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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17120 | # Kruskal-Katona type question for union-closed families of sets
Question : Let $n,k$ be two positive integers with $n \geq k$. Let $\mathcal{F}$ be a family of $C(n,k)$ sets, each of size $k$, and let $<\mathcal{F}>$ denote the union-closed family generated by $\mathcal{F}$, i.e.: $<\mathcal{F}>$ consists of all those sets which can be expressed as a union of members of $\mathcal{F}$. Must it be the case that $$|<\mathcal{F}>| \geq \sum_{j=k}^{n} C(n,j),$$ with equality if and only if $\mathcal{F}$ consists of all $k$-element subsets of an $n$-set ?
It is easy to see that if the inequality holds (whatever about uniqueness), then it implies that, for any union-closed family $\mathcal{G}$ and non-negative integer $m$, if $|\mathcal{G}| \geq 2^{m}$ then the average size, let's denote it $w(\mathcal{G})$, of a member of $\mathcal{G}$ is at least $m/2$. This is, in turn, a special case of a result of Reimer [1] that, for any union-closed family $\mathcal{G}$ one has $w(\mathcal{G}) \geq \frac{1}{2} \log_{2} |\mathcal{G}|$. Indeed I had conjectured the same result and in thinking about it was led to the above question, before I recently became aware of Reimer's proof, which is a beautiful piece of work !
One can obviously try to generalise my question to an arbitrary number of generating $k$-sets, perhaps along the lines of the Kruskal-Katona theorem for shadows ?
[1] D. Reimer, An average set size theorem, Combin. Probab. Computing 12 (2003), 89-93.
-
You might make it clear that union F is potentially greater than n. Gerhard "Ask Me About System Design" Paseman, 2011.09.16 – Gerhard Paseman Sep 16 '11 at 16:07
What about the underlying set being of size C(n,k), and each member of F is that set minus a singleton? I then get < F > having one set more than F. Or is there something I am missing? Gerhard "Ask Me About System Design" Paseman, 2011.09.16 – Gerhard Paseman Sep 16 '11 at 16:11
I apologize for not meeting all of the problem requirements, especially the condition of each set being of size k. Nevertheless, I have a feeling that some example similar to the above is possible, and that one can find F which do not satisfy the inequality because of getting several pairs of elements to have the same union. Gerhard "Back To The Drawing Board" Paseman, 2011.09.16 – Gerhard Paseman Sep 16 '11 at 17:52
But, intuiutively, it seems to me that the best way to get most overlap in the unions is to start off by "packing" the sets as close as you can, i.e: as k-subsets of an n-set. Not that this proves anything :) – Peter Hegarty Sep 17 '11 at 15:04
I sympathize with the intuition. So instead let's consider k-sets of N >> n. Perhaps it can be shown that there is a partition of [k,N] into intervals [k_i, k_{i+1}) such that C(N,k+i) is less than the sets of < F > whose sizes fall into the ith interval, and that the inequality can be established that way. It would be interesting to know this even in the case N=n+1. Gerhard "Ask Me About System Design" Paseman, 2011.09.17 – Gerhard Paseman Sep 17 '11 at 19:05 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17121 | ## anonymous one year ago Let f(x) = 4x2 + x + 1 and g(x) = x2 – 2. Find g(f(x)). Show each step of your work. Help please
1. anonymous
@marihelenh can you help me with this one too?
2. anonymous
@freckles
3. marihelenh
You would replace what you have for f(x) into the x position of g(x). $\left( 4x ^{2}+x+1 \right)^{2}-2$ Does that make sense?
4. anonymous
but where does the x^2-2 go
5. anonymous
the -2 is there but the x^2 i only see the ^2 added but not the x
6. marihelenh
The f(x) becomes a function of g(x), so you just replace what f was equal to into wherever the x is in the g equation.
7. anonymous
oh i think i get it
8. anonymous
yes i get it :D
9. marihelenh
Ok great! It is a little difficult trying to explain on here sometimes.
10. anonymous
4x^4+x^3-1 would this be the final answer?
11. anonymous
@marihelenh
12. marihelenh
Just one sec, let me check.
13. marihelenh
No, the answer is actually even longer.
14. marihelenh
To find the answer you would have to do, $(4x ^{2}+x+1)(4x ^{2}+x+1) -2$
15. anonymous
This is so hard, i thought i understood but now i don't.
16. marihelenh
Do you have to simplify it?
17. anonymous
It says Find g(f(x)) so I think I have to but maybe not
18. marihelenh
Well, if you don't have to, you already have your answer, but I will help you simplify it. Just hang in with me. It will seem confusing, but we will get through it and to the answer. OK?
19. anonymous
ok, thank you
20. marihelenh
The first thing you should do, is multiply the first term of the first set by the second set. |dw:1436743172709:dw|
21. anonymous
16x^5?
22. marihelenh
It would be 16x^4 + 4x^3 + 4x^2
23. marihelenh
Then the same thing with the second term |dw:1436743368652:dw| 4x^3 + x^2 + x
24. anonymous
4 times 4 = 16 then 16^4 times x = 16x^5 or the x doesnt add ^1?
25. marihelenh
No, it doesn't. You add the current exponents, so it would be 2 + 2 = 4.
26. anonymous
oh ok
27. marihelenh
Then you would do it with the third term. |dw:1436743540161:dw| 4x^2 + x + 1
28. anonymous
and what happens to the -2?
29. marihelenh
That will get added on in the end when we put all of the terms together.
30. anonymous
so final answer is this 16x^4 + 4x^3 + 4x^2 - 2 ?
31. marihelenh
That is how you multiply the two trinomials out, so then add them all together. $16x ^{4}+4x ^{3}+4x ^{2}+4x ^{3}+x ^{2}+x+4x ^{2}+x+1-2$ (I added in the -2)
32. anonymous
^that last part is so confusing
33. marihelenh
Add like terms $16x ^{4}+8x ^{3}+9x ^{2}+2x-1$
34. marihelenh
Do you understand how I got the three different terms on the pictures by multiplying them out?
35. anonymous
yes that part i understand
36. anonymous
The f(x) becomes a function of g(x) so you just replace what f was equal to into wherever the x is in the g equation. Then you get: (4x^2+x+1)^2(4x^2+x+1)^2-2 then multiply 4x^2 by all terms of second equation to get 16x^4 now multiply x by all terms of second equation to get 4x^3 repeat now with last term to get 4x^2 you get 16x^4 + 4x^3 + 4x^2 now to add the -2 you add like terms to get 16x^4 + 8x^3 + 9x^2 + 2x - 1 would this be good for the answer ?
37. marihelenh
I just put the nine different terms together and then added the -2.
38. marihelenh
Do you just have to explain how to get the answer?
39. anonymous
yes
40. anonymous
41. marihelenh
Ok, well then let me just fix what you have, you were pretty close.
42. anonymous
ok
43. marihelenh
then multiply 4x^2 by all terms of second equation to get 16x^4+4x^3+4x^2 now multiply x by all terms of second equation to get 4x^3+x^2+x repeat now with last term to get 4x^2+x+1 you get 16x^4+4x^3+4x^2+4x^3+x^2+x+4x^2+x+1 now to add the -2 and add like terms to get 16x^4+8x^3+9x^2+2x-1
44. marihelenh
The first three lines were good. You just had to add a little extra into the next ones, but it was all looking good.
45. anonymous
yeah i missed a couple of things, thank you so much for your time :D
46. marihelenh
It's not a problem. I am soooo sorry for how long it took and how confusing it was.
47. anonymous
I don't care as long as I get it, thank you.
48. marihelenh
You're welcome |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17122 | # If $A=\begin{bmatrix}1&2\\3&-5\end{bmatrix}$ then $A^{-1}$=
$\begin{array}{1 1}(A)\;\begin{bmatrix}-5&-2\\-3&1\end{bmatrix}\\(B)\;\begin{bmatrix}5/11&2/11\\3/11&-1/11\end{bmatrix}\\(C)\;\begin{bmatrix}-5/11&-2/11\\-3/11&1/11\end{bmatrix}\\(D)\;\begin{bmatrix}5&2\\3&-1\end{bmatrix}\end{array}$ |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17123 | # Buying pants at the mall
Originally posted on What's new:
The classical inverse function theorem reads as follows:
Theorem 1 ($latex {C^1}&fg=000000$ inverse function theorem) Let $latex {\Omega \subset {\bf R}^n}&fg=000000$ be an open set, and let $latex {f: \Omega \rightarrow {\bf R}^n}&fg=000000$ be an continuously differentiable function, such that for every $latex {x_0 \in \Omega}&fg=000000$, the derivative map $latex {Df(x_0): {\bf R}^n \rightarrow {\bf R}^n}&fg=000000$ is invertible. Then $latex {f}&fg=000000$ is a local homeomorphism; thus, for every $latex {x_0 \in \Omega}&fg=000000$, there exists an open neighbourhood $latex {U}&fg=000000$ of $latex {x_0}&fg=000000$ and an open neighbourhood $latex {V}&fg=000000$ of $latex {f(x_0)}&fg=000000$ such that $latex {f}&fg=000000$ is a homeomorphism from $latex {U}&fg=000000$ to $latex {V}&fg=000000$.
It is also not difficult to show by inverting the Taylor expansion
View original
Originally posted on Team Python:
Okay, so you have “registered an interest” or pre-ordered or even (lucky few) ordered an RPi. It’s time to get together the stuff you’ll need.
1. A screen. The most obvious choice is a TV with an HDMI input.
2. A usb mouse. If you don’t have a spare around, why not treat yourself to a new one – you’re worth it!
View original |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17124 | It is currently 11 Dec 2017, 18:03
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# n-x) + (n-y) + (n-c) + (n-k) What is the value of the
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(n-x) + (n-y) + (n-c) + (n-k)
What is the value of the expression above?
(1) The average (arithmetic mean) of x, y, c, and k is n.
(2) x, y, c, and k are consecutive integers.
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18 Mar 2006, 22:32
A. the value of this expression is 0.
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18 Mar 2006, 22:40
oa is A
cant beleive i overlooked this stupid q
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19 Mar 2006, 09:24
Can someone explain this?
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22 Mar 2006, 01:59
(n-x) + (n-y) + (n-c) + (n-k) = n-x+n-y+n-c+n-k
= 4n -(x+y+c+k)............(i)
now if (1) The average (arithmetic mean) of x, y, c, and k is n.
means (x+y+c+k)/4 = n
means (x+y+c+k) = 4n
substitute this in (i)
4n - 4n
=0
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07 May 2013, 15:17
Angela780 wrote:
Can someone explain this?
Can you explain this problem for me, I'm not understanding it.
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07 May 2013, 20:04
Quote:
(n-x) + (n-y) + (n-c) + (n-k)
What is the value of the expression above?
(1) The average (arithmetic mean) of x, y, c, and k is n.
(2) x, y, c, and k are consecutive integers.
laythesmack23 wrote:
Angela780 wrote:
Can someone explain this?
Can you explain this problem for me, I'm not understanding it.
It asks for the value of (n-x) + (n-y) + (n-c) + (n-k) .......or 4n-(x+y+c+k) .....so that means we need to know the value of n
& x+y+c+k .....
statement:: 1 says ..... The average (arithmetic mean) of x, y, c, and k is n..... that means ....
$$\frac{x+y+c+k}{4}= n$$ ... Theforefore, 4n = x+y+c+k ...
when plugging in the value in the expression given .... 4n-4n .. Therefore, 0.
The value of above expression is 0. Sufficient.
Statement :: 2 says x, y, c, and k are consecutive integers. ..lets say for ex. the consecutive integers x,y,c,k are 1,2,3,4, respectively...
therefore, we have (n-1)(n-2)(n-3)(n-4)..... or 4n-10 .. so here we need to know the value of n in order to know the value of the
expression above........ so, Insufficient.
Hence, A ............
Hope it Helps !! & let me know if there is any problem ..............
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Re: n-x) + (n-y) + (n-c) + (n-k) What is the value of the [#permalink]
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03 May 2017, 08:21
Statement 1: (x+y+c+k)/4 = n
or x+y+c+k = 4n = n+n+n+n. Now take all the terms from left hand side to right hand side
It gives (n-x)+(n-y)+(n-c)+(n-k) = 0. Sufficient.
Statement 2: It doesn't tell us anything about n. OR even about the differences between n and any of the four variables x,y,c or k. Insufficient.
Hence A.
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Re: n-x) + (n-y) + (n-c) + (n-k) What is the value of the [#permalink]
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03 May 2017, 10:04
joemama142000 wrote:
(n-x) + (n-y) + (n-c) + (n-k)
What is the value of the expression above?
(1) The average (arithmetic mean) of x, y, c, and k is n.
(2) x, y, c, and k are consecutive integers.
OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/what-is-the- ... 03146.html
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Re: n-x) + (n-y) + (n-c) + (n-k) What is the value of the [#permalink] 03 May 2017, 10:04
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17125 | # 11. Consider the titration of 25.00 mL of 0.200 M methyl amine (CH3NH2). The titrant is...
###### Question:
11. Consider the titration of 25.00 mL of 0.200 M methyl amine (CH3NH2). The titrant is 0.120 M HCI. Calculate each of the following: a. the initial pH of the base. b. the pH at 5.00 mL added acid c. the pH at the HCI needed to reach the equivalence point d. the volume of added acid required to reach the equivalence point. e. the pH at the equivalence point.
Questions Use correct units throughout all calculations, including those involving the equation of the line from the graph. 1. Use the Clausius-Clapeyron equation and the graph, as described in the Introduction, to determine the entropy and enthalpy of vaporization. Show where the units for AH and AS originate. In PH₂o Asuap Huap po A RT. -2.26068 AS S. Ag = 221.825 in R 5/molk SH° = 8.05x10 J/mol 9680.4 = -AH J/molok R/melk 2. Use the equation of the line from the graph to determine the vapor pressure of water at 65 °C. y=-96604x+26.08 PH₂O 3380K 4= -9 680..1 (338x) +20.08 -1.90 In PH₂O po = -1.90.. 3. Use a table in your text to determine the vapor pressure of water at its normal boiling point. Compare this value to the vapor pressure calculated in 2. Why is one value lower than the other?
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17126 | # Moving the branch cut of the complex logarithm
The complex logarithm is defined as $\log z:=\operatorname{Log} |z|+i\arg z$ , with the branch cut on the non-negative real axis.
Determine a branch of $f(z)=\log(z^3-2)$ that is analytic at $z=0$ and find $f(0)$ and $f'(0)$.
So, I would need to choose a branch of the logarithm such that it is analytic at $g(0)=-2$ (where $g(z)=z^3-2)$. I am having difficulty knowing which branch can be chosen. The book I am using claims that $\mathcal{L_{-\pi/4}}$ works, and shows that the branch cut has moved from the negative real axis to somewhere in quadrant $4$. Why is this? I would have thought that the branch cut would have been moved to somewhere in quadrant $2$ if I moved it by $-\pi/4$.
• Taking a single-valued branch is always annoying to me. +1 – Yai0Phah Apr 19 '14 at 14:25
• It works if we rewrite $f(z)=\log(z-\sqrt[3]2)+\log(z-\sqrt[3]2\omega)+\log(z-\sqrt[3]2\omega^2)$, but I want to obtain good insight on how Riemann surfaces work. For example, what's the exact meaning of $\log(fg)=\log f+\log g$ w.r.t. Riemann surfaces when $f,g$ are multi-valued functions. – Yai0Phah Apr 19 '14 at 14:28 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17127 | # Suppose x and y are two arbitrary, distinct (unequal) real numbers. Prove that there exists a rational number q between x and y.
I thought this proof was much simpler than it actually is. I used an Axiom that states, "if $p$ and $q$ are real numbers, then there is a number between them, i.e: $$\frac{(p + q)}{2}$$ My attempt at the proof was:
By the above Axiom, there exists a number, $q$, between $x$ and $y$, i.e: $$\frac{(x+y)}{2}$$ We know $q$ is rational because it is in the form $\frac ab$, where $a = x + y$ and $b = 2$. Thus, there exists a rational number, $q$ such that $q$ is between $x$ and $y$.
However, I realized that if $x=\pi$ and $y = e$, then $\frac{(x+y)}{2}$ will not be rational.
I know that to prove this, I need to use the Lemma: If $y-x>1$, ∃ n ∈ $\mathbb{Z}$ such that $x<n<y$.
I really don't know how to approach this differently because I did think it would be simpler.
• x and y are distinct real numbers. It cannot be that both are the same. One concrete example could be pi and pi + e. Then both are real and distinct – Icycarus Feb 3 '17 at 1:35
• We actually don't know whether $\pi+e$ is rational. It could be. – Matt Samuel Feb 3 '17 at 2:48
• Find the minimum integer value of $k$ for which $$|x-y| < 10^k$$
• Round $\min(x,y)$ to the smallest multiple of $10^{k-1}$ greater than $\min(x,y)$.
• Add $10^{k-2}$. The result is a rational number strictly between $x$ and $y$.
Since $x$ and $y$ are different you can choose an integer $N$ large enough so that $1/N$ is less than their (positive) difference. Now imagine a ruler (number line) with marks every $1/N$th. One of those marks will have to fall strictly between $x$ and $y$. It's rational since it looks like $k/N$ for some integer $k$.
• So I wouldn't necessarily have to use the lemma? – Mathgirl Feb 3 '17 at 1:45
• @Mathgirl No, you wouldn't. – Ethan Bolker Feb 3 '17 at 11:27
If the difference between two real numbers is greater than 1, then there will be an integer between the numbers. If the difference is less than , take the difference of first decimal number from where the numbers start to differ. And that decimal number at its position has to exist between the real numbers.
Eg: $0.12345......$ and $0.123378.....$ are such real numbers then, $0.1234$ has to exist between them which is a rational number.
Similar trick Can be applied for negative numbers also.
• I understand the first part, but I am still a bit confused about the case when the difference is less than 1. Since $x$ and $y$ are arbitrary, is there a way to prove this would be true for all real numbers? Or would we use proof by example? – Mathgirl Feb 3 '17 at 1:55
• Any arbitrary numbers will have decimal expansion similar to the example I gave. From there only required decimal numbers can be used to prove the existence of rational number. Try cooking up your own example and getting a decimal number between them. You will get the idea – jnyan Feb 3 '17 at 2:09 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17128 | • # question_answer Among $[Ni{{(CO)}_{4}}],{{[NiC{{l}_{4}}]}^{2-}},[Co{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]Cl,$ $N{{a}_{3}}[Co{{F}_{6}}],N{{a}_{2}}{{O}_{2}}$ and $Cs{{O}_{2}}$ the total number of paramagnetic compounds is A) 2 B) 3 C) 4 D) 5
Compound/ Ion of compound Magnetic nature 1. $[Ni{{(CO)}_{4}}]$ Diamagnetic 2. ${{[NiC{{l}_{4}}]}^{2-}}$ Paramagnetic 3. $a[Co{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]Cl~$ Diamagnetic 4. $Na[Co{{F}_{6}}]$ Paramagnetic 5. $N{{a}_{2}}{{O}_{2}}$ Diamagnetic 6. $Cs{{O}_{2}}$ Paramagnetic |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17129 | author = "Abreu, L. D."
1
Found 7 documents, displaying page 1 of 1
## Completeness criteria for sequences of special functions
Author(s) : Abreu, L. D.
Description : Two Lp completeness criteria for sequences of functions written in the form f(lnx), where ln is the nth zero of another function g, are obtained. The results are illustrated with applications to sequences of special functions , Fundação para a Ciência e Tecnologia; Centro de Matemática da Universidade ...
Repository : Estudo Geral
Language(s) : English
## A q-sampling theorem related to the q-Hankel transform
Author(s) : Abreu, L. D.
Description : A q-version of the sampling theorem is derived using the q-Hankel transform introduced by Koornwinder and Swarttouw. The sampling points are the zeros of the third Jackson q-Bessel function , Fundação para a Ciência e Tecnologia; Centro de Matemática da Universidade de Coimbra
Repository : Estudo Geral
Language(s) : English
## The roots of the third Jackson q-Bessel function
Description : We derive analytic bounds on the roots of the third Jackson Function. This bounds prove a conjecture of M. E. H. Ismail concerning the asymptotic behaviour of the roots.
Repository : Estudo Geral
Language(s) : English
## Hardy-type theorem for orthogonal functions with respect to their zeros. The Jacobi weight case
Description : Motivated by the G.H. Hardy's 1939 results [G.H. Hardy, Notes on special systems of orthogonal functions II: On functions orthogonal with respect to their own zeros, J. London Math. Soc. 14 (1939) 37-44] on functions orthogonal with respect to their real zeros [lambda]n, , we will consider, under th...
Repository : Estudo Geral
Language(s) : English
## Hardy-type theorem for functions orthogonal with respect to their zeros. The Jacobi weight case
Description : Motivated by G. H. Hardy's 1939 results \cite{Hardy} on functions orthogonal with respect to their real zeros $\lambda_{n}, n=1,2,...$, we will consider, within the same general conditions imposed by Hardy, functions satisfying an orthogonality with respect to their zeros with Jacobi weights on t...
Repository : arXiv
Language(s) : Undetermined
## Bilinear biorthogonal expansions and the Dunkl kernel on the real line
Description : We study an extension of the classical Paley-Wiener space structure, which is based on bilinear expansions of integral kernels into biorthogonal sequences of functions. The structure includes both sampling expansions and Fourier-Neumann type series as special cases, and it also provides a bilinear...
Repository : arXiv
Language(s) : Undetermined
## Pointwise convergent expansions in q-Fourier-Bessel series
Description : We define q-analogues of Fourier-Bessel series, by means of complete q- orthogonal systems constructed with the third Jackson q-Bessel function. Sufficient conditions for pointwise convergence of these series are obtained, in terms of a general convergence principle valid for other Fourier series on...
Repository : Estudo Geral
Language(s) : English
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17130 | # Differentiating $90 + 45 \cos(t^2/18)$ using the chain rule?
Can someone please explain to me how to differentiate this function?
$$90 + 45 \cos(t^2/18)$$
I know you use the chain rule, but I can't get the final result correct.
-
$$\zeta(t)=90+45\cos\left(\frac{t^2}{18}\right)$$ \require{cancel}\begin{align} \frac{d\zeta}{dt}&=\cancelto{0}{\frac{d}{dt}(90)}+\frac{d}{dt}\left(45\cos\left(\frac{t^2}{18}\right)\right)\\&= \frac{d}{dt}\left(45\cos\left(\frac{t^2}{18}\right)\right) \\&=45\left(-\sin\left(\frac{t^2}{18}\right)\right)\underbrace{\left(\frac{d}{dt}\left(\frac{t^2}{18}\right)\right)}_{\frac{t}{9}}\\& =-5t\sin\left(\frac{t^2}{18}\right)\end{align}
Because: $$\frac{d}{dx}\cos(\varphi)=-\sin(\varphi)\cdot\frac{d\varphi}{dx}$$
-
thank you this made so much sense! Splitting up the equation helped me see what i had to differentiate. The 1/9 was were I went wrong and messed up. Thanks!!! – sloth1111 May 7 '14 at 0:24
@sloth1111 No problem. Good luck. – Shahar May 7 '14 at 0:25
Here by using the chain rule:
$\frac{d}{dt}(90+45\cos(\frac{t^2}{18})=\frac{d}{dt}(90)+45\frac{d}{dt}(\frac{t^2}{18})\cos'(\frac{t^2}{18})=0-45\frac{2t}{18}\sin(\frac{t^2}{18})=-5t\sin(\frac{t^2}{18}).$
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Well the $90$ becomes simply zero. So you have $$45\left(f\left(g(t)\right)\right)'$$ where $g(t)=t^2/18$ and cos $f(x)=\cos(x)$. The chain rule goes: $$\left(f\left(g(t)\right)\right)'=f'(g(t))\cdot g'(t)$$ And $f'(x)=-\sin(x)$ so $f'(t^2/18)=-\sin(t^2/18)$. and $g'(t)=t/9$, that's clear, right?
So $$45\cdot f'(g(t))\cdot g'(t)=-45\cdot\sin(t^2/18)\cdot t/9$$
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$\frac{d}{dt}90 = 0$ so we concentrate on your second term.
$\displaystyle\frac{d}{dt} \cos(t^2/18) = \frac{dt^2}{dt}\frac{d}{dt^2} \cos(t^2/18) = 2t \times (-\frac{1}{18}\sin(t^2/18))= -\frac{1}{9}t\sin(t^2/18)$
And then multiply by 45.
When you take the derivative of the $\cos()$, think of $t^2$ as a variable in its own right (which it really is, after all). If it helps, replace it with another symbol, like $t^2 = u$, and then, when you have your result, put the $t^2$ back in where you have $u$.
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Remark that $cos'(t)=-sin( t)$ and $\frac{d}{dt}(\frac{t^2}{18})=\frac{t}{9}$. Then $$\frac{d}{dt}(90+45cos(\frac{t^2}{18}))=\frac{d}{dt}(90)+\frac{d}{dt}(45cos(\frac{t^2}{18}))=0+45\frac{d}{dt}(cos(\frac{t^2}{18}))$$ But setting $s=s(t)=\frac{t^2}{18}$, $$\frac{d}{dt}(cos(\frac{t^2}{18}))=\frac{d}{ds}(cos(s))\frac{d}{dt}(s(t))=-sin(s).s'(t)=-sin(\frac{t^2}{18})\frac{t}{9}.$$ Thus $$\frac{d}{dt}(90+45cos(\frac{t^2}{18}))=45(-sin(\frac{t^2}{18})\frac{t}{9})=-5t.sin(\frac{t^2}{18})$$
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You need to see the nesting:
$$f(t) = 90 + 45\cos\left(\frac{t^2}{18}\right) \\ g(x) = 90 + 45\cos(x) \rightarrow g\left(x = \frac{t^2}{18}\right) = f(t) = 90 + 45\cos\left(\frac{t^2}{18}\right) \\ g'(x) = 90 - 45\sin(x) \rightarrow g'\left(x = \frac{t^2}{18}\right) = 90 - 45\sin\left(\frac{t^2}{18}\right) \\ x'(t) = \frac{d}{dt}\left(\frac{t^2}{18}\right) = \frac{t}{9} \\ \left.\left.\frac{d}{dt}\right(f(t)\right) = \left.\left.\frac{d}{dt}\right (g\left(x(t)\right)\right) = \frac{dx}{dt}\cdot\left(\frac{dg}{dx} \circ x(t)\right) \\ \frac{df}{dt} = \frac{t}{9}\left(90 - 45\sin\left(\frac{t^2}{18}\right)\right)$$
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17131 | # Start a new discussion
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## Discussion Tag Cloud
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeJul 11th 2014
• (edited Jul 11th 2014)
at normed division algebra it used to say that “A normed field is either $\mathbb{R}$ or $\mathbb{C}$. ” I have changed that to “a normed field over $\mathbb{R}$ is…” and changed normed field from being a redirect to “normed division algebra” to instead being an entry on its own.
• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeJul 11th 2014
I added more examples to normed field.
• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeJul 11th 2014
• (edited Jul 11th 2014)
Thanks for the edits!
Regarding the edits on the links:
I am thinking that “archimedean” and “non-archimedean” should eventually point to something other than archimedean field, an entry both more focused – in that it immediately mentions the ultrametric inequality – and more general – in that it includes not just normed fields but also normed vector spaces more generally.
Wikipediae has “Archimedean property”. Hm.
• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeDec 14th 2016
• (edited Dec 15th 2016)
at normed division algebra in the Classification section it suddenly says “…exactly four finite-dimensional normed division algebras…”. This makes it sound as if there might be further normed division algebras that evade the classification by being infinite-dimensional. Should the “finite dimensional” qualifier not just be removed in this sentence?
• CommentRowNumber5.
• CommentAuthorDavidRoberts
• CommentTimeDec 14th 2016
John Baez has some stuff about this that we discussed once. I can track it down. Possibly it’s on MathOverflow.
• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeDec 14th 2016
Re #4: I think that’s right, but to rule out infinite-dimensional examples might require more functional analysis than is currently discussed in the relevant nLab articles. (Not much, but some.)
• CommentRowNumber7.
• CommentAuthorDavidRoberts
• CommentTimeDec 14th 2016
• (edited Dec 14th 2016)
OK, you were right. See this MO answer and JB’s comment. I’ve edited the entry to include.
EDIT: one has to have characteristic 2, and the algebra needs to be a purely inseparable extension of the base field, in order to get an infinite-dimensional example in the unital case – or else one needs a non-unital algebra (examples are known over the reals in this case).
• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeDec 15th 2016
• CommentRowNumber9.
• CommentAuthorDavid_Corfield
• CommentTimeDec 15th 2016
• (edited Dec 15th 2016)
Over the complex numbers, the only normed division algebra is the algebra of complex numbers themselves.
Does this have implications if and when you try to take the superpoint story (superspacetime and branes) in intergeometric fashion over to the complex analytic world?
What happens if you try to find maximal invariant central extensions of $\mathbb{C}^{0|2}$, say?
• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeDec 15th 2016
• (edited Dec 15th 2016)
What happens if you try to find maximal invariant central extensions of $\mathbb{C}^{0\vert 2}$, say?
By the same simple argument as for the real case (here) the maximal central extension of $\mathbb{C}^{0\vert 2}$ is $\mathbb{C}^{3\vert 2}$, with the odd-odd bracket the one that takes two 2-component complex vectors $\psi$, $\phi$ to the symmetric complex matrix $\psi \phi^t + \phi \psi^t$.
This does not seem to have any good spacetime meaning. The reason why the same simple step works magic over the reals is that over the reals symmetric matrices are hermitian (trivially, but crucially), and it is hermitian $2 \times 2$ matrices over the real normed division algebras $\mathbb{K}$ which are identified with Minkowski spacetimes of dimension $2 + dim_{\mathbb{R}}(\mathbb{K})$ (here).
• CommentRowNumber11.
• CommentAuthorDavid_Corfield
• CommentTimeDec 15th 2016
This does not seem to have any good spacetime meaning.
Which takes me back to the g+ question I posed to you about what counts as a ’valuable’ departure from a good spacetime meaning, as in F-theory. I guess there could be all kinds of ’weird’ prequantum theories which possess equivalent quantizations to more ’meaningful’ ones, as at duality in physics.
• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeDec 15th 2016
The F-theory super Lie algebra as in section 7 of arXiv:1611.06536 just so happens to be a useful correspondence space between two actual spacetimes, a “non-perturbative T-fold”. One should not read more into this than what it is: a useful book-keeping tool, for T-duality and, in this case, also for S-duality. Generally, these “doubled” T-fold spacetimes are by design not meant to be actual spacetimes, but correspondence spaces witnessing duality between actual spacetimes. They are not part of the bouquet, they are an auxiliary construction surrounding it.
• CommentRowNumber13.
• CommentAuthorDavid_Corfield
• CommentTimeDec 15th 2016
Not a task for now, but maybe you could tell us one day what’s so special about an “actual spacetime”. At spacetime there is
A spacetime is a manifold that models space and time in physics.
and this is said to require a smooth Lorentzian space equipped with a time orientation.
Can we ’transcendentally’ deduce from what we experience the need for a fundamental physical theory to refer to such a spacetime?
As I say, not something for the working day, but rather late night speculative chat.
• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeDec 2nd 2018
• (edited Dec 2nd 2018)
There is a gap in the entry normed division algebra:
Ever since rev 1 it claims that a normed division algebra is an algebra that is both a division algebra and a normed algebra.
But a priori normed algebras satisfy only
$(1)\;\; {\vert a \cdot b\vert} \leq C {\vert a \vert} \cdot {\vert b \vert} \,.$
wheras all texts I checked define “normed division algebra” by directly imposing instead the stronger clause
$(2)\;\;{\vert a \cdot b\vert} = {\vert a \vert} \cdot {\vert b \vert}$
I gather the idea is that with the division algebra property (1) implies (2).
[edit: oh, of course rev 1 claims just that ]
But this seems a bit fiddly. Or maybe I am missing something.
• CommentRowNumber15.
• CommentAuthorTodd_Trimble
• CommentTimeDec 2nd 2018
I think the necessary repairs would probably be at hand provided that we are dealing with the finite-dimensional case only. But it would take some time, which I don’t have at the moment, to go through it all. You’re right that there is some messiness here that needs cleaning up (and some of it might be non-trivial).
• CommentRowNumber16.
• CommentAuthorUrs
• CommentTimeDec 2nd 2018
Thanks, Todd! No rush. Maybe we could for the moment just add a line to the entry that confirms that the intended implication exists.
Myself, have to rush off now.
• CommentRowNumber17.
• CommentAuthorUrs
• CommentTimeDec 3rd 2018
expanded the first lines as follows – this still needs attention:
A normed division algebra is a not-necessarily associative algebra, over some ground field, that is
1. a division algebra $\big( \text{i.e.}\, (a \cdot b = 0) \Rightarrow (a = 0 \,\text{or}\, b = 0) \big)$
2. a multiplicatively normed algebra $\big( \text{i.e.}\, {\Vert a \cdot b\Vert} \leq C \cdot {\Vert a\Vert} \cdot {\Vert b\Vert} \big)$.
It should be the case (at least maybe for finite-dimensional algebras) that the division property (1) implies that the norm property (2) holds in the stronger form
${\vert a \cdot b\vert} \;=\; {\vert a \vert} \cdot {\vert b \vert}$
and this is how most (or all) authors actually define normed division algebras, and that’s what we assume to be meant now.
• CommentRowNumber18.
• CommentAuthorTodd_Trimble
• CommentTimeDec 4th 2018
I mean to get back to this, but here are a few notes.
I think it’s not that property 2. gets strengthened in this way assuming property 1. All that 2. is really trying to say is that multiplication is (uniformly) continuous with respect to the topology induced by the norm. It’s more that we always have an option to rescale the norm by putting ${|a|} = C {\|a\|}$, so that at least ${|a \cdot b|} \leq {|a|} \cdot {|b|}$ (called submultiplicativity).
In the associative unital case, we can further enforce ${|\mathbf{1}|} = 1$, by pulling back the operator norm on the space of continuous linear maps $\hom(A, A)$ along the map $A \to \hom(A, A)$ that you get by currying the multiplication map on $A$. If $A$ has zero divisors, then the pulled-back pseudonorm does not satisfy separability (${|a|} = 0 \Rightarrow a = 0$), but with property 1. you do get a norm.
In that case, I believe the Frobenius theorem that characterizes finite-dimensional Banach algebras over $\mathbb{R}$ as $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ does give you strict multiplicativity over submultiplicativity, i.e., the equality that everyone else assumes. I’m afraid I don’t see an easy argument for getting that otherwise, and I don’t know what to make of the situation if associativity is not assumed.
Another thing is that the article seems to goof up later on, inadvertently assuming associativity when it begins talking about Banach algebras. I assume that was unintended.
It may be that all this fuss about starting with weaker assumptions than strict multiplicativity is just not worth it in the end, but I’m curious whether I’ve overlooked something. Maybe the author (Toby?) sees how to derive that after all. For that reason, I haven’t implemented any changes.
• CommentRowNumber19.
• CommentAuthorTodd_Trimble
• CommentTimeDec 4th 2018
Obviously in the 4th paragraph I forgot to use the word “division”. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17132 | # If a definite integral produces a finite value, does that mean it's convergent?
$\int{_0^5\frac{x}{x-2}dx}$
This integral produces a finite value of 5+ln(9/4). However, according to Wolfram Alpha, it diverges (http://www.wolframalpha.com/input/?i=integral+of+x%2F%28x-2%29+from+0+to+5). How can I tell that this integral is divergent without using Wolfram Alpha? Or is it actually convergent from 0 to 5?
• How did you get the value $5+\ln(9/4)$? – Tim Raczkowski Feb 6 '15 at 3:27
• Did you scroll down to the bottom of the Wolfram Alpha result? It provides the same principal value for the integral which you claimed. – David H Feb 6 '15 at 3:32
• @DavidH Yes, I did. Nevertheless, I was still confused by "integral does not converge". I assumed the Cauchy principle value is similar to the complex value solutions that Wolfram Alpha often gives. – Leo Jiang Feb 6 '15 at 3:35
We must be very careful with what we mean by "converges" when talking about improper integrals. The usual definition of $\int_a^b f(x)dx$ converging is that $\int_a^b f^+(x) dx$ and $\int_a^b f^-(x)dx$ are both finite, where $f^+(x) = \max\{f(x),0\}$ and $f^-(x) = \max\{-f(x),0\}$.
In the case of $\mathbb{R}$ (as opposed to $\mathbb{R}^n$) if $a$ or $b$ is infinite it is common to say an integral converges if, e.g. the case where $b$ is infinite $\lim_{M \to \infty} \int_a^M f(x)dx$ exists in $\mathbb{R}$.
• By "first definition", you mean when $\int_a^b f^+(x) dx$ and $\int_a^b f^-(x)dx$ are both finite right? Aren't they both finite in my example? If a=0 and b=5, then the integral produces 5+ln(9/4) – Leo Jiang Feb 6 '15 at 3:57
• The integral does not produce $5 + \log (9/4)$. The integrand does not satisfy the hypotheses of the fundamental theorem of calculus, and so the calculation that you probably did to get that answer is not valid. – nullUser Feb 6 '15 at 4:10 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17133 | # 13.6 Binomial theorem (Page 3/6)
Page 3 / 6
## Key equations
Binomial Theorem ${\left(x+y\right)}^{n}=\sum _{k-0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){x}^{n-k}{y}^{k}$ $\left(r+1\right)th\text{\hspace{0.17em}}$ term of a binomial expansion $\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}$
## Key concepts
• $\left(\begin{array}{c}n\\ r\end{array}\right)\text{\hspace{0.17em}}$ is called a binomial coefficient and is equal to $C\left(n,r\right).\text{\hspace{0.17em}}$ See [link] .
• The Binomial Theorem allows us to expand binomials without multiplying. See [link] .
• We can find a given term of a binomial expansion without fully expanding the binomial. See [link] .
## Verbal
What is a binomial coefficient, and how it is calculated?
A binomial coefficient is an alternative way of denoting the combination $\text{\hspace{0.17em}}C\left(n,r\right).\text{\hspace{0.17em}}$ It is defined as $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ r\end{array}\right)=\text{\hspace{0.17em}}C\left(n,r\right)\text{\hspace{0.17em}}=\frac{n!}{r!\left(n-r\right)!}.$
What role do binomial coefficients play in a binomial expansion? Are they restricted to any type of number?
What is the Binomial Theorem and what is its use?
The Binomial Theorem is defined as $\text{\hspace{0.17em}}{\left(x+y\right)}^{n}=\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){x}^{n-k}{y}^{k}\text{\hspace{0.17em}}$ and can be used to expand any binomial.
When is it an advantage to use the Binomial Theorem? Explain.
## Algebraic
For the following exercises, evaluate the binomial coefficient.
$\left(\begin{array}{c}6\\ 2\end{array}\right)$
15
$\left(\begin{array}{c}5\\ 3\end{array}\right)$
$\left(\begin{array}{c}7\\ 4\end{array}\right)$
35
$\left(\begin{array}{c}9\\ 7\end{array}\right)$
$\left(\begin{array}{c}10\\ 9\end{array}\right)$
10
$\left(\begin{array}{c}25\\ 11\end{array}\right)$
$\left(\begin{array}{c}17\\ 6\end{array}\right)$
12,376
$\left(\begin{array}{c}200\\ 199\end{array}\right)$
For the following exercises, use the Binomial Theorem to expand each binomial.
${\left(4a-b\right)}^{3}$
$64{a}^{3}-48{a}^{2}b+12a{b}^{2}-{b}^{3}$
${\left(5a+2\right)}^{3}$
${\left(3a+2b\right)}^{3}$
$27{a}^{3}+54{a}^{2}b+36a{b}^{2}+8{b}^{3}$
${\left(2x+3y\right)}^{4}$
${\left(4x+2y\right)}^{5}$
$1024{x}^{5}+2560{x}^{4}y+2560{x}^{3}{y}^{2}+1280{x}^{2}{y}^{3}+320x{y}^{4}+32{y}^{5}$
${\left(3x-2y\right)}^{4}$
${\left(4x-3y\right)}^{5}$
$1024{x}^{5}-3840{x}^{4}y+5760{x}^{3}{y}^{2}-4320{x}^{2}{y}^{3}+1620x{y}^{4}-243{y}^{5}$
${\left(\frac{1}{x}+3y\right)}^{5}$
${\left({x}^{-1}+2{y}^{-1}\right)}^{4}$
$\frac{1}{{x}^{4}}+\frac{8}{{x}^{3}y}+\frac{24}{{x}^{2}{y}^{2}}+\frac{32}{x{y}^{3}}+\frac{16}{{y}^{4}}$
${\left(\sqrt{x}-\sqrt{y}\right)}^{5}$
For the following exercises, use the Binomial Theorem to write the first three terms of each binomial.
${\left(a+b\right)}^{17}$
${a}^{17}+17{a}^{16}b+136{a}^{15}{b}^{2}$
${\left(x-1\right)}^{18}$
${\left(a-2b\right)}^{15}$
${a}^{15}-30{a}^{14}b+420{a}^{13}{b}^{2}$
${\left(x-2y\right)}^{8}$
${\left(3a+b\right)}^{20}$
$3,486,784,401{a}^{20}+23,245,229,340{a}^{19}b+73,609,892,910{a}^{18}{b}^{2}$
${\left(2a+4b\right)}^{7}$
${\left({x}^{3}-\sqrt{y}\right)}^{8}$
${x}^{24}-8{x}^{21}\sqrt{y}+28{x}^{18}y$
For the following exercises, find the indicated term of each binomial without fully expanding the binomial.
The fourth term of $\text{\hspace{0.17em}}{\left(2x-3y\right)}^{4}$
The fourth term of $\text{\hspace{0.17em}}{\left(3x-2y\right)}^{5}$
$-720{x}^{2}{y}^{3}$
The third term of $\text{\hspace{0.17em}}{\left(6x-3y\right)}^{7}$
The eighth term of $\text{\hspace{0.17em}}{\left(7+5y\right)}^{14}$
$220,812,466,875,000{y}^{7}$
The seventh term of $\text{\hspace{0.17em}}{\left(a+b\right)}^{11}$
The fifth term of $\text{\hspace{0.17em}}{\left(x-y\right)}^{7}$
$35{x}^{3}{y}^{4}$
The tenth term of $\text{\hspace{0.17em}}{\left(x-1\right)}^{12}$
The ninth term of $\text{\hspace{0.17em}}{\left(a-3{b}^{2}\right)}^{11}$
$1,082,565{a}^{3}{b}^{16}$
The fourth term of $\text{\hspace{0.17em}}{\left({x}^{3}-\frac{1}{2}\right)}^{10}$
The eighth term of $\text{\hspace{0.17em}}{\left(\frac{y}{2}+\frac{2}{x}\right)}^{9}$
$\frac{1152{y}^{2}}{{x}^{7}}$
## Graphical
For the following exercises, use the Binomial Theorem to expand the binomial $f\left(x\right)={\left(x+3\right)}^{4}.$ Then find and graph each indicated sum on one set of axes.
Find and graph $\text{\hspace{0.17em}}{f}_{1}\left(x\right),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{1}\left(x\right)\text{\hspace{0.17em}}$ is the first term of the expansion.
Find and graph $\text{\hspace{0.17em}}{f}_{2}\left(x\right),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{2}\left(x\right)\text{\hspace{0.17em}}$ is the sum of the first two terms of the expansion.
${f}_{2}\left(x\right)={x}^{4}+12{x}^{3}$
Find and graph $\text{\hspace{0.17em}}{f}_{3}\left(x\right),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{3}\left(x\right)\text{\hspace{0.17em}}$ is the sum of the first three terms of the expansion.
Find and graph $\text{\hspace{0.17em}}{f}_{4}\left(x\right),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{4}\left(x\right)\text{\hspace{0.17em}}$ is the sum of the first four terms of the expansion.
${f}_{4}\left(x\right)={x}^{4}+12{x}^{3}+54{x}^{2}+108x$
Find and graph $\text{\hspace{0.17em}}{f}_{5}\left(x\right),\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}{f}_{5}\left(x\right)\text{\hspace{0.17em}}$ is the sum of the first five terms of the expansion.
## Extensions
In the expansion of $\text{\hspace{0.17em}}{\left(5x+3y\right)}^{n},\text{\hspace{0.17em}}$ each term has the form successively takes on the value $\text{\hspace{0.17em}}0,1,2,\text{\hspace{0.17em}}...,\text{\hspace{0.17em}}n.$ If $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k\end{array}\right)=\left(\begin{array}{c}7\\ 2\end{array}\right),\text{\hspace{0.17em}}$ what is the corresponding term?
$590,625{x}^{5}{y}^{2}$
In the expansion of $\text{\hspace{0.17em}}{\left(a+b\right)}^{n},\text{\hspace{0.17em}}$ the coefficient of $\text{\hspace{0.17em}}{a}^{n-k}{b}^{k}\text{\hspace{0.17em}}$ is the same as the coefficient of which other term?
Consider the expansion of $\text{\hspace{0.17em}}{\left(x+b\right)}^{40}.\text{\hspace{0.17em}}$ What is the exponent of $b$ in the $k\text{th}$ term?
$k-1$
Find $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k-1\end{array}\right)+\left(\begin{array}{c}n\\ k\end{array}\right)\text{\hspace{0.17em}}$ and write the answer as a binomial coefficient in the form $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k\end{array}\right).\text{\hspace{0.17em}}$ Prove it. Hint: Use the fact that, for any integer $\text{\hspace{0.17em}}p,\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}p\ge 1,\text{\hspace{0.17em}}p!=p\left(p-1\right)!\text{.}$
$\left(\begin{array}{c}n\\ k-1\end{array}\right)+\left(\begin{array}{l}n\\ k\end{array}\right)=\left(\begin{array}{c}n+1\\ k\end{array}\right);\text{\hspace{0.17em}}$ Proof:
$\begin{array}{}\\ \\ \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ k-1\end{array}\right)+\left(\begin{array}{l}n\\ k\end{array}\right)\\ =\frac{n!}{k!\left(n-k\right)!}+\frac{n!}{\left(k-1\right)!\left(n-\left(k-1\right)\right)!}\\ =\frac{n!}{k!\left(n-k\right)!}+\frac{n!}{\left(k-1\right)!\left(n-k+1\right)!}\\ =\frac{\left(n-k+1\right)n!}{\left(n-k+1\right)k!\left(n-k\right)!}+\frac{kn!}{k\left(k-1\right)!\left(n-k+1\right)!}\\ =\frac{\left(n-k+1\right)n!+kn!}{k!\left(n-k+1\right)!}\\ =\frac{\left(n+1\right)n!}{k!\left(\left(n+1\right)-k\right)!}\\ =\frac{\left(n+1\right)!}{k!\left(\left(n+1\right)-k\right)!}\\ =\left(\begin{array}{c}n+1\\ k\end{array}\right)\end{array}$
Which expression cannot be expanded using the Binomial Theorem? Explain.
• $\left({x}^{2}-2x+1\right)$
• ${\left(\sqrt{a}+4\sqrt{a}-5\right)}^{8}$
• ${\left({x}^{3}+2{y}^{2}-z\right)}^{5}$
• ${\left(3{x}^{2}-\sqrt{2{y}^{3}}\right)}^{12}$
The expression $\text{\hspace{0.17em}}{\left({x}^{3}+2{y}^{2}-z\right)}^{5}\text{\hspace{0.17em}}$ cannot be expanded using the Binomial Theorem because it cannot be rewritten as a binomial.
#### Questions & Answers
(1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$)
Ajay Reply
hatdog
Mark
how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching
Shahid Reply
bsc F. y algebra and trigonometry pepper 2
Aditi Reply
given that x= 3/5 find sin 3x
Adamu Reply
4
DB
remove any signs and collect terms of -2(8a-3b-c)
Joeval Reply
-16a+6b+2c
Will
is that a real answer
Joeval
(x2-2x+8)-4(x2-3x+5)
Ayush Reply
sorry
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
(X2-2X+8)-4(X2-3X+5)=0 ?
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
Y
master
X2-2X+8-4X2+12X-20=0 (X2-4X2)+(-2X+12X)+(-20+8)= 0 -3X2+10X-12=0 3X2-10X+12=0 Use quadratic formula To find the answer answer (5±Root11i)/3
master
Soo sorry (5±Root11* i)/3
master
x2-2x+8-4x2+12x-20 x2-4x2-2x+12x+8-20 -3x2+10x-12 now you can find the answer using quadratic
Mukhtar
2x²-6x+1=0
Ife
explain and give four example of hyperbolic function
Lukman Reply
What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?
Racelle Reply
y/y+10
Mr
Find nth derivative of eax sin (bx + c).
Anurag Reply
Find area common to the parabola y2 = 4ax and x2 = 4ay.
Anurag
y2=4ax= y=4ax/2. y=2ax
akash
A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden
Jhovie Reply
to find the length I divide the area by the wide wich means 1125ft/25ft=45
Miranda
thanks
Jhovie
What do you call a relation where each element in the domain is related to only one value in the range by some rules?
Charmaine Reply
A banana.
Yaona
given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
Snalo Reply
what are you up to?
Mark Reply
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
Propessor Reply
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
sita Reply
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17134 | Home » Math Theory » Numbers » Cardinal Numbers
# Cardinal Numbers
## Introduction
In mathematics, we come across different types of sets of numbers that have their own characteristics. One such type of number is the cardinal numbers. So, what are these numbers and how are they used? Let us find out.
## History of Cardinal Numbers
George Cantor in 1874 – 1884 originated set theory and established the concept of the cardinality of a set. According to him, cardinal numbers were infinite sets and they could also be called as transfinite cardinal numbers. Cantor replaced the elementary concept of a number with a derived and abstract one based on sets (for what it is worth, the informal counting conception is closer to his ordinals than to his cardinals, the notions are equivalent in scope in the finite case but diverge dramatically for infinities). Another mathematician, Traski in 1924 proposed that every set has an association with a cardinal number. Two other mathematicians, Dana Scott and A P Morse later gave the definition of cardinal numbers by considering a set and calling the magnitude of a set as cardinality.
## What are Cardinal Numbers?
Cardinal numbers are the numbers that are used as counting numbers. In other words, the numbers that we use for counting are called cardinal numbers. Another name by which cardinal numbers are known as is natural numbers. In fact, cardinal numbers are the generalisation of natural numbers. The word “ Cardinal” means “how many” of anything is existing in a group. Like if we want to count the number of oranges that are present in the basket, we will have to make use of these numbers, such as 1, 2, 3, 4, 5….and so on. Let us understand the cardinal numbers by an example.
Suppose we want to tell how many students are present in the class. The answer could be any number such as 8, 20, 45 and so on. These numbers that tell the count of the number of students present in the class are cardinal numbers. Some other examples of cardinal numbers are –
1. There are 6 clothes in the bag.
2. 3 cars are driving in a lane.
3. Peter has 2 dogs and 1 cat as pets in his house.
In the above three examples, the numbers 6, 3, 2 and 1 are the cardinal numbers. So basically it denotes the quantity of something, irrespective of their order. It defines the measure of the size of a set but does not take account of the order.
### How are Cardinal Numbers Represented in English?
We now know that Cardinal numbers define how many things or people are there, for example, five women standing under a tree. In this sentence, the word “five “ represents the cardinal number “ 5 “.
### How many Cardinal Numbers are there?
Since we know that cardinal numbers are used as counting numbers and counting can be of less number of things or more number of things, therefore, this means that there is no end to the list of cardinal numbers. In other words, the number of cardinal numbers is infinite. But an important point to note here is that although the counting of numbers can go on to infinity, but the digits that are used to count the numbers are fixed. In fact, there are only 10 digits that are used for counting of numbers or we can say to represent cardinal numbers. These 10 digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. It is just the combination of these numbers which make different cardinal numbers such as 36, 43, 75, 89 and so on. The order of these numbers is called counting of numbers.
This means that the cardinal numbers, since being used as counting numbers are used in the order, 1 ,2 , 3 , 4 , 5 , 6 , 7 , 8 , 9. After 9, the numbers are formed in two digits i.e. the cardinal numbers after 9 will be 1 0, 1 1, 1 2, 1 3 , 1 4 , 1 5 , 1 6 , 1 7 , 1 8 , 1 9. It can be clearly seen here that the digit in tens place remains constant while the digit in one place follows the series. Once the series in one’s place reaches the end, i.e. it reaches 9, the digit in ten’s place is changed to a number of higher-order and the pattern is followed. Therefore, after 1 9 we will have 2 0 , 2 1 , 2 2 and so on. After all the numbers in two digits are over, three-digit numbers are followed. The three-digit number starts from 100 and goes on till 999. Three-digit numbers are followed by four-digit numbers and so on. Therefore, there is no limit to the cardinal numbers which means cardinal numbers are infinite.
## Cardinal Numbers From 1 to 100
Below is the list of cardinal numbers from 1 to 100
## Properties of Cardinal Numbers
Following are the properties of cardinal numbers –
1. Cardinal numbers help us to count the number of things or people in or around a place or a group.
2. The collection of all the ordinal numbers can be denoted by the cardinal.
3. Cardinal numbers can be written as words such as one, two, three, etc.
4. Cardinal numbers define the measure of the size of a set but do not take account of the order.
5. Cardinal numbers are also called counting numbers and natural numbers
6. A group of ordinal numbers can be represented by cardinal numbers
7. Cardinal numbers are always used to count and are stated by ‘how many’
8. Fractions and decimals are not cardinal numbers
9. Zero (0) is not a cardinal number, since it means nothing
10. The cardinality of a set represents how many objects or elements are there in the set
## Cardinal Numbers vs Other Numbers
As we know that there are other types of sets as well that possesses their own properties. For example, apart from cardinal numbers, there are two other numbers that are ordinal numbers and nominal numbers. So, how similar are cardinal numbers with ordinal numbers and nominal numbers? Or, rather how different all the three types of numbers are from each other. Let us find out.
Nominal numbers are numbers in numeric form. For instance, 25 is the nominal number for the number “ Twenty Five “. Hence, the nominal numbers and cardinal numbers are different representations of the same numbers. We can also say that the nominal numbers are another type of number, different from cardinals and ordinals, used to name an object or a thing in a set of groups. They are used for the identification of something. It is not for representing the quantity or the position of an object.
Ordinal number describes the position of things. In other words, an ordinal number is a number that denotes the position or place of an object. For example, 1 st, 2 nd, 3 rd etc. are all ordinal numbers. Ordinal numbers are used for the purpose of ranking.
Let us compare cardinal numbers vs ordinal numbers
### Cardinal Numbers vs Ordinal Numbers
Let us consider an example. Suppose in a race there were 10 athletes who participated. The athlete who came first was awarded a gold medal, while a silver medal was given to the candidate who stood second and a bronze medal was given to the athlete who came third. In this case, the number 10 which represents the number of athletes that participated in the race is the cardinal number. On the other hand, the positions first ( 1 st ), second (2 nd ) and third (3 rd ) are ordinal numbers as they represent the position.
Let us summarise the difference between cardinal numbers and ordinal numbers
Let us now compare some cardinal numbers with their nominal numbers and ordinal counterparts
### What is Cardinality of a Set?
The cardinality of a group (set) tells how many objects or terms are there in that set or group. In other words, the cardinality of a finite set is a natural number: the number of elements in the set. For example, the set A = { 1 , 3 , 5 , 7 , 9 } has a cardinality of 3 as there are 3 elements in the set.
## Solved Examples
Example 1 What is the cardinal number of set A = { 3, 5, 7, 9, 10, 11, 4, 19 } ? What is the ordinal number of the number 7 in the set?
Solution We have been given the set A = { 3, 5, 7, 9, 10, 11, 4, 19 }. We need to find
1. The cardinal number of set A
2. The ordinal number of the number 7 in the set
Let us find these one by one.
First, let us count the number of elements in the set A. The total number of elements in the given set is 8.
Therefore, the cardinal number of the set A is 8.
Now, let us check the position of the element 7 in the given set. It is at the third position in the set.
Therefore, the ordinal number of the number 7 in the set A is 3 rd ( Third ).
Example 2 These are the first 10 English letters given in order. Express them in ordinal numbers as well as cardinal numbers where D is the fourth letter at the number 4 in the set.
{ A, B, C, D, E, F, G, H, I, J }
Solution We have been given the set { A, B, C, D, E, F, G, H, I, J }. It has also been given that D is the fourth letter at the number 4 in the set.
Let us write the ordinal and the cardinal numbers for element of the given set. We will have,
# Key Facts and Summary
1. Cardinal numbers are the numbers that are used as counting numbers
2. Cardinal numbers are the generalisation of natural numbers.
3. Cardinal numbers help us to count the number of things or people in or around a place or a group.
4. The collection of all the ordinal numbers can be denoted by the cardinal.
5. Cardinal numbers can be written as words such as one, two, three, etc.
6. Cardinal numbers define the measure of the size of a set but do not take account of the order.
7. Cardinal numbers are also called counting numbers and natural numbers
8. A group of ordinal numbers can be represented by cardinal numbers
9. Cardinal numbers are always used to count and are stated by ‘how many’
10. Fractions and decimals are not cardinal numbers
11. Zero (0) is not a cardinal number, since it means nothing
12. Cardinality of a set represents how many objects or elements are there in the set
13. The nominal numbers and cardinal numbers are different representations of the same numbers.
14. The number of cardinal numbers is infinite.
15. Although the counting of numbers can go on to infinity, but the digits that are used to count the numbers are fixed. In fact, there are only 10 digits that are used for counting of numbers or we can say to represent cardinal numbers.
16. Ordinal number describes the position of things. In other words, an ordinal number is a number that denotes the position or place of an object. Ordinal numbers are used for the purpose of ranking.
17. The cardinality of a group (set) tells how many objects or terms are there in that set or group. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_17135 | ### Home > PC > Chapter 5 > Lesson 5.1.2 > Problem5-22
5-22.
Find a function that has a vertical asymptote at $x=−7$ and a horizontal asymptote at $y=5$.
Your function should be in the form:
$y=\frac{1}{x-h}+k$ |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20208 | # Combinatorics using averages
How many solutions exist for $$x+2y+4z=100$$ in non-negative integers?
The author, Martin Erickson, in his book, Aha! Solutions, published by MAA, gives the following brief solution :
There are $$26$$ choices for $$z$$, namely, all integers from $$0$$ to $$25$$. Among these choices, the average value of $$4z$$ is $$50$$. So, on average, $$x+2y=50$$. In this equation, there are $$26$$ choices for $$y$$, namely, all integers from $$0$$ to $$25$$. The value of $$x$$ is determined by the value of $$y$$. Hence, altogether there are $$26^2=676$$ solutions to the original equation.
Can somebody explain to me why this mindblowing solution works using averages?
• Just because you got the numerical answer, doesn't necessarily mean that the solution must have a correct reasoning. Sep 12 '21 at 19:23
• @CalvinLin Is there no reasonable explanation for given solution? Sep 12 '21 at 19:26
• I didn't say that. There could be. But this requires more of an explanation/details than just what i written. Sep 12 '21 at 19:27
• @CalvinLin Unfortunately the author gives only this much solution to above problem. Sep 12 '21 at 19:28
There are $$26$$ choices for $$z$$ in $$\{0,1,\ldots,25\}$$.
Given $$z$$, we have $$x+2y=100-4z$$, which leaves $$\frac{1}{2}(100-4z)+1=51-2z$$ choices for $$y$$ in $$\{0,1,\ldots,\frac{1}{2}(100-4z)\}$$. Having fixed $$y$$, we must have $$x=100-4z-2y$$.
The total number of solutions is therefore $$\sum_{z=0}^{25} (51-2z) = 1 + 3 + \cdots + 49 + 51.$$ Note that the average value of the addends here is $$26$$. Because the addends form an arithmetic sequence, you can replace the addends with $$26$$ via $$1 + 3 + \cdots + 49 + 51 = 26 + 26 + \cdots + 26 + 26 = 26 \cdot 26.$$ This can be seen by noting that $$1+51=26+26$$ and $$3 + 49 = 26 + 26$$, and so on.
Although the solution in the book is correct, it omits justification of several steps.
• Thank you for your explanation. It seems ultimately it boils down to what are the coefficients of the equation, $1,2,4$ and their relation with the sum , $100$. Sep 13 '21 at 16:15
Consider the cartesian product $$S=\{0,1,\ldots, 50\}\times\{0,1,\ldots, 25\}$$, which represents choosing $$y$$ to be at most $$50$$, and $$z$$ to be at most $$25$$. We call elements of $$S$$ solutions, though they may not actually give solutions to the equation. We say $$(y,z)\in P$$ is valid if $$2y+4z\leq 100$$, since we can then choose $$x$$ to be $$100-2y-4z$$, and invalid otherwise. I'll show a way to match invalid solutions with valid ones. If $$(y,z)$$ is an invalid solution, then $$(50-y, 25-z)$$ is a valid solution. This is because $$\color{red}{2y+4z}+\color{blue}{2(50-y)+4(25-z)}=200$$, so if one of the two colored terms is greater than $$100$$, the other is less than $$100$$. Two invalid solutions are never matched this way, and if two valid solutions are matched this way, it's because $$2y+4z=100$$. There are $$26$$ solutions for which this can occur.
Hence, there are $$51\cdot 26$$ solutions in $$S$$. Of these, $$26$$ of these solutions are valid and are matched with another valid solution. All other $$51\cdot 26-26$$ solutions can be grouped into pairs, consisting of one valid solution, and one invalid one. Hence, the number of valid solutions is $$\frac{51\cdot 26-26}{2}+26=26\cdot 26$$.
• Thanks, you have detailed what Calvin Lin said in one sentence. Sep 13 '21 at 16:13
• @MyMolecules this is distinct from my solution. It's a great solution IMO. However, it doesn't truly convey the "average value of 4z" in the original solution. $\quad$ Kevin, you can modify this slightly to solve for $2y+4 + 2 (51-y)+4(25-z ) = 202$ which allows for a bijection of valid and invalid solutions. Sep 13 '21 at 18:23
Claim: The number of solutions to $$x + 2y = 50-2k$$ plus the number of solutions to $$x + 2y = 50+2k$$ is the constant 52.
We can prove this via counting: The number of solutions to $$x + 2y = K$$ is $$\lfloor \frac{K}{2} \rfloor + 1$$.
IMO the "on average" part is dubious. I don't see this as an "Aha!" solution, unless there's more of an explanation like that above.
• So essentially the author is taking the advantage of coefficients, $1,2,4$? If the coefficients were say $1,2,3$ in the equation, ($3$ being coprime to $100$), then the method using averages would fail? Sep 12 '21 at 19:49
• @MyMolecules The main coefficient we are using is the "1". I believe this extends to solving $x + Ay + Bz = k$ (but should still check through the details). EG For $x+2y+3z = 100$, because we're studying $x+2y = 1, 4, \ldots 97$, the number of solutions to $x+2y=49 \pm 3k$ is $\lfloor (49 - 3k) / 3 \rfloor + \lfloor (49+3k) / 3 \rfloor + 2 = 34$, which is the same as $2\times ( \lfloor (49) / 3 \rfloor + 1 )$. Sep 13 '21 at 18:19
• Thank you. That is great insight! Sep 13 '21 at 18:38
This answer was written to give you another tricky approach. It may help you for expanding your perspective.
Lets use generating functions.
Generating function for $$x$$ is equal to $$\frac{1}{1-x}= x^0 +x + x^2 +x^3+...$$
Generating function for $$2y$$ is equal to $$\frac{1}{1-x^2}= x^0 +x^2 +x^4+..$$
Generating function for $$4z$$ is equal to $$\frac{1}{1-x^4}= x^0 +x^4 +x^8+..$$
Then find the coefficient of $$x^{100}$$ in the expansion of $$\frac{1}{1-x} \times \frac{1}{1-x^2} \times \frac{1}{1-x^4}$$ such that https://www.wolframalpha.com/input/?i=expanded+form+of+%281+%2F+%281-x%29%29%281+%2F+%281-x%5E2%29%29%281+%2F+%281-x%5E4%29%29
So , answer is $$676$$ |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20209 | Question
# Which of the following images is that of a desert?
Solution
## The correct option is A Deserts are large dry areas which have very little vegetation. They are usually covered with sand. The below given image is that of a desert.
Suggest corrections |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20210 | # How do you solve 1/2x + 1/3y = 5 and 1/4x + y = 10 using substitution?
May 26, 2018
$x = 4 , y = 9$
#### Explanation:
From the second equation we get
$y = 10 - \frac{1}{4} x$
plugging this in the first one
$\frac{1}{2} x + \frac{1}{3} \left(10 - \frac{1}{4} x\right) = 5$
$\frac{1}{2} x + \frac{10}{3} - \frac{1}{12} x = 5$
Multplying by $12$
$6 x + 40 - x = 60$
$5 x = 20$
$x = 4$
so
$y = 9$ |
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• CommentRowNumber1.
• CommentAuthorHarry Gindi
• CommentTimeJun 20th 2010
• (edited Jun 20th 2010)
I posted this as a question over at MO, but I figured I’d post it here as well:
## Question:
Let $D:A\to (X\downarrow C)$ be a $\kappa$-good $S$-tree rooted at $X$ for a collection of morphisms $S$ in $C$, where $\kappa$ is a fixed uncountable regular cardinal. Then according to the proof of Lemma A.1.5.8 of Higher Topos Theory by Lurie, for any $\kappa$-small downward-closed $B\subseteq A$, the colimit of the restricted diagram, $\varinjlim D|_B$ is $\kappa$-compact in $(X\downarrow C)$.
Why is this true? (It is stated without proof.)
## Definitions:
For your convenience, here are the definitions:
Recall that an object $X$ in $C$ is called $\kappa$-compact if $h^X (-) := Hom(X,-)$ preserves all $\kappa$-filtered colimits (where $\kappa$-filtered means “< $\kappa$“-filtered, since the terminology is different depending on the source).
Recall that an $S$-tree rooted at $X$ for a collection of morphisms $S$ in $C$ consists of the following data:
• An object $X$ in C (the root)
• A partially ordered set $A$ whose order structure is well-founded (the index)
• A diagram $D:A\to (X\downarrow C)$ such that given any element $\alpha\in A$, the canonical map
$\varinjlim D|_{\{\beta:\beta < \alpha\}}\to D(\alpha)$
is the pushout of some map $U_\alpha\to V_\alpha\in S$.
We say that an $S$-tree is $\kappa$-good if for all of the morphisms $U_\alpha\to V_\alpha$ above, $U_\alpha$ and $V_\alpha$ are $\kappa$-compact, and such that for any $\alpha\in A$, the subset $\{\beta: \beta$ < $\alpha \}\subseteq A$ is $\kappa$-small.
Edit: It’s easy to reduce the proof to showing that $D(\alpha)$ is $\kappa$-compact, since projective limits of diagrams $B\to Set$ are $|Arr(B)|$-accessible (and therefore $\kappa$-accessible since $B$ is $\kappa$-small), we perform the computation for $I$ a $\kappa$-filtered poset, and $F:I\to C$, assuming that $D(\alpha)$ is $\kappa$-compact for all $\alpha\in B$:
$\begin{matrix} \varinjlim_I Hom_C(\varinjlim_B D, F)&\cong& \varinjlim_I\varprojlim_{B^{op}} Hom_C(D,F)\\ &\cong& \varprojlim_{B^{op}} \varinjlim_I Hom_C(D,F)\\ &\cong& \varprojlim_{B^{op}} Hom_C(D,\varinjlim_IF)\\ &\cong& Hom_C(\varinjlim_B D,\varinjlim_IF) \end{matrix}$
Edit 2: I think the above reduction actually won’t work, since it doesn’t use the hypothesis that B is downward-closed.
nForum Edit 1: For some reason, \varinjlim, \varprojlim, and \cdot don’t render. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20212 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 4.5: The Binomial Probability Distribution
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Know the characteristics of a binomial random variable.
• Understand a binomial probability distribution.
• Know the definitions of the mean, the variance, and the standard deviation of a binomial random variable.
• Identify the type of statistical situation to which a binomial distribution can be applied.
• Use a binomial distribution to solve statistical problems.
## Introduction
Many experiments result in responses for which there are only two possible outcomes, such as either 'yes' or 'no', 'pass' or 'fail', 'good' or 'defective', 'male' or 'female', etc. A simple example is the toss of a coin. Say, for example, that we toss the coin five times. In each toss, we will observe either a head, \begin{align*}H\end{align*}, or a tail, \begin{align*}T\end{align*}. We might be interested in the probability distribution of \begin{align*}X\end{align*}, the number of heads observed. In this case, the possible values of \begin{align*}X\end{align*} range from 0 to 5. It is scenarios like this that we will examine in this lesson
### Binomial Experiments
Example: Suppose we select 100 students from a large university campus and ask them whether they are in favor of a certain issue that is being debated on their campus. The students are to answer with either a 'yes' or a 'no'. Here, we are interested in \begin{align*}X\end{align*}, the number of students who favor the issue (a 'yes'). If each student is randomly selected from the total population of the university, and the proportion of students who favor the issue is \begin{align*}p\end{align*}, then the probability that any randomly selected student favors the issue is \begin{align*}p\end{align*}. The probability of a selected student who does not favor the issue is \begin{align*}1 - p\end{align*}. Sampling 100 students in this way is equivalent to tossing a coin 100 times. This experiment is an example of a binomial experiment.
### Characteristics of a Binomial Experiment
• The experiment consists of \begin{align*}n\end{align*} independent, identical trials.
• There are only two possible outcomes on each trial: \begin{align*}S\end{align*} (for success) or \begin{align*}F\end{align*} (for failure).
• The probability of \begin{align*}S\end{align*} remains constant from trial to trial. We will denote it by \begin{align*}p\end{align*}. We will denote the probability of \begin{align*}F\end{align*} by \begin{align*}q\end{align*}. Thus, \begin{align*}q = 1 - p\end{align*}.
• The binomial random variable \begin{align*}X\end{align*} is the number of successes in \begin{align*}n\end{align*} trials.
Example: In the following two examples, decide whether \begin{align*}X\end{align*} is a binomial random variable.
Suppose a university decides to give two scholarships to two students. The pool of applicants is ten students: six males and four females. All ten of the applicants are equally qualified, and the university decides to randomly select two. Let \begin{align*}X\end{align*} be the number of female students who receive the scholarship.
If the first student selected is a female, then the probability that the second student is a female is \begin{align*}\frac{3}{9}\end{align*}. Here we have a conditional probability: the success of choosing a female student on the second trial depends on the outcome of the first trial. Therefore, the trials are not independent, and \begin{align*}X\end{align*} is not a binomial random variable.
A company decides to conduct a survey of customers to see if its new product, a new brand of shampoo, will sell well. The company chooses 100 randomly selected customers and asks them to state their preference among the new shampoo and two other leading shampoos on the market. Let \begin{align*}X\end{align*} be the number of the 100 customers who choose the new brand over the other two.
In this experiment, each customer either states a preference for the new shampoo or does not. The customers’ preferences are independent of each other, and therefore, \begin{align*}X\end{align*} is a binomial random variable.
Let’s examine an actual binomial situation. Suppose we present four people with two cups of coffee (one percolated and one instant) to discover the answer to this question: “If we ask four people which is percolated coffee and none of them can tell the percolated coffee from the instant coffee, what is the probability that two of the four will guess correctly?” We will present each of four people with percolated and instant coffee and ask them to identify the percolated coffee. The outcomes will be recorded by using \begin{align*}C\end{align*} for correctly identifying the percolated coffee and \begin{align*}I\end{align*} for incorrectly identifying it. A list of the 16 possible outcomes, all of which are equally likely if none of the four can tell the difference and are merely guessing, is shown below:
Number Who Correctly Identify Percolated Coffee Outcomes, \begin{align*}C\end{align*} (correct), \begin{align*}I\end{align*} (incorrect) Number of Outcomes
0 \begin{align*}IIII\end{align*} 1
1 \begin{align*}ICII \ IIIC \ IICI \ CIII\end{align*} 4
2 \begin{align*}ICCI \ IICC \ ICIC \ CIIC \ CICI \ CCII\end{align*} 6
3 \begin{align*}CICC \ ICCC \ CCCI \ CCIC\end{align*} 4
4 \begin{align*}CCCC\end{align*} 1
Using the Multiplication Rule for Independent Events, you know that the probability of getting a certain outcome when two people guess correctly, such as \begin{align*}CICI\end{align*}, is \begin{align*}(\frac{1}{2}) \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)=\left(\frac{1}{16}\right)\end{align*}. The table shows six outcomes where two people guessed correctly, so the probability of getting two people who correctly identified the percolated coffee is \begin{align*}\frac{6}{16}\end{align*}. Another way to determine the number of ways that exactly two people out of four people can identify the percolated coffee is simply to count how many ways two people can be selected from four people:
\begin{align*}_4C_2 = \frac{4 !}{2!2!}=\frac{24}{4}=6\end{align*}
In addition, a graphing calculator can also be used to calculate binomial probabilities.
By pressing [2ND][DISTR], you can enter 'binompdf (4,0.5,2)'. This command calculates the binomial probability for \begin{align*}k\end{align*} (in this example, \begin{align*}k = 2\end{align*}) successes out of \begin{align*}n\end{align*} (in this example, \begin{align*}n = 4\end{align*}) trials, when the probability of success on any one trial is \begin{align*}p\end{align*} (in this example, \begin{align*}p = 0.5\end{align*}).
A binomial experiment is a probability experiment that satisfies the following conditions:
• Each trial can have only two outcomes\begin{align*}-\end{align*}one known as a success, and the other known as a failure.
• There must be a fixed number, \begin{align*}n\end{align*}, of trials.
• The outcomes of the trials must be independent of each other. The probability of each success doesn’t change, regardless of what occurred previously.
• The probability, \begin{align*}p\end{align*}, of a success must remain the same for each trial.
The distribution of the random variable \begin{align*}X\end{align*}, where \begin{align*}x\end{align*} is the number of successes, is called a binomial probability distribution. The probability that you get exactly \begin{align*}x = k\end{align*} successes is as follows:
\begin{align*}P(x=k) = \binom{n}{k} p^k (1-p)^{n-k}\end{align*}
where:
\begin{align*}\binom{n}{k} = \frac{n!}{k!(n-k)!}\end{align*}
Let’s return to the coffee experiment and look at the distribution of \begin{align*}X\end{align*} (correct guesses):
\begin{align*}k\end{align*} \begin{align*}P(x=k)\end{align*}
0 \begin{align*}\frac{1}{16}\end{align*}
1 \begin{align*}\frac{4}{16}\end{align*}
2 \begin{align*}\frac{6}{16}\end{align*}
3 \begin{align*}\frac{4}{16}\end{align*}
4 \begin{align*}\frac{1}{16}\end{align*}
The expected value for the above distribution can be calculated as follows:
\begin{align*}E(x) &= (0)\left(\frac{1}{16}\right) + (1)\left(\frac{4}{16}\right)+(2)\left(\frac{6}{16}\right) +(3)\left(\frac{4}{16}\right) + (4)\left(\frac{1}{16}\right)\\ E(x) &= 2\end{align*}
In other words, you would expect half of the four to guess correctly when given two equally-likely choices. \begin{align*}E(x)\end{align*} can be written as \begin{align*}(4)\left(\frac{1}{2}\right)\end{align*}, which is equivalent to \begin{align*}np\end{align*}.
For a random variable \begin{align*}X\end{align*} having a binomial distribution with \begin{align*}n\end{align*} trials and a probability of success of \begin{align*}p\end{align*}, the expected value (mean) and standard deviation for the distribution can be determined by the following formulas:
\begin{align*}E(x)=\mu_X=np\end{align*} and \begin{align*}\sigma_X=\sqrt{np(1-p)}\end{align*}
To apply the binomial formula to a specific problem, it is useful to have an organized strategy. Such a strategy is presented in the following steps:
• Identify a success.
• Determine \begin{align*}p\end{align*}, the probability of success.
• Determine \begin{align*}n\end{align*}, the number of experiments or trials.
• Use the binomial formula to write the probability distribution of \begin{align*}X\end{align*}.
Example: According to a study conducted by a telephone company, the probability is 25% that a randomly selected phone call will last longer than the mean value of 3.8 minutes. What is the probability that out of three randomly selected calls:
a. Exactly two last longer than 3.8 minutes?
b. None last longer than 3.8 minutes?
Using the first three steps listed above:
• A success is any call that is longer than 3.8 minutes.
• The probability of success is \begin{align*}p = 0.25\end{align*}.
• The number of trials is \begin{align*}n = 3\end{align*}.
Thus, we can now use the binomial probability formula:
\begin{align*}p(x)=\binom{n}{x} p^x (1-p)^{n-x}\end{align*}
Substituting, we have: \begin{align*}p(x)=\binom{3}{x} (0.25)^x (1-0.25)^{3-x}\end{align*}
a. For \begin{align*}x = 2\end{align*}:
\begin{align*}p(x)&=\binom{3}{2} (0.25)^2 (1-0.25)^{3-2}\\ &= (3)(0.25)^2 (1-0.25)^1\\ &= 0.14\end{align*}
Thus, the probability is 0.14 that exactly two out of three randomly selected calls will last longer than 3.8 minutes.
b. Here, \begin{align*}x = 0\end{align*}. Again, we use the binomial probability formula:
\begin{align*}p(x=0) &= \binom{3}{0} (0.25)^0 (1-0.25)^{3-0}\\ &= \frac{3!}{0!(3-0)!}(0.25)^0(0.75)^3\\ &= 0.422\end{align*}
Thus, the probability is 0.422 that none of the three randomly selected calls will last longer than 3.8 minutes.
Example: A car dealer knows from past experience that he can make a sale to 20% of the customers who he interacts with. What is the probability that, in five randomly selected interactions, he will make a sale to:
a. Exactly three customers?
b. At most one customer?
c. At least one customer?
Also, determine the probability distribution for the number of sales.
A success here is making a sale to a customer. The probability that the car dealer makes a sale to any customer is \begin{align*}p = 0.20\end{align*}, and the number of trials is \begin{align*}n = 5\end{align*}. Therefore, the binomial probability formula for this case is:
\begin{align*}p(x)=\binom{5}{x} (0.2)^x(0.8)^{5-x}\end{align*}
a. Here we want the probability of exactly 3 sales, so \begin{align*}x =3\end{align*}.
\begin{align*}p(x)=\binom{5}{3} (0.2)^3 (0.8)^{5-3}=0.051\end{align*}
This means that the probability that the car dealer makes exactly three sales in five attempts is 0.051.
b. The probability that the car dealer makes a sale to at most one customer can be calculated as follows:
\begin{align*}p(x \le 1) &= p(0)+p(1)\\ &= \binom{5}{0} (0.2)^0 (0.8)^{5-0} + \binom{5}{1} (0.2)^1 (0.8)^{5-1}\\ &= 0.328 + 0.410 = 0.738\end{align*}
c. The probability that the car dealer makes at least one sale is the sum of the probabilities of him making 1, 2, 3, 4, or 5 sales, as is shown below:
\begin{align*}p(x \ge 1)=p(1)+p(2)+p(3)+p(4)+p(5)\end{align*}
We can now apply the binomial probability formula to calculate the five probabilities. However, we can save time by calculating the complement of the probability we're looking for and subtracting it from 1 as follows:
\begin{align*}p(x \ge 1) &= 1-p(x < 1) = 1-p(x = 0)\\ 1- p(0) &= 1- \binom{5}{0} (0.2)^0 (0.8)^{5-0}\\ &= 1-0.328 = 0.672\end{align*}
This tells us that the salesperson has a probability of 0.672 of making at least one sale in five attempts.
We are also asked to determine the probability distribution for the number of sales, \begin{align*}X\end{align*}, in five attempts. Therefore, we need to compute \begin{align*}p(x)\end{align*} for \begin{align*}x = 1, 2, 3, 4\end{align*}, and 5. We can use the binomial probability formula for each value of \begin{align*}X\end{align*}. The table below shows the probabilities.
\begin{align*}x\end{align*} \begin{align*}p(x)\end{align*}
0 0.328
1 0.410
2 0.205
3 0.051
4 0.006
5 0.00032
Figure: The probability distribution for the number of sales.
Example: A poll of twenty voters is taken to determine the number in favor of a certain candidate for mayor. Suppose that 60% of all the city’s voters favor this candidate.
a. Find the mean and the standard deviation of \begin{align*}X\end{align*}.
b. Find the probability of \begin{align*}x \le 10\end{align*}.
c. Find the probability of \begin{align*}x>12\end{align*}.
d. Find the probability of \begin{align*}x=11\end{align*}.
a. Since the sample of twenty was randomly selected, it is likely that \begin{align*}X\end{align*} is a binomial random variable. Of course, \begin{align*}X\end{align*} here would be the number of the twenty who favor the candidate. The probability of success is 0.60, the percentage of the total voters who favor the candidate. Therefore, the mean and the standard deviation can be calculated as shown:
\begin{align*}\mu &= np = (20)(0.6)=12\\ \sigma^2 &= np (1-p)=(20)(0.6)(0.4)=4.8\\ \sigma &=\sqrt{4.8}=2.2\end{align*}
b. To calculate the probability that 10 or fewer of the voters favor the candidate, it's possible to add the probabilities that 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10 of the voters favor the candidate as follows:
\begin{align*}p(x \le 10) = p(0)+p(1)+p(2)+ \ldots + p(10)\end{align*}
or
\begin{align*}p(x \le 10) = \sum^{10}_{x=0} p(x) = \sum^{10}_{x=0} \binom{20}{x} (0.6)^x (0.4)^{20-x}\end{align*}
As you can see, this would be a very tedious calculation, and it is best to resort to your calculator. See the technology note at the end of the section for more information.
c. To find the probability that \begin{align*}x > 12\end{align*}, it's possible to add the probabilities that 13, 14, 15, 16, 17, 18, 19, or 20 of the voters favor the candidate as shown:
\begin{align*}p(x > 12)=p(13)+p(14)+ \ldots + p(20)=\sum^{20}_{x=13} p(x)\end{align*}
Alternatively, using the Complement Rule, \begin{align*}p(x > 12)=1-p(x \le 12)\end{align*}.
Using a calculator (see the technology note below) with \begin{align*}n = 20, p = 0.6,\end{align*} and \begin{align*}k = 12\end{align*}, we get a probability of 0.584 that \begin{align*}x \le 12\end{align*}. Thus, \begin{align*}p(x > 12) = 1 - 0.584 = 0.416.\end{align*}
d. To find the probability that exactly 11 voters favor the candidate, it's possible to subtract the probability that less than or equal to 10 voters favor the candidate from the probability that less than or equal to 11 voters favor the candidate. These probabilities can be found using a calculator. Thus, the probability that exactly 11 voters favor the candidate can be calculated as follows:
\begin{align*}p(x=11)=p(x \le 11) - p(x \le 10)=0.404 - 0.245=0.159\end{align*}
A graphing calculator will now be used to graph and compare different versions of a binomial distribution. Each binomial distribution will be entered into two lists and then displayed as a histogram. First, we will use the calculator to generate a sequence of integers, and next, we will use it to generate a corresponding list of binomial probabilities.
To generate a sequence of integers, press [2ND][LIST], go to OPS, select '5:seq', enter '(X, X, 0, \begin{align*}n\end{align*}, 1)', where \begin{align*}n\end{align*} is the number of independent binomial trials, and press [STO][2ND][L1].
To enter the binomial probabilities associated with this sequence of integers, press [STAT] and select '1:EDIT'.
Clear out L2 and position the cursor on the L2 list name.
Press [2ND][DISTR] to bring up the list of distributions.
Select 'A:binompdf(' and enter '\begin{align*}n, p)\end{align*}', where \begin{align*}n\end{align*} is the number of independent binomial trials and \begin{align*}p\end{align*} is the probability of success.
To graph the histogram, make sure your window is set correctly, press [2ND][STAT PLOT], turn a plot on, select the histogram plot, choose L1 for Xlist and L2 for Freq, and press [GRAPH]. This will display the binomial histogram.
Horizontally, the following are examples of binomial distributions where \begin{align*}n\end{align*} increases and \begin{align*}p\end{align*} remains constant. Vertically, the examples display the results where \begin{align*}n\end{align*} remains fixed and \begin{align*}p\end{align*} increases.
\begin{align*}n=5 \ \text{and} \ p=0.1 && n=10 \ \text{and} \ p=0.1 && n=20 \ \text{and} \ p=0.1\end{align*}
For a small value of \begin{align*}p\end{align*}, the binomial distributions are skewed toward the higher values of \begin{align*}X\end{align*}. As \begin{align*}n\end{align*} increases, the skewness decreases and the distributions gradually move toward being more normal.
\begin{align*}n=5 \ \text{and} \ p=0.5 && n=10 \ \text{and} \ p=0.5 && n=20 \ \text{and} \ p=0.5\end{align*}
As \begin{align*}p\end{align*} increases to 0.5, the skewness disappears and the distributions achieve perfect symmetry. The symmetrical, mound-shaped distribution remains the same for all values of \begin{align*}n\end{align*}.
\begin{align*}n=5 \ \text{and} \ p=0.75 && n=10 \ \text{and} \ p=0.75 && n=20 \ \text{and} \ p=0.75\end{align*}
For a larger value of \begin{align*}p\end{align*}, the binomial distributions are skewed toward the lower values of \begin{align*}X\end{align*}. As \begin{align*}n\end{align*} increases, the skewness decreases and the distributions gradually move toward being more normal.
Because \begin{align*}E(x)=np=\mu_X\end{align*}, the expected value increases with both \begin{align*}n\end{align*} and \begin{align*}p\end{align*}. As \begin{align*}n\end{align*} increases, so does the standard deviation, but for a fixed value of \begin{align*}n\end{align*}, the standard deviation is largest around \begin{align*}p = 0.5\end{align*} and reduces as \begin{align*}p\end{align*} approaches 0 or 1.
Technology Note: Calculating Binomial Probabilities on the TI-83/84 Calculator
Press [2ND][DIST] and scroll down to 'A:binompdf('. Press [ENTER] to place 'binompdf(' on your home screen. Type values of \begin{align*}n, p\end{align*}, and \begin{align*}k\end{align*}, separated by commas, and press [ENTER].
Use the 'binomcdf(' command to calculate the probability of at most \begin{align*}x\end{align*} successes. The format is 'binomcdf\begin{align*}(n, p, k)\end{align*}' to find the probability that \begin{align*}x \le k\end{align*}. (Note: It is not necessary to close the parentheses.)
Technology Note: Using Excel
In a cell, enter the function =binomdist(\begin{align*}x,n,p\end{align*},false). Press [ENTER], and the probability of \begin{align*}x\end{align*} successes will appear in the cell.
For the probability of at least \begin{align*}x\end{align*} successes, replace 'false' with 'true'.
## Lesson Summary
Characteristics of a Binomial Experiment:
• A binomial experiment consists of \begin{align*}n\end{align*} identical trials.
• There are only two possible outcomes on each trial: \begin{align*}S\end{align*} (for success) or \begin{align*}F\end{align*} (for failure).
• The probability of \begin{align*}S\end{align*} remains constant from trial to trial. We denote it by \begin{align*}p\end{align*}. We denote the probability of \begin{align*}F\end{align*} by \begin{align*}q\end{align*}. Thus, \begin{align*}q = 1 - p\end{align*}.
• The trials are independent of each other.
• The binomial random variable \begin{align*}X\end{align*} is the number of successes in \begin{align*}n\end{align*} trials.
The binomial probability distribution is: \begin{align*}p(x)=\binom{n}{x} p^x (1-p)^{n-x}=\binom{n}{x} p^x q^{n-x}\end{align*}.
For a binomial random variable, the mean is \begin{align*}\mu = np\end{align*}.
The variance is \begin{align*}\sigma^2 = npq = np(1-p)\end{align*}.
The standard deviation is \begin{align*}\sigma=\sqrt{npq}=\sqrt{np(1-p)}\end{align*}.
On the Web
http://tinyurl.com/268m56r Simulation of a binomial experiment. Explore what happens as you increase the number of trials.
http://tinyurl.com/299hsjo Explore a binomial distribution as you change \begin{align*}n\end{align*} and \begin{align*}p\end{align*}.
For an explanation of binomial distribution and notation used for it (4.0)(7.0), see ExamSolutions, A-Level Statistics: Binomial Distribution (Introduction) (10:30).
For an explanation on using tree diagrams and the formula for finding binomial probabilities (4.0)(7.0), see ExamSolutions, A-Level Statistics: Binomial Distribution (Formula) (14:19).
For an explanation of using the binomial probability distribution to find probabilities (4.0), see patrickJMT, The Binomial Distribution and Binomial Probability Function (6:45).
## Review Questions
1. Suppose \begin{align*}X\end{align*} is a binomial random variable with \begin{align*}n = 4\end{align*} and \begin{align*}p = 0.2\end{align*}. Calculate \begin{align*}p(x)\end{align*} for each of the following values of \begin{align*}X\end{align*}: \begin{align*}0, 1, 2, 3, 4\end{align*}. Give the probability distribution in tabular form.
2. Suppose \begin{align*}X\end{align*} is a binomial random variable with \begin{align*}n = 5\end{align*} and \begin{align*}p = 0.2\end{align*}. Display \begin{align*}p(x)\end{align*} in tabular form. Compute the mean and the variance of \begin{align*}X\end{align*}.
3. Over the years, a medical researcher has found that one out of every ten diabetic patients receiving insulin develops antibodies against the hormone, thus, requiring a more costly form of medication.
1. Find the probability that in the next five patients the researcher treats, none will develop antibodies against insulin.
2. Find the probability that at least one will develop antibodies.
4. According to the Canadian census of 2006, the median annual family income for families in Nova Scotia is $56,400. [Source: Stats Canada. www.statcan.ca ] Consider a random sample of 24 Nova Scotia households. 1. What is the expected number of households with annual incomes less than$56,400?
2. What is the standard deviation of households with incomes less than $56,400? 3. What is the probability of getting at least 18 out of the 24 households with annual incomes under$56,400?
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20213 | ## Second derivative of an integral
Good day,
I don't understand the following:
$$\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\phi''(t)$$
All I know is:
$$\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\frac{d^{2}}{dt^{2}}\int_{0}^{t}t \cdot \phi (\epsilon)d\epsilon-\frac{d^{2}}{dt^{2}}\int_{0}^{t}\epsilon \cdot \phi (\epsilon)d\epsilon$$
Is it allowed to say:
$$\frac{d^{2}}{dt^{2}}\int_{0}^{t}t \cdot \phi (\epsilon)d\epsilon=\frac{d^{2}}{dt^{2}}\cdot t\int_{0}^{t}\phi (\epsilon)d\epsilon$$?
And if so, why is this correct? This is only correct when $$t\neq f(\epsilon )$$, right? But I am not sure whether this is the case...
Thank you in advance!
PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
Recognitions: Homework Help The result (if correct; I haven't checked) will follow from the Liebniz rule: $$\frac{d}{dt}\int_{\alpha(t)}^{\beta(t)} du~f(u,t) = \int_{\alpha(t)}^{\beta(t)}du~ \frac{\partial}{\partial t}f(u,t) + f(t,t)\frac{d\beta}{dt} - f(t,t)\frac{d\alpha}{dt}$$ Then you have to do the derivative again, so you would have to use the Liebniz rule on the integral term again.
Quote by Pietair Good day, I don't understand the following: $$\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\phi''(t)$$
I think it's
$$\frac{d^{2}}{dt^{2}}\int_{0}^{t}(t-\epsilon )\phi (\epsilon)d\epsilon=\phi(t)$$
Similar discussions for: Second derivative of an integral Thread Forum Replies Calculus & Beyond Homework 5 Calculus & Beyond Homework 7 Calculus & Beyond Homework 6 Calculus 4 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20214 | ## Ramanujan Misterious PI formula..
Is there any mathematical explanation to the incredible fast converging formula by Ramanujan?:
$$\frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}=\frac{1}{\pi}$$
or simply "ocurred to him" and put it on a paper.
PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
Recognitions: Homework Help There most likely is a mathematical explanation to the series, but from arguments that are far more advanced than my knowledge. There is however a very small chance it just luckily occurred to him, just as this interesting approximation did (he got it in a dream apparently) : $$\sqrt[4]{\frac{2143}{22}}$$ Thats accurate to 9 digits, and came from a dream with no mathematical basis, so obviously Ramanujan was extremely proficient in his numeracy. I can only offer 2 ideas : The first is the following expression for pi, which looks like it may be somehow related to the series and had been transformed : $$\frac{\sqrt2}2 \cdot \frac{\sqrt{2+\sqrt2}}2 \cdot \frac{\sqrt{2+\sqrt{2+\sqrt2}}}2 \cdot \cdots = \frac2\pi$$ The 2nd idea is to send an email to the Chudnovsky brothers, because I know that the series you ask about is in fact the basis for this faster series: $$\frac{1}{\pi} = 12 \sum^\infty_{k=0} \frac{(-1)^k (6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 640320^{3k + 3/2}}$$ Maybe they can help you.
Recognitions: Homework Help O just in case there was any confusion over my last part of the post, the Chudnovsky brothers discovered that series. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20215 | # High School Math : How to find the volume of a prism
## Example Questions
### Example Question #1 : How To Find The Volume Of A Prism
Find the volume of the following triangular prism.
Explanation:
The formula for the volume of a triangular prism is:
Where is the length of the triangle, is the width of the triangle, and is the height of the prism
Plugging in our values, we get:
### Example Question #2 : How To Find The Volume Of A Prism
Find the volume of the following triangular prism.
Explanation:
The formula for the volume of a triangular prism is:
Where is the length of the base, is the width of the base, and is the height of the prism
Plugging in our values, we get:
### Example Question #3 : How To Find The Volume Of A Prism
Find the volume of the following triangular prism:
Explanation:
The formula for the volume of an equilateral, triangular prism is:
Where is the length of the triangle side and is the length of the height.
Plugging in our values, we get:
### Example Question #4 : How To Find The Volume Of A Prism
What is the volume?
Explanation:
The volume is calculated using the equation:
### Example Question #91 : Prisms
A rectangular box has two sides with the following lengths:
and
If it possesses a volume of , what is the area of its largest side?
28
49
12
16
21
28
Explanation:
The volume of a rectangular prism is found using the following formula:
If we substitute our known values, then we can solve for the missing side.
Divide both sides of the equation by 12.
We now know that the missing length equals 7 centimeters.
This means that the box can have sides with the following dimensions: 3cm by 4cm; 7cm by 3cm; or 7cm by 4cm. The greatest area of one side belongs to the one that is 7cm by 4cm. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20216 | Hot Topics
# IL state football champs to be crowned this weekend
#### Voxitatis has applied a modified version of its mathematical model to predict all eight football titles in Illinois, but, like last year, we are waiting until after the Class 8A game is played to publish those predictions.
Our reason for doing this is that the overwhelming majority of hits on that page, which come from Google searches, involve gambling or betting on the games, and we don’t encourage this type of activity for high school athletics.
We’re not going to change our answers based on the actual outcomes—the computer has already printed the story. We just need to delay the publication of our predictions.
The games take place today and tomorrow at Huskie Stadium on the campus of Northern Illinois University. Classes 1A through 4A play Friday, and classes 5A through 8A play Saturday.
1A: Lena-Winslow vs Tuscola
2A: Gibson City-Melvin-Sibley vs Maroa-Forsyth
3A: Immaculate Conception vs Pleasant Plains
4A: Morris vs Rochester
5A: Phillips vs Dunlap
6A: Prairie Ridge vs Nazareth
7A: Batavia vs Lake Zurich
8A: Lincoln-Way East vs Loyola
### Our basic mathematical model
What we did in the past was to compute a statistic that would tell us about how good a team was at winning games, which comes from scoring more points than your opponents in the games. The general form of the model we used is as follows:
$\sigma_t = (Af_t - Bd_o) - (Cd_t - Df_o)$
where σ represents a statistic that is directly proportional to a team’s overall strength. In our model, f represents the offense (average number of points scored in every game up to that point in the season), d represents the defense (average number of points given up in every game so far), and the subscripts have t for the team for which the statistic is being computed and o for that team’s opponents.
In other words, a team that gives up as many points as they make will have a σ somewhere close to 0. We adjust the average points scored by the team by subtracting the average number of points given up by that team’s opponents. What this does is something akin to giving teams that play tough opponents a higher rating, because a lower number of points will be subtracted if the team faced all tough opponents than if the team faced only very weak opponents all season.
Likewise, the number of points given up is adjusted by subtracting the average number of points the team’s opponents earned across all the games that season. This effectively means that if the team played opponents who didn’t score very many points anyway, it’s not such a big deal if this team held them to a shutout.
Over the years, we have played with the coefficients on each of the terms, but A always equals C and B and D have always been less than A and C, respectively. Two years ago, for example, we used:
• A = C = 1.00
• B = 0.50
• D = 0.25
Using those values, we predicted the winners in seven of the eight games for a “batting average” of 0.875. This year, we have put considerably more computing power behind the predictions, and the coefficients are going to be adjusted based on the results in each team’s semifinal games. That is, whichever coefficients best predict the margin of victory in the team’s semifinal game—the same values of A, B, C, and D have to be used for the team and its semifinal opponent—those are the coefficients we’ll use to get the σ statistic for the team in the final game.
Our prediction is simple: Whichever team has the higher σ will be predicted to win the final game. The greater the difference between the σ statistic for the two teams, the more likely our computer model predicts that team will win the game. This is not a “spread” or anything related to a point differential: it is a probability statistic used for comparing two teams playing against each other in a single game at a certain point of the season. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20217 | <meta http-equiv="refresh" content="1; url=/nojavascript/"> Whole Number Exponents ( Read ) | Arithmetic | CK-12 Foundation
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# Whole Number Exponents
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Practice Whole Number Exponents
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Whole Number Exponents
Have you ever been hiking with a group of other teens?
On the first day of Teen Adventure, Kelly thought they would be hiking, but when the group assembled at the Lafayette Place Campground she realized that there was a lot to do before they could begin hiking. First, the leaders organized each group into 10 hikers with 2 leaders each. Then the leaders split off with their groups to do some training.
There was a lot to learn. The leaders of Kelly’s group, Scott and Laurel began by having the hikers introduce themselves and share a little about their hiking experience. They learned that the group would be taking it easy the first week while everyone got into shape and had a chance to get to know each other. The hiking would get more strenuous as the time went on.
After introductions, Scott and Laurel gave the campers two tents. Since there were five boys and five girls in each group, the team would need two tents. There would be times when they would be sleeping in cabins, but there also would be times where tents would be necessary.
Their first task was to set up the tent and figure out the square footage of the floor. The girls and boys were each given a Kelty Trail Dome 6.
Kelly and the other girls took one tent and began to take it out of its package. They were so excited that they did not pay attention and almost lost the directions. Luckily, Kara saw this and caught them before the wind did. Kelly looked at the directions. The tent was sized to sleep six so it would be perfect for them and one of the leaders.
Dimensions of the floor $= 120^2$ inches
Kelly and Jessica looked at the dimensions. Who would have thought that they would be solving math problems when hiking! Jessica took out a piece of paper and began working on the problem.
$120^2$ inches is a measurement that has an exponent. To figure out the dimensions of the floor of the tent you will need to know how to work with exponents. In this Concept, you will learn all about exponents. By the end, you will know how to figure out the area of floor of the tent.
### Guidance
Sometimes, we have to multiply the same number several times. We can say that we are multiplying the number by itself in this case.
$4 \times 4 \times 4$ is 4 multiplied by itself three times.
When we have a situation like this, it is helpful to use a little number to show how many times to multiply the number by itself. That little number is called an exponent.
If we were going to write $4 \times 4 \times 4$ with an exponent, we would write $4^3$ . This lesson is all about exponents. By the end of it, you will how and when to use them and how helpful this shortcut is for multiplication.
Using exponents has an even fancier name too. We can say that we use exponential notation when we express multiplication in terms of exponents.
We can use exponential notation to write an expanded multiplication problem into a form with an exponent, we write $4 \times 4 \times 4$ with an exponent $= 4^3$
We can work the other way around too. We can write a number with an exponent as a long multiplication problem and this is called expanded form.
The base is the number being multiplied by itself in this case the base is 4.
The exponent tells how many times to multiply the base by itself in this case, it is a 3.
Using an exponent can also be called “raising to a power.” The exponent represents the power.
Here $4^3$ would be read as “Four to the third power.”
Write the following in exponential notation: $6 \times 6 \times 6 \times 6$
Exponential Notation means to write this as a base with an exponent.
Six times itself four times $= 6^4$
Write the following in expanded form: $5^3$
Expanded form means to write this out as a multiplication problem.
$5 \times 5 \times 5$
We can also evaluate expressions with variables.
$4^3$
Our first step is to write it out into expanded form.
$4 \times 4 \times 4$
Now multiply.
$4 \times 4 = 16 \times 4 = 64$
Now it's time for you to try a few on your own.
#### Example A
Write the following in exponential form: $3 \times 3 \times 3 \times 3 \times 3$
Solution: $3^5$
#### Example B
Write the following in expanded form and evaluate the expression: $6^3$
Solution: $6 \times 6 \times 6$
#### Example C
Evaluate: $4^3-5^2$
Solution: 39
Now let's go back to the tent dilemma from the beginning of the Concept.
Kelly and the other girls took one tent and began to take it out of its package. They were so excited that they did not pay attention and almost lost the directions. Luckily, Kara saw this and caught them before the wind did. Kelly looked at the directions. The tent was sized to sleep six so it would be perfect for them and one of the leaders.
Dimensions of the floor $= 120^2$ inches
Kelly and Jessica looked at the dimensions. Who would have thought that they would be solving math problems when hiking! Jessica took out a piece of paper and began working on the problem.
First, notice that the measurement is in inches not feet. Our final answer needs to be in square footage, so after figuring out these dimensions, the girls will need to convert the measurement to feet.
The area of a square is one place where we use exponents all the time. The square has side $x$ side, so we can write $s^2$ to find the area of a square. Since the tent floor is square, the dimensions have been written in square inches.
$120^2$ inches
To start, the girls need to multiply this out.
$120 \times 120$
Next, they can covert each inch dimension to feet.
There are 12 inches in 1 foot, so we divide each measurement by 12. 120 divided by $12 = 10$ .
Now we multiply to find the area in square feet.
$10 \ ft \times 10 \ ft = 100$ square feet
Exponents are very useful when working with area!
### Vocabulary
Exponent
a little number that tells you how many times to multiply the base by itself.
Base
the big number in a variable expression with an exponent.
Exponential Notation
writing long multiplication using a base and an exponent
Expanded Form
taking a base and an exponent and writing it out as a long multiplication problem.
### Guided Practice
Here is one for you to try on your own.
$2^3+4^2$
To evaluate this expression write it out in expanded form.
$(2)(2)(2) + (4)(4)$
Now multiply each part of the expression.
$& 8 + 16\\& 24$
### Practice
Directions: Name the base and exponent in the following examples. Then write each in expanded form.
1. $4^5$
2. $3^2$
3. $5^8$
4. $4^3$
5. $6^3$
6. $2^5$
7. $1^{10}$
8. $2^{5}$
9. $3^{4}$
10. $5^{2}$
11. $4^{4}$
12. $8^{10}$
13. $9^{3}$
14. $12^{2}$
15. $13^{3}$ |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20218 | ## Chapter 14 # 107
$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$
Tim Foster 2A
Posts: 73
Joined: Fri Sep 29, 2017 7:07 am
### Chapter 14 # 107
Why do we use E(naught)=lnK for this problem, instead of E=E(naught)-lnQ in which Enaught would be zero (PH7)
I paraphrased the equations because I'm lazy |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20219 | A calculus of the absurd
10.4 Planes
This is a Further Maths topic.
TODO
In the meantime, http://www.mit.edu/~hlb/1802/pdf/MIT18_02SC_notes_6.pdf was pretty helpful. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20220 | An ideal in a Noetherian ring is nilpotent if each element of the ideal is nilpotent.
Lemma 10.32.5. Let $R$ be a Noetherian ring. Let $I, J$ be ideals of $R$. Suppose $J \subset \sqrt{I}$. Then $J^ n \subset I$ for some $n$. In particular, in a Noetherian ring the notions of “locally nilpotent ideal” and “nilpotent ideal” coincide.
Proof. Say $J = (f_1, \ldots , f_ s)$. By assumption $f_ i^{d_ i} \in I$. Take $n = d_1 + d_2 + \ldots + d_ s + 1$. $\square$
Comment #851 by Bhargav Bhatt on
Suggested slogan: An ideal in a noetherian ring is nilpotent if each element of the ideal is nilpotent.
There are also:
• 2 comment(s) on Section 10.32: Locally nilpotent ideals
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20221 | Of Two Time Indices
In the appendix to a paper I am currently co-authoring, I recently wrote the following within a parenthetical excursus:
When talking of dynamical systems, our probability assignments really carry two time indices: one for the time our betting odds are chosen, and the other for the time the bet concerns.
A parenthesis in an appendix is already a pretty superfluous thing. Treating this as the jumping-off point for further discussion merits the degree of obscurity which only a lengthy post on a low-traffic blog can afford.
The Book of Numbers Erratum
Among the fun math books I have on my overburdened bookshelves is John Conway and Richard Guy’s fascinating volume, The Book of Numbers (1996). In following up one of the topics discussed in its very last chapter, I discovered that Conway and Guy had made a bibliographic error, which in the interests of scholarship should be publicly noted. While I could give the correction in a line and be done with it, the topic and its background are curious enough to merit a few paragraphs. To wit:
Anybody who has had a brush with calculus is familiar with taking derivatives of a function. The derivative of a function is a whole new function which gives the rate of change of the original; plug a function into the machine, and out comes a new one, which is also just as “complicated” as the one you started with. If your initial function was something like
$$f(x) = x^2,$$
which maps each real number to a real number, then the derivative will be something like
$$f^\prime(x) = 2x,$$
which also maps elements of $$\mathbb{R}$$ to elements of $$\mathbb{R}$$. That’s a whole lot of mappings! If we were so inclined, we could also represent the “growth rate” of functions by numbers, instead of by functions. The operation of “finding the growth rate” would then be a functional, mapping functions to numbers — though the sort of numbers we find ourselves using are a little out of the ordinary.
Continue reading The Book of Numbers Erratum
Categories and Surreals: Disordered Thoughts
I’ve been going through André Joyal’s category-theoretic construction of the surreal numbers, futzing around to see how restricting the construction to yield only the real numbers affects things. (See my earlier remarks on this and/or Mark Chu-Carroll’s discussion for background information.) If I were an actual mathematician, I’d probably be done by now. Instead, I have a headful of half-baked notions, which I’d like to spill out into the Intertubes (mixing my metaphors as necessary).
Why would a physics person even care about the surreal numbers? Well, ultimately, my friends and I were going through Baez and Dolan’s “From Finite Sets to Feynman Diagrams” (2000), which touches upon the issue that subtraction can be a real pain to interpret categorically. If you interpret the natural numbers as the decategorification of FinSet, then addition is easy: you’re just talking about a coproduct. But subtraction and negation — oh, la!
Continue reading Categories and Surreals: Disordered Thoughts
No seminar today (Sorry!)
Due to an illness among the teaching staff, today’s statistical-mechanics session will be postponed until later in the week. In the meantime, check out science writer Carl Zimmer and his experience being quote-mined by global warming denialists. To cheer up after that study in human folly, try Mark Chu-Carroll’s ongoing series on surreal numbers. Based on his prior habits, I’m curious to see if he takes a crack at explaining the relation between surreals and category theory (definitely one of the branches of mathematics you need to study if you want to be like the guy in Pi).
One point should be raised about the surreals which has not yet appeared in Mark Chu-Carroll’s exposition. You can get the reals — that ordinary, familiar number line — from the surreals by imposing an extra condition on the construction. It’s exercise 17 in the back of Knuth’s Surreal Numbers (a problem originally suggested to Knuth by John Conway). A number x is defined to be real if -n < x < n for some integer n, and if x falls in the same equivalence class as the surreal number
({x – 1, x – 1/2, x – 1/4, …}, {x + 1, x + 1/2, x + 1/4, …)}.
This topic is also discussed in chapter 2 of Conway’s On Numbers and Games. Theorem 13 proves that dyadic rationals are real numbers, and Conway then deduces that each real number not a dyadic rational is born on day ω (“Aleph Day” in Knuth’s book).
The practical upshot of all this is that surreals may provide a better pathway to understanding the real numbers than the standard way of teaching real analysis! Dealing with Dedekind cuts, for example, leads you to an explosion of special cases and general irritation. Conway says:
This discussion should convince the reader that the construction of the real numbers by any of the standard methods is really quite complicated. Of course the main advantage of an approach like that of the present work is that there is just one kind of number, so that one does not spend large amounts of time proving the associative law in several different guises. I think that this makes it the simplest so far, from a purely logical point of view.
Nevertheless there are certain disadvantages. One that can be dealt with quickly is that it is quite difficult to make the process stop after constructing the reals! We can cure this by adding to the construction the proviso that if L is non-empty but with no greatest member, then R is non-empty with no least member, and vice versa. This happily restricts us exactly to the reals.
The remaining disadvantages are that the dyadic rationals receive a curiously special treatment, and that the inductive definitions are of an unusual character. From a purely logical point of view these are unimportant quibbles (we discuss the induction problems later in more detail), but they would predispose me against teaching this to undergraduates as “the” theory of real numbers.
There is another way out. If we adopt a classical approach as far as the rationals Q, and then define the reals as sections of Q with the definitions of addition and multiplication given in this book, then all the formal laws have 1-line proofs and there is no case-splitting. The definition of multiplication seems complicated, but is fairly easy to motivate. Altogether, this seems the easiest possible approach. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20222 | # Example 410 x r n with x y iff x i y i i 1 n is a
• Notes
• 119
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Chapter 6 / Exercise 124
College Algebra: Real Mathematics, Real People
Larson
Expert Verified
Example 4.10 . ( X, - ) = ( R n , ) with x y iff x i y i , i = 1 , . . . , n is a lattice with x y = (min( x 1 , y 1 ) , . . . , min( x n , y n )) and x y = (max( x 1 , y 1 ) , . . . , max( x n , y n )) . X is not a complete lattice, but ([0 , 1] n , ) is, with inf S = (inf { x 1 : x S } , inf { x 2 : x S } ) and sup S = (sup { x 1 : x S } , sup { x 2 : x S } ) . 35
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Chapter 6 / Exercise 124
College Algebra: Real Mathematics, Real People
Larson
Expert Verified
Chapter 4.5 Example 4.11 . Suppose that X 6 = , let P ( X ) denote the class of subsets of X , and for A, B ∈ P ( X ) , define A - B if A B . ( P ( X ) , - ) is a complete lattice, with inf S = T { E : E ∈ S} for S ⊂ P ( X ) and sup S = S { E : E ∈ S} . We can turn lattice orderings around and keep the lattice structure. Example 4.12 . Suppose that ( X, - ) is a lattice. Define x - - y if y - x . Show that ( X, - - ) is a lattice. [ Thus, ( R n , - ) , i.e. ( R n , ) , is a lattice, as is ( P ( X ) , ) . ] Example 4.13 . ( P ( N ) , ) is a lattice with many noncomparable pairs of elements, but the class of sets {{ 1 } , { 1 , 2 } , . . . , { 1 , 2 , . . . , n } , . . . } is a chain in P ( N ) . Problem 4.2 . Show that if A B and B is totally ordered, then A is totally ordered. As an application, show that, in ( R 2 , ) , any subset of the graph of a non-decreasing function from R to R is a chain. There are partially ordered sets that are not lattices. Example 4.14 . Let T 1 R 2 be the triangle with vertices at (0 , 0) , (1 , 0) and (0 , 1) . ( T 1 , ) is partially ordered, but is not a lattice because (0 , 1) (1 , 0) is not in T 1 . On the other hand, ( T 2 , ) is a lattice when T 2 is the square with vertices at (1 , 1) , (1 , 0) , (0 , 1) , and (0 , 0) . 5. Monotone comparative statics We now generalize in two directions: (1) we now allow for X and T to have more general properties, they are not just linearly ordered in what follows, and (2) we now also allow for the set of available points to vary with t , not just the utility function. 5.1. Product Lattices. Suppose that ( X, - X ) and ( T, - T ) are lattices. Define the order - X × T on X × T by ( x 0 , t 0 ) % X × T ( x, t ) iff x 0 % X x and t 0 % T t . (This is the unanimity order again.) Lemma 4.15 . ( X × T, - X × T ) is a lattice. Proof : ( x 0 , t 0 ) ( x, t ) = (max { x 0 , x } , max { t 0 , t } ) X × T . ( x 0 , t 0 ) ( x, t ) = (min { x 0 , x } , min { t 0 , t } ) X × T . 5.2. Supermodular Functions. Definition 4.16 . For a lattice ( L, - ) , f : L R is supermodular if for all ‘, ‘ 0 L , f ( 0 ) + f ( 0 ) f ( ) + f ( 0 ) , equivalently, f ( 0 ) - f ( 0 ) f ( ) - f ( 0 ) . Example 4.17 . Taking 0 = ( x 0 , t ) and = ( x, t 0 ) recovers Definition 4.2 . Problem 4.3 . Show that a monotonic convex transformation of a monotonic supermodular function is supermodular. 36
Chapter 4.5 Problem 4.4 . Let ( L, - ) = ( R n , ) . Show that f : L R is supermodular iff it has increasing differences in x i and x j for all i 6 = j . Show that a twice continuously differentiable f : L R is supermodular iff 2 f/∂x i ∂x j 0 for all i 6 = j . 5.3. Ordering Subsets of a Lattice. Definition 4.18 . For A, B L , L a lattice, the strong set order is defined by A - Strong B iff ( a, b ) A × B , a b A and a b B . Recall that interval subsets of R are sets of the form ( -∞ , r ), ( -∞ , r ], ( r, s ), ( r, s ], [ r, s ), [ r, s ], ( r, ), or [ r, ). |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_20223 | # Newbie
• August 15th 2012, 10:03 AM
Rpuxster
Newbie
So glad I found this site .... I have alot of questions, so please be patient with me :P ... so here goes, how do you denote the square root cubed?
Found this .... though.
$\sqrt{y=1}$
• August 15th 2012, 12:28 PM
HallsofIvy
Re: Newbie
Quote:
Originally Posted by Rpuxster
So glad I found this site .... I have alot of questions, so please be patient with me :P ... so here goes, how do you denote the square root cubed?
Found this .... though.
$\sqrt{y=1}$
The "square root of x, cubed" could be written $(\sqrt{x})^2$. As long as x is positive, that is the same as "the cube of the square root of x", $\sqrt{x^3}$. And both of those can be written as $x^{3/2}$.
Where did you find $\sqrt{y=1}$? That makes no sense. You take the square root of a number, not an expression. Perhaps you mean $\sqrt{y+1}$ and just missed the "shift" key! But I don't see what that has to do with your question.
• August 15th 2012, 01:54 PM
Rpuxster
Re: Newbie
I was not clear in my question, I was looking for the format (LaTeX) to show $\sqrt[3]{y+1}$
sticky fingers, Yes it should be $\sqrt{y+1}$
Thank you, you have been most helpful - Roux |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_384 | # Testing $H_0: \beta_1 = \beta_2$ in a model $Y = \alpha + \beta_1x_1+ \beta_2x_2 + \varepsilon$
Consider a linear regression model $$Y = \alpha + \beta_1x_1+ \beta_2x_2 + \varepsilon$$, and suppose we are interested in testing $$H_0: \beta_1 = \beta_2$$. These slopes are two population slopes.
This means that under $$H_0$$, we have $$Y = \alpha + \beta(x_1 + x_2) + \varepsilon$$ where $$\beta$$ is the common value of $$\beta_1$$ and $$\beta_2$$.
My question is, can we just use the incremental F-test, where you compare the incremental sum of squares?
Also, is this test meaningful? I feel like it doesn't make sense to perform this test if $$x_1$$ and $$x_2$$ have different units.
EDIT: I just found in the post
t test of individual coefficient and wald test of euqality of two coefficients
that I should use the Wald test. But how do you compute the test statistic? The one in Wikipedia is for comparing a parameter obtained by MLE with a constant.
• If you are worried about incommensurable units, then you hypothesis makes no sense. However, such hypotheses can make sense in many applications. For instance, you might postulate that the rates at which the growth of a plant changes with respect to standardized amounts of fertilizer (mass) and sunlight (energy) are the same. As far as the computing goes, you could simply reformulate your model as $Y=\alpha + \gamma_1(x_1+x_2)+\gamma_2(x_1-x_2)+\varepsilon,$ where the gammas equal $(\beta_1\pm\beta_2)/2,$ and let the software automatically test $\gamma_2=0.$ – whuber Apr 29 at 13:45 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_385 | # zbMATH — the first resource for mathematics
## Akahira, Masafumi
Compute Distance To:
Author ID: akahira.masafumi Published as: Akahira, Masafumi; Akahira, M.
Documents Indexed: 130 Publications since 1975, including 9 Books Reviewing Activity: 54 Reviews Biographic References: 1 Publication
all top 5
#### Co-Authors
52 single-authored 36 Takeuchi, Kei 11 Ohyauchi, Nao 7 Takahashi, Kunihiko 5 Hida, Eisuke 5 Koike, Ken-ichi 5 Sato, Michikazu 4 Kawai, Shinichi 3 Nishihira, Yuji 3 Tanaka, Hidekazu 3 Torigoe, Norio 2 Kashima, Hiroyuki 1 Hashimoto, Shintaro 1 Hirakawa, Fumiko 1 Kim, Hyo Gyeong 1 Maihara, Hirosuke 1 Puri, Madan Lal
all top 5
#### Serials
26 RIMS Kokyuroku 20 Annals of the Institute of Statistical Mathematics 11 Journal of the Japan Statistical Society 7 Sequential Analysis 7 Communications in Statistics. Theory and Methods 6 Metron 4 The Australian Journal of Statistics 3 Metrika 3 Statistics & Decisions 3 Communications in Statistics. Simulation and Computation 3 Statistical Papers 2 The Annals of Statistics 2 Reports of Statistical Application Research 2 Statistics 2 Lecture Notes in Statistics 2 İstatistik 1 Publications de l’Institut de Statistique de l’Université de Paris 1 Statistica 1 Statistica Neerlandica 1 Sūgaku 1 SpringerBriefs in Statistics
#### Fields
128 Statistics (62-XX) 11 General and overarching topics; collections (00-XX) 10 Probability theory and stochastic processes (60-XX) 3 Numerical analysis (65-XX) 2 History and biography (01-XX)
#### Citations contained in zbMATH Open
57 Publications have been cited 223 times in 129 Documents Cited by Year
Asymptotic efficiency of statistical estimators: concepts and higher order asymptotic efficiency. Zbl 0463.62026
Akahira, Masafumi; Takeuchi, Kei
1981
Non-regular statistical estimation. Zbl 0833.62024
Akahira, Masafumi; Takeuchi, Kei
1995
On asymptotic deficiency of estimators in pooled samples in the presence of nuisance parameters. Zbl 0561.62023
Akahira, Masafumi; Takeuchi, Kei
1982
Third order asymptotic efficiency of the sequential maximum likelihood estimation procedure. Zbl 0689.62065
Akahira, Masafumi; Takeuchi, Kei
1989
Characterizations of prediction sufficiency (adequacy) in terms of risk functions. Zbl 0333.62010
Takeuchi, Kei; Akahira, Masafumi
1975
Asymptotic theory for estimation of location in non-regular cases. I: Order of convergence of consistent estimators. Zbl 0326.62022
Akahira, Masafumi
1975
Asymptotic theory for estimation of location in non-regular cases. II: Bounds of asymptotic distributions of consistent estimators. Zbl 0331.62024
Akahira, Masafumi
1975
Loss of information of a statistic for a family of non-regular distributions. Zbl 0878.62004
Akahira, Masafumi
1996
Asymptotic optimality of the generalized Bayes estimator in multiparameter cases. Zbl 0451.62029
Takeuchi, Kei; Akahira, Masafumi
1979
Bhattacharyya bound of variances of unbiased estimators in nonregular cases. Zbl 0588.62040
Akahira, Masafumi; Puri, Madan L.; Takeuchi, Kei
1986
The structure of asymptotic deficiency of estimators. Zbl 0634.62024
Akahira, Masafumi
1986
Asymptotic optimality of estimators in non-regular cases. Zbl 0485.62029
Akahira, Masafumi
1982
Discretized likelihood methods. Asymptotic properties of discretized likelihood estimators (DLE’s). Zbl 0453.62024
Akahira, Masafumi; Takeuchi, Kei
1979
A note on minimum variance. Zbl 0588.62039
Takeuchi, K.; Akahira, M.
1986
The lower bound for the variance of unbiased estimators for one- directional family of distributions. Zbl 0656.62034
Akahira, Masafumi; Takeuchi, Kei
1987
Loss of information associated with the order statistics and related estimators in the double exponential distribution case. Zbl 0744.62040
Akahira, Masafumi; Takeuchi, Kei
1990
The 3/2th and 2nd order asymptotic efficiency of maximum probability estimators in non-regular cases. Zbl 0782.62031
Akahira, Masafumi
1991
Asymptotics on the statistics for a family of non-regular distributions. Zbl 0857.62011
Akahira, Masafumi
1993
An information inequality for the Bayes risk. Zbl 0867.62014
Sato, Michikazu; Akahira, Masafumi
1996
Second order asymptotic efficiency in terms of the asymptotic variance of sequential estimation procedures in the presence of nuisance parameters. Zbl 0721.62078
Akahira, Masafumi; Takeuchi, Kei
1991
Second order asymptotic properties of the generalized Bayes estimators for a family of non-regular distributions. Zbl 0737.62026
Akahira, Masafumi
1988
Second order asymptotic efficiency in terms of asymptotic variances of the sequential maximum likelihood estimation procedures. Zbl 0737.62068
Takeuchi, Kei; Akahira, Masafumi
1988
On the asymptotic efficiency of estimators in an autoregressive process. Zbl 0362.62092
Akahira, Masafumi
1976
On the second order asymptotic efficiencies of estimators. Zbl 0398.62027
Takeuchi, Kei; Akahira, Masafumi
1976
A higher order approximation to a percentage point of the non-central $$t$$-distribution. Zbl 0850.62189
Akahira, M.
1995
Second-order asymptotic comparison of the MLE and MCLE for a two-sided truncated exponential family of distributions. Zbl 1348.62058
Akahira, M.; Hashimoto, S.; Koike, K.; Ohyauchi, N.
2016
A Bayesian view of the Hammersley-Chapman-Robbins-type inequality. Zbl 1117.62026
Akahira, M.; Ohyauchi, N.
2007
Estimation of a common parameter for pooled samples from the uniform distributions. Zbl 0607.62022
Akahira, Masafumi; Takeuchi, Kei
1985
Locally minimum variance unbiased estimator in a discontinuous density function. Zbl 0608.62030
Akahira, M.; Takeuchi, K.
1987
Second and third order asymptotic completeness of the class of estimators. Zbl 0641.62021
Akahira, Masafumi; Hirakawa, Fumiko; Takeuchi, Kei
1988
On the bound of the asymptotic distribution of estimators when the maximum order of consistency depends on the parameter. Zbl 0651.62014
Akahira, Masafumi; Takeuchi, Kei
1986
Second order asymptotic comparison of estimators of a common parameter in the double exponential case. Zbl 0622.62033
Akahira, Masafumi
1987
Remarks on the asymptotic efficiency and inefficiency of maximum probability estimators. Zbl 0479.62023
Akahira, Masafumi; Takeuchi, Kei
1979
Third order efficiency implies fourth order efficiency: A resolution of the conjecture of J. K. Ghosh. Zbl 0874.62023
Akahira, Masafumi
1996
Second order asymptotic sufficiency for a family of distributions with one-directionality. Zbl 0758.62004
Akahira, M.
1991
On the second order asymptotic efficiency of unbiased confidence intervals. Zbl 0459.62026
Akahira, Masafumi; Takeuchi, Kei
1979
The amount of information and the bound for the order of consistency for a location parameter family of densities. Zbl 0852.62027
Akahira, Masafumi
1995
Information inequalities in a family of uniform distributions. Zbl 0989.62014
Akahira, Masafumi; Takeuchi, Kei
2001
Confidence intervals for the difference of means: Application to the Behrens-Fisher type problem. Zbl 1017.62025
Akahira, Masafumi
2002
The information inequality in sequential estimation for the uniform case. Zbl 1048.62078
Akahira, Masafumi; Takeuchi, Kei
2003
On a family of distributions attaining the Bhattacharyya bound. Zbl 1049.62025
Tanaka, Hidekazu; Akahira, Masafumi
2003
The asymptotic bound by the Kiefer-type information inequality and its attainment. Zbl 1124.62008
Akahira, M.; Ohyauchi, N.
2007
An information inequality bound for the asymptotic variance of sequential estimation procedures of a linearly combined parameter and its attainment. Zbl 0879.62070
Akahira, Masafumi
1997
A generalized binomial distribution determined by a two-state Markov chain and a distribution by the Bayesian approach. Zbl 0892.62005
Akahira, M.; Kashima, H.; Takahashi, K.
1997
On the consistency of the maximum likelihood estimator through its uniform consistency. Zbl 0811.62032
Akahira, Masafumi; Kashima, Hiroyuki
1994
Second order asymptotic optimality of estimators for a density with finite cusps. Zbl 0669.62005
Akahira, Masafumi
1988
On asymptotic deficiency of estimators. Zbl 0473.62028
Akahira, Masafumi
1981
Asymptotic deficiency of the jackknife estimator. Zbl 0505.62020
Akahira, Masafumi
1983
Sequential analysis and statistical inference. Proceedings of a symposium held at the Research Institute for Mathematical Sciences, Kyoto University, Kyoto, Japan, March 9–11, 1993. Zbl 0938.62500
Akahira, Masafumi (ed.)
1993
Information inequalities for the Bayes risk for a family of non-regular distributions. Zbl 1051.62025
Akahira, Masafumi; Ohyauchi, Nao
2002
On a new approximation to non-central $$t$$-distributions. Zbl 0845.62015
Akahira, Masafumi; Sato, Michikazu; Torigoe, Norio
1995
The Bhattacharyya type bound for the asymptotic variance and the sequential discretized likelihood estimation procedure. Zbl 0838.62068
Akahira, Masafumi
1995
Sequential interval estimation of a location parameter with the fixed width in the uniform distribution with an unknown scale parameter. Zbl 1061.62121
Akahira, Masafumi; Koike, Ken-ichi
2005
Randomized confidence intervals of a parameter for a family of discrete exponential type distributions. Zbl 0925.62126
Akahira, Masafumi; Takahashi, Kunihiko; Takeuchi, Kei
1997
On the application of the Minkowski-Farkas theorem to sampling designs. Zbl 0778.62007
Akahira, M.; Takeuchi, K.
1993
Completeness for sequential sampling plans. Zbl 0793.62042
Koike, Ken-ichi; Akahira, Masafumi
1993
An approximation to the generalized hypergeometric distribution. Zbl 1050.62022
Hida, Eisuke; Akahira, Masafumi
2003
Second-order asymptotic comparison of the MLE and MCLE for a two-sided truncated exponential family of distributions. Zbl 1348.62058
Akahira, M.; Hashimoto, S.; Koike, K.; Ohyauchi, N.
2016
A Bayesian view of the Hammersley-Chapman-Robbins-type inequality. Zbl 1117.62026
Akahira, M.; Ohyauchi, N.
2007
The asymptotic bound by the Kiefer-type information inequality and its attainment. Zbl 1124.62008
Akahira, M.; Ohyauchi, N.
2007
Sequential interval estimation of a location parameter with the fixed width in the uniform distribution with an unknown scale parameter. Zbl 1061.62121
Akahira, Masafumi; Koike, Ken-ichi
2005
The information inequality in sequential estimation for the uniform case. Zbl 1048.62078
Akahira, Masafumi; Takeuchi, Kei
2003
On a family of distributions attaining the Bhattacharyya bound. Zbl 1049.62025
Tanaka, Hidekazu; Akahira, Masafumi
2003
An approximation to the generalized hypergeometric distribution. Zbl 1050.62022
Hida, Eisuke; Akahira, Masafumi
2003
Confidence intervals for the difference of means: Application to the Behrens-Fisher type problem. Zbl 1017.62025
Akahira, Masafumi
2002
Information inequalities for the Bayes risk for a family of non-regular distributions. Zbl 1051.62025
Akahira, Masafumi; Ohyauchi, Nao
2002
Information inequalities in a family of uniform distributions. Zbl 0989.62014
Akahira, Masafumi; Takeuchi, Kei
2001
An information inequality bound for the asymptotic variance of sequential estimation procedures of a linearly combined parameter and its attainment. Zbl 0879.62070
Akahira, Masafumi
1997
A generalized binomial distribution determined by a two-state Markov chain and a distribution by the Bayesian approach. Zbl 0892.62005
Akahira, M.; Kashima, H.; Takahashi, K.
1997
Randomized confidence intervals of a parameter for a family of discrete exponential type distributions. Zbl 0925.62126
Akahira, Masafumi; Takahashi, Kunihiko; Takeuchi, Kei
1997
Loss of information of a statistic for a family of non-regular distributions. Zbl 0878.62004
Akahira, Masafumi
1996
An information inequality for the Bayes risk. Zbl 0867.62014
Sato, Michikazu; Akahira, Masafumi
1996
Third order efficiency implies fourth order efficiency: A resolution of the conjecture of J. K. Ghosh. Zbl 0874.62023
Akahira, Masafumi
1996
Non-regular statistical estimation. Zbl 0833.62024
Akahira, Masafumi; Takeuchi, Kei
1995
A higher order approximation to a percentage point of the non-central $$t$$-distribution. Zbl 0850.62189
Akahira, M.
1995
The amount of information and the bound for the order of consistency for a location parameter family of densities. Zbl 0852.62027
Akahira, Masafumi
1995
On a new approximation to non-central $$t$$-distributions. Zbl 0845.62015
Akahira, Masafumi; Sato, Michikazu; Torigoe, Norio
1995
The Bhattacharyya type bound for the asymptotic variance and the sequential discretized likelihood estimation procedure. Zbl 0838.62068
Akahira, Masafumi
1995
On the consistency of the maximum likelihood estimator through its uniform consistency. Zbl 0811.62032
Akahira, Masafumi; Kashima, Hiroyuki
1994
Asymptotics on the statistics for a family of non-regular distributions. Zbl 0857.62011
Akahira, Masafumi
1993
Sequential analysis and statistical inference. Proceedings of a symposium held at the Research Institute for Mathematical Sciences, Kyoto University, Kyoto, Japan, March 9–11, 1993. Zbl 0938.62500
Akahira, Masafumi
1993
On the application of the Minkowski-Farkas theorem to sampling designs. Zbl 0778.62007
Akahira, M.; Takeuchi, K.
1993
Completeness for sequential sampling plans. Zbl 0793.62042
Koike, Ken-ichi; Akahira, Masafumi
1993
The 3/2th and 2nd order asymptotic efficiency of maximum probability estimators in non-regular cases. Zbl 0782.62031
Akahira, Masafumi
1991
Second order asymptotic efficiency in terms of the asymptotic variance of sequential estimation procedures in the presence of nuisance parameters. Zbl 0721.62078
Akahira, Masafumi; Takeuchi, Kei
1991
Second order asymptotic sufficiency for a family of distributions with one-directionality. Zbl 0758.62004
Akahira, M.
1991
Loss of information associated with the order statistics and related estimators in the double exponential distribution case. Zbl 0744.62040
Akahira, Masafumi; Takeuchi, Kei
1990
Third order asymptotic efficiency of the sequential maximum likelihood estimation procedure. Zbl 0689.62065
Akahira, Masafumi; Takeuchi, Kei
1989
Second order asymptotic properties of the generalized Bayes estimators for a family of non-regular distributions. Zbl 0737.62026
Akahira, Masafumi
1988
Second order asymptotic efficiency in terms of asymptotic variances of the sequential maximum likelihood estimation procedures. Zbl 0737.62068
Takeuchi, Kei; Akahira, Masafumi
1988
Second and third order asymptotic completeness of the class of estimators. Zbl 0641.62021
Akahira, Masafumi; Hirakawa, Fumiko; Takeuchi, Kei
1988
Second order asymptotic optimality of estimators for a density with finite cusps. Zbl 0669.62005
Akahira, Masafumi
1988
The lower bound for the variance of unbiased estimators for one- directional family of distributions. Zbl 0656.62034
Akahira, Masafumi; Takeuchi, Kei
1987
Locally minimum variance unbiased estimator in a discontinuous density function. Zbl 0608.62030
Akahira, M.; Takeuchi, K.
1987
Second order asymptotic comparison of estimators of a common parameter in the double exponential case. Zbl 0622.62033
Akahira, Masafumi
1987
Bhattacharyya bound of variances of unbiased estimators in nonregular cases. Zbl 0588.62040
Akahira, Masafumi; Puri, Madan L.; Takeuchi, Kei
1986
The structure of asymptotic deficiency of estimators. Zbl 0634.62024
Akahira, Masafumi
1986
A note on minimum variance. Zbl 0588.62039
Takeuchi, K.; Akahira, M.
1986
On the bound of the asymptotic distribution of estimators when the maximum order of consistency depends on the parameter. Zbl 0651.62014
Akahira, Masafumi; Takeuchi, Kei
1986
Estimation of a common parameter for pooled samples from the uniform distributions. Zbl 0607.62022
Akahira, Masafumi; Takeuchi, Kei
1985
Asymptotic deficiency of the jackknife estimator. Zbl 0505.62020
Akahira, Masafumi
1983
On asymptotic deficiency of estimators in pooled samples in the presence of nuisance parameters. Zbl 0561.62023
Akahira, Masafumi; Takeuchi, Kei
1982
Asymptotic optimality of estimators in non-regular cases. Zbl 0485.62029
Akahira, Masafumi
1982
Asymptotic efficiency of statistical estimators: concepts and higher order asymptotic efficiency. Zbl 0463.62026
Akahira, Masafumi; Takeuchi, Kei
1981
On asymptotic deficiency of estimators. Zbl 0473.62028
Akahira, Masafumi
1981
Asymptotic optimality of the generalized Bayes estimator in multiparameter cases. Zbl 0451.62029
Takeuchi, Kei; Akahira, Masafumi
1979
Discretized likelihood methods. Asymptotic properties of discretized likelihood estimators (DLE’s). Zbl 0453.62024
Akahira, Masafumi; Takeuchi, Kei
1979
Remarks on the asymptotic efficiency and inefficiency of maximum probability estimators. Zbl 0479.62023
Akahira, Masafumi; Takeuchi, Kei
1979
On the second order asymptotic efficiency of unbiased confidence intervals. Zbl 0459.62026
Akahira, Masafumi; Takeuchi, Kei
1979
On the asymptotic efficiency of estimators in an autoregressive process. Zbl 0362.62092
Akahira, Masafumi
1976
On the second order asymptotic efficiencies of estimators. Zbl 0398.62027
Takeuchi, Kei; Akahira, Masafumi
1976
Characterizations of prediction sufficiency (adequacy) in terms of risk functions. Zbl 0333.62010
Takeuchi, Kei; Akahira, Masafumi
1975
Asymptotic theory for estimation of location in non-regular cases. I: Order of convergence of consistent estimators. Zbl 0326.62022
Akahira, Masafumi
1975
Asymptotic theory for estimation of location in non-regular cases. II: Bounds of asymptotic distributions of consistent estimators. Zbl 0331.62024
Akahira, Masafumi
1975
all top 5
#### Cited by 109 Authors
28 Akahira, Masafumi 14 Takeuchi, Kei 10 Koike, Ken-ichi 7 Ohyauchi, Nao 5 Taniguchi, Masanobu 4 Kumon, Masayuki 4 Miao, Yu 4 Tanaka, Hidekazu 4 Yoshida, Nakahiro 3 Hashimoto, Shintaro 3 Hayashi, Masahito 3 Kakizawa, Yoshihide 3 Lim, Wooi Khai 3 Ogasawara, Haruhiko 3 Pal, Nabendu 3 Takada, Yoshikazu 3 Takemura, Akimichi 2 Liang, Hua 2 Mohtashami Borzadaran, Gholam Reza 2 Roknabadi, Abdol Hamid Rezaei 2 Takagi, Yoshiji 2 Tamaki, Kenichiro 2 Venetiaan, Shanti A. 2 Wefelmeyer, Wolfgang 1 Amari, Shun-ichi 1 Antonelli, Sabrina 1 Arcones, Miguel A. 1 Bai, Xuchao 1 Bentler, Peter M. 1 Bhattacharjee, Debanjan 1 Bingham, Nicholas Hugh 1 Björk, Tomas 1 Bosq, Denis 1 Carro, Jesus M. 1 Chafaï, Djalil 1 Cheng, Ping 1 Concordet, Didier 1 Eguchi, Shinto 1 Fenoy, Mar 1 Ferguson, Helaman 1 Friedlander, Benjamin 1 Fujikoshi, Yasunori 1 Gammerman, Alex 1 Gunasekera, Sumith 1 Heijmans, Risto D. H. 1 Hosoya, Yuzo 1 Ibarrola, Pilar 1 Jiang, Yiwen 1 Johansson, Björn 1 Jurečková, Jana 1 Kabaila, Paul V. 1 Kaniovska, I. 1 Kano, Yutaka 1 Kanter, Marek 1 Kashima, Hiroyuki 1 Kato, Shogo 1 Kawai, Shinichi 1 Khorashadizadeh, Mohammad 1 Kim, Hyo Gyeong 1 Klaassen, Chris A. J. 1 Knight, Keith 1 Krishnamoorthy, Kalyanam 1 Kunitomo, Naoto 1 Liu, Bin 1 Liu, Yiming 1 Maekawa, Koichi 1 Magnus, Jan R. 1 Milhaud, Xavier 1 Mukhopadhyay, Nitis 1 Müller, Ursula U. 1 Naing, May Thu 1 Nair, K. R. Muraleedharan 1 Nayak, Tapan Kumar 1 Nayeban, Samira 1 Nouretdinov, Ilia 1 Obayashi, Chie 1 Ochi, Yoshimichi 1 O’Neill, Terence J. 1 Paris, Matteo G. A. 1 Patil, H. S. 1 Perakis, Michael 1 Picard, Dominique B. 1 Podraza-Karakulska, A. 1 Prakasa Rao, B. L. S. 1 Puri, Madan Lal 1 Puza, Borek D. 1 Qin, Min 1 Rattihalli, Ranganath Narayanacharya 1 Ravishanker, Nalini 1 Regoli, Giuliana 1 Reid, Nancy M. 1 Sakamoto, Yuji 1 Seoane-Sepúlveda, Juan Benigno 1 Seveso, Luigi 1 Shen, Luming 1 Shen, Si 1 Shi, Yimin 1 Smitha, S. 1 Söderström, Torsten 1 Stoica, Petre Gheorghe ...and 9 more Authors
all top 5
#### Cited in 31 Serials
27 Annals of the Institute of Statistical Mathematics 22 Communications in Statistics. Theory and Methods 12 Sequential Analysis 7 Statistics & Probability Letters 6 Journal of Multivariate Analysis 6 Journal of Statistical Planning and Inference 6 Journal of Time Series Analysis 5 Metrika 5 Statistics 4 Journal of Econometrics 3 Probability Theory and Related Fields 3 Communications in Statistics. Simulation and Computation 3 Statistical Papers 2 The Canadian Journal of Statistics 2 The Annals of Statistics 1 International Journal of Control 1 Lithuanian Mathematical Journal 1 Psychometrika 1 Results in Mathematics 1 Applied Mathematics Letters 1 Applied Mathematical Modelling 1 Journal of Statistical Computation and Simulation 1 Stochastic Processes and their Applications 1 Bulletin des Sciences Mathématiques 1 Australian & New Zealand Journal of Statistics 1 Econometric Theory 1 Brazilian Journal of Probability and Statistics 1 Entropy 1 International Journal of Quantum Information 1 Journal of Statistical Theory and Practice 1 Chilean Journal of Statistics
all top 5
#### Cited in 13 Fields
126 Statistics (62-XX) 23 Probability theory and stochastic processes (60-XX) 3 Differential geometry (53-XX) 2 Numerical analysis (65-XX) 1 Several complex variables and analytic spaces (32-XX) 1 Dynamical systems and ergodic theory (37-XX) 1 Computer science (68-XX) 1 Quantum theory (81-XX) 1 Statistical mechanics, structure of matter (82-XX) 1 Operations research, mathematical programming (90-XX) 1 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 1 Systems theory; control (93-XX) 1 Information and communication theory, circuits (94-XX) |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_386 | # What is the axis of symmetry and vertex for the graph y= -7x^2 +2x?
Jul 7, 2018
$x = \frac{1}{7} , \text{ vertex } = \left(\frac{1}{7} , \frac{1}{7}\right)$
#### Explanation:
$\text{calculate the zeros by letting y = 0}$
$- 7 {x}^{2} + 2 x = 0$
$x \left(- 7 x + 2\right) = 0$
$x = 0 , x = \frac{2}{7} \leftarrow \textcolor{b l u e}{\text{are the zeros}}$
$\text{the vertex lies on the axis of symmetry which is}$
$\text{situated at the midpoint of the zeros}$
$\text{axis of symmetry } x = \frac{0 + \frac{2}{7}}{2} = \frac{1}{7}$
$\text{substitute this value into the equation for y-coordinate}$
$y = - 7 {\left(\frac{1}{7}\right)}^{2} + 2 \left(\frac{1}{7}\right) = - \frac{1}{7} + \frac{2}{7} = \frac{1}{7}$
$\textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{1}{7} , \frac{1}{7}\right)$
graph{-7x^2+2x [-10, 10, -5, 5]} |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_387 | # Applying Initial Conditions
#### MathematicalPhysics
I need to find all the separated solns of
$$x^2 \frac{\partial^2 u}{\partial x^2} + x\frac{\partial u}{\partial x} + \frac{\partial^2 u}{\partial y^2} = 0$$
in the strip $${(x,y) : 0 < y < a, -\infty < x < \infty }$$
the separated solns must also satisfy u = 0 on both the edges, that is, on y=0 and y=a for all values of x.
Iv got the general solutions to be..
$$X(x) = Dlnx + C , (k = 0)$$
$$X(x) = Dx^{n} + Cx^{-n} , (k \neq 0)$$
and
$$Y(y) = A\cos{ky} + B\sin{ky} , (k \neq 0)$$
$$Y(y) = Ay + B , (k = 0)$$
where k is just the constant iv let the two bits equal when I separated the variables. (well -k^2 actually).
I just need help interpreting the conditions to sort out the constants..I think!
Related Differential Equations News on Phys.org
#### Galileo
Homework Helper
I haven't checked your answer, but if it is correct then, since $u(x,y)=X(x)Y(y)$, the boundary conditions say:
$$u(x,0)=X(x)Y(0)=0$$
and
$$u(x,a)=X(x)Y(a)=0$$
So $Y(0)=Y(a)=0$
For example: if k=0, then applying the boundary condition at y=0 gives:
$$Y(0)=B=0$$
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_388 | # 78 of 100: Visualizing Algebra
Algebra Level 1
What equation does this image correspond to?
The same setup can also be used to give a general formula using $n.$
× |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_389 | # How do you combine like terms in (6+ 8i ) + ( 1- 9i )?
May 18, 2018
$\left(7 - i\right)$
#### Explanation:
$\left(6 + 8 i\right) + \left(1 - 9 i\right)$
$\left(6 + 8 i + 1 - 9 i\right)$
$\left(6 + 1 + 8 i - 9 i\right)$
$\left(7 - i\right)$ |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_390 | # Coefficients from Stone Weierstrass versus Fourier Transform
Usually one shows the density of the functions $\sin(kx)$ in $L^2([0,1])$ using the Fourier transform. This in fact comes from the Stone-Weirstrass theorem however and then uses the density of continuous functions in $L^2([0,1])$.
However, the Stone Weirstrass theorem can be used to show, for example, that the functions $e^{ikx}$ are dense in $C([0,1])$ and hence dense in $L^1([0,1])$ as well. So we obtain (not-necessarily-unique) coefficients $c_k$ such that $f_k(x) =c_ke^{ikx}$ converge to any given $f \in L^1([01])$. How should I think about these coefficients? How do they relate to the Fourier series of $f$ (with basis $e^{ikx})$?
-
Your question might be; in which $L^1$? All your functions are $2\pi$-periodic, so you cannot approximate anything that isn't. – Thierry Zell Aug 24 '10 at 13:27
What you're saying is false I think unless you're misunderstanding me. I'm working on say L^1[0,1] (any compact set will suffice). I can approximate $1$ as well as I'd like for instance by functions $\sum c_k sin(kx)$ for instance on $[0,2pi]$. I'm not talking about pointwise convergence just $L^1$ convergence. – Dorian Aug 24 '10 at 13:38
you need to change your exponents to $e^{ikx}$ so they display properly.. – Otis Chodosh Aug 24 '10 at 13:47
The coefficients are the same. If you work in $L^2([0,1])$ instead, its easier to see. It all comes from the fact that Fourier coefficients are unique. If $\sum c_n e^{inx} = \sum d_n e^{inx}$ then $\sum (c_n - d_n) e^{inx} = 0$. Inner product with $e^{imx}$ and you get $c_n = d_n$. Its just another way to prove that Fourier series converge in $L^2$ (and thus $L^1$) – Otis Chodosh Aug 24 '10 at 13:54
The set $\{ e^{inx} \}$ is trivially not dense in $L^1(0,1)$. What you probably mean is that their linear combinations (trigonometric polynomials) are. Still, this does not give a priori a development as a Fourier series, since the coefficients of the trigonometric polynomial approximation to a given function $f$ may not stabilize. – Andrea Ferretti Aug 24 '10 at 16:18
Just a comment if you choose coefficients $c_{k,n}$ such that $$\lim_{n\to\infty} \left(\sum_{k=-n}^{n} c_{k,n} e^{2\pi i n x}\right) \to f (x)$$ in some sense, e.g. $L^1$, then these are not unique. It is even known that the obvious choice $c_{k,n} = \hat{f}(k) = \int e^{-2\pi i n x} f(x) dx$ is not the best. It's much better to choose $$c_{k,n} = \left(1 - \frac{|k|}{n}\right) \hat{f}(k).$$ Then one Cesaro sums the Fourier series, and this is known to converge.
As pointed out by Zen Harper below, I should mention that with the choice $c_{k,n} = \hat{f}(k)$ for $-n \leq k \leq n$, the Fourier series of a $L^1$ function must not converge. In fact it can diverge almost-everywhere.
Having said these things, the obvious advantage of this is, that everything is explicit and does not rely on any abstract hocus pocus.
I realized one more thing: Consider the case $f \in L^2$. Then the choice $c_{k,n} = \hat{f}(k)$ for $-n \leq k \leq n$ is optimal. This follows from easy Hilbert space theory!
-
Is this the best you can do in $L^1$? This is very interesting! – Otis Chodosh Aug 24 '10 at 20:34
Not sure. Certainly not for functions having some regularity, one can check that the "normal" Fourier series become superior at one point ($C^1$ should be sufficient). – Helge Aug 24 '10 at 22:14
Apologies to Helge and everyone else who knows this, but it's worth stressing that the partial sums of the Fourier series DO NOT always converge to $f \in L^1$, either pointwise or in $L^1$, which is why the Cesaro averages are needed. (Although they do, by Cauchy-Schwarz, if you start off with $f \in L^2$). So, when Helge says "...the obvious choice...is not the best...", not only is it not the best, but in fact it may not even work at all! – Zen Harper Aug 25 '10 at 2:48
@Zen: I edited my post to include your comment. Furthermore, I added a comment on the L^2 case. – Helge Aug 25 '10 at 11:00
The Fourier transform is bounded from $L^1([0,1])$ to $\ell^\infty(Z)$ (with norm 1 if you expand in terms of $e^{2\pi ikx}$). If we take a sequence of finite sums of the form $f_n = \sum_k c_{n,k} e^{2\pi ikx}$ where for each $n$ there are only finitely many terms which are not zero ("trigonometric polynomials"), and $\|f_n -f\|_{L^1}\to 0$, then for all $k$, we have $c_{n,k} \to \hat f(k)$ as $n\to \infty$.
It may help to think of the kind of stuff that doesn't happen in $L^2$: Namely, Kolmogorov famously showed in his first publication that there exist $L^1$ functions $f$ whose Fourier series diverge almost everywhere (and I believe one can arrange for the divergence to be everywhere as well). If such a function $f$ can be written as an infinite sum $f(x) = \sum c_k e^{2\pi ikx}$, with the sum converging in $L^1$ and the coefficients not necessarily the Fourier coefficients, then the sum must converge almost everywhere, so it can't be the Fourier series. By the previous paragraph, it follows that no such representation as a convergent infinite sum is possible.
So what must happen as one uses density of $C(X)$ and Stone-Weierstrass to approximate such a function $f$ by trigonometric polynomials? If $\|f_n - f\|_{L^1} < \epsilon$, then for all $k$, $|\hat f_n(k) - \hat f(k)| < \epsilon$. One way to think about it is that $\hat f_n(k)$ is forced to be like $\hat f(k)$ whenever the latter is large compared to $\epsilon$, while there's potentially up to an $\epsilon$ of leeway in every coefficient. It is clear that $f_n$ can't just match $f$ in all the Fourier coefficients which are larger than $\epsilon$, though, or else the $L^1$ norm of the resulting sequence would diverge. So density and Stone-Weierstrass are somehow smart enough to use the (less than) $\epsilon$ of room to carry out the $L^1$ approximation.
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Hey Mike, What you're saying sounds reasonable but I"m a bit confused by something. Isn't saying that there is some $f_n = \sum_k c_{n,k} e^{2 \pi i k x}$ where $||f_n-f||_{L^1} \to 0$ the same as saying that the span of trigonometric polynomials is dense in $L^1$? – Dorian Aug 24 '10 at 23:31
Oh yes, their span is certainly dense in $L^1$, like you said in your post. It is dense in $C([0,1])$ in the sup norm by Stone-Weierstrass, which implies that it's dense in $L^1$ norm as well, from which follows density in all of $L^1$. Since $L^1$ is a metric space, this means there is a sequence of trig polynomials converging to $f$ in $L^1$. – Mike Hall Aug 25 '10 at 2:42
Note that it does not follow that $f$ can be written as an infinite trigonometric series in the $L^1$ sense. In the $L^2$ case you can use Hilbert space geometry to prove that the best $L^2$ approximation to $f$ by an $n$th degree trig polynomial is given by the $n$th partial Fourier series. Stone-Weierstrass implies that trig polynomials are dense in $L^2$, and hence we have $L^2$ convergence of Fourier series. $L^1$, on the other hand, isn't so nice geometrically (there's no inner product which gives the norm). – Mike Hall Aug 25 '10 at 2:44
In answer to Dorian's comment question: yes, dense span is exactly the same as saying existence of $f_n$. However, just knowing a set with dense span in general DOESN'T tell you anything about which coefficients you should actually use! But Fourier series have very special additional properties which can be exploited, as Mike Hall says. (Basically because the Hilbert space $L^2$ is densely embedded in $L^1$, so we can use Hilbert space arguments and orthogonality). – Zen Harper Aug 25 '10 at 2:53
Per @Andrea Ferretti's comments, you have to be careful to distinguish between $\{e^{inx}\}$ and $span \ \{e^{inx}\}$. You certainly are interested the latter. Sorry if my comments were sloppy and confusing above.
So, I think that the it goes like this:
From some corollary of Stone-Weirstrauss you can show that $span \ \{e^{inx}\}$ is dense in $C(\mathbb{S}^1)$ with the supremum norm. Because we know that $C(\mathbb{S}^1)\hookrightarrow L^2([0,1])$ has its image a dense subset of $L^2([0,1])$ and we know that if $f_n \to f$ in the supremum topology on $C(\mathbb{S}^1)$, then the images also converge in $L^2([0,1])$.
Thus, by this reasoning, for $f\in L^2([0,1])$ we can find $f_n \in span \ \{e^{inx}\}$ such that $f_n = L^2([0,1])$. Lets write $$f_n = \sum_{k\in \mathbb{Z}} c_k^{(n)} e^{ikx}$$ where all but finitely many of the $c_k^{(n)}$ are zero (this is because in the span of infinitely many objects we only take a finite number of them to add together)
Now, what I think you are asking is: what can we say about the coefficients $c_k^{(n)}$? The answer is that they converge to the $k$-th Fourier coefficient of $f$ as $n\to\infty$ because $$\hat f(k) = \langle f, e^{ikx} \rangle = \lim_{n\to\infty} \langle f_n ,e^{ikx}\rangle = \lim_{n\to\infty} c_k^{(n)}$$
In fact if $c_k^{(n)}$ are arbitrary complex numbers, defining $f_n$ as above, we see that $$\Vert f - f_n \Vert_{L^2} = \sum_{k\in \mathbb{Z}} |\hat f(k) - c_k^{(n)}|^2$$ assuming convergence. Thus, if $(c_k^{(n)})_k \to (\hat f(k))_k$ as $n\to\infty$ in $\ell^2(\mathbb{Z})$ then $f_n\to f$ in $L^2$, which is a pretty weak condition.
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You hae misunderstood my question. I'm concerned with $L^1([0,1])$. Of course we have a fourier series in $L^2([0,1])$ but my point is that we get coefficients regardless and we should get convergence in $L^1([0,1])$ as well for certain coefficients (not the fourier coefficients though a priori if our function is not $L^2([0,1])$). – Dorian Aug 24 '10 at 18:59
By Holder's inequality on $[0,1]$, $\Vert f \Vert_{L^1} \leq \Vert f \Vert_{L^2}$ so what I have shown is that if you started with a f that was actually in L2 , then you can replace the L2 convergence statements with L1 convergence. – Otis Chodosh Aug 24 '10 at 19:38
I'm concerned with $f \in L^1([0,1])$ but not in $L^2([0,1])$ so that you can't just use the $L^2$ structure. – Dorian Aug 24 '10 at 23:20 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_391 | # Change of basis matrix verification
Let $B$ and $C$ be two bases. To find the change of basis matrix $\phi_{B,C}$, I compute $\phi_{SB,B}$ and $\phi_{SB,C}$. Create the new matrix $T=[\phi_{SB,B}|\phi_{SB,C}]$. Reducing it to reduced row echelon form, should yield $T=[I|\phi_{B,C}$, right? Is there a way to verify that the new basis indeed maps $[v]_B$ to $[v]_C$?
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What are $\phi_{SB, B}$ and $\phi_{SB, C}$? – student Jan 19 '12 at 13:21
@Leandro The change of basis from the standard basis to B, respectively C. – Andrew Jan 19 '12 at 13:23
## 2 Answers
To check that the change of basis is correct, we can multiply $B$ times $\phi_{B,C}$ and we should get the $C$ basis. This can tell you that $\phi_{B,C}$ is correct.
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The matrix $\phi_{B,C}$ should have columns that are the coordinates of the $B$-basis vectors in terms of the $C$-basis vectors (i.e., old basis in terms of new basis). Let your matrix $T$ have left block whose columns are the $C$ basis vectors, and right block whose columns are the $B$-basis vectors. Row reduce $T$. The right hand block will be $\phi_{B,C}$.
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I dont think this works – Olivia Irving Mar 26 '13 at 3:59 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_392 | ## College Algebra (6th Edition)
$y=\displaystyle \frac{1}{3}x-2$ $y=\displaystyle \frac{1}{3}x+2$
A graphical solution to a system of linear equations is a point of intersection of two lines. If the solution set is empty, the lines do not intersect. They are parallel. From the graph, the pair of parallel lines have equations $x-3y=6\qquad$and$\quad x+3y=-6$ Slope-intercept form: solve each equation for $y$: $\left[\begin{array}{lll} x-3y=6 & & x+3y=-6\\ -3y=-x+6 & & -3y=-x-6\\ y=\frac{1}{3}x-2 & & y=\frac{1}{3}x+2 \end{array}\right]$ The system: $y=\displaystyle \frac{1}{3}x-2$ $y=\displaystyle \frac{1}{3}x+2$ |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_393 | Atomic Number The mass number is the total number of an atom’s protons and neutrons. Which isotope of hydrogen is present in heavy... What is the most common isotope of silicon? Learn what an ion is and how it forms. for any atom with a + or - charge, # protons will not = # electrons. The atomic number of Pb from the periodic table is 82, indicating that it has 82 protons. it is uneven. that's true. Since it has a charge of +2 , this ion must contain 80 electrons. The four isotopes of Pb are given below. Atomic Symbol: Atomic number. Similarly, A = 82 + 125 = 207 and A = 82 + 126 = 208 for the second and third isotopes, respectively. Protons, Neutrons and Electrons of all the Elements: Shell Diagram: 1: Hydrogen has 1 proton, 0 neutron and 1 electron: 2: Helium has 2 protons, 2 neutrons and 2 electrons: 3: Lithium has 3 protons, Is it harmful? Gibbs Free Energy: Definition & Significance. An atom is built with a combination of three distinct particles: electrons, protons, and neutrons. That's the number of protons about 36. What is Atomic Mass? What is the most abundant isotope of boron? Determine the number of protons, neutrons, and electrons in a 210 Pb 2 + ion. Atomic Mass Unit (AMU): Definition, Standard & Conversion. Magic Numbers of Protons and Neutrons The nuclear notation #""_82^206"Pb"# gives us the mass number, 206, for this isotope of lead, and the atomic number of lead, 82.. Halogens Finally, the neutrons depends on the isoptope (sum of protons plus neutrons) you're talking about! Have you ever been in a room where someone has put on perfume or scented lotion and a few minutes later you are able to smell it? Lithium has more protons, neutrons, and electrons. So any atom of lead-206 has #82# protons. A glucose molecule has 96 protons, 84 neutrons and 96 electrons. This article discusses atomic mass and how it is calculated. Formula Mass of a Compound: Definition & Formula. Have you ever asked yourself why salt dissolves in water? In this lesson, we will discuss the examples and types of isotopes. 125 neutrons. You can't put a molecule on a scale to determine its mass, so how can you figure it out? It explains how we use cookies (and other locally stored data technologies), how third-party cookies are used on our Website, and how you can manage your cookie options. Not all atoms of an element are identical - atoms of the same element can have different numbers of neutrons. Report. A glucose molecule has 96 protons, 84 neutrons and 96 electrons. See, we have 90. In this lesson, we are going to use our knowledge of the electron configurations and quantum numbers to see what goes on during the creation of light. {eq}^{204}Pb, ^{206}Pb,... See full answer below. Alkaline Earth Metals Explore this lesson to learn more about atomic mass units (amus), carbon-12, and amu to kilogram conversion! Lithium has only two naturally occuring isotopes: lithiumi-6 and lithium-7. What is the difference between an isotope and an... What is the most common isotope of bromine? Take the quiz at the end to see how well you understand the lesson. Date of Discovery Using information from the periodic table, you'll learn how to use the definitions to calculate either quantity. B For the first isotope, A = 82 protons + 124 neutrons = 206. How many protons and electrons does it have? There are over 1,000 known radioactive isotopes of elements in the periodic table. Calculate the average atomic mass of chromium... How to find fractional abundance of an isotope. Sciences, Culinary Arts and Personal That's are we looking at our atomic number? Neutrons. https://www.wikihow.com/Find-the-Number-of-Protons,-Neutrons,-and-Electrons In this lesson, we will discuss electronegativity, its trends in the periodic table, and bonding. Of the four quantum numbers, our focus in this lesson is the principal quantum number. They are electrically neutral. This is because the number of protons and electrons is always equal. That's 86 electrons saying month same amount and neutrons is 22 to minus 86 which is equal to 146"} How do scientists represent very large and very small quantities? And, how do the atoms that make up the elements in the periodic table differ from one another? The particles in the nucleus i.e the protons and the neutrons are known as nucleons. Putting it all together, given that the atomic number of lead is 82, the number of protons a … So that's equal to 54. Protons. Pb (lead) 82 protons 82electrons 125 neutrons. Lead is a chemical element with atomic number 82 which means there are 82 protons in its nucleus.Total number of protons in the nucleus is called the atomic number of the atom and is given the symbol Z.The total electrical charge of the nucleus is therefore +Ze, where e (elementary charge) equals to 1,602 x 10-19 … Become a Study.com member to unlock this Number of neutrons present in lead = 126 . Do you know where that energy and power is coming from? Lithium-6 has a mass of 6.01512 and makes up 7.42 percent of all naturally occuring lithium. Our experts can answer your tough homework and study questions. Non-Metals FTCE Physics 6-12 (032): Test Practice & Study Guide, ILTS Music (143): Test Practice and Study Guide, CSET Business Subtest I (175): Practice & Study Guide, CSET Business Subtest II (176): Practice & Study Guide, CSET Business Subtest III (177): Practice & Study Guide, CSET Science Subtest I - General Science (215): Practice & Study Guide, CSET Social Science Subtest I (114): Practice & Study Guide, FTCE Business Education 6-12 (051): Test Practice & Study Guide, NYSTCE English Language Arts (003): Practice and Study Guide, ILTS Science - Physics (116): Test Practice and Study Guide, ILTS Social Science - History (246): Test Practice and Study Guide, CSET English Subtest IV (108): Practice & Study Guide, ILTS School Counselor (181): Test Practice and Study Guide, Praxis Marketing Education (5561): Practice & Study Guide, Praxis Health Education (5551): Practice & Study Guide, Praxis Mathematics - Content Knowledge (5161): Practice & Study Guide, ILTS Social Science - Psychology (248): Test Practice and Study Guide, FTCE Music K-12 (028): Study Guide & Test Practice, ILTS School Psychologist (237): Test Practice and Study Guide, Biological and Biomedical In this lesson, we will learn how to find the average atomic mass of the elements from its isotopes. These different versions of the same element are called isotopes. The third column shows the masses of the three subatomic particles in "atomic mass units." Create your account. Electrons do not have mass. number Protons. Melting Point: 327.5 °C (600.65 K, 621.5 °F) Boiling Point: 1740.0 °C (2013.15 K, 3164.0 °F) Number of Protons/Electrons: 82. © copyright 2003-2021 Study.com. Significant Figures and Scientific Notation. Rare Earth Elements, Basic Information | Atomic Structure | Isotopes | Related Links | Citing This Page. In this lesson, you will understand the meaning of the Gibbs free energy and how it helps us predict if a chemical reaction will happen spontaneously. We know all matter is made of atoms, but how is the mass of an atom expressed? In this lesson, we will learn all about radioactive isotopes and their uses. The 35 would be the isotope number, which is the sum of protons plus neutrons. The numbers of subatomic particles in an atom can … An atomic mass unit ($$\text{amu}$$) is defined as one-twelfth of the mass of a carbon-12 atom. Atomic Pb 82 125 82 Zn 30 35 30 O 2-8 8 10 Na + 11 12 11 Mass number: total number of protons and neutrons How do we figure out how many neutrons? The chemical behavior of the isotopes is nearly the same. The atomic number of lead is 82, which means this ion contains 82 protons.Since it has a charge of +2 , this ion must contain 80 electrons.. Also know, what isotope has 14 protons and 15 neutrons? This lesson will give you the steps you need to calculate the formula mass of a compound and then will briefly discuss mass spectroscopy. Learn about the most common kinds of chemical bonds: ionic, covalent, polar covalent, and metallic. All other types of atomic nuclei are composed of two or more protons and various numbers of neutrons. The atomic weight of lead (pb) is is 207, meaning that there is a total of 207 protons and neutrons in its nucleus. Tune into this lesson to find out what matters about matter. We are given Pb. Quantum numbers were developed to characterize electrons - its electron configuration, movement and position in an atom. Use of this web site is restricted by this site's license Atoms are the basic building blocks of everything around you. What's the difference? 80 electrons. The atomic number of any isotope of any element is the number of protons in its atomic nucleus. and # electrons = # protons for a neutral atom. 28 protons and 25 electrons charge. You would not believe the hassle it took to make this little video. We first need to determine the number of protons and neutrons that are present in the nucleus of the isotope lead-206. The most common nuclide of the common chemical element lead, 208 Pb, has 82 protons and 126 neutrons, for example. Copyright © 1996-2012 Yinon Bentor. Protons, Neutrons, and Electrons Practice Worksheet Calculating the number of each particle in an atom: # Protons = Atomic Number # Electrons = Protons # Neutrons = Atomic Mass – Atomic Number OR Big # - Small # Use the periodic table to find the numbers of protons, neutrons, and electrons for atoms of the following elements. Pb: Atomic Number: 82: Atomic Mass: 207.2 atomic mass units: Number of Protons: 82: Number of Neutrons: 125: Number of Electrons: 82: Melting Point: 327.5° C: Boiling Point: 1740.0° C: Density: 11.34 grams per cubic centimeter: Normal Phase: Solid: Family: Other Metals: Period Number: 6: Cost: Unavailable Q. In nuclear reactors, promethium equilibrium exists in power operation. This page was created by Yinon Bentor. Since its atomic number is 82, there are 82 protons in its nucleus. When you drink a glass of water, you are actually drinking a combination of heavy water and light water. Lead is a heavy metal that is denser than most common materials. Did you know there are three types of oxygen? Boiling Point Actually, the atomic number of Cl is 17, so that equals the number of protons and the number of electrons. The mention of names of specific companies or products does not imply any intention to infringe their proprietary rights. The nucleus of an atom contains protons and neutrons. Atomic Mass - Definition & Examples. Symbol: Pb. Number of Neutrons 40 Ca and 208 Pb, have magic numbers of both neutrons and protons; due to that factor, these nuclei have exceptional stability and are called "doubly magic." We will look at some common ions and get to know the terms for the two types of ions. Protons, Neutrons, and Electrons Practice Worksheet Calculating the number of each particle in an atom: # Protons = Atomic Number # Electrons = Protons # Neutrons = Atomic Mass – Atomic Number OR Big # - Small # Use the periodic table to find the numbers of protons, neutrons, and electrons for atoms of the following elements. and Fe, Ni, Cu, Zn, I, Ag, Sn, Au, W, Hg, Pb, U. protons and neutrons are composed of, Protons have a positive charge. Protons carry a positive electrical change, while electrons are negatively charged, and neutrons are neutral. Question: Complete The Following Chart: Atomic # # Of Protons Mass # # Of Neutrons # Of Electrons Isotope Name 92 Uranium- 235 92 Uranium- 238 5 Boron-10 5 Boron-11 Atomic Mass 24 89 Complete The Following Chart: Atomic Atomic Symbol Protons Neutrons Electrons Number B 6 11 31 37 39 29 35 43 Pb 102 70 89 Μο 53 81 100 159 No Yb 106 159 100 207 225 206 261 … ***** by the way, the mass number is actually equal to the mass of the protons + neutrons + electrons. Are 7.5 grams and 7.50 grams the same? Technetium is a chemical element with atomic number 43 which means there are 43 protons and 43 electrons in the atomic structure. The electrons equal the number of protons. Comments In nuclear reactors, promethium equilibrium exists in power operation. Atomic number Isotopes Isotopes: atoms with the same number of protons and different number of neutrons o Ex: Potassium always has 19 p + & 19 e-but it can have 20, 21, or 22 neutrons. In this lesson, we are going to use the kinetic molecular theory of gases to explain some of their behaviors and determine how we can compare the speeds of different gases. In this lesson, we will learn about the three isotopes of hydrogen. answer! This degree of attraction is measured by the element's electronegativity. In this lesson, we'll be learning about the three types of radioactivity: alpha, beta and gamma radiation. 120 seconds . What exactly is an atom? All Rights Reserved. Question: Atomic Symbol Atomic Number Protons Neutrons Electrons Mass Number B 5 6 11 24 31 37 39 89 29 36 43 100 Pb 207 54 131 Mo 53 . For example, 116 Sn atoms has a magic number of 50 protons, and the atom 54 Fe has a magic number of 28 neutrons. To find the number of Neutrons … Electrons. symbol Atomic. Atomic mass is a characteristic of an atom that is a measure of its size. All rights reserved. Identify one element that has the same number of protons and neutrons its nucleus. Anonymous. Number of Neutrons: 125. 3) Identify a proton, electron and neutron from a drawing of an atom. The attraction of electrons is different depending on the element. By the end of the lesson, you'll be able to explain what makes these three types unique and how humans use them. Alkali Metals … In this lesson, we'll dissect atoms so we can see just what really goes into those little building blocks of matter. Platinum is used in catalytic converters, laboratory equipment, electrical contacts and electrodes, platinum resistance thermometers, dentistry equipment, and jewelry. It is plays a major role in the chemical properties of elements. Helium has more protons, neutrons, and electrons. - Definition, Types & Examples, Working Scholars® Bringing Tuition-Free College to the Community. The "210" is the mass number. The mentioned isotopes are isotopes of elements Uranium (U) and Lead (Pb). A. actual. So, because this atom has 7 protons, its atomic number is 7 which is Nitrogen (N). This chemistry video tutorial explains the subatomic particles found inside an atom such as protons, neutrons, and electrons. This is the highest atomic number for the stable elements. Metalloids Atomic no. Find out the answers to these questions in this video. Neutrons. Likewise, how many protons neutrons and electrons are there in 90sr2+? Therefore, there are 125 (207 – 82) neutrons in its nucleus. About This Site 10. Mass-Energy Conversion, Mass Defect and Nuclear Binding Energy. The most common isotope of lead has a mass number (number of protons + number of neutrons) of 208. The negative and positive charges cancel each other. The atomic number of lead is 82, which means this ion contains 82 protons. Nuclear binding energy is the minimum energy that would be required to disassemble the nucleus of an atom into its component parts. Electron Configuration A The element with 82 protons (atomic number of 82) is lead: Pb. agreement. A neutral atom has the same number of protons and electrons (charges cancel each other out). Transition Metals The chemical symbol for Lead is Pb . don't expect # protons = # neutrons. Crystal Structure, Element Groups: Discover how they form and why they hold together. When we are looking at the atomic number of an element in the periodic table, we may not know it, but these elements may have isotopes. Earn Transferable Credit & Get your Degree, Get access to this video and our entire Q&A library. What isotope of lead (Pb) has 126 neutrons? There are many applications of radioactive isotopes in various fields. Tags: ... (Pb) Tags: Question 29 . Isotope is defined as the different forms of the same element that have the same atomic number but different atomic masses. Melting Point Platinum is used in catalytic converters, laboratory equipment, electrical contacts and electrodes, platinum resistance thermometers, dentistry equipment, and jewelry. Other Metals An … The mass number is given as 204, which means: 204 = protons... See full answer below. This depends on the number of their neutrons. Protons are located outside the nucleus of an atom. SURVEY . Indicate their: - charge (positive/negative/neutral) - mass (1 amu or negligible) - role in an atom 4) Identify the following characteristics of an element: - Number of protons - Number of electrons - Number of neutrons These component parts are neutrons and protons, which are collectively called nucleons.The binding energy is always a positive number, as we need to spend energy in moving these nucleons, attracted to each other by the strong nuclear force, away … Name And lastly we have 222 r n its atomic number. Song as above by The Cat Empire. Table $$\PageIndex{1}$$ gives the properties and locations of electrons, protons, and neutrons. When an isotope is reported as U-238 for example, 238 is the mass number, which is the sum of the number of protons and neutrons. In this lesson, we are going to zoom in on the nucleus of a helium atom to explain how something as small as a nucleus can produce an extremely large amount of energy. Links, Show Table With: This lesson introduces the definitions of molecular mass and molar mass. Atomic. Protons: Neutrons: Electrons: The isotopes of an element have the same number of protons but a different number of neutrons. Molecular Mass: Definition, Formula & Calculation. mass B 6 11 24 31 37 39 89 29 35 43 100 Pb 207 102 70 89 225 Mo 53 81 206 100 159 No 261 Yb 172 106 159 Solutions for the Protons, Neutrons, and Electrons Practice Worksheet: Atomic. Determine the number of protons, neutrons, and electrons in a 210 Pb 2+ ion. 82 protons. Take a quiz and see how much knowledge you've held onto. Average Atomic Mass: Definition & Formula. Atomic Number: 82. When you hear the term 'nuclear power,' what comes to mind? Our lot you're on the is the same in our new John is 90 minus 36. The atomic number of U and Pb are 92 and 82, respectively. Some nuclei, for e.g. SURVEY . Tags: Question 4 . Atoms of elements have different versions of each other called isotopes. … A glucose molecule's chemical formula is: C 6 H 1 C 6 H 1 ; simple: if a nucleus has an imbalance between the number of protons and neutrons it contains, leading it to become unstable, then one or more protons or neutrons will transform into the other to redress the balance. Atoms are made of protons, neutrons, and electrons. Also Know, how many protons neutrons and electrons are there in 90sr2+? This means it has 82 Protons and a neutral atom also has 82 electrons. I love the Cat Empire. Atomic Mass: 207.2 amu. Protons have a positive charge; electrons have a negative charge; and neutrons are electrically neutral. Protons, neutrons and electrons of all elements are mentioned in the table below. What causes you to be able to smell something from so far away? 208 - 82 = 126, so this isotope has 126 neutrons in its nucleus. In order to really understand how atoms combine to form molecules, it's necessary to be familiar with their structure. » 28 protons and 25 electrons charge. Atomic Number – Protons, Electrons and Neutrons in Lead. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_394 | Approximating erf by tanh - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T19:14:24Z http://mathoverflow.net/feeds/question/117735 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/117735/approximating-erf-by-tanh Approximating erf by tanh Aryeh Kontorovich 2012-12-31T15:18:47Z 2012-12-31T17:57:10Z <p>It appears to be well-known that $\tanh(x)\le \mathrm{erf}(x)$ on $[0,\infty)$. It's off-handedly mentioned <a href="http://web.cs.dal.ca/~jborwein/tanh-sinh.pdf" rel="nofollow">here</a>, for example. Where can I find a formal proof? On the one hand, it's hard to imagine that a "classic" like this wouldn't have been proven already. On the other hand, the Taylor expansions are somewhat involved (tanh involves Bernoulli numbers) and unfortunately, the inequality does not hold termwise in the expansions -- so it's certainly far from obvious.</p> http://mathoverflow.net/questions/117735/approximating-erf-by-tanh/117743#117743 Answer by Alexandre Eremenko for Approximating erf by tanh Alexandre Eremenko 2012-12-31T17:17:41Z 2012-12-31T17:17:41Z <p>Let $f(x)={\mathrm{erf}}(x)-\tanh(x)$. It can be easily seen from Taylor series at $0$ and from asymptotics at $\infty$ that $f(x)>0$ for small $x$ and for large $x$.</p> <p>Let us prove that $f(x)>0$ by contradiction. Suppose that $f(x)$ is negative for some $x$, then $f'$ must have at least $3$ positive zeros, by Rolle's theorem. This means that the equation $$g(x):=e^{-x^2}(e^{2x}+2+e^{-2x})=2\sqrt{\pi}$$ has at least $3$ positive solutions. But this is not the case because the LHS is monotone. Indeed, differentiating $g$, dividing by $e^{-x^2}$ and replacing $2x$ with $y$ we obtain $$g'(x)=\sinh(y)-y\cosh(y)-y<0,$$ because $\sinh(y) < y \cosh(y)$ as you can see from their Taylor series.</p> http://mathoverflow.net/questions/117735/approximating-erf-by-tanh/117745#117745 Answer by Gerald Edgar for Approximating erf by tanh Gerald Edgar 2012-12-31T17:37:42Z 2012-12-31T17:57:10Z <p>First, \begin{align} 1-\mathrm{erf}(x) &= \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}dt, \cr 1-\tanh(x) &= \int_x^\infty \mathrm{sech}^2 t\;dt . \end{align} Subtract: $$\mathrm{erf}(x)-\mathrm{tanh}(x) = \int_x^\infty \left(\mathrm{sech}^2 t - \frac{2}{\sqrt{\pi}}e^{-t^2}\right)dt$$ So it suffices to show that this integrand is positive. It is positive for $t>1$ (proof needed), so we establish $\mathrm{erf}(x) > \mathrm{tanh}(x)$ for $x > 1$.</p> |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_395 | # How do you find the value of 16^(1/4)?
May 16, 2018
2
#### Explanation:
${16}^{\frac{1}{4}} =$
$\sqrt[4]{16} =$
$\sqrt[4]{2 \cdot 2 \cdot 2 \cdot 2} =$
$2$ |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_396 | # Can you find the solution?
Algebra Level 5
The number of solutions (x, y, z) to the system of equations $x+2y+4z=9$ $4yz+2xz+xy=13$ $xyz=3$ such that at least two of x, y, z are integers is $Note-\:\:Its\:\:an\:\:old\:\:KVPY\:\:question$
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_397 | # Ngô Quốc Anh
## September 9, 2016
### Benefits of “complete” and “compact” for analysis on Riemannian manifolds
Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 10:08
When working on Riemannian manifolds, it is commonly assumed that the manifold is complete and compact. (The case of non-compactness is also of interest too.) In this entry, let us review the role of completeness and compactness in this setting.
How important the completeness is? Let us recall that for given a Riemannian manifold $(M,g)$, what we have is a nice structure as well as an appropriate analysis on any tangent space $T_pM$. For a $C^1$-curve $\gamma : [a,b] \to M$ on $M$, the length of $\gamma$ is
$\displaystyle L(\gamma) = \int_a^b \sqrt{g(\gamma (t)) \langle \partial_t \gamma \big|_t, \partial_t \gamma \big|_t\rangle} dt$
where $\partial_t \gamma\big|_t \in T_{\gamma (t)}M$ a tangent vector. (Note that by using curves, the tangent vector $\partial_t \gamma\big|_t$ is being understood as follows
$\displaystyle \partial_t \gamma\big|_t (f) = (f \circ \gamma)'(t)$
for any differentiable function $f$ at $\gamma(t)$.) Length of piecewise $C^1$ curves can be defined as the sum of the lengths of its pieces. From this a distance on $M$ whose topology coincides with the old one on $M$ is given as follows
$\displaystyle d_g(x,y) = \inf_\gamma L(\gamma)$
where the infimum is taken on all over the set of all piecewise $C^1$ curves connecting $x$ and $y$.
## April 20, 2016
### Stereographic projection, 6
Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 1:08
I want to propose an alternative way to calculate the Jacobian of the stereographic projection $\mathcal S$. In Cartesian coordinates $\xi=(\xi_1, \xi_2,...,\xi_{n+1})$ on the sphere $\mathbb S^n$ and $x=(x_1,x_2,...,x_n)$ on the plane, the projection and its inverse are given by the formulas
$\displaystyle\xi _i = \begin{cases} \dfrac{{2{x_i}}}{{1 + {{\left| x \right|}^2}}},&1 \leqslant i \leqslant n, \hfill \\ \dfrac{{{{\left| x \right|}^2} - 1}}{{1 + {{\left| x \right|}^2}}},&i = n + 1. \hfill \\ \end{cases}$
and
$\displaystyle {x_i} = \frac{{{\xi _i}}}{{1 - {\xi _{n + 1}}}}, \quad 1 \leqslant i \leqslant n$.
It is well-known that the Jacobian of the stereographic projection $\mathcal S: \xi \mapsto x$ is
$\displaystyle \frac{\partial \xi}{\partial x} = {\left( {\frac{2}{{1 + {{\left| x \right|}^2}}}} \right)^n}.$
The way to calculate its Jacobian is to compare the ratio of volumes. First pick two arbitrary points $x, y \in \mathbb R^n$ and denote $\xi = \mathcal S(x)$ and $\eta = \mathcal S(y)$.
The Euclidean distance between $\xi$ and $\eta$ is
$\displaystyle |\xi -\eta|^2 = \sum_{i=1}^{n+1} |\xi_i - \eta_i|^2 =\sum_{i=1}^n |\xi_i - \eta_i|^2+|\xi_{n+1} - \eta_{n+1}|^2.$
## August 9, 2015
### The third fundamental lemma in the method of moving spheres
Filed under: Uncategorized — Ngô Quốc Anh @ 4:03
In this post, we proved the following result (appeared in a paper by Y.Y. Li published in J. Eur. Math. Soc. (2004))
Lemma 1. For $n \geqslant 1$ and $\nu \in \mathbb R$, let $f$ be a function defined on $\mathbb R^n$ and valued in $[-\infty, +\infty]$ satisfying
$\displaystyle {\left( {\frac{\lambda }{{|y - x|}}} \right)^\nu }f\left( {x + {\lambda ^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) \leqslant f(y), \quad \forall |y - x| > \lambda > 0.$
Then $f$ is constant or $\pm \infty$.
Later, we considered the equality case in this post and proved the following result:
Lemma 2. Let $n\geqslant 1$, $\nu \in \mathbb R$ and $f \in C^0(\mathbb R^n)$. Suppose that for every $x \in \mathbb R^n$ there exists $\lambda(x)>0$ such that
$\displaystyle {\left( {\frac{\lambda(x) }{{|y - x|}}} \right)^\nu }f\left( {x + {\lambda(x) ^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) =f(y), \quad \forall |y - x| > 0.$
Then for some $a \geqslant 0$, $d>0$ and $\overline x \in \mathbb R^n$
$\displaystyle f(x) = \pm a{\left( {\frac{1}{{d + {{\left| {x - \overline x } \right|}^2}}}} \right)^{\frac{\nu }{2}}}.$
In this post, we consider the third result which can be stated as follows:
## April 28, 2015
### On the simplicity of the first eigenvalue of elliptic systems with locally integrable weight
Filed under: Uncategorized — Ngô Quốc Anh @ 0:52
Of interest in this note is the simplicity of the first eigenvalue of the following problem
$\begin{array}{rcl}-\text{div}(h_1 |\nabla u|^{p-2}\nabla u) &=& \lambda |u|^{\alpha-1} |v|^{\beta-1}v \quad \text{ in } \Omega\\-\text{div}(h_2 |\nabla v|^{q-2}\nabla v) &=& \lambda |u|^{\alpha-1} |v|^{\beta-1}u \quad \text{ in } \Omega\\u &=&0 \quad \text{ on } \partial\Omega\\v &=&0 \quad \text{ on } \partial\Omega\end{array}$
where $1 \leqslant h_1, h_2 \in L_{\rm loc}^1 (\Omega)$ and $\alpha, \beta>0$ satisfy
$\displaystyle \frac \alpha p + \frac \beta q = 1$
with $p,q >1$. A simple variational argument shows that $\lambda$ exists and can be characterized by
$\lambda = \inf_{\Lambda} J(u,v)$
where
$\displaystyle J(u,v)=\frac \alpha p \int_\Omega h_1 |\nabla u|^p dx + \frac \beta q \int_\Omega h_2 |\nabla v|^q dx$
and
$\Lambda = \{(u,v) \in W_0^{1,p} (\Omega) \times W_0^{1,q} (\Omega) : \Lambda (u,v) = 1\}$
with
$\displaystyle \Lambda (u,v)= \int_\Omega |u|^{\alpha-1}|v|^{\beta-1} uv dx.$
## April 10, 2015
### Existence of antiderivative of discontinuous functions
Filed under: Uncategorized — Ngô Quốc Anh @ 0:17
It is well-known that every continuous functions admits antiderivative. In this note, we show how to prove existence of antiderivative of some discontinuous functions.
A typical example if the following function
$f(x)=\begin{cases}\sin \frac 1x, & \text{ if } x \ne 0,\\ 0, & \text{ if }x=0.\end{cases}$
By taking to different sequences $x_k = 1/(2k\pi)$ and $y_k = 1/(\pi/2 + 2k\pi)$ we immediately see that $f$ is discontinuous at $x=0$. However, we will show that $f$ admits $F$ as its antiderivative.
To this end, we first consider the following function
$G(x)=\begin{cases}x^2\cos \frac 1x, & \text{ if } x \ne 0,\\ 0, & \text{ if }x=0.\end{cases}$
First we show that $G$ is differentiable. Clearly whenever $x \ne 0$, we obtain
$\displaystyle G'(x)=\sin \frac 1x + 2x \cos \frac 1x.$
## April 5, 2015
### The set of continuous points of Riemann integrable functions is dense
Filed under: Uncategorized — Ngô Quốc Anh @ 15:03
In this note, we prove that the set of continuous point of Riemann integrable functions $f$ on some interval $[a,b]$ is dense in $[a,b]$. Our proof start with the following simple observation.
Lemma: Assume that $P=\{t_0=a,...,t_n=b\}$ is a partition of $[a,b]$ such that
$\displaystyle U(f,P)-L(f,P)<\frac{b-a}m$
for some $m$; then there exists some index $i$ such that $M_i-m_i < \frac 1m$ where $M_i$ and $m_i$ are the supremum and infimum of $f$ over the subinterval $[t_{i-1},t_i]$.
We now prove this result.
Proof of Lemma: By contradiction, we would have $M_i-m_i \geqslant \frac 1m$ for all $i$; hence
$\displaystyle \frac{b-a}m = \sum_{i} \frac{t_i-t_{i-1}}{m}\leqslant \sum_{i} \big(M_i-m_i\big)(t_i-t_{i-1})=U(f,P)-L(f,P),$
We now state our main result:
Theorem. Let $f$ be Riemann integrable over $[a,b]$. Define
$\displaystyle \Gamma = \{ x\in [a,b] : f \text{ is continuous at } x\}$
Then $\Gamma$ is dense in $[a,b]$.
## March 13, 2015
### Comparing topologies of normed spaces: The equivalency of norms and the convergence of sequences
Filed under: Uncategorized — Ngô Quốc Anh @ 23:01
The aim of this note is to derive some connections between topologies of normed spaces in terms of the equivalency of norms and the convergence of sequences.
Topological space and its topology: First, we start with a topological space, call $X$. Its topology, say $\mathcal T$ is the collection of subsets of $X$ which satisfies certain conditions. In the literature, each member of the collection $\mathcal T$ is called an open set.
Regarding to topologies we have the following basic facts:
• Given two topologies $\mathcal T_1$ and $\mathcal T_2$ on $X$, we say that $\mathcal T_1$ is stronger (or finer or richer) than $\mathcal T_2$ if $\mathcal T_2 \subset \mathcal T_1$.
• Given a sequence $(x_n)_n$ in $X$, we say that $x_n$ converges to $x$ in topology $\mathcal T$ of $X$ if for any neighborhood $V$ of $x$, there exists some large number $N$ such that $x_n \in V$ for all $n \geqslant N$. (Here by the neighborhood $V$ of $x$ we mean that there exists an open set $O$ of $X$, i.e. $O$ is a member of the topology $\mathcal T$, such that $x \in O \subset V$.)
The key ingredient to compare topologies is to make use of the identity map. In the following part, we state a result which shall be used frequently in this note.
Topologies under the identity map: Given two topologies $\mathcal T_1$ and $\mathcal T_2$ on a topological space $X$, we are interested in comparing $\mathcal T_1$ and $\mathcal T_2$ in terms of the identity map $\rm id : (X, \mathcal T_1) \to (X, \mathcal T_2)$.
Lemma 1. The identity map $\rm id : (X, \mathcal T_1) \to (X, \mathcal T_2)$ is continuous if and only if $\mathcal T_1$ is stronger than $\mathcal T_2$.
Proof.
The proof is relatively easy. Indeed, if the map $\rm id$ is continuous, then the preimage of any $O_2 \in \mathcal T_2$ is also a member of $\mathcal T_1$ which immediately implies that $\mathcal T_1$ includes $\mathcal T_2$.
Having Lemma 1 in hand, we now try to compare topologies using norms.
## February 25, 2015
### Continuous functions on subsets can be extended to the whole space: The Kirzbraun-Pucci theorem
Filed under: Uncategorized — Ngô Quốc Anh @ 1:22
Let $f$ be a continuous function defined on a set $E \subset \mathbb R^N$ with values in $\mathbb R$ and with modulus of continuity
$\displaystyle \omega_f (s) := \sup_{|x-y|\leqslant s,x,y\in E} |f(x) - f(y)| \quad s>0.$
Obviously, the function $s \mapsto \omega_f(s)$ is nonnegative and nondecreasing in $[0,+\infty)$.
Our first assumption is that $\omega_f$ is bounded from above in $[0, \infty)$ by some increasing, affine function; that is to say there exists some $a,b \in \mathbb R^+$ such that
$\displaystyle \omega_f (s) \leqslant a s +b \quad \forall s \geqslant 0$.
Associated with $\omega_f$ having the above first assumption is the concave modulus of continuity of $f$, i.e. some smallest concave function $c_f$ lies above $\omega_f$. Such the function $c_f$ can be easily constructed using the following
$\displaystyle c_f (s) = \inf_\ell \{\ell(s) : \ell \text{ is affine and } \ell \geqslant \omega_f \text{ in } [0,+\infty)\}.$
As can be easily seen, once $\omega_f$ can be bounded from above by some affine function, the concave modulus of continuity of $f$ exists and is well-defined.
By definition and the monotonicity of $\omega_f$, we obtain
$\displaystyle |f(x)-f(y)| \leqslant \omega_f (|x-y|) \leqslant c_f (|x-y|).$
In this note, we prove the following extension theorem.
Theorem (Kirzbraun-Pucci). Let $f$ be a real-valued, uniformly continuous function on a set $E \subset \mathbb R^N$ with modulus of continuity $\omega_f$ satisfying the first assumption. There exists a continuous function $\widetilde f$ defined on $\mathbb R^N$ that coincides with $f$ on $E$. Moreover, $f$ and $\widetilde f$ have the same concave modulus of continuity $c_f$ and
$\displaystyle \sup_{\mathbb R^N} \widetilde f = \sup_E f, \quad \inf_{\mathbb R^N} \widetilde f = \inf_E f.$
## February 22, 2015
### The conditions (NN), (P), (NN+) and (P+) associated to the Paneitz operator for 3-manifolds
Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 18:54
Of recent interest is the prescribed Q-curvature on closed Riemannian manifolds since it involves high-order differential operators.
In a previous post, I have talked about prescribed Q-curvature on 4-manifolds. Recall that for 4-manifolds, this question is equivalent to finding a conformal metric $\widetilde g =e^{2u}g$ for which the Q-curvature of $\widetilde g$ equals the prescribed function $\widetilde Q$? That is to solving
$\displaystyle P_gu+2Q_g=2\widetilde Q e^{4u},$
where for any $g$, the so-called Paneitz operator $P_g$ acts on a smooth function $u$ on $M$ via
$\displaystyle {P_g}(u) = \Delta _g^2u - {\rm div}\left( {\frac{2}{3}{R_g} - 2{\rm Ric}_g} \right)du$
which plays a similar role as the Laplace operator in dimension two and the Q-curvature of $\widetilde g$ is given as follows
$\displaystyle Q_g=-\frac{1}{12}(\Delta\text{Scal}_g -\text{Scal}_g^2 +3|{\rm Ric}_g|^2).$
Sometimes, if we denote by $\delta$ the negative divergence, i.e. $\delta = - {\rm div}$, we obtain the following formula
$\displaystyle {P_g}(u) = \Delta _g^2u + \delta \left( {\frac{2}{3}{R_g} - 2{\rm Ric}_g} \right)du.$
Generically, for $n$-manifolds, we obtain
$\displaystyle Q_g=-\frac{1}{2(n-1)} \Big(\Delta\text{Scal}_g - \frac{n^3-4n^2+16n-16}{4(n-1)(n-2)^2} \text{Scal}_g^2+\frac{4(n-1)}{(n-2)^2} |{\rm Ric}_g|^2 \Big)$
and
$\displaystyle {P_g}(u) = \Delta _g^2u + {\rm div}\left( { a_n {R_g} + b_n {\rm Ric}_g} \right)du + \frac{n-4}{2} Q_g u,$
where $a_n = -((n-2)^2+4)/2(n-1)(n-2)$ and $b_n =4/(n-2)$.
## January 24, 2015
### Reversed Gronwall-Bellman’s inequality
Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 23:01
In mathematics, Gronwall’s inequality (also called Grönwall’s lemma, Gronwall’s lemma or Gronwall–Bellman inequality) allows one to bound a function that is known to satisfy a certain differential or integral inequality by the solution of the corresponding differential or integral equation. The differential form was proven by Grönwall in 1919. The integral form was proven by Richard Bellman in 1943. A nonlinear generalization of the Gronwall–Bellman inequality is known as Bihari’s inequality.
First, we consider the Gronwall inequality.
Type 1. Bounds by integrals based on lower bound $a$.
Let $\beta$ and $u$ be real-valued continuous functions defined on $[a,b]$. If $u$ is differentiable in $(a,b)$ and satisfies the differential inequality
$\displaystyle u'(t) \leqslant \beta(t) u(t),$
then $u$ is bounded by the solution of the corresponding differential equation $y'(t) = \beta (t)y(t)$, that is to say
$\displaystyle \boxed{u(t) \leqslant u(a) \exp\biggl(\int_a^t \beta(s) ds\biggr)}$
for all $t \in [a,b]$.
Older Posts » |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_398 | ## The smoothness of solutions to nonlinear variational inequalities.(English)Zbl 0278.49011
Consider a variational inequality: $u\in K\quad\text{and}\quad \int_\Omega Au(v-u)\,dx\geq \int_\Omega f(v-u)\quad \forall v\in K$ where $$K=\{v: v\geq \psi \;\text{in}\;\Omega\;\text{and}\;v=0\;\text{on}\;\partial\Omega\}$$ and $$Au=-\sum D_ja_j(Du)$$, $$a_j$$ is $$C^2$$ and satisfies $$a(p)-a(q)\cdot p-q\geq \nu\,| p-q|^2$$ uniformly on bounded sets. Under some conditions, it is proved that $$u\in C^{1,1}(\Omega)$$ and that $$Au$$ is of bounded variation.
Reviewer: H. Brézis
### MSC:
49J40 Variational inequalities
Full Text: |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_399 | ### Body-centered cubic problems
Problem #1: The edge length of the unit cell of Ta, is 330.6 pm; the unit cell is body-centered cubic. Tantalum has a density of 16.69 g/cm3.
(a) calculate the mass of a tantalum atom.
(b) Calculate the atomic weight of tantalum in g/mol.
Solution:
1) Convert pm to cm:
330.6 pm x 1 cm/1010 pm = 330.6 x 10¯10 cm = 3.306 x 10¯8 cm
2) Calculate the volume of the unit cell:
(3.306 x 10¯8 cm)3 = 3.6133 x 10¯23 cm3
3) Calculate mass of the 2 tantalum atoms in the body-centered cubic unit cell:
16.69 g/cm3 times 3.6133 x 10¯23 cm3 = 6.0307 x 10¯22 g
4) The mass of one atom of Ta:
6.0307 x 10¯22 g / 2 = 3.015 x 10¯22 g
5) The atomic weight of Ta in g/mol:
3.015 x 10¯22 g times 6.022 x 1023 mol¯1 = 181.6 g/mol
Problem #2a: Chromium crystallizes in a body-centered cubic structure. The unit cell volume is 2.583 x 10¯23 cm3. Determine the atomic radius of Cr in pm.
Solution:
1) Determine the edge length of the unit cell:
$\sqrt[3]{\mathrm{2.583 x 10¯23cm3}}$ = 2.956 x 10¯8 cm
2) Examine the following diagram:
The triangle we will use runs differently than the triangle used in fcc calculations.d is the edge of the unit cell, however d√2 is NOT an edge of the unit cell. It is adiagonal of a face of the unit cell. 4r is a body diagonal. Since it is a right triangle,the Pythagorean Theorem works just fine.
We wish to determine the value of 4r, from which we will obtain r, the radius of the Cr atom. Using the Pythagorean Theorem, we find:
d2 + (d√2)2 = (4r)2
3d2 = (4r)2
3(2.956 x 10¯8 cm)2 = 16r2
r = 1.28 x 10¯8 cm
3) The conversion from cm to pm is left to the student.
Problem #2b: Chromium crystallizes with a body-centered cubic unit cell. The radius of a chromium atom is 128 pm . Calculate the density of solid crystalline chromium in grams per cubic centimeter.
Solution:
1) Convert pm to cm:
(125 pm) (1 cm / 1010 pm) = 1.25 x 10¯8 cm
2) Use the Pythagorean theorem to calculate the unit cell edge length:
d2 + (d√2)2 = (4r)2
3d2 = (4r)2
3d2 = 16r2
3d2 = (16) (1.25 x 10¯8 cm)2
d = 2.8868 x 10¯8 cm
3) Calculate volume of the unit cell
(2.8868 x 10¯8 cm)3 = 2.4056 x 10¯23 cm3
4) Determine mass of two atoms in body-centered unit cell:
51.996 g/mol / 6.022 x 1023 atoms/mol = 8.63434 x10¯23 g/atom
8.63434 x10¯23 g/atom times 2 = 1.726868 x 10¯22 g
5) Determine the density:
1.726868 x 10¯22 g / 2.4056 x 10¯23 cm3 = 7.18 g/cm3 (to three sig figs)
Book value is 7.15.
Problem #3: Barium has a radius of 224 pm and crystallizes in a body-centered cubic structure. What is the edge length of the unit cell? (This is the reverse of problem #4.)
Solution:
1) Calculate the value for 4r (refer to the above diagram):
radius for barium = 224 pm
4r = 896 pm
2) Apply the Pythagorean Theorem:
d2 + (d√2)2 = (896)2
3d2 = 802816
d2 = 267605.3333. . .
d = 517 pm
Problem #4: Metallic potassium has a body-centered cubic structure. If the edge length of unit cell is 533 pm, calculate the radius of potassium atom. (This is the reverse of problem #3.)
Solution:
1) Solve the Pythagorean Theorem for r (with d = the edge length):
d2 + (d √2)2 = (4r)2
d2 + 2d2 = 16r2
3d2 = 16r2
r2 = 3d2 / 16
r = √3 (d / 4)
2) Solve the problem:
√3 (533 / 4)
r = 231 pm
Problem #5: Sodium has a density of 0.971 g/cm3 and crystallizes with a body-centered cubic unit cell. (a) What is the radius of a sodium atom? (b) What is the edge length of the cell? Give answers in picometers.
Solution:
1) Determine mass of two atoms in a bcc cell:
22.99 g/mol divided by 6.022 x 1023 mol-1 = 3.81767 x 10-23 g (this is the average mass of one atom of Na)
3.81767 x 10-23 g times 2 = 7.63534 x 10-23 g
2) Determine the volume of the unit cell:
7.63534 x 10-23 g divided by 0.971 g/cm3 = 7.863378 x 10-23 cm3
3) Determine the edge length (the answer to (b)):
$\sqrt[3]{\mathrm{7.863378 x 10-23cm3}}$ = 4.2842 x 10-8 cm
4) Use the Pythagorean Theorem (refer to above diagram):
d2 + (d√2)2 = (4r)2
3d2 = 16r2
r2 = 3(4.2842 x 10-8)2 / 16
r = 1.855 x 10-8 cm
The radius of the sodium atom is 185.5 pm. The edge length is 428.4 pm. The manner of these conversions are left to the reader.
Problem #6: At a certain temperature and pressure an element has a simple body-centred cubic unit cell. The corresponding density is 4.253 g/cm3 and the atomic radius is 1.780 Å. Calculate the atomic mass (in amu) for this element.
Solution:
1) Convert 1.780 Å to cm:
1.780 Å = 1.780 x 10-8 cm
2) Use the Pythagorean Theorem to calculate d, the edge length of the unit cell:
d2 + (d√2)2 = (4r)2
3d2 = 16r2
d2 = (16/3) (1.780 x 10-8 cm)2
d = 4.11 x 10-8 cm
3) Calcuate the volume of the unit cell:
(4.11 x 10-8 cm)3 = 6.95 x 10-23 cm3
4) Calcuate the mass inside the unit cell:
6.95 x 10-23 cm3 times 4.253 g/cm3 = 2.95 x 10-22 g
Use a ratio and proportion to calculate the atomic mass:
2.95 x 10-22 g is to two atoms as 'x' is to 6.022 x 1023 mol-1
x = 88.95 g/mol (or 88.95 amu)
Problem #7: Mo crystallizes in a body-centered cubic arrangement. Calculate the radius of one atom, given the density of Mo is 10.28 g /cm3.
Solution:
1) Determine mass of two atoms in a bcc cell:
95.96 g/mol divided by 6.022 x 1023 mol-1 = 1.59349 x 10-22 g (this is the average mass of one atom of Mo)
1.59349 x 10-22 g times 2 = 3.18698 x 10-22 g
2) Determine the volume of the unit cell:
3.18698 x 10-22 g divided by 10.28 g/cm3 = 3.100175 x 10-23 cm3
3) Determine the edge length:
$\sqrt[3]{\mathrm{3.100175 x 10-23cm3}}$ = 3.14144 x 10-8 cm
4) Use the Pythagorean Theorem (refer to above diagram):
d2 + (d√2)2 = (4r)2
3d2 = 16r2
r2 = 3(3.14144 x 10-8)2 / 16
r = 1.3603 x 10-8 cm (or 136.0 pm, to four sig figs)
Problem #8: Sodium crystallizes in body-centered cubic system, and the edge of the unit cell is 430. pm. Calculate the dimensions of a cube that would contain one mole of Na.
Solution:
A cube that is bcc has two atoms per unit cell.
6.022 x 1023 atoms divided by 2 atoms/cell = 3.011 x 1023 cells required.
430. pm = 4.30 x 10-8 cm <--- I'm going to give the answer in cm3 rather than pm3
(4.30 x 10-8 cm)3 = 7.95 x 10-23 cm3 <--- vol. of unit cell in cm3
(3.011 x 1023 cell) (7.95 x 10-23 cm3/cell) = 23.9 cm3
23.9 cm3 would be a cube 2.88 cm on a side (2.88 being the cube root of 23.9)
Problem #9: Vanadium crystallizes with a body-centered unit cell. The radius of a vanadium atom is 131 pm. Calculate the density of vanadium. (in g/cm3)
Solution:
1) We are going to use the Pythagorean Theorem to determine the edge length of the unit cell. That edge length will give us the volume.
131 pm times ( 1 cm / 1010 pm) = 131 x 10-10 cm = 1.31 x 10-8 cm
The right triangle for Pythagorean Theorem is here. The image is in problem #2.
3d2 = (4 * 1.31 x 10-8 cm)2
d2 = (4 * 1.31 x 10-8 cm)2 / 3
d = 3.0253 x 10-8 cm <--- this is the edge length
Cube the edge length to give the volume:
2.7689 x 10-23 cm3
2) We will use the average mass of one V atom and the two atoms in bcc to determine the mass of V inside the unit cell.
50.9415 g/mol divided by 6.022 x 1023 mol-1 = 8.459 x 10-23 g <--- average mass of one atom
8.459 x 10-23 g times 2 = 1.6918 x 10-22 g <--- mass of V in unit cell
3) Step 2 divided by step 1 gives the density.
1.6918 x 10-22 g / 2.7689 x 10^-23 cm3 = 6.11 g/cm3
Problem #10: In modeling solid-state structures, atoms and ions are most often modeled as spheres. A structure built using spheres will have some empty space in it. A measure of the empty (also called void) space in a particular structure is the packing efficiency, defined as the volume occupied by the spheres divided by the total volume of the structure.
Given that a solid crystallizes in a body-centered cubic structure that is 3.05 Å on each side, please answer the following questions. (The ChemTeam formatted this question while in transit through the Panama Canal, Nov. 7, 2010.)
Solution:
a. How many atoms are there in each unit cell?
2
b. What is the volume of one unit cell in Å3?
(3.05 Å)3 = 28.372625 Å3
c. Assuming that the atoms are spheres and the radius of each sphere is 1.32 Å, what is the volume of one atom in Å3?
(4/3) (3.141592654) (1.32)3 = 9.63343408 Å3
I used the key for π on my calculator, so there were some internal digits in addition to that last 4 (which is actually rounded up from the internal digits).
d. Therefore, what volume of atoms are in one unit cell?
(9.63343408 Å3 times 2) = 19.26816686 Å3
e. Based on your results from parts b and d, what is the packing efficiency of the solid expressed as a percentage?
19.26816686 Å3 / 28.372625 Å3 = 0.679
67.9% |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3472 | # What is a primitive point on an elliptic curve?
While working with elliptic curves for cryptography reasons, I found the notion of a primitive point, but no definition.
For example, $P(0,6)$ is a primitive point on the elliptic curve $y^2\equiv x^3+2x+2 \mod 17$. What does that mean? How can I tell if a point is primitive or not?
-
It seems to mean that the point generates the group. I guess there are algorithms for checking this. – Qiaochu Yuan Feb 9 '11 at 16:40
The points on an elliptic curve (plus a 'point at infinity') form a group under a certain addition law, explained in this Wikipedia article. (You probably know this already.) A primitive point $P$ is simply a generator of this group: all elements of the group can be expressed as $P+P+...+P$ ($k$ times) for some $k$. If the elliptic curve has a prime number of points, then all its points (except the point at infinity) are primitive; but in general, the elliptic curve may or may not have a primitive point.
-
I have a question from Tonyk... you mentioned that "If the elliptic curve has a prime number of points, then all its points (except the point at infinity) are primitive;" Can anyone give a reason why is that so. I mean what difference does it make if the order of the group is a prime number? thanks – user9462 Apr 11 '11 at 22:36
Because a point on the curve always generates a subgroup of the full elliptic curve group - just think of P, (-P), P+P, etc. with the obvious group structure; it's easy enough to see that this is a subgroup. (Indeed, this is true for any group, not just for elliptic curves). The order of a subgroup divides the order of a group, so if the order of the group is prime then any subgroups are either trivial or the full group. – Steven Stadnicki Apr 11 '11 at 22:54
How can you find the primitive point of a curve? – curious Dec 3 '13 at 15:16
@curious This might be a good, separate question (rather than a comment). – Niels Abildgaard May 12 at 17:52 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3473 | ### The equation of a tangent which is parallel to a given line
Thứ sáu - 31/01/2020 11:06
This article shows how to write the equation of a tangent which is parallel to a given line.
The following facts are regularly used.
$(i)$ If $\left( d \right)$ has its equation of form $\left( d \right):y = kx + b$, then its slope is $k$. $(ii)$ The slope of the tangent $\Delta$ of the graph of function $y=f\left( x \right)$ at $x_0$ is $f'\left( {{x_0}} \right)$. $(iii)$ Every two parallel lines have the same slope. $(iv)$ The equation of the targent $\Delta$ of the fuction $y=f\left( x \right)$ at $M_0(x_0;y_0)$ is $$y = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) + {y_0}.\;\;$$
Problem. Write the equation of the tangent $(\Delta)$ of the graph $\left( C \right):y = f\left( x \right)$ paralleling to the strange line $\left( d \right):y = kx + b$.
Solution. Let $M_0(x_0;y_0)$ be the point of tangency. It follows from $(i)$ and $(ii)$ that the slope of $\Delta$ and $d$ are $f'\left( {{x_0}} \right)$ and $k$ respectively.
Since $\Delta \parallel d$, the statement $(iii)$ implies that $f'\left( {{x_0}} \right) = k$, from which it follows that $x_0$ is a solution of the equation $f'\left( {{x}} \right) = k$.
Therefore, to find equation of the tangent $(\Delta)$, we can do the following steps.
Step 1. Solve the equation $f'\left( {{x}} \right) = k$. Its solution $x_0$, say, is the $x$-coordinate of the point of tangency. Step 2. Calculate ${y_0} := f\left( {{x_0}} \right)$ to obtain $M_0(x_0;y_0)$. Step 3. Write the equation of the tangent of the graph $\left( C \right)$ with the point of tangency $M_0(x_0;y_0)$ according to $(iv)$.
Example 1. Write the equation of the tangent $\Delta$ of $(C): y = {x^2} - 2x - 1$ paralleling to $\left( d \right):y = 2x - 1$.
Solution. Step 1. We have $f\left( x \right) = {x^2} - 2x - 1 \Rightarrow f'\left( x \right) = 2x - 2.$ The $x$-coordinate of the point of tangency is the solution of the equation $$f'\left( x \right) = 2 \Leftrightarrow 2x - 2 = 2 \Leftrightarrow x = 2.$$
Step 2. Substitute $x_0=2$ into $(C)$ to obtain $y_0=-1$. It follows that the coordinate of the point of tangency is ${M_0}\left( {2; - 1} \right).$
Step 3. We also have $f'\left( {{x_0}} \right) = 2$. The equation of the tangent at ${M_0}\left( {2; - 1} \right)$ is
$$y = f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) + {y_0} \Leftrightarrow y = 2\left( {x - 2} \right) - 1 \Leftrightarrow y = 2x - 5.$$
Example 2. Write the equaion of the tangent $\Delta$ of $\left( C \right): y = {x^3} + 3x - 1$ paralleling to $\left( d \right):y = 6x - 1$.
Solution. The $x$-coordinate of the point of tangency is the solution of the following equation.
$$f'\left( x \right) = 6 \Leftrightarrow 3{x^2} + 3 = 6 \Leftrightarrow \left[ \begin{array}{l} {x_1} = 1 \Rightarrow {y_1} = f\left( 1 \right) = 3\\ {x_2} = - 1 \Rightarrow {y_2} = f\left( 1 \right) = - 5 \end{array} \right.$$
Since there are two solutions, there exist two points of tangency ${M_1}\left( {1;3} \right),{M_2}\left( { - 1; - 5} \right)$.
The equation of the tangent at ${M_1}\left( {1;3} \right)$ is $$\left( {{\Delta_1}} \right):\;\;\;\;y = 6\left( {x - {x_1}} \right) + {y_1} \Leftrightarrow y = 6\left( {x - 1} \right) + 3 \Leftrightarrow y = 6x - 3.$$
Similarly, the equation of the tangent at ${M_2}\left( { - 1; - 5} \right)$ is $$\left( {{\Delta_2}} \right):\;\;\;\;y = 6\left( {x - {x_2}} \right) + {y_2} \Leftrightarrow y = 6\left( {x + 1} \right) - 5 \Leftrightarrow y = 6x + 1.$$
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3474 | Tuesday
March 3, 2015
# Homework Help: Calculus
Posted by John on Monday, March 4, 2013 at 12:30am.
Given that \displaystyle \int_0^4 x^3\sqrt{9+x^2} dx = a, what is the value of \lfloor a \rfloor?
• Calculus - Steve, Monday, March 4, 2013 at 4:44am
I get 14
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3475 | ## Sunday, 31 January 2016
### On This Day in Math - January 31
Joost Bürgi nich at Kepler monument
The capacity to blunder slightly is the real marvel of DNA.
Without this special attribute, we would still be anaerobic bacteria
and there would be no music.
The 31st day of the year; 31 = 22 + 33, i.e., The eleventh prime, and third Mersenne prime, it is also the sum of the first two primes raised to themselves. *Number Gossip (Is there another prime which is the sum of consecutive primes raised to themselves? A note from Andy Pepperdine of Bath who informed me that $2^2 + 3^3 +5^5 + 7^7 = 826699$, a prime.)
Jim Wilder @wilderlab offered, The sum of digits of the 31st Fibonacci number (1346269) is 31.
If you like unusual speed limits, the speed limit in downtown Trenton, a small city in northwestern Tennessee, is 31 miles per hour.
And the little teapot on the sign? Well, Trenton also bills itself as the teapot capital of the nation. The 31 mph road sign seems to come from a conflict between Trenton and a neighboring town which I will not name ,...but I will tell you they think of themselves as the white squirrel capital.
31 is also the smallest integer that can be written as the sum of four positive squares in two ways 1+1+4+25; 4+9+9+9.
31 is an evil math teacher number. The sequence of the maximum number of regions obtained by joining n points around a circle by straight lines begins 2, 4, 8, 16... but for five points, it is 31.
And 31 is also the minimal number of moves to solve the Towers of Hanoi problem with five disks.
EVENTS
1599 During an observation of the lunar eclipse, Tycho Brahe discovers that his predictive theory about the movement of the Moon is wrong since the eclipse started 24 minutes before his calculations predicted: he improves on his theory. On March 21 he sent a letter to Longomontanus, in which he reports his revised theory.*Wik
1802 Gauss elected a corresponding member of the St. Petersburg Academy of Science. *VFR
1834 Felix Klein declines to be the successor of J. J. Sylvester at John's Hopkins. Klein had been offered the position on December 13th of the previous year, but had demanded a salary equal to the departing Sylvester and some form of security for his family which Johns Hopkins did not meet. By October he would send notes to his family, "Gottingen is beginning to make noises." In the spring of 1836 he took over as Professor at Gottingen (he had been their second choice). *Constance Reid, The Road Not Taken, Mathematical Intelligencer, 1978
1839, Fox Talbot read a paper before the Royal Society, London, to describe his photographic process using solar light, with an exposure time of about 20 minutes: Some Account of the Art of Photogenic Drawing or the Process by which Natural Objects may be made to Delineate Themselves without the Aid of the Artist's Pencil. He had heard that Daguerre of Paris was working on a similar process. To establish his own priority, Fox Talbot had exhibited "such specimens of my process as I had with me in town," the previous week at a meeting of the Royal Institution, before he had this more detailed paper ready to present.*TIS
1939 Hewlett-Packard founded. Their calculators use the “reverse Polish notation” devised by Jan L Lukasiewicz (see here, 1878). *VFR
1939 Joseph Ehrenfried Hofmann began his academic career as a professor of the history of mathematics at the University of Berlin. He is noted for his work on Leibniz, especially the book Leibniz in Paris, 1672–1676: His Growth to Mathematical Maturity. *VFR Leibniz in Paris 1672-1676: His Growth to Mathematical Maturity
1958 Explorer 1 was launched on January 31, 1958 at 22:48 Eastern Time (equal to February 1, 03:48 UTC because the time change goes past midnight). It was the first spacecraft to detect the Van Allen radiation belt, returning data until its batteries were exhausted after nearly four months. It remained in orbit until 1970, and has been followed by more than 90 scientific spacecraft in the Explorer series. *Wik
Actually the Van Allen radiation was detectable by the Russian’s first satellite, Sputnik. Because the signals were sent in a secret code, it’s signal could not be received by the Russians when it was detecting the radiation of the belt. *Frederich Pohl, Chasing Science, pg 85
1995 AT&T Bell Laboratories and VLSI Technology announce plans to develop strategies for protecting communications devices from eavesdroppers. The goal would be to prevent problems such as insecure cellular phone lines and Internet transmissions by including security chips in devices. *CHM
2016, Since the year is a leap year beginning on a Friday, the typical calendar page for January takes six lines. Such months are called perverse months. (The same months will be perverse in a year starting on Saturday. 2016 has three such months, Jan, July and October. 2012 had only two. It is possible for there to be four in a single year. When will that year be? Is it possible for there to be a year with no perverse months?
There is an inverse relationship between Friday-the-thirteenths and perverse months; so what is good for the calendar makers is bad for the superstitious., so 2016 has only one Friday the 13th
Image credit: NASA/JPL-Caltech/MSSS/TAMU
2014 The Mars rover's view of its original home planet even includes our moon, just below Earth.
The images, taken about 80 minutes after sunset during the rover's 529th Martian day (Jan. 31, 2014) are available for a broad scene of the evening sky, and a zoomed-in view of Earth and the moon.
The distance between Earth and Mars when Curiosity took the photo was about 99 million miles (160 million kilometers). * NASA
BIRTHS
1715 Giovanni Francesco Fagnano dei Toschi (31 Jan 1715 in Sinigaglia, Italy - 14 May 1797 in Sinigaglia, Italy) He proved that the triangle which has as its vertices the bases of the altitudes of any triangle has those altitudes as its bisectors. *VFR Of all the triangles that could be inscribed in a given triangle, the one with the smallest perimeter is the orthic triangle. This has sometimes been called Fagnano's Problem since it was first posed and answered by Giovanni Francesco Fagnano dei Toschi. Fagnano also was the first to show that the altitudes of the original triangle are the angle bisectors of the orhtic triangle, so the incenter of the orthic triangle is the orthocenter of the original triangle.*pb
He was the son of the mathematician
Giulio Carlo Fagnano. He calculated the integral of the tangent and also proved the reduction formula $\int x^n \sin {x} dx = -x^n \cos{x}+n\int x^{n-1} \cos{x} dx$
*VFR
1763 The Rt. Rev. John Mortimer Brinkley D.D. (ca. 1763 (Baptized 31 Jan,1763, Woodbridge, Suffolk – 14 September 1835, Dublin) was the first Royal Astronomer of Ireland and later Bishop of Cloyne.
He graduated B.A. in 1788 as senior wrangler and Smith's Prizeman, was elected a fellow of the college and was awarded M.A. in 1791. He was ordained at Lincoln Cathedral in the same year, and in 1792 became the second Andrews Professor of Astronomy in the University of Dublin, which carried the new title of Royal Astronomer of Ireland. Together with John Law, Bishop of Elphin, he drafted the chapter on "Astronomy" in William Paley's Natural Theology. His main work concerned stellar astronomy and he published his Elements of Plane Astronomy in 1808. In 1822 he was elected a Foreign Honorary Member of the American Academy of Arts and Sciences. He was awarded the Copley Medal by the Royal Society in 1824. Brinkley's observations that several stars shifted their apparent place in the sky in the course of a year were disproved at Greenwich by his contemporary John Pond, the Astronomer Royal. In 1826, he was appointed Bishop of Cloyne in County Cork, a position he held for the remaining nine years of his life. Brinkley was elected President of the Royal Astronomical Society in 1831, serving in that position for two years.
He died in 1835 at Leeson Street, Dublin and was buried in Trinity College chapel. He was succeeded at Dunsink Observatory by Sir William Rowan Hamilton. *Wik
1841 Samuel Loyd (31 Jan 1841 ; died 10 Apr 1911) was an American puzzlemaker who was best known for composing chess problems and games, including Parcheesi, in addition to other mathematically based games and puzzles. He studied engineering and intended to become a steam and mechanical engineer but he soon made his living from his puzzles and chess problems. Loyd's most famous puzzle was the 14-15 Puzzle which he produced in 1878. The craze swept America where employers put up notices prohibiting playing the puzzle during office hours. Loyd's 15 puzzle is the familiar 4x4 arrangement of 15 square numbered tiles in a tray that must be reordered by sliding one tile at a time into the vacant space. *TIS When he offered a cash prize to anyone who could solve the puzzle with 14&15 reversed, it swept the country. To show it impossible requires only a little group theory; see W. E. Story, “Note on the ‘15’ puzzle,” American Journal of Mathematics, 2, 399–404. For samples of Loyd’s many puzzles, see Mathematical Puzzles of Sam Loyd, edited by Martin Gardner, Dover 1959 [p. xi]. *VFR
Although Lloyd popularized the puzzle in his books and articles, he most certainly did not invent it. Loyd's first article about the puzzle was published in 1886 and it wasn't until 1891 that he first claimed to have been the inventor. The article mentioned by Story(1878) was dated prior to Loyd's first mention of the puzzle) Here is the history of the puzzle as related by Wikipedia:The puzzle was "invented" by Noyes Palmer Chapman, a postmaster in Canastota, New York, who is said to have shown friends, as early as 1874, a precursor puzzle consisting of 16 numbered blocks that were to be put together in rows of four, each summing to 34. Copies of the improved Fifteen Puzzle made their way to Syracuse, New York by way of Noyes' son, Frank, and from there, via sundry connections, to Watch Hill, RI, and finally to Hartford (Connecticut), where students in the American School for the Deaf started manufacturing the puzzle and, by December 1879, selling them both locally and in Boston, Massachusetts. Shown one of these, Matthias Rice, who ran a fancy woodworking business in Boston, started manufacturing the puzzle sometime in December 1879 and convinced a "Yankee Notions" fancy goods dealer to sell them under the name of "Gem Puzzle". In late-January 1880, Dr. Charles Pevey, a dentist in Worcester, Massachusetts, garnered some attention by offering a cash reward for a solution to the Fifteen Puzzle.
The game became a craze in the U.S. in February 1880, Canada in March, Europe in April, but that craze had pretty much dissipated by July. Apparently the puzzle was not introduced to Japan until 1889.
Noyes Chapman had applied for a patent on his "Block Solitaire Puzzle" on February 21, 1880. However, that patent was rejected, likely because it was not sufficiently different from the August 20, 1878 "Puzzle-Blocks" patent (US 207124) granted to Ernest U. Kinsey.*Wik
Play with an online version here.
1886 George Neville Watson (31 Jan 1886 in Westward Ho!, Devon, England - 2 Feb 1965 in Leamington Spa, Warwickshire, England) studied at Cambridge, and then taught at Cambridge and University College London before becoming Professor at Birmingham. He is best known as the joint author with Whittaker of one of the standard text-books on Analysis. Titchmarsh wrote of Watson's books, "Here one felt was mathematics really happening before one's eyes. ... the older mathematical books were full of mystery and wonder. With Professor Watson we reached the period when the mystery is dispelled though the wonder remains." *SAU
1914 Lev Arkad'evich Kaluznin (31 Jan 1914 in Moscow, Russia - 6 Dec 1990 in Moscow, Russia) Kaluznin is best known for his work in group theory and in particular permutation groups. He studied the Sylow p-subgroups of symmetric groups and their generalisations. In the case of symmetric groups of degree pn, these subgroups were constructed from cyclic groups of order p by taking their wreath product. His work allowed computations in groups to be replaced by computations in certain polynomial algebras over the field of p elements. Despite the fact that the earliest applications of wreath products of permutation groups was due to C Jordan, W Specht and G Polya, it was Kaluznin who first developed special computational tools for this purpose. Using his techniques, he was able to describe the characteristic subgroups of the Sylow p-subgroups, their derived series, their upper and lower central series, and more. These results have been included in many textbooks on group theory. *SAU
1928 Heinz Bauer (31 January 1928 – 15 August 2002) was a German mathematician.
Bauer studied at the University of Erlangen-Nuremberg and received his PhD there in 1953 under the supervision of Otto Haupt and finished his habilitation in 1956, both for work with Otto Haupt. After a short time from 1961 to 1965 as professor at the University of Hamburg he stayed his whole career at the University of Erlangen-Nuremberg. His research focus was the Potential theory, Probability theory and Functional analysis
Bauer received the Chauvenet Prize in 1980 and became a member of the German Academy of Sciences Leopoldina in 1986. Bauer died in Erlangen. *Wik
1929 Rudolf Ludwig Mössbauer (31 Jan 1929 - 14 September 2011) German physicist and co-winner (with American Robert Hofstadter) of the Nobel Prize for Physics in 1961 for his researches concerning the resonance absorption of gamma-rays and his discovery in this connection of the Mössbauer effect. The Mössbauer effect occurs when gamma rays emitted from nuclei of radioactive isotopes have an unvarying wavelength and frequency. This occurs if the emitting nuclei are tightly held in a crystal. Normally, the energy of the gamma rays would be changed because of the recoil of the radiating nucleus. Mössbauer's discoveries helped to prove Einstein's general theory of relativity. His discoveries are also used to measure the magnetic field of atomic nuclei and to study other properties of solid materials. *TIS
Rudolf Mössbauer was an excellent teacher. He gave highly specialized lectures on numerous courses, including Neutrino Physics, Neutrino Oscillations, The Unification of the Electromagnetic and Weak Interactions and The Interaction of Photons and Neutrons With Matter. In 1984, he gave undergraduate lectures to 350 people taking the physics course. He told his students: “Explain it! The most important thing is, that you are able to explain it! You will have exams, there you have to explain it. Eventually, you pass them, you get your diploma and you think, that's it! – No, the whole life is an exam, you'll have to write applications, you'll have to discuss with peers... So learn to explain it! You can train this by explaining to another student, a colleague. If they are not available, explain it to your mother – or to your cat!” *Wik
1945 Persi Warren Diaconis (January 31, 1945; ) is an American mathematician and former professional magician. He is the Mary V. Sunseri Professor of Statistics and Mathematics at Stanford University. He is particularly known for tackling mathematical problems involving randomness and randomization, such as coin flipping and shuffling playing cards.
Diaconis left home at 14 to travel with sleight-of-hand legend Dai Vernon, and dropped out of high school, promising himself that he would return one day so that he could learn all of the math necessary to read William Feller's famous two-volume treatise on probability theory, An Introduction to Probability Theory and Its Applications. He returned to school (City College of New York for his undergraduate work graduating in 1971 and then a Ph.D. in Mathematical Statistics from Harvard University in 1974), and became a mathematical probabilist.
According to Martin Gardner, at school Diaconis supported himself by playing poker on ships between New York and South America. Gardner recalls that Diaconis had "fantastic second deal and bottom deal".
Diaconis is married to Stanford statistics professor Susan Holmes. *Wik
DEATHS
1632 Joost Bürgi (28 Feb 1552, 31 Jan 1632) Swiss watchmaker and mathematician who invented logarithms independently of the Scottish mathematician John Napier. He was the most skilful, and the most famous, clockmaker of his day. He also made astronomical and practical geometry instruments (notably the proportional compass and a triangulation instrument useful in surveying). This led to becoming an assistant to the German astronomer Johannes Kepler. Bürgi was a major contributor to the development of decimal fractions and exponential notation, but his most notable contribution was published in 1620 as a table of antilogarithms. Napier published his table of logarithms in 1614, but Bürgi had already compiled his table of logarithms at least 10 years before that, and perhaps as early as 1588.
*TIS I posted about Burgi and his work w/ "proto" logarithms here if you would like more detail.
1903 Norman Macleod Ferrers; (11 Aug 1829 in Prinknash Park, Upton St Leonards, Gloucestershire, England - 31 Jan 1903 in Cambridge, England) John Venn wrote of him,.. ,
the Master, Dr Edwin Guest, invited Ferrers, who was by far the best mathematician amongst the fellows, to supply the place. His career was thus determined for the rest of his life. For many years head mathematical lecturer, he was one of the two tutors of the college from 1865. As lecturer he was extremely successful. Besides great natural powers in mathematics, he possessed an unusual capacity for vivid exposition. He was probably the best lecturer, in his subject, in the university of his day.
It was as a mathematician that Ferrers acquired fame outside the university. He made many contributions of importance to mathematical literature. His first book was "Solutions of the Cambridge Senate House Problems, 1848 - 51". In 1861 he published a treatise on "Trilinear Co-ordinates," of which subsequent editions appeared in 1866 and 1876. One of his early memoirs was on Sylvester's development of Poinsot's representation of the motion of a rigid body about a fixed point. The paper was read before the Royal Society in 1869, and published in their Transactions. In 1871 he edited at the request of the college the "Mathematical Writings of George Green" ... Ferrers's treatise on "Spherical Harmonics," published in 1877, presented many original features. His contributions to the "Quarterly Journal of Mathematics," of which he was an editor from 1855 to 1891, were numerous ... They range over such subjects as quadriplanar co-ordinates, Lagrange's equations and hydrodynamics. In 1881 he applied himself to study Kelvin's investigation of the law of distribution of electricity in equilibrium on an uninfluenced spherical bowl. In this he made the important addition of finding the potential at any point of space in zonal harmonics (1881).
Ferrers proved the proposition by Adams that "The number of modes of partitioning (n) into (m) parts is equal to the number of modes of partitioning (n) into parts, one of which is always m, and the others (m) or less than (m). " with a graphic transformation that is named for him. *SAU
1934 Duncan MacLaren Young Sommerville (24 Nov 1879 in Beawar, Rajasthan, India - 31 Jan 1934 in Wellington, New Zealand) Sommerville studied at St Andrews and then had a post as a lecturer there. He left to become Professor of Pure and Applied mathematics at Victoria College, Wellington New Zealand. He worked on non-Euclidean geometry and the History of Mathematics. He became President of the EMS in 1911. *SAU
1966 Dirk Brouwer (1 Sep 1902; 31 Jan 1966) Dutch-born U.S. astronomer and geophysicist known for his achievements in celestial mechanics, especially for his pioneering application of high-speed digital computers for astronomical computations. While still a student he determined the mass of Titan from its influence on other Saturnian moons. Brouwer developed general methods for finding orbits and computing errors and applied these methods to comets, asteroids, and planets. He computed the orbits of the first artificial satellites and from them obtained increased knowledge of the figure of the earth. His book, Methods of Celestial Mechanics, taught a generation of celestial mechanicians. He also redetermined astronomical constants.*TIS
1973 Noel Bryan Slater, often cited NB Slater, (29 July 1912 in Blackburn, Lancashire, England - January 31 1973 in Hull, England) was a British mathematician and physicist who worked on including statistical mechanics and physical chemistry, and probability theory.*Wik
1995 George Robert Stibitz (30 Apr 1904, 31 Jan 1995) U.S. mathematician who was regarded by many as the "father of the modern digital computer." While serving as a research mathematician at Bell Telephone Laboratories in New York City, Stibitz worked on relay switching equipment used in telephone networks. In 1937, Stibitz, a scientist at Bell Laboratories built a digital machine based on relays, flashlight bulbs, and metal strips cut from tin-cans. He called it the "Model K" because most of it was constructed on his kitchen table. It worked on the principle that if two relays were activated they caused a third relay to become active, where this third relay represented the sum of the operation. Also, in 1940, he gave a demonstration of the first remote operation of a computer.*TIS
Credits :
*CHM=Computer History Museum
*FFF=Kane, Famous First Facts
*NSEC= NASA Solar Eclipse Calendar
*RMAT= The Renaissance Mathematicus, Thony Christie
*SAU=St Andrews Univ. Math History
*TIA = Today in Astronomy
*TIS= Today in Science History
*VFR = V Frederick Rickey, USMA
*Wik = Wikipedia
*WM = Women of Mathematics, Grinstein & Campbell |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3476 | This is an old version, view current version.
## 11.1 Suffix Marks Type of Function
The suffix is determined by the type of function according to the following table.
function outcome suffix
log probability mass function discrete _lpmf
log probability density function continuous _lpdf
log cumulative distribution function any _lcdf
log complementary cumulative distribution function any _lccdf
random number generator any _rng
For example, normal_lpdf is the log of the normal probability density function (pdf) and bernoulli_lpmf is the log of the bernoulli probability mass function (pmf). The log of the corresponding cumulative distribution functions (cdf) use the same suffix, normal_lcdf and bernoulli_lcdf. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3477 | Language: Search: Contact
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Zbl 1175.34035
Zhou, Jianwen; Li, Yongkun
Existence and multiplicity of solutions for some Dirichlet problems with impulsive effects.
(English)
[J] Nonlinear Anal., Theory Methods Appl. 71, No. 7-8, A, 2856-2865 (2009). ISSN 0362-546X
The authors consider the following Dirichlet boundary value problem with impulses \left\{\aligned &-u''(t)+g(t)u(t)=f(t,u(t)) \quad \text{a.e.}\,\, t\in [0,T]\\ &u(0)=u(T)=0,\\ & \Delta u'(t_j)=u'(t_j^+)-u'(t_j^-)=I_j(u(t_j)), \,\,\, j=1,2,\dots, p, \endaligned \right. where $t_0=0<t_1<t_2<\dots<t_p<t_{p+1}=T,$ $g\in L^{\infty}[0,T],$ $f: [0,T]\times {\Bbb R}\to {\Bbb R}$ is continuous and $I_j: {\Bbb R}\to {\Bbb R},$ $j=1,2,\dots,p$ are continuous. Existence and multiplicity results are obtained via Lax-Milgram theorem and critical points theorems. The main results are illustrated by examples.
[Sotiris K. Ntouyas (Ioannina)]
MSC 2000:
*34B37 Boundary value problems with impulses
58E05 Abstract critical point theory
Keywords: Dirichlet problems; impulse; Lax-Milgram theorem; critical points
Highlights
Master Server |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3478 | # Bianchi Identity-where did I go wrong?
zn5252
hello ,
In the Bianchi Identities of the second kind, we have ∇a Rbcde + ∇b Rcade + ∇c Rabde ≡ 0
but since ∇c Rabde = - ∇c Rbade
we get :
in the last term, we exchange the c and the b indices and we would arrive at:
∇a Rbcde = 0 .
but this incorrect ? did I do something wrong ?
I would like actually to arrive at the formula :
2 ∇b Racde − ∇a Rbcde ≡ 0.
see formula 9 here from original Bel's article from 1938:
http://gallica.bnf.fr/ark:/12148/bp...ction+d'un+tenseur+du+quatrième+ordre;.langEN
and
(see formula 5 here : https://docs.google.com/viewer?a=v&...9JJLhB&sig=AHIEtbRUONCiZUSV_8erdxK9YSMiouUrjA)
Thanks,
cheers,
Last edited:
in the last term, we exchange the c and the b indices
Only "dummy" indices can be freely changed like this, i.e. indices that are being summed over.
zn5252
Thanks I see , I must have been on a rush to get to the formula.
My problem is then : how do i get rid of the last variable on which the divergence is acting I mean c ? is there a formula I must have have overlooked perhaps ?
I don't think what you're trying to prove is true. Bel's result follows only after he's multiplied by another factor of the Riemann tensor.
zn5252
I see, now I think I found it . I have tried it out indeed with the additional tensor factor. it goes like this :
we have :
a Rbcde + ∇b Rcade + ∇c Rabde ≡ 0
multiply out by Rbcdf , we get :
Rbcdfa Rbcde + Rbcdfb Rcade + Rbcdfc Rabde ≡ 0
=>in the last term , we exchange the b and the c since now they are dummy indices I presume and in the second term, we use the antisymmetry of a and c:
Rbcdfa Rbcde - Rbcdfb Racde + Rcbdfb Racde = Rbcdfa Rbcde - 2 Rbcdfb Racde
Thank you.
PS : Sorry the Bel's article above is from 1958. I must have confused it with the article of Lanczos .
Last edited: |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3479 | # How do you factor x^2+ 11x = 180?
May 24, 2015
First subtract $180$ from both sides to get:
${x}^{2} + 11 x - 180 = 0$
Noticing that the coefficient of ${x}^{2}$ is $1$ and the signs of the other two coefficients, this may factorize as:
$\left(x + a\right) \left(x - b\right) = {x}^{2} + \left(a - b\right) x - a b$ with $a > 0$, $b > 0$, $a - b = 11$ and $a b = 180$.
$180 = 20 \times 9$ and $11 = 20 - 9$
So we can let $a = 20$ and $b = 9$ to get:
${x}^{2} + 11 x - 180 = \left(x + 20\right) \left(x - 9\right)$
Hence ${x}^{2} + 11 x = 180$ has solutions $x = - 20$ and $x = 9$ |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3480 | # Lesson 21
One Hundred and Eighty
## 21.1: What Went Wrong? (10 minutes)
### Warm-up
The purpose of this activity is for students to analyze a proof and decide whether each step makes sense. By finding the error in this proof students reinforce their understanding that corresponding angles are only congruent when the given lines are parallel.
### Launch
Arrange students in groups of 4 and assign a different statement to each student in a group. Give students quiet work time to decide if their statement is true. Next, invite students to share their responses with their group before discussing the circumstances that make corresponding angles congruent. Follow with whole-class discussion.
### Student Facing
Here are 2 lines $$\ell$$ and $$m$$ that are not parallel that have been cut by a transversal.
Tyler thinks angle $$EBF$$ is congruent to angle $$BCD$$ because they are corresponding angles and a translation along the directed line segment from $$B$$ to $$C$$ would take one angle onto the other. Here are his reasons.
• The translation takes $$B$$ onto $$C$$, so the image of $$B$$ is $$C$$.
• The translation takes $$E$$ somewhere on ray $$CB$$ because it would need to be translated by a distance greater than $$BC$$ to land on the other side of $$C$$.
• The image of $$F$$ has to land somewhere on line $$m$$ because translations take lines to parallel lines and line $$m$$ is the only line parallel to $$\ell$$ that goes through $$B’$$.
• The image of $$F$$, call it $$F’$$, has to land on the right side of line $$BC$$ or else line $$FF’$$ wouldn’t be parallel to the directed line segment from $$B$$ to $$C$$.
1. Your teacher will assign you one of Tyler’s statements to think about. Is the statement true? Explain your reasoning.
2. In what circumstances are corresponding angles congruent? Be prepared to share your reasoning.
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
The purpose of discussion is to highlight the fact that the corresponding angle theorem is only true when the given lines are parallel. Ask students to pinpoint exactly where the argument breaks down in the situation where the two given lines are not parallel (Tyler’s third statement). Ask students to share what circumstances ensure corresponding angles are congruent and explain their reasoning. (Only parallel lines with a transversal produce congruent corresponding angles as we saw with rotation and transformation proofs.)
## 21.2: Triangle Angle Sum One Way (15 minutes)
### Activity
In grade 8, students used informal arguments to prove the Triangle Angle Sum Theorem. This activity revisits that work and takes it further by using more rigorous definitions and careful reasoning. Students should make use of their reference charts.
Making dynamic geometry software available gives students an opportunity to choose appropriate tools strategically (MP5).
### Launch
Remind students of the meaning of the Triangle Angle Sum Theorem by displaying a large triangle on tracing paper with angle measures labeled $$a^\circ$$, $$b^\circ$$, and $$c^\circ$$ for all to see. Tear off the angles and rearrange them to form a straight line. Ask what the sum of the three angle measures is, given that they form a straight line. (180 degrees.)
Representation: Internalize Comprehension. Begin the activity with concrete or familiar contexts. For example, provide students with a triangle that has known angle measurements such as $$50^\circ$$, $$60^\circ$$, and $$70^\circ$$. Ask students to complete the task with this triangle. Then, ask students to consider the relationship between the angles formed by the parallel lines and each transversal.
Supports accessibility for: Conceptual processing; Memory
### Student Facing
1. Use technology to create a triangle. Use the Text tool to label the 3 interior angles as $$a$$, $$b$$, and $$c$$.
2. Mark the midpoints of 2 of the sides.
3. Extend the side of the triangle without the midpoint in both directions to make a line.
4. Use what you know about rotations to create a line parallel to the line you made that goes through the opposite vertex.
5. What is the value of $$a+b+c$$? Explain your reasoning.
### Student Response
For access, consult one of our IM Certified Partners.
### Launch
Remind students of the meaning of the Triangle Angle Sum Theorem by displaying a large triangle on tracing paper with angle measures labeled $$a^\circ$$, $$b^\circ$$, and $$c^\circ$$ for all to see. Tear off the angles and rearrange them to form a straight line. Ask what the sum of the three angle measures is, given that they form a straight line. (180 degrees.)
Representation: Internalize Comprehension. Begin the activity with concrete or familiar contexts. For example, provide students with a triangle that has known angle measurements such as $$50^\circ$$, $$60^\circ$$, and $$70^\circ$$. Ask students to complete the task with this triangle. Then, ask students to consider the relationship between the angles formed by the parallel lines and each transversal.
Supports accessibility for: Conceptual processing; Memory
### Student Facing
1. Use a straightedge to create a triangle. Label the 3 angle measures as $$a^\circ$$, $$b^\circ$$, and $$c^\circ$$.
2. Use paper folding to mark the midpoints of 2 of the sides.
3. Extend the side of the triangle without the midpoint in both directions to make a line.
4. Use what you know about rotations to create a line parallel to the line you made that goes through the opposite vertex.
5. What is the value of $$a+b+c$$? Explain your reasoning.
### Student Response
For access, consult one of our IM Certified Partners.
### Anticipated Misconceptions
Some students may get stuck finding the measures of the missing angles. Direct those students to their reference charts.
### Activity Synthesis
The important idea is that, no matter what triangle students started with, the sum of the measures of the three angles is always 180 degrees. Select several student responses to display for all to see. Here are some questions for discussion:
• “What is the same in these figures? What is different?” (The particular labelings of the points and angle measures may be different, but the result is always the same. The angles form a straight line and thus the angle measures sum to 180 degrees.)
• “How did you create a second line that is parallel to the first?” (Rotating by 180 degrees around either of the midpoints created takes the first line to the second line.)
• “Why does rotating 180 degrees around one midpoint create the same line as rotating around the other?” (Both lines are parallel to the given line, and both lines go through the vertex opposite the given line. The Parallel Postulate says that there is one unique line that is parallel to the given line that goes through the point in question, so the two lines must be the same line.)
Speaking, Representing: MLR 8 Discussion Supports. Use this routine to support whole-class discussion. After each student shares, provide the class with the following sentence frames to help them respond: "I agree because ….” or "I disagree because ….” If necessary, revoice student ideas to demonstrate mathematical language use by restating a statement as a question in order to clarify, apply appropriate language, and involve more students.
Design Principle(s): Support sense-making
## 21.3: Triangle Angle Sum Another Way (10 minutes)
### Activity
In this activity, students prove the Triangle Angle Sum theorem using translations.
Making dynamic geometry software available gives students an opportunity to choose appropriate tools strategically (MP5).
### Launch
Instruct students to create a triangle with sides extended to a line using this image as a guide.
Here is triangle $$ABC$$ with angle measures $$a^\circ$$, $$b^\circ$$, and $$c^\circ$$. Each side has been extended to a line.
Action and Expression: Internalize Executive Functions. Begin with a small-group or whole-class demonstration and think aloud of the first question to remind students how to translate a triangle along a directed line segment. Keep the worked-out translation on display for students to reference as they work.
Supports accessibility for: Memory; Conceptual processing
### Student Facing
1. Translate triangle $$ABC$$ along the directed line segment from $$B$$ to $$C$$ to make triangle $$A’B’C’$$. Label the measures of the angles in triangle $$A’B’C’$$.
2. Translate triangle $$A’B’C’$$ along the directed line segment from $$A’$$ to $$C$$ to make triangle $$A’’B’’C’’$$. Label the measures of the angles in triangle $$A’’B’’C’’$$.
3. Label the measures of the angles that meet at point $$C$$. Explain your reasoning.
4. What is the value of $$a+b+c$$? Explain your reasoning.
### Student Response
For access, consult one of our IM Certified Partners.
### Launch
Action and Expression: Internalize Executive Functions. Begin with a small-group or whole-class demonstration and think aloud of the first question to remind students how to translate a triangle along a directed line segment. Keep the worked-out translation on display for students to reference as they work.
Supports accessibility for: Memory; Conceptual processing
### Student Facing
Here is triangle $$ABC$$ with angle measures $$a^\circ$$, $$b^\circ$$, and $$c^\circ$$. Each side has been extended to a line.
1. Translate triangle $$ABC$$ along the directed line segment from $$B$$ to $$C$$ to make triangle $$A’B’C’$$. Label the measures of the angles in triangle $$A’B’C’$$.
2. Translate triangle $$A’B’C’$$ along the directed line segment from $$A’$$ to $$C$$ to make triangle $$A’’B’’C’’$$. Label the measures of the angles in triangle $$A’’B’’C’’$$.
3. Label the measures of the angles that meet at point $$C$$. Explain your reasoning.
4. What is the value of $$a+b+c$$? Explain your reasoning.
### Student Response
For access, consult one of our IM Certified Partners.
### Student Facing
#### Are you ready for more?
One reason mathematicians like to have rigorous proofs even when conjectures seem to be true is that it can help reveal what assertions were used. This can open up new areas to explore if we change those assumptions. For example, both of our proofs that the measures of the angles of a triangle sum to 180 degree were based on rigid transformations that take lines to parallel lines. If our assumptions about parallel lines changed, so would the consequences about triangle angle sums. Any study of geometry where these assumptions change is called non-Euclidean geometry. In some non-Euclidean geometries, lines in the same direction may intersect while in others they do not. In spherical geometry, which studies curved surfaces like the surface of Earth, lines in the same direction always intersect. This has amazing consequences for triangles. Imagine a triangle connecting the north pole, a point on the equator, and a second point on the equator one quarter of the way around Earth from the first. What is the sum of the angles in this triangle?
### Student Response
For access, consult one of our IM Certified Partners.
### Anticipated Misconceptions
Some students may have difficulty drawing a reasonably accurate image of the figure under the translation. Remind them of the tools in their geometry toolkits, such as tracing paper, straightedges, and compasses.
Some students may get stuck finding the measures of the missing angles. Direct those students to their reference charts.
### Activity Synthesis
Here are some questions for discussion:
• “How did you find the other angle measures at point $$C$$?” (Some of the angles are images of the original triangle after a rigid transformation. They must have the same measure as the original triangle. I can find the other angle measures using properties of vertical angles.)
• “How do you know the point $$C’’$$ lands on line $$AC$$?” (Translations takes lines to parallel lines and $$A’$$ is taken to $$C$$, so the segment $$A’C’’$$ has to be taken to a parallel line that goes through $$C$$. There’s only one such line, which is line $$AC$$. So $$C’’$$ lands somewhere on line $$AC$$. In order for $$C’C’’$$ to be parallel to line $$A’C$$, it would have to land on line $$AC$$ on the right side of $$C$$.)
## Lesson Synthesis
### Lesson Synthesis
In this lesson, students explored two different proofs of the Triangle Angle Sum Theorem. Ask students to add this theorem to their reference charts as you add it to the class reference chart:
Triangle Angle Sum Theorem: The three angle measures of any triangle always sum to 180 degrees.
(Theorem)
Here are some questions for discussion:
• “Look back at the two proofs of the Triangle Angle Sum Theorem. Are there any parts of the argument that depend on the particular measurements of your triangle, or would the same arguments have worked with other kinds of triangles?” (The same argument would work for any triangle. The particular measurements didn't matter.)
• “How are the 2 proofs of the Triangle Angle Sum Theorem different? How are they the same?” (Both activities use transformations to place the three angles adjacent to each other to make a straight line. The proofs are different because the first proof used 180 degree rotations whereas the second proof used translations.)
## 21.4: Cool-down - Triangle Angle Sum a Third Way (5 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners.
## Student Lesson Summary
### Student Facing
Using rotations and parallel lines, we can understand why the angles in a triangle always add to 180 degrees. Here is triangle $$ABC$$.
Rotate triangle $$ABC$$ 180 degrees around the midpoint of segment $$AB$$ and label the image of $$C$$ as $$D$$. Then rotate triangle $$ABC$$ 180 degrees around the midpoint of segment $$AC$$ and label the image of $$B$$ as $$E$$.
Note that each 180 degree rotation takes line $$BC$$ to a parallel line. So line $$DA$$ is parallel to $$BC$$ and line $$AE$$ is also parallel to $$BC$$. There is only one line parallel to $$BC$$ that goes through point $$A$$, so lines $$DA$$ and $$AE$$ are the same line. Since line $$DE$$ is parallel to line $$BC$$, we know that alternate interior angles are congruent. That means that angle $$BAD$$ also measures $$b^\circ$$ and angle $$CAE$$ also measures $$c^\circ$$.
Since $$DE$$ is a line, the 3 angle measures at point $$A$$ must sum to 180 degrees. So $$a+b+c=180$$. This argument does not depend on the triangle we started with, so that proves the sum of the 3 angle measures of any triangle is always 180 degrees. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3481 | # How to characterize the properties, as opposed to predicates expressing such properties, of mathematical objects?
More Abstract Background for my Question
Assuming a sort of realism about mathematical objects, there is an intuitive notion of the properties a mathematical object instantiates which is non-linguistic. Predicates express these properties but, assuming the language is countable, there may be more properties than can be expressed by predicates of the language. While I'm somewhat familiar with how to characterize the predicates the objects of a theory satisfy (in terms of various hierarchies, e.g., arithmetical hierarchy, analytical hierarchy, and projective hierarchy), I am less clear on how to characterize properties these same objects have when there may be more properties than formulas in the language.
More Concrete Background for my Question
For example, first-order arithmetic as given by $\mathsf{PA}$ is not categorical and so admits non-standard models. There is, however, a standard model captured by True Arithmetic -- the set of sentences in the language of first-order arithmetic that are true in the structure given by $\mathsf{PA}$. These can captured in second-order arithmetic as the sentences that are arithmetical -- either $\Sigma^0_n$ or $\Pi^0_n$.
Since, however, there are only countably many formulas and a fortiori only countably many sentences. So, formulas that are $\Sigma^0_n$ or $\Pi^0_n$ with $n$ free variables can characterize $n$-place arithmetical predicates, but these don't obviously express every "truly arithmetic" property of the natural numbers. (Maybe there are only countably many "truly arithmetic" properties, but then the same issue arises for the properties of the reals as described by the (categorical) axioms characterizing the reals as a Dedekind-complete ordered field. Since there are uncountably many reals, if there is at least one distinct property per real then there are uncountably many properties of the reals.)
The Question
Since properties can typically be identified with sets of sorts (e.g., unary properties with sets of their instances and $n$-ary relations with sets of $n$-tuples instantiating them), is the relevant notion of "arithmetical property" that doesn't force the countability of the set of properties given by the sets classified by some relativized version of the arithmetical hierarchy? Or something like the Borel hierarchy? Is there a general way of pairing any categorical theory with the sets that correspond to properties possessed by objects of that theory? (In a way that doesn't count every set in the powerset of the domain as a "legitimate" property.)
In lieu of at least a vague definition of "legitimate property," this question is difficult to answer. However, one reasonable response is to say a bit about transfinite hierarchies of definability in general, in the hope that one or more of these is of interest.
Let's begin by focusing on $\mathbb{N}$. There is no obvious notion of "arithmetic mod parameters" which will work here: either our parameters are elements of $\mathbb{N}$ in which case (a) they're countable and (b) we gain nothing since each natural number is definable, or our parameters are elements of $\mathcal{P}(\mathbb{N})$ in which case every set corresponds to a property.
Instead of relativizing things, one possible line of attack is extending them: continue the arithmetic hierarchy "past $\omega$" in some reasonable way. The hyperarithmetic hierarchy is of course the best known way to do this, but it still reaches only countably many sets. Can we do better?
The answer is yes. Recall the definition of the constructible universe:
• $L_0=\emptyset$
• $L_\lambda=\bigcup_{\alpha<\lambda} L_\alpha$ for $\lambda$ limit.
• $L_{\alpha+1}$ is the set of subsets of $L_\alpha$ which are definable in the structure $(L_\alpha, \in)$ (this is usually denoted "$\mathcal{P}_{def}(L_\alpha)$").
Let $\mathcal{P}^L(\mathbb{N})$ denote the collection of subsets of $\omega$ in $L$. I suspect that this will be fairly satisfying to you, based on the following:
• $L$ is definable, and if $V$ is the set-theoretic universe and $W$ is an "inner model" of $V$ (= a class containing all ordinals and satisfying ZFC), then $L$ is definable in $W$ via the same definition defining it in $V$. So $L$, and its theory and the theory of its levels, are "absolute" in a precise sense.
• It is consistent that there are uncountably many, but still fewer than continuum, subsets of $\mathbb{N}$ in $L$ - this will happen whenever we have $\omega_1^L=\omega_1<2^{\aleph_0}$ (and this can be gotten by e.g. adding $\aleph_2$-many Cohen reals to $L$). So I think this may be what you are looking for.
• $\mathcal{P}^L(\omega)$ comes equipped with a hierarchy, namely $(L_\alpha\cap\mathcal{P}^L(\omega))_{\alpha<\omega_1^L}$, and this jibes nicely with the idea that legitimate properties should fall into a complexity hierarchy like the arithmetic hierarchy. Two quick notes on this:
• Every set $s$ of naturals in $L$ shows up at some countable-in-$L$ level: letting $M$ be the minimal elementary substructure of $L$ containing $s$, the Mostowski collapse of $M$ alsocontains $x$ and is isomorphic to some $L_\gamma$ with $\gamma<\omega_1$ by the condensation lemma).
• We can find even finer hierarchies - e.g. by looking at master codes - and these reveal tight connections with the Turing jump.
Now what about generalizing past $\mathbb{N}$?
For an arbitrary structure $\mathcal{A}$, we can form an analogue $L^\mathcal{A}$ of $L$ - this time a model of ZF + "urelements" corresponding to the elements of $\mathcal{A}$ - built with $\mathcal{A}$ on the "ground floor." We can then define $\mathcal{P}^{L^\mathcal{A}}(\mathcal{A})$ to be the collection of subsets of $\mathcal{A}$ which are in $L^\mathcal{A}$. A similar argument shows that $\mathcal{P}^{L^\mathcal{A}}(\mathcal{A})$ has size at most $\vert\mathcal{A}\vert^+$, which consistently is $<\vert\mathcal{P}(\mathcal{A})\vert$, so again we may can get an interesting hierarchy of sets.
Incidentally, going back to the countable, generalizations of the hyperarithmetic hierarchy (and related hierarchies) on arbitrary structures have been studied by Moschovakis and others.
We can also generalize past $L$ - under certain hypotheses, there are other "canonical" inner models which yield similar hierarchies - but that seems of secondary interest for now.
• I didn't know Moschovakis' excellent book was reprinted by Dover! Awesome! – Nagase Oct 30 '17 at 22:40
• @Nagase Yup! It's excellent! – Noah Schweber Oct 30 '17 at 22:49
• Once again, exactly what I was after. You have a knack for sussing out the question I would have asked were I able to express it. The Moschovakis book is especially helpful. It had been sitting in my database for a while -- clearly I thought it was relevant to my interests at some point -- and looking at it now I wish I had started to work through it earlier. – Dennis Oct 31 '17 at 0:32 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3482 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# Yesterday Diana spent a total of 240 minutes attending a
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Yesterday Diana spent a total of 240 minutes attending a [#permalink]
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07 Sep 2010, 15:37
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Yesterday Diana spent a total of 240 minutes attending a training class, responding to E-mails, and talking on the phone. If she did no two of these three activities at the same time, how much time did she spend talking on the phone?
(1) Yesterday the amount of time that Diana spent attending the training class was 90 percent of the amount of time that she spent responding to E-mails.
(2) Yesterday the amount of time that Diana spent attending the training class was 60 percent of the total amount of time that she spent responding to E-mails and talking on the phone.
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Joined: 02 Sep 2009
Posts: 47946
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07 Sep 2010, 15:48
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mehdiov wrote:
Yesterday Diana spent a total of 240 minutes attending a training class, responding to E-
mails, and talking on the phone. If she did no two of these three activities at the same
time, how much time did she spend talking on the phone?
(1) Yesterday the amount of time that Diana spent attending the training class was
90 percent of the amount of time that she spent responding to E-mails.
(2) Yesterday the amount of time that Diana spent attending the training class was
60 percent of the total amount of time that she spent responding to E-mails and
talking on the phone.
Given: $$C+E+P=240$$, where C is th time she spent on training class, E is the time she spent on E-mail and P is the time she spent on phone. Question: $$P=?$$
(1) $$C=0.9E$$ --> $$C+E+P=0.9E+E+P=1.9E+P=240$$. Not sufficient to calculate $$P$$ (one equation, two variables).
(2) $$C=0.6(E+P)$$ --> $$C+E+P=0.6(E+P)+E+P=1.6E+1.6P=240$$. Not sufficient to calculate $$P$$ (one equation, two variables).
(1)+(2) $$1.9E+P=240$$ and $$1.6E+1.6P=240$$ --> we have two distinct linear equations with two variables hence we can calculate each of them. Sufficient.
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22 Oct 2013, 07:02
Bunuel wrote:
mehdiov wrote:
Yesterday Diana spent a total of 240 minutes attending a training class, responding to E-
mails, and talking on the phone. If she did no two of these three activities at the same
time, how much time did she spend talking on the phone?
(1) Yesterday the amount of time that Diana spent attending the training class was
90 percent of the amount of time that she spent responding to E-mails.
(2) Yesterday the amount of time that Diana spent attending the training class was
60 percent of the total amount of time that she spent responding to E-mails and
talking on the phone.
Given: $$C+E+P=240$$, where C is th time she spent on training class, E is the time she spent on E-mail and P is the time she spent on phone. Question: $$P=?$$
(1) $$C=0.9E$$ --> $$C+E+P=0.9E+E+P=1.9E+P=240$$. Not sufficient to calculate $$P$$ (one equation, two variables).
(2) $$C=0.6(E+P)$$ --> $$C+E+P=0.6(E+P)+E+P=1.6E+1.6P=240$$. Not sufficient to calculate $$P$$ (one equation, two variables).
(1)+(2) $$1.9E+P=240$$ and $$1.6E+1.6P=240$$ --> we have two distinct linear equations with two variables hence we can calculate each of them. Sufficient.
Bunuel, for statement 2 we can rearrange to get E+P=240/1.6. If we have E+P, cant we plug this into C+E+P=240, which gives us C+(240/1.6)=240, which gives us C=240-(240/1.6)? Hence 2 would be sufficient as you rearrange to get P?
Math Expert
Joined: 02 Sep 2009
Posts: 47946
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22 Oct 2013, 07:56
bulletpoint wrote:
Bunuel wrote:
mehdiov wrote:
Yesterday Diana spent a total of 240 minutes attending a training class, responding to E-
mails, and talking on the phone. If she did no two of these three activities at the same
time, how much time did she spend talking on the phone?
(1) Yesterday the amount of time that Diana spent attending the training class was
90 percent of the amount of time that she spent responding to E-mails.
(2) Yesterday the amount of time that Diana spent attending the training class was
60 percent of the total amount of time that she spent responding to E-mails and
talking on the phone.
Given: $$C+E+P=240$$, where C is th time she spent on training class, E is the time she spent on E-mail and P is the time she spent on phone. Question: $$P=?$$
(1) $$C=0.9E$$ --> $$C+E+P=0.9E+E+P=1.9E+P=240$$. Not sufficient to calculate $$P$$ (one equation, two variables).
(2) $$C=0.6(E+P)$$ --> $$C+E+P=0.6(E+P)+E+P=1.6E+1.6P=240$$. Not sufficient to calculate $$P$$ (one equation, two variables).
(1)+(2) $$1.9E+P=240$$ and $$1.6E+1.6P=240$$ --> we have two distinct linear equations with two variables hence we can calculate each of them. Sufficient.
Bunuel, for statement 2 we can rearrange to get E+P=240/1.6. If we have E+P, cant we plug this into C+E+P=240, which gives us C+(240/1.6)=240, which gives us C=240-(240/1.6)? Hence 2 would be sufficient as you rearrange to get P?
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Re: Yesterday Diana spent a total of 240 minutes attending a [#permalink]
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03 Oct 2017, 23:11
mehdiov wrote:
Yesterday Diana spent a total of 240 minutes attending a training class, responding to E-mails, and talking on the phone. If she did no two of these three activities at the same time, how much time did she spend talking on the phone?
(1) Yesterday the amount of time that Diana spent attending the training class was 90 percent of the amount of time that she spent responding to E-mails.
(2) Yesterday the amount of time that Diana spent attending the training class was 60 percent of the total amount of time that she spent responding to E-mails and talking on the phone.
Let T denotes the time spent attending a training class.
Let E denotes the time spent responding to E-mails.
Let P denotes the time spent talking on the phone.
T +E + P = 240
DS: Find P ?
Statement 1 : T = 0.9 E .......(i)
Also, T +E + P = 240
0.9 E + E + P = 240
1.9 E + P = 240 .........(ii)
NOT SUFFICIENT
Statement 2:
T = 0.6 (E+P) ........(iii)
Also, T +E + P = 240
T + T /0.6 = 240 ..........(iv)
OR
(0.6+1) (E+P) = 240.......(v)
we can find T but not P.
NOT SUFFICIENT
Combined : From i & iii
0.9 E = 0.6 (E+P)
0.3 E = 0.6 P
E = 2P
From (v) 1.6 (2P+P) = 240
P = 240 / 4.8 = 50
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Re: Yesterday Diana spent a total of 240 minutes attending a &nbs [#permalink] 03 Oct 2017, 23:11
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3483 | Short-time dynamics of finite-size mean-field systems
# Short-time dynamics of finite-size mean-field systems
Celia Anteneodo Departmento de Física, PUC-Rio and National Institute of Science and Technology for Complex Systems, Rua Marquês de São Vicente 225, Gávea, CEP 22453-900 RJ, Rio de Janeiro, Brazil Ezequiel E. Ferrero Facultad de Matemática, Astronomía y Física, Universidad Nacional de Córdoba and Instituto de Física Enrique Gaviola (IFEG-CONICET), Ciudad Universitaria, 5000 Córdoba, Argentina Sergio A. Cannas Facultad de Matemática, Astronomía y Física, Universidad Nacional de Córdoba and Instituto de Física Enrique Gaviola (IFEG-CONICET), Ciudad Universitaria, 5000 Córdoba, Argentina
###### Abstract
We study the short-time dynamics of a mean-field model with non-conserved order parameter (Curie-Weiss with Glauber dynamics) by solving the associated Fokker-Planck equation. We obtain closed-form expressions for the first moments of the order parameter, near to both the critical and spinodal points, starting from different initial conditions. This allows us to confirm the validity of the short-time dynamical scaling hypothesis in both cases. Although the procedure is illustrated for a particular mean-field model, our results can be straightforwardly extended to generic models with a single order parameter.
Mean-field models, Short-time dynamics, Fokker-Planck equation
###### pacs:
64.60.Ht, 05.10.Gg, 64.60.My, 64.60.an
## I Introduction
Universal scaling behavior appears to be an ubiquitous property of critical dynamic systems. While initially believed to hold only in the long time limit, it was realized during the last decade that the dynamical scaling hypothesis can be extended to the short-time limit JaScSc1989 (). This is accomplished by assuming that, close to the critical point, the moment of the order parameter obeys the homogeneity relation
m(n)(t,τ,L,m0)=b−nβ/νm(n)(b−zt,b1/ντ,L/b,bμm0), (1)
where is time, is the reduced temperature , is the linear system size, is the initial value of the order parameter and is a spatial rescaling parameter. is a universal exponent that describes the short-time behavior, while , , and are the usual critical exponents. When we recover the usual dynamic scaling relation, from which a power law relaxation at the critical point (for instance, in the magnetization ) when and follows. This is the critical slowing down.
On the other hand, the short-time dynamics (STD) scaling properties of the system depend on the initial preparation, i.e., on the scaling field . Setting , from Eq. (1) one obtains for small (but non-null) values of in the large limit
m(t,τ,m0)∼m0tθF(t1/νzτ),θ=μ−β/νz. (2)
Hence, at the critical point an initial increase of the magnetization is observed. For the second moment the dependency on can be neglected when . Since in the large limit ( is the spatial dimension), one obtains at the critical point
m(2)(t)∼N−1td/z−2β/zν. (3)
The short-time universal scaling behavior has been verified in a large variety of critical systems both by renormalization group (RG) calculations JaScSc1989 (); PrPrKaTs2008 () and Monte Carlo (MC) numerical simulations Zh1998 (); Zh2006 (); PrPrKrVaPoRy2010 (). The hypothesis also applies when the system starts in the completely ordered state, i.e., . In this case it is assumed that the homogeneity relation
m(n)(t,τ,L)=b−nβ/νm(n)(b−zt,b1/ντ,L/b) (4)
holds even for short (macroscopic) time scales. Hence, in the large limit we have
m(t)=t−β/νzG(t1/νzτ), (5)
and taking the derivative of ,
∂logm(t,τ)∂τ∣∣∣τ=0∼t1/νz. (6)
While the scaling hypothesis starting from the disordered state is supported both by numerical simulations and RG, its validity for an initial ordered state relies up to now only on numerical simulations.
Recently, numerical simulations have shown that the short-time scaling hypothesis (1) holds not only close to a critical point, but also close to spinodal points in systems exhibiting a first-order phase transition, both for mean-field and short-range interactions models LoFeCaGr2009 (). This is particularly interesting, because it suggests the existence of some kind of diverging correlation length associated to a spinodal point. Since the proper concept of spinodal in short-range interactions systems is still a matter of debate (see Ref. LoFeCaGr2009 () and references therein), a deeper understanding of the microscopic mechanisms behind the observed short time scaling could shed some light on this problem. One way of achieving this goal is to look for exact solutions of particular models. A first step in that direction is to analyze mean-field (i.e., infinite-range interactions) models, for which the concept of spinodal is well defined LoFeCaGr2009 (). That is the objective of the present work: we analyze the exact STD behavior of far from equilibrium mean-field systems with non-conserved order parameter.
Non-equilibrium phenomena in physics and other fields are commonly studied through Fokker-Planck equations (FPEs). In particular, non-equilibrium dynamical aspects of phase transitions can be analyzed by means of the FPE associated to the master equation describing the microscopic dynamics binder (); hanggi (); munoz (). In fact, this tool was proved to be useful in the description of the relaxation of metastable states binder (), finite-size effects ruffo () or the impact of fluctuations in control parameters politi (), and have been considered for mean-field spin models binder (); mori () and coupled oscillators ruffo (), amongst many others.
As soon as the degrees of freedom of the system can be reduced to a few relevant ones, a low-dimensional FPE can be found. Although this description is suitable for properties that do not depend on the details of the dynamics, or for mean-field kinetics, many conclusions are expected to hold in more general instances.
For a single order parameter , the FPE for its probability is
∂tP=[−∂mD1(m)+∂mmD2(m)]P≡LFP(m)P, (7)
where the drift and diffusion coefficients are determined by the Hamiltonian and the particular dynamics (e.g., Glauber or Metropolis).
Following this stochastic approach, here we study the scaling of the short-time relaxational dynamics in the vicinity of critical and spinodal points. In first approximation, the drift () is generically linear in the vicinity of a critical point and quadratic in the spinodal, following the quadratic and cubic behavior of the drift potential , respectively. Meanwhile, typically in various models, the noise intensity scales as binder (); ruffo (). Therefore, although we will present the STD for a particular spin model, our results can be straightforwardly extended to more general mean-field ones.
## Ii Formal FPE solution and moment expansions
The formal solution of the FPE (7), for the initial condition , is risken ()
P(m,t|m0,0)=etLFP(m)δ(m−m0).
The average of an arbitrary quantity can be derived directly from the FPE, by means of the adjoint Fokker-Plank operator , as follows
⟨Q⟩(m0,t) = ∫Q(m)P(m,t|m0,0)dm=∫Q(m)etLFP(m)δ(m−m0)dm (8) = ∫δ(m−m0)etL†FP(m)Q(m)dm=etL†FP(m0)Q(m0)= = ∑k≥0[L†FP(m0)]kQ(m0)tk/k!.
Therefore, the first two moments of the order parameter are
⟨m⟩ = m0+D1t+12[D1D′1+D2D′′1]t2+…, ⟨m2⟩ = ⟨m⟩2+2D2t+[2D2D′1+D1D′2+D2D′′2]t2+…, (9)
where and their derivatives are evaluated in . Notice that if and are not state-dependent, the expansion up to first order is exact.
Alternatively, evolution equations for moments can be obtained by integration of Eq. (7), after multiplying each member of the equation by the quantity to be averaged, that is
d⟨mn⟩dt=n⟨mn−1D1(m)⟩+n(n−1)⟨mn−2D2(m)⟩. (10)
For we have
d⟨m⟩dt=⟨D1(m)⟩. (11)
Eqs.(10) lead in general to a hierarchy of coupled equations for the moments. Only for a few special cases ( and polynomials in of degree smaller or equal than one and two respectively) these equations decouple. Otherwise, one has to rely on approximated methods to solve their dynamics.
Let us exhibit our STD analysis for the paradigmatic system of fully connected Ising spins (Curie-Weiss model), subject to a magnetic field , ruled by the mean-field Hamiltonian
H=−J2NM2−HM. (12)
Since the Hamiltonian depends only on the total magnetization , the master equation for this model can be written in closed form for binder (); mori (). In the large limit, when the magnetization per spin can be taken as a continuous variable, an expansion of the master equation up to first order in the perturbative parameter leads for the Glauber dynamics to a FP equation (7) with mori ()
D1(m) = −m+tanh[m′]−ϵβJmsech2[m′], D2(m) = ϵ(1−mtanh[m′]), (13)
where we have defined , with .
In the next sections we derive asymptotic solutions of the FPE with these coefficients, both close to the critical point ( and ) and to spinodal points for . Analytical results are compared against Monte Carlo simulation ones using Glauber algorithm. Time was adimensionalized with the characteristic time of the transition rate . The unit of time in theoretical expressions corresponds to one MC step in simulations. We also performed several checks using Metropolis algorithm. The outcomes were indistinguishable from the Glauber ones, except for a trivial time rescaling factor 2 close to the critical point, as expected binder ().
## Iv STD near the critical point
In the vicinity of the critical point (at and ), the coefficients (13) can be approximated for small (i.e., ) respectively by
D1(m) = −ω(λ,ϵ)m−κ(λ,ϵ)m3+O(m5), D2(m) = ϵ([1−(1−λ)m2]+O(m4)), (14)
where and , with .
Within the domain of validity of these approximations and therefore Concerning , its linear term dominates, that is,
D1(m)≃−ω(λ,ϵ)m, (15)
if
|ω|>>κm2. (16)
This implies a parabolic approximation of the drift potential , whose shape is plotted in Fig. 1 for different values of , found from the integration of in Eq. (13) and of the linearized expression(15), for comparison. For , one has a confining quadratic potential, while for the parabolic potential is inverted, with an unstable point at .
### iv.1 Ornstein-Ulhenbeck approximation
Now, for linear and constant , the exact solution of Eq. (7) reads risken ()
P(m,t|m0,0)=1√2πσ2(t)exp(−[m−m0exp(−ωt)]22σ2(t)), (17)
where . This solution applies for (Ornstein-Uhlenbeck (OU) process) as well as for , and is valid as long as the probability distribution remains strongly picked so that the inequality (16) holds for any value of with non-negligible probability.
Performing the average with Eq. (17) gives
⟨m⟩=m0exp(−ωt). (18)
Therefore, for , that is , the average magnetization decays (grows) exponentially, with characteristic time . Then, for time scales , it remains . Since in the large limit , then the magnetization scales as . This is consistent with Eq. (2), provided that and , in agreement with the mean-field exponents and . The same exponents are displayed by the Gaussian model JaScSc1989 (). For higher-order moments with even , one has
m(n)=Γ(n+12)√π[2ϵω−1(1−exp[−2ωt])]n2. (19)
Then, for short times ,
m(n)∼[ϵt]n/2. (20)
Hence, , consistently with Eq. (3) (), provided that we choose , the upper critical dimension.
The characteristic time scale for STD behavior is then with
τSTD≈1|λ+ϵ|=N|1+Nλ|. (21)
If we have , while for we have .
Fig. 2 displays the comparison between numerical simulations and the approximate OU solutions Eqs. (18)-(19), for , and different values of , such that . The OU approximation gives an excellent agreement for time scales up to ( for the present parameter values). Averages were taken over 1000 independent MC runs. The main differences between the theoretical and numerical results appear for and , where finite-size effects shift the equilibrium value of both the average magnetization and its variance.
Fig. 2 also shows the performance of Eq. (22), which reproduces the simulation results for longer times than Eq. (18), predicting the transient steady state. The lower saturation level observed in simulation outcomes for is due to the presence of fluctuations that drive some trajectories to the equilibrium state with negative magnetization, while the deterministic equation rules the stabilization at the level of the local minimum. Also notice that this discrepancy decreases as departs from the critical value because of the consequent increase of the potential barrier height, which makes such events less probable. For , the system evolves quickly towards the vicinity of the equilibrium state and the saturation level of the second moment is very close to the value given by the (bimodal) steady state distribution . In any case, finite-size higher order corrections can be neglected as far as the STD behavior is concerned.
### iv.2 Quartic approximation of the drift potential
When (16) does not apply, one can not discard the cubic contribution to . For such case we show in Appendix A that the inclusion of the cubic correction in the drift coefficient Eq. (14) leads for to
⟨m⟩=m0e−ωt√1+m20κ(1−e−2ωt)/ω. (22)
This solution is exact in the thermodynamic limit , as can be verified by direct integration of the deterministic version of Eq. (11binder (), i.e.,
d⟨m⟩dt=D1(⟨m⟩). (23)
Notice that the expansion of Eq. (22) up to first order in reproduces Eq. (18). The case (), can also be drawn from Eq. (22) by taking the limit , yielding
⟨m⟩=m0√1+2m20κt. (24)
In Appendix A we additionally show that finite-size corrections do not change the STD scaling of . For the second moment we obtain
m(2)≡⟨m2⟩−⟨m⟩2=2ϵt(1+z)(1+2z+2z2)(1+2z)3+O(ϵ2,ϵω), (25)
where . Notice that up to a typical time scale , the approximation holds. For , a crossover to a second linear (hence normal diffusive) regime but with a different diffusion constant is predicted, namely , although it typically falls beyond the STD region.
### iv.3 Other initial conditions
To investigate the scaling behavior for other initial conditions, we analyzed the STD behavior when . As can be seen in the inset of Fig. 1, the cubic approximation still holds close to . Hence, the thermodynamic-limit expression (22) is expected to apply too, as verified in Fig. 3a. In comparison with the initial condition of Fig. 2, here trajectories get more trapped around the positive minimum, hence the agreement with deterministic Eq. (22) is still better. For finite systems, the intensity of the fluctuations is state dependent following Eq. (13). Therefore, the finite-size corrections derived by assuming do not hold. However, for very short times one still expects , according to Eq. (9), as in fact verified in numerical simulations illustrated in Fig. 3. From Eq. (22) we have that for , in agreement with Eq. (5). The excellent accord between Eq. (22) and numerical simulation outcomes displayed in Fig. 3 when confirms our previous assumptions. Numerical simulations for other values of also verify the above scaling. For the equilibrium (final steady state) values of both mean and variance are quickly approached as in Fig.2. However, when , we see from Fig.3b that all the curves lie below the critical curve, at variance with the behavior observed when (compare with Fig.2b). This is because when almost all the trajectories get trapped in the positive minimum. Thus, the variance stabilizes in a value corresponding to the fluctuations in a single potential minimum. At long enough times, both minima in a finite size system get equally populated and therefore the equilibrium value of will be higher. However, the time scales needed to observe this effect fall outside the STD regime. On the contrary, when , a relatively large number of trajectories cross the barrier between minima and approaches the equilibrium value (which is larger than the steady one), even at very short times, as can be verified by comparing the numerical plateaux in Fig.2b with the equilibrium value
m(2)eq=∫1−1m2e−V(m)ϵdm∫1−1e−V(m)ϵdm. (26)
## V STD near the spinodal
When the model has a line of first-order transitions at and metastable stationary solutions for a range of values of . Without loss of generality we will restrict hereafter to the metastable solutions with positive magnetization, that is, those analytic continuations of the equilibrium magnetization from positive to negative values of . Defining , the metastable state exists as long as , where the spinodal field is given by
hSP = −βJmSP+12ln1+mSP1−mSP mSP = √1−1βJ.
where is the magnetization at the spinodal point LoFeCaGr2009 ().
Suppose now that we start the system evolution from the completely ordered state with and and let us define and . Considering as an order parameter, numerical simulations using Metropolis dynamics LoFeCaGr2009 () showed that close enough to the spinodal point () its moments obey the scaling form (4) with . For temperatures far enough from the spinodal magnetization is close to one and we can expand and in powers of and . Moreover, close to the spinodal we can neglect mori () the finite-size correction of . Then, from Eqs. (13) one has at first order in and second order in :
D1(m) ≃ ΔhβJ−2mSPΔmΔh−βJmSP(Δm)2, D2(m) ≃ ϵ(1βJ−2mSPΔm+(βJ−2)(Δm)2 (27) −mSPβJΔh+(2−3βJ)ΔmΔh).
In Fig. 4 we plot the shape of for different values of in the vicinity of , obtained both from integration of in Eq. (13) and of the approximate quadratic polynomial (27), for comparison.
The moments of can be calculated by means of Eq. (8), namely
⟨(Δm)n⟩=∑k≥0[D1∂x+D2∂xx]kxntk/k!, (28)
where we have defined .
For we can neglect in a first approximation the diffusion term, that is, at least for short times we can disregard finite-size effects. Then, from Eq. (28) using , with and one has (see Appendix B)
⟨Δm⟩=√γu+tanh(√γAt)1+utanh(√γAt)−Aα, (29)
where and . For (hence ), Eq. (29) becomes
⟨Δm⟩=√|γ|u−tan(√|γ|At)1+utan(√|γ|At)−Aα. (30)
Alternatively, Eqs. (29)-(30) can be obtained by integrating Eq. (23), and are in good agreement with numerical simulations, as illustrated in Fig. (5). One observes the following asymptotic behaviors:
(i) For (), a constant level is reached. In fact, since the potential presents a local minimum, the plateau occurs at a level associated to that minimum. This is in accord with numerical simulations (Fig. 5), notice that the local minimum of the potential is at , then , in agreement with the observed level.
(ii) For : (), Eq. (30) yields a rapid decay towards zero attained at finite . This is because the potential is tilted towards the absolute minimum (without local minimum).
In the limit , from Eq. (29) it follows
⟨Δm⟩=x1+Axt. (31)
Hence, at the spinodal point one has for , consistently with Eq. (5) with and , in agreement with previous numerical results LoFeCaGr2009 (). This behavior corresponds to the relaxation towards the saddle point . While in an infinite system such point is an stationary state, finite-size fluctuations destabilize it, with the subsequent exponential relaxation towards the equilibrium value at longer times, as depicted in Fig. 5. Finite-size corrections to Eq. (31), that we compute for , can be obtained by including the diffusion term in Eq. (28). When , following Eq. (27), we have , with , , and . In Appendix B we obtain Eq. (44), furnishing corrected at first order in , that for leads to
⟨Δm⟩∼1At[1−ϵcA210t3+O(ϵt2,ϵ2)]. (32)
Hence, finite-size effects will become relevant only when , with
t∗=(10βJϵA2)1/3=(10N−λ)1/3, (33)
in agreement with the scaling proposed in Ref. LoFeCaGr2009 (): , with and .
Finally, let us consider the second moment. In Appendix B we obtain Eq. (46), that gives the -correction to . It allows to compute , Eq. (47), that at short times leads to
Δm(2)∼2ϵ(ax2+bx+c)t≃2D(x)t, (34)
in accord with Eq. (9).
Meanwhile, for , Eq. (46) behaves as
⟨(Δm)2⟩∼1(At)2[1+ϵcA25t3+O(ϵt2,ϵ2)]. (35)
Hence from Eqs. (32) and (35) one gets
Δm(2)∼2ϵct5. (36)
Notice that in this regime, the prefactor of given by Eq. (36) is generically different from that obtained in the very short-time regime following Eq. (34). Fig. (6) illustrates this cross-over for different values of and fixed temperature. The prefactor at small times, varies with (panel a), while at intermediate times the prefactor becomes independently of , which is evident in the linear scale (panel b).
In any case the behavior up to is consistent with the STD scaling hypothesis for the set of mean-field exponents , , and in agreement with numerical outcomes LoFeCaGr2009 ().
We studied the short-time dynamical behavior of finite-size mean-field models (infinite-range interactions) with non conserved order parameter dynamics. By solving the associated Fokker-Plank equation we obtained closed expressions for the first moments of the order parameter, in the vicinity of both the critical and spinodal points. This allowed us to confirm the STD scaling hypothesis in both situations, as well as to determine the dynamical ranges of its validity. In particular, we confirmed analytically its validity when the system starts from an ordered state. Moreover, we found that a diffusion-like scaling behavior of the second moment appears for any initial value of the order parameter, but the associated diffusion coefficient presents a crossover between two different values, for short and intermediate times within the STD regime.
We found in general that the scaling behavior of the first moment is mainly determined by the shape of the potential and therefore by the equilibrium generalized free energy , which has the same extrema structure as binder () . The scaling behavior of higher moments, on the other hand, has its origin on the Gaussian nature of finite-size fluctuations close to the singular points. Although our results were obtained for a particular model, it is worth to stress that the above facts are characteristic of mean-field systems, since they depend only on the shape of and on the proportionality . This makes the analysis quite general and independent of the particular mean-field model.
Acknowledgments: The authors would like to thank T. S. Grigera and E. S. Loscar for sharing with us their simulation codes for Metropolis dynamics, as well as for useful discussions. This work was supported by CNPq and Faperj (Brazil), CONICET, Universidad Nacional de Córdoba, and ANPCyT/FONCyT (Argentina).
## Appendix A Quartic potential approximation near the critical point
To investigate the effect of including the cubic correction in the drift coefficient Eq. (14), we evaluate the particular setting of Eq. (8)
⟨mn⟩=∑k≥0[(−ωm0−κm30)∂m0+ϵ∂m0m0]kmn0tkk!. (37)
In the limit , we can neglect in a first approximation the diffusion term and compute
⟨m⟩≈∑k≥0[(−ωm0−κm30)∂m0]km0tkk!.
By iterating the operator times and identifying the general form of the coefficients of , with the aid of symbolic manipulation programs, we obtain
⟨m⟩ ≈ (38) = m0e−ωt∑j≥0(2jj)(−m20κ4ω(1−e−2ωt))j = m0e−ωt√1+m20κ(1−e−2ωt)/ω,
that coincides with the exact deterministic solution (22).
Fluctuations can be neglected as long as . However, while remains of order one (except for extreme temperatures), typically . Then, a finite-size correction can be included by keeping only the terms of order and in each coefficient of in Eq. (37). This procedure yields the correction term,
C1(ϵ) = −ϵκm0t2∑k≥0(2kk)(2k2+6k+3)(−z/2)k = −ϵκm0t23+4z+2z2(1+2z)5/2,
where . Then, it results
⟨m⟩ = m0(1−ωt)(1+2z)1/2+m0ωtz(1+2z)3/2 −ϵκm0t23+4z+2z2(1+2z)5/2+O(ϵ2,ϵω,ω2).
Notice that the first two terms in the right-hand side come from the expansion of the deterministic Eq. (38) up to first order in .
In particular, exactly at the critical point we have and (hence ). Therefore, as in the case of the OU approximation, one concludes that the magnetization remains up to a characteristic time .
Similarly, for , one obtains the correction
C2(ϵ) = ϵt∑k≥0(k+1)(k+2)(−2z)k=2ϵt(1+2z)3,
⟨m2⟩=m20(1+2z)−2ωz(1+z)κ(1+2z)2+2ϵt(1+2z)3+O(ϵ2,ϵω,ω2). (40)
Since in the deterministic limit , then the first two terms in the right-hand side come from the expansion of the squared Eq. (38) up to first order in .
In the computation of the centered second moment, using Eqs. (A) and (40), the purely deterministic terms cancels out to yield Eq. (25).
## Appendix B Moment calculation near the spinodal
If , from Eq. (8) using , and , the average magnetization is given by
⟨Δm⟩=∑k≥0[−(x2+2Aαx−α)∂x]kx(At)k/k!, (41)
where . Completing squares and making the change of variables with we obtain
⟨Δm⟩=√γ∑k≥0[(1−u2)∂u]ku(√γAt)k/k!−Aα. (42)
Considering the generating function for tangent, with the change of variable , one has tangent ()
∑n≥0[(1−u2)∂u]nuτn/n! = ∑n≥0[∂z]ntanhzτn/n! = (u+tanhτ)/(1+utanhτ),
from where Eqs. (29)-(30) follow.
To include finite-size effects we have to consider the complete expression
⟨Δm⟩=∑k≥0[D1∂x+D2∂xx]kxtkk!. (43)
When , from Eq. (27), we have , with , , and . The contributions of order associated to each coefficient of the quadratic approximation of are
C1a=−ϵa3A∑k≥2(k2−1)(−y)k=−ϵay2(3+y)3A(1+y)3,
C1b=ϵbt12∑k≥2(k+1)(3k−2)(−y)k−1=−ϵbty(6+4y+y2)6(1+y)3,
C1c=−ϵcAt2(1+110∑k≥2(k+2)(2k+1)(−y)k−1)=−ϵcAt2(10+10y+5y2+y3)10(1+y)3,
where . Summing the -corrections together with the deterministic one, given by Eq. (31), yields
⟨Δm⟩=x1+y−(c10Ax2(10+10y+5y2+y3)+ +b6Ax(6+4y+y2)+a3A(3+y))ϵy2(1+y)3. (44)
Likewise, we calculate
⟨(Δm)2⟩=∑k≥0[D1∂x+D2∂xx]kx2tkk!. (45)
In this case, the contributions of order are
C2a=−2axA∑k≥1(k+23)(−y)k=2axyA(1+y)4,
C2b=−b12A∑k≥1(k+1)(k+2)(3k+1)(−y)k=by(12+6y+4y2+y3)6A(1+y)4,
Summing up the corrections , together with the deterministic term (given by the squared Eq. (31)), yields
⟨(Δm)2⟩ = x2(1+y)2+(c5x(10+10y+10y2+5y3+y4) (46) +b6(12+6y+4y2+y3)+2ax)ϵyA(1+y)4.
Finally, the second moment is obtained through . The purely deterministic terms cancel out and at first order in it remains
Δm(2)=ϵy30Ax(1+y)4(20ax2(3+3y+y2) (47) + 15bx(2+y)(2+2y+y2)+ + 12c(5+10y+10y2+5y3+y4))+O(ϵ2).
## References
• (1) H.K. Janssen, B. Schaub, and B. Schmittmann, Z. Phys. B - Condensed Matter 73, 539 (1989).
• (2) V. V. Prudnikov, P. V. Prudnikov, I. A. Kalashnikov, and S. S. Tsirkin, Journal of Experimental and Theoretical Physics 106, 1095 (2008).
• (3) B. Zheng, Int. J. Mod. Phys. B 12, 1419 (1998).
• (4) B. Zheng, in Computer Simulation Studies in Condensed Matter Physics, edited by D. P. Landau, S. P. Lewis and H. B. Schüttler (Springer, New York, 2006).
• (5) V. V. Prudnikov, P. V. Prudnikov, A. S. Krinitsyn, A. N. Vakilov, E. A. Pospelov, and M, V. Rychkov, Phys. Rev. E 81 011130 (2010).
• (6) E.S. Loscar, E.E. Ferrero, T.S. Grigera and S.A. Cannas, J. Chem. Phys. 131, 024120 (2009).
• (7) W. Paul, D.W. Heermann and K. Binder, J. Phys A: Math. Gen. 22 3325 (1989).
• (8) P. Hanggi, H. Grabert, P. Talkner and H. Thomas, Phys. Rev. A 29, 371 (1994).
• (9) M. Munoz and P.L. Garrido, J. Phys A: Math. Gen. 28 2637 (1995).
• (10) A. Pikovsky and S. Ruffo, Phys. Rev. E 59, 1633 (1999).
• (11) F.T. Arecchi and A. Politi, Optics communications 29, 362 (1979).
• (12) T. Mori, S. Miyashita and P.A. Rikvold, Phys. Rev. E 81, 011135 (2009).
• (13) H. Risken, The Fokker-Planck Equation: Methods of Solution and Applications (Springer-Verlag, Berlin, 1984).
• (14) M.E. Hoffman, Electron. J. Combin. 6, R21 (1999); D. Cvijović, Appl. Math. and Computation 215, 3002 (2009).
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3484 | # Higher Derivatives of Exponential Function
## Theorem
Let $\exp x$ be the exponential function.
Then:
$\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$
### Corollary
Let $c$ be a constant.
Then:
$\map {\dfrac {\d^n} {\d x^n} } {\map \exp {c x} } = c^n \map \exp {c x}$
## Proof
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
$\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$
### Basis for the Induction
$\map P 1$ is true, as this is the case proved in Derivative of Exponential Function:
$\map {\dfrac \d {\d x} } {\exp x} = \exp x$
This is our basis for the induction.
### Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
$\map {\dfrac {\d^k} {\d x^k} } {\exp x} = \exp x$
Then we need to show:
$\map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\exp x} = \exp x$
### Induction Step
This is our induction step:
$\ds \map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\exp x}$ $=$ $\ds \map {\dfrac \d {\d x} } {\dfrac {\d^k} {\d x^k} } {\exp x}$ Definition of Higher Derivatives $\ds$ $=$ $\ds \map {\dfrac \d {\d x} } {\exp x}$ Induction Hypothesis $\ds$ $=$ $\ds \exp x$ Basis for the Induction
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
$\forall n \in \N_{>0}: \map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$
$\blacksquare$ |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3485 | How much lift does the wing attach fitting actually carry?
How do I calculate the lift force that the the bolt that join the wings to the fuselage of a strutted or braced airplane carry?
I know in a cantilever airplane,the bolts have to carry all of the lift load on a wing but how about a strutted airplane - how do the wing attach bolts share the lift with the wing braces or struts, i.e. is the lift load shared between them, the bolts carrying half and the bracing wires or struts carrying the other half?
• The wings carry themselves, any engines attached to them and the fuel inside. Only the remaining load e.g. the weight of the fuselage is carried through the connection of the fuselage and wing. – Jan Jun 13 '19 at 13:12
• This only applies to aircraft with conventional spar designs. Some aircraft, like the homebuilt BD-5, have a spar that slides over a center carry-through spar and uses a small taper bolt to prevent movement in torsion. Works well as long you remember to inspect the hardware on a regular basis. – Juan Jimenez Jun 13 '19 at 14:40
• On an aircraft with flexible wire-based wings and significant washout, I've seen the lower wires go completely slack during high-speed flight, due to the torque generated by the downlifting wingtips during high-speed flight. Food for thought when formulating the answers-- – quiet flyer Jun 13 '19 at 19:27
• Instead of the lift loads being carried by the bolts you might consider bolted flanges. You basicly make this part stronger than the wing and spar, not weaker. Like Peter K said, it is also important to know the strength properties of the components of the structure. Find the weakest point. That is where improvement is possible, provided weight increase is not too extreme. – Robert DiGiovanni Jun 15 '19 at 0:42
The answer depends on the elastic deformation of wing and strut. Extreme example: Make the strut a rubber band and all lift needs to be carried by the wing and its fittings. Other extreme: Make the wing spar (or attachment points - this works either way) from rubber - now the strut needs to take all the loads.
In real life, stiffness attracts tension and compression, and in order to know the load distribution between wing spar and strut you need to look at them with a slight deformation from loading. How much resistance does each show (in isolation) against that deformation? This will tell you in what ratio the forces will split between wing and strut. Make sure to use the stiffness of the real structure - the material property called Young's modulus is not what I mean here.
If you want to be safe, you need to consider any single point failure and still have a flyable airplane. Normally, you design your structure with a safety factor (usually 1.5, but higher for fittings). Now assume that in turn each of the structural members is missing and calculate the loads on the remaining structure with a safety factor of 1.0. Dimension each part at least for those loads and you should be pretty sure that the intact structure will work out fine.
• Anonymous down voter: If you don't understand structural mechanics please ask in the comments or your own question. Sprinkling downvotes on answers you don't understand isn't helping anybody. – Peter Kämpf Jun 13 '19 at 18:40
• Deformation is caused by the forces acting on the structural members. Determine forces first, then deformation follows from the material properties. I wasn't the downvoted, by the way. – Koyovis Jun 14 '19 at 8:08
• @Koyovis And I didn't downvote your answer, either. The deformation is to determine stiffness of the part. The ratio in stiffness is inverse to the ratio of the forces in an overdetermined structure. The forces first approach only works if there are enough degrees of freedom (Cremona plan etc.). – Peter Kämpf Jun 14 '19 at 10:32
• The diagonal strut transforms the inner wing construction to a kinematically and statically determined system, with the triangle experiencing compression and tension forces. Within the triangle, both upper and lower skin experience stress in the same direction (tension or compression). Only outside the strut is the wing still a cantilever beam, with upper skin under compression and lower skin under tension. Point is: adding the strut does not create a statically overdetermined structure, which your answer addresses. – Koyovis Jun 19 '19 at 7:53
• @Koyovis: Depends entirely on the wing root mount. If that is a hinge, you are correct. If it has a central spar running all the span (which is how it is mostly done), it is indeed an overdetermined structure. – Peter Kämpf Jun 19 '19 at 9:07
In flight, the fuselage hangs off of the wing. For dimensioning purposes, consider:
• the intersection points to exert no moments and to behave as hinges;
• weight W of the fuselage to be concentrated in the Centre of Gravity;
• lift of each wing to be concentrated in its Centre of Lift.
If we dimension the construction in this way, we over-dimension which is never a bad idea with primary construction bolts. In reality the following factors alleviate the loads:
• The bolts are modelled as hinges which cannot exert a moment, but actually they do exert torque.
• The wing lift is distributed loading, with most of it near the wing root.
Bolts 1, 2 and 3 experience lift and gravity forces from the construction, and exert equal and opposite forces in order for everything to stay in one piece. Fuselage weight is transferred to bar 2-3 which distributes the load evenly over bolts 2 and 3. Remove the bolts and the fuselage falls from the assembly.
Force equilibrium in point 1 from this answer:
• $$F_{13} \cdot sinψ = ½L => F_{13} = \frac{L}{2sinψ}$$
• $$F_{12} = F_{13} \cdot cosψ = \frac{L}{2tanψ}$$
In point 2, bolt reaction force in green:
* $$F_{V} = ¼W$$ $$\tag{Vertical}$$ * $$F_{H} = \frac{L}{2tanψ}$$ $$\tag{Horizontal}$$
In point 3, bolt reaction force in green:
* $$F_{V} = ¼W - F_{13} \cdot sinψ = ¼W - ½L$$ $$\tag{Vertical}$$ * $$F_{H} = \frac{L}{2tanψ}$$ $$\tag{Horizontal}$$
• The forces on the strut are not in equilibrium – Gypaets Jun 14 '19 at 8:26
• @Gypaets Strut 2 - 3 can be removed, all forces to be absorbed by the bolt supports. – Koyovis Jun 14 '19 at 8:29
• I mean the vertical forces on strut 1-3. Vertical reaction force on the upper hinge (2) should also be 0. – Gypaets Jun 14 '19 at 8:30
• This appears to ignore that a portion of created lift must lifts the wing itself, not the fuselage. – J Walters Jun 14 '19 at 12:44
• +1 just for the picture. Try making a bookshelf without the "strut". Shown here is the positive G case. The bolts 1 and 3 would have to be torn from their mountings. – Robert DiGiovanni Jun 15 '19 at 0:15 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3486 | # Thread: 15.2.63 Reverse the order of integration in the following integral
1. ## 15.2.63 Reverse the order of integration in the following integral
$\textsf{Reverse the order of integration in the following integral and find the value of }$
\begin{align*}\displaystyle
I&=\int_0^3 \int_y^{3} x^2 e^{xy} dxdy
\end{align*}
ok apparently $0 \le x \le 3$ and $y \le y \le 3$
but ???
2. ## Re: 15.2.63 Reverse the order of integration in the following integral
The integration area is a triangle with top border being $y=x$, right border being the line x=3, and bottom border being the x-axis
Thus the new integration is
$I = \displaystyle \int_0^3 \int_0^y~x^2 e^{xy} ~dy~dx$
3. ## Re: 15.2.63 Reverse the order of integration in the following integral
Originally Posted by bigwave
$\textsf{Reverse the order of integration in the following integral and find the value of }$
\begin{align*}\displaystyle
I&=\int_0^3 \int_y^{3} x^2 e^{xy} dxdy
\end{align*}
ok apparently $0 \le x \le 3$ and $y \le y \le 3$
but ???
NO. This is $\int_0^3\left(\int_{y}^3 x^2e^{xy}dx\right) dy$.
The "outer integral" is with respect to y so $0\le y\le 3$. The "inner integral" is with respect to x and goes from y to 3: $y\le x\le 3$. (You should have seen that "$y\le y\le 3$" made no sense. Perhaps that was a typo.)
Draw a graph with the horizontal lines y= 0 (x-axis) and y= 3 drawn. Draw the vertical line x= 3 and the line $x= y^3$. The region of integration is the triangle bounded by those four lines. Notice that $x= y$ crosses y= 3 at (3, 3). That is a diagonal that divides the square into two triangles. Since the "inner integral" is from y= x on the left to y= 3 on the right, the region of integration is the lower triangle. To cover that triangle, x must range from 0 to 3 and, for each x, y must go from x to 3. The integral is $\int_0^3\int_x^3 x^2e^{xy}dydx$. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_3487 | # Chapter 4 - Section 4.1 - Angles and Radian Measure - Exercise Set - Page 532: 30
#### Work Step by Step
To convert degrees to radians, multiply degrees by $\frac{\pi radians}{180°}.$ So 76° = 76$^{\circ}$ x $\frac{3.142}{180°}$ = 1.33 radians.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14160 | # Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$
Notation 5.1.1.11. Let $q: X \rightarrow S$ be a right fibration of simplicial sets, and let $e: s \rightarrow s'$ be an edge of the simplicial set $S$. We will often use the symbol $e^{\ast }$ to denote a morphism of Kan complexes $X_{s'} \rightarrow X_{s}$ which is given by covariant transport along $e$. By virtue of Proposition 5.1.1.10, such a morphism exists and is uniquely determined up to homotopy. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14161 | # In Q. No. 14, write the sign of c.
Question:
In Q. No. 14, write the sign of c.
Solution:
The parabola $y=a x^{2}+b x+c$ cuts $y$-axis at point P which lies on $\mathrm{y}$-axis. Putting $x=0$ in $y=a x^{2}+b x+c$, we get $y=c$. So the coordinates of P are $(0, c)$. Clearly, P lies on OY. Therefore $c>0$ |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14162 | # Homework Help: Photoelectric effect and Rydberg's formula
1. Sep 29, 2009
### iuchem16
A peak in the energy spectrum is seen at 134.2 keV when using a germanium detector. In coincidence with this, an x-ray corresponding to the 2s to 1s electronic transition in germanium is detected. Find the energy (in keV) of the gamma ray that ejected the electron.
Use Rydberg's formula for spectral lines of hydrogenic atoms:
1/lambda = R Z2 ( ( 1/n12) - ( 1/n22) )
where R, the Rydberg constant, is such that Rhc = 13.61 eV
(c is the speed of light, h is Planck's constant).
not really sure where to begin, do I solve for the wavelength and use that in the energy equation E=hc/lambda??
2. Sep 29, 2009
### mgb_phys
You might want to substitute the above equation into the equation for energy that you mentioned. Especiay since you are given R in terms of Rhc
3. Sep 29, 2009
### iuchem16
nevermind..i got it...thanks! |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14163 | # Question No.4
#### Topic : Solving Pipe Networks using Quantity Balance Method
If in the network shown in the following figure the flow to C is regulated by a valve to 100 lit/sec, calculate the effect on the flows to the other reservoirs; determine the head loss to be provided by the valve.
Data :
Pipe Length (m) Diameter (mm) AJ 10,000 450 BJ 2,000 350 CJ 3,000 300 DJ 3,000 250
Roughness size of all pipes = 0.06 mm
Given :
Steps of solution :
Step (1) - Estimate ZJ (pressure head elevation at J) = 150 m (Note that the elevation of the pipe junction itself does not affect the solution.)
Step (2) - Since the flow in CJ is prescribed, it is simply treated as an external outflow at J.
For automatic computer analysis, Darcy-Colebrook-White combination can be used :
$Q=−2A2gDhfLlogk/D3.7+2.51νD2gDhfL$
For each pipe (friction head loss) is initialized to :
$ZI−ZJ$
Step (3) – Calculate the correction value for the assumed pressure head elevation. The calculations proceed in tabular form. Note that Q is written in liter/sec simply for convenience; all computations are based on Q in m3/sec.
To calculate the correction required in ZJ value :
$ΔZJ=2(ΣQIJ−FJ)ΣQIJhL,IJ$
For this problem, the flow in CJ is considered as an external outflow at junction J, therefore :
Therefore :
$ΔZJ=2(ΣQIJ−0.1)ΣQIJhL,IJ$
For the first iteration :
Correction to ZJ :
For the second iteration :
Correction to ZJ :
For the third iteration :
Correction to ZJ :
For the fourth iteration :
Correction to ZJ :
For the fifth iteration :
Correction to ZJ :
Final Discharges :
Final discharges (after 5 iterations) :
Pipe Q (l/sec) Flow Direction AJ 340.565 A to J BJ 125.365 J to B CJ 100.000 J to C DJ 115.201 J to D
To calculate the head loss provided by the valve :
$ZJ−ZC=KCJQCJ2+hvalve$
To calculate Reynolds number :
$Re=vDν=1.415∗0.31.011∗10−6=419,795$
To calculate the friction coefficient, Barr's equation is used :
$1f=−2logk/D3.7+5.1286Re0.89$
$1f=−2log0.06/3003.7+5.1286419,7950.89$
$f=0.0158$
To calculate the head loss coefficient K for pipe CJ :
$KCJ=8fCJLCJπ2gDCJ5=8∗0.0158∗3,000π2∗9.81∗(0.3)5=1,610.271$
$ZJ−ZC=KCJQCJ2+hvalve$ $→$ $127.576−100=1,610.271∗0.12+hvalve$
Finally, the head loss provided by the valve is : |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14164 | # Mirror Image
## Deriving Gibbs distribution from stochastic gradients
Stochastic gradients is one of the most important tools in optimization and machine learning (especially for Deep Learning – see for example ConvNet). One of it’s advantage is that it behavior is well understood in general case, by application of methods of statistical mechanics.
In general form stochastic gradient descent could be written as
$w_{n+1} = w_n - \alpha \bigtriangledown E(w_n) + F_n$
where $F_n$ is a random variable with expectation zero.
To apply methods of statistical mechanics we can rewrite it in continuous form, as stochastic gradient flow
$\partial J/\partial t = - \bigtriangledown_J E(J) +F(t)$
and random variable F(t) we assume to be white noise for simplicity.
In that moment most of textbooks and papers refer to “methods of statistical mechanics” to show that
stochastic gradient flow has invariant probability distribution, which is called Gibbs distribution
$p(x) = \frac {exp(-\beta E(x))}{Z}$
and from here derive some interesting things like temperature and free energy.
The question is – how Gibbs distribution derived from stochastic gradient flow?
First we have to understand what stochastic gradient flow really means.
It’s not a partial differential equation (PDE), because it include random variable, which is not a function $\mathbb{R} \to \mathbb{R}$. In fact it’s a stochastic differential equation (SDE) . SDE use some complex mathematical machinery and relate to partial differential equations, probability/statistics, measure theory and ergodic theory. However they used a lot in finance, so there is quite a number of textbooks on SDE for non-mathematicians. For the short, minimalistic and accessible book on stochastic differential equation I can’t recommend highly enough introduction to SDE by L.C. Evans
SDE in question is called Ito diffusion. Solution of that equation is a stochastic process – collection of random variables parametrized by time. Sample path of stochastic process in question as a function of time is nowhere differentiable – it’s difficult to talk about it in term of derivatives, so it is defined through it’s integral form.
First I’ll notice that integral of white noise is actually Brownian motion, or Wiener process.
Assume that we have stochastic differential equation written in informal manner
$\partial X / \partial t = b(X, t) + g(X, t) F(t)$ with X -stochastic process and F(t) – white noise
It’s integral form is
$X(T) - X(0) = \int_{0}^{T} b(X, t) dt + \int_{0}^{T} g(X, t) dW(t) \ \ (1)$
where W(t) is a Wiener process
$W(t) = \int F(t) dt$
This equation is usually written in the form
$dX = b(X,t)dt + g(X,t)dW \ \ (2)$
This is only a notation for integral equation, d here is not a differential.
Returning to (1)
$\int_{0}^{T} b(X, t) dt$ is an integral along sample path, it’s meaning is obvious, or it can be defined as limit of Riemann sums with respect to time.
The most notable thing here is
$\int_{0}^{T} g(X, t) dW(t)$ – integral with respect to Wiener process (3)
It’s a stochastic integral, and it’s defined in the courses of stochastic differential equation as the limit of Riemann sums of random variables, in the manner similar to definition of ordinary integral.
Curiously, stochastic integral is not quite well defined. Depending on the form of the sum it produce different results, like Ito integral:
$\int_0^t g \,d W =\lim_{n\rightarrow\infty} \sum_{[t_{i-1},t_i]\in\pi_n}g_{t_{i-1}}(W_{t_i}-W_{t_{i-1}})$
Different Riemann sums produce different integral – Stratonovich integral:
$\int_0^t g \,d W =\lim_{n\rightarrow\infty} \sum_{[t_{i-1},t_i]\in\pi_n}(g_{t_i} + g_{t_{i-1}})/2(W_{t_i}-W_{t_{i-1}})$
Ito integral used more often in statistics because it use $g_{t_{i-1}}$ – it don’t “look forward”, and Stratonovich more used in theoretical physics.
Returning to Ito integral – Ito integral is stochastic process itself, and it has expectation zero for each t.
From definition of Ito integral follow one of the most important tools of stochastic calculus – Ito Lemma (or Ito formula)
Ito lemma states that for solution of SDE (2)
$dX = b(X, t)dt + g(x, t)dW$ X, b, W – vectors, g – matrix
were W is Wiener process (actually some more general process) and b and g are good enough
$du(\mathbf{X}, t) = \frac{\partial u}{\partial t} dt + (\nabla_\mathbf{X}^{\mathsf T} u) d\mathbf{X} + \tfrac{1}{2} \sum_{i,j} (gg^\mathsf{T})_{i,j}\tfrac{ \partial^2 u}{\partial x_i \partial x_j} dt$
where $\nabla_\mathbf{X} u$ is the gradient.
From Ito lemma follow Ito product rule for scalar processes: applying Ito formula to process $V = { X \choose Y}$ combined from two processes X and Y to function u(V) = XY
$d(XY) = XdY + YdX + g_x \cdot g_y dt$
Using Ito formula and Ito product rule it is possible to get Feynman–Kac formula (derivation could be found in the wikipedia, it use only Ito formula, Ito product rule and the fact that expectation of Ito integral (3) is zero):
for partial differential equation (PDE)
$\frac{\partial u}{\partial t} + (\nabla_\mathbf{X}u)^{\mathsf T} b + \tfrac{1}{2} \sum_{i,j} (gg^\mathsf{T})_{i,j}\tfrac{ \partial^2 u}{\partial x_i \partial x_j} - vu = f$
with terminal condition
$u(x, T) = \psi(x) \ \ (4)$
solution can be written as conditional expectation:
$u(x,t) = E\left[ \int_t^T e^{- \int_t^r v(X_\tau,\tau)\, d\tau}f(X_r,r)dr + e^{-\int_t^T v(X_\tau,\tau)\, d\tau}\psi(X_T) \Bigg| X_t=x \right]$
Feynman–Kac formula establish connection between PDE and stochastic process.
From Feynman–Kac formula taking $v\equiv 0$ and $f \equiv 0$ we get Kolmogorov backward equation :
for
$Lu = (\nabla_\mathbf{X}u)^{\mathsf T} b + \tfrac{1}{2} \sum_{i,j} (gg^\mathsf{T})_{i,j}\tfrac{ \partial^2 u}{\partial x_i \partial x_j}$
equation
$\frac{\partial u}{\partial t} + Lu = 0 \ \ (5)$
with terminal condition (4) have solution as conditional expectation
$u(x, t) = E(\psi(X_T) | X_t = x) \ \ (6)$
From Kolmogorov backward equation we can obtain Kolmogorov forward equation, which describe evolution of probability density for random process X (2)
In SDE courses it’s established that (2) is a Markov process and has transitional probability P and transitional density p:
p(x, s, y, t) = probability density at being at y in time t, on condition that it started at x in time s
taking u – solution of (5) with terminal condition (6)
$u(x, s) = E(\psi(X_T) | X_s = x) = \int p(x, s, z, T) \psi(z) dz$
From Markov property
$p(x, s, z, T) = \int p(x, s, y, t)p(y, t, z, T) dz$
from here
$u(x, s) = \int \int p(x, s, y, t) p(y, t, z, T) \psi(z) dz dy$
$u(x, s) = \int p(x, s, y, t) u(y, t) dy$
form here
$0 = \frac{\partial u}{\partial t} = \int \frac {\partial p(x, s, y, t)}{\partial t}u(y, t) + p(x, s, y, t) \frac {\partial u(y, t)}{\partial t} dy$
from (5)
$\int \frac{\partial p(x, s, y, t)}{\partial t}u(y, t) - (Lu) p(x, s, y, t) dy = 0 \ \ (7)$
Now we introduce dual operator $L^*$
$\int p (Lu) dy = \int (L^{*}p) u dy$
By integration by part we can get
$L^*u = -\nabla_\mathbf{X}^{\mathsf T}(ub) + \tfrac{1}{2} \sum_{i,j} \tfrac{ \partial^2 }{\partial x_i \partial x_j}(gg^\mathsf{T}u)_{i,j}$
and from (7)
$\int (\frac{\partial p(x, s, y, t)}{\partial t} - L^{*}p) u(y, t) dy = 0$
for t=T
$\int (\frac{\partial p(x, s, y, T)}{\partial t} - L^{*}p(x, s, y, T)) \psi(y) dy = 0$
This is true for any $\psi$, wich is independent from p
And we get Kolmogorov forward equation for p. Integrating by x we get the same equation for probability density at any moment T
$\frac{\partial p}{\partial t} - L^{*}p = 0$
$\tfrac{\partial J}{\partial t} = - \bigtriangledown_J E(J) + \sqrt{\tfrac{2}{\beta}}F(t)$
Stochastic gradient flow in SDE notation
$dJ = - \bigtriangledown_J E(J)dt + \sqrt{\tfrac{2}{\beta}} dW$ – integral of white noise is Wiener process
We want to find invariant probability density $p_G$. Invariant – means it doesn’t change with time, $\tfrac{\partial p_G}{\partial t} = 0$
so from Kolmogorov forward equation
$L^*p_G = 0$
or
$\nabla^{\mathsf T}(p_G\nabla E)+ \tfrac{1}{\beta} \nabla^{\mathsf T} \nabla p_G = 0$
$p_G\nabla E + \tfrac{1}{\beta} \nabla p_G = C$
C = 0 because we want $p_G$ integrable
$\nabla(E + \tfrac{1}{\beta} log(p_G)) = 0$
and at last we get Gibbs distribution
$p_G = \frac{exp(-\beta E)}{Z}$
Recalling again the chain of reasoning:
Wiener process
SDE + Ito Lemma + Ito product rule + zero expecation of Ito integral →
Kolmogorov forward equation for probability density →
Gibbs invariant distribution
13, November, 2013 Posted by | Uncategorized | Comments Off on Deriving Gibbs distribution from stochastic gradients
## Meaning of conditional probability
Conditional probability was always baffling me. Empirical, frequentists meaning is clear, but the abstract definition, originating from Kolmogorov – what was its mathematical meaning? How it can be derived? It’s a nontrivial definition and is appearing in the textbooks out the air, without measure theory intuition behind it.
Here I mostly follow Chang&Pollard paper Conditioning as disintegartion. Beware that the paper use non-standard notation, but this post follow more common notation, same as in wikipedia.
Here is example form Chang&Pollard paper:
Suppose we have distribution on $\mathbb{R}^2$ concentrated in two straight lines $L_1$ and $L_2$ with respective density $g_i(x)$ and angles with X axis $\alpha_i$. Observation (X,Y) taken, giving $X=x_0$, what is probability that point lies on the line $L_1$ ?
Standard approach would be approximate $x_0$ with $[ x_0, x_0 +\Delta ]$ and take limit with $\Delta \to 0$
$\frac{\int_{x_0}^{x_0+ \Delta}g_1/cos(\alpha_1)}{\int_{x_0}^{x_0+ \Delta}g_1/cos(\alpha_1) + \int_{x_0}^{x_0+ \Delta}g_2/cos(\alpha_2)}$
Not only taking this limit is kind of cumbersome, it’s also not totally obvious that it’s the same conditional probability that defined in the abstract definition – we are replacing ratio with limit here.
Now what is “correct way” to define conditional probabilities, especially for distributions?
For simplicity we will first talk about single scalar random variable, defined on probability space. We will think of random variable X as function on the sample space. Now condition $X=x_0$ define fiber – inverse image of $x_0$.
Disintegration theorem say that probability measure on the sample space can be decomposed into two measures – parametric family of measures induced by original probability on each fiber and “orthogonal” measure on $\mathbb{R}$ – on the parameter space of the first measure. Here $\mathbb{R}$ is the space of values of X and serve as parameter space for measures on fibers. Second measure induced by the inverse image of the function (random variable) for each measurable set on $\mathbb{R}$. This second measure is called Pushforward measure. Pushforward measure is just for measurable set on $\mathbb{R}$ (in our case) taking its X inverse image on sample space and measuring it with μ.
Fiber is in fact sample space for conditional event, and measure on fiber is our conditional distribution.
Full statement of the theorem require some term form measure theory. Following wikipedia
Let P(X) is collection of Borel probability measures on X, P(Y) is collection of Borel probability measures on Y
Let Y and X be two Radon spaces. Let μ ∈ P(Y), let $\pi$: Y → X be a Borel- measurable function, and let ν ∈ P(X) be the pushforward measure $\mu \circ \pi^{-1}$.
* Then there exists a ν-almost everywhere uniquely determined family of probability measures $\mu_x$ ⊆ P(Y) such that
* the function $x \mapsto \mu_{x}$ is Borel measurable, in the sense that $x \mapsto \mu_{x} (B)$ is a Borel-measurable function for each Borel-measurable set B ⊆ Y;
* $\mu_x$ “lives on” the fiber $\pi^{-1}(x)$ : for ν-almost all x ∈ X,
$\mu_x(Y \setminus \pi^{-1} (x))=0$
and so $\mu_x(E)=\mu_x(E \cap \pi^{-1}(x))$
* for every Borel-measurable function $f$ : Y → [0, ∞],
$\int_{Y} f(y) \, \mathrm{d} \mu (y) = \int_{X} \int_{\pi^{-1} (x)} f(y) \, \mathrm{d} \mu_{x} (y) \mathrm{d} \nu (x)$
From here for any event E form Y
$\mu (E) = \int_{X} \mu_{x} \left( E \right) \, \mathrm{d} \nu (x)$
This was complete statement of the disintegration theorem.
Now returning to Chang&Pollard example. For formal derivation I refer you to the original paper, here we will just “guess” $\mu_x$, and by uniqueness it give us disintegration. Our conditional distribution for $X=x$ will be just point masses on the intersections of lines $L_1$ and $L_2$ with axis $X=x$
Here $\delta$ is delta function – point mass.
$d\mu_x(y) = \frac{\delta (L_1( x )-y) g_1( x ) / cos(\alpha_1) + \delta (L_2( x )-y) g_2( x ) / cos(\alpha_2)}{Z(x)}dy$
$Z(x) = g_1(x)/cos(\alpha_1) + g_2(x)/cos(\alpha_2)$
and for $\nu$
$d\nu = Z(x)dx$
Our conditional probability that event lies on $L_1$ with condition $X = x_0$, form conditional density $d\mu_{x_0}/dy$ thus
$\frac{g_1(x_0)/cos(\alpha_1)}{g_1(x_0)/cos(\alpha_1) + g_2(x_0)/cos(\alpha_2)}$
Another example from Chen&Pollard. It relate to sufficient statistics. Term sufficient statistic used if we have probability distribution depending on some parameter, like in maximum likelihood estimation. Sufficient statistic is some function of sample, if it’s possible to estimate parameter of distribution from only values of that function in the best possible way – adding more data form the sample will not give more information about parameter of distribution.
Let $\mathbb{P}_{\theta}$ be uniform distribution on the square $[0, \theta]^2$. In that case M = max(x, y) is sufficient statistics for $\theta$. How to show it?
Let take our function $(x,y) \to m$ , $m = max(x,y)$ and make disintegration.
$d\mu_m(x,y) = s_m(x,y)dxdy$ is a uniform distribution on edges where x and y equal m and $d\nu = (2m / \theta^2 )dm$
$s_m(x,y)$ is density of conditional probability $P(\cdot | M = m)$ and it doesn’t depend on $\theta$
For any $f \ \ \mathbb{P}_{\theta}(f| m) = \int f s_m(x,y) \ , \ \mathbb{P}_{\theta, m}(\cdot) = \mathbb{P}_{m}(\cdot)$ – means that M is sufficient.
It seems that in most cases disintegration is not a tool for finding conditional distribution. Instead it can help to guess it and form uniqueness prove that the guess is correct. That correctness could be nontrivial – there are some paradoxes similar to Borel Paradox in Chang&Pollard paper.
7, November, 2013 Posted by | Math | , , | Comments Off on Meaning of conditional probability |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14165 | # Why/When we need the axiom schema of replacement?
In a quite informal way the axiom schema of replacement says that
Let $X$ be a set.
Let $\phi(x,y)$ a formula s.t. for each $x$ in $X$ there exists an unique set $y$ satisying $\phi$.
Then there exists the family $Y$ composed by all such $y$.
Therefore even if I know that any single set $y$ exists, without this axiom I cannot conclude that the family $Y$ exists.
In the finite case this should not be true, because from single sets $a$, $b$ and $c$ I can build $\{a,b,c\}$ by means the axiom of pair and the union axiom.
Therefore I suppose that we need this axiom schema just for the infinite case.
Am I wrog?
That depends very much on your axiomatization of $\sf ZF$.
Specifically, the axiom of pairing is a consequence of replacement, power set, and infinity (one could use "empty set" instead of infinity, but that would be redundant, since infinity implies the empty set exists directly).
If minimality is what your heart desires, then pairing is a theorem, not an axiom, and then I don't see why things work out in the finite case either.
But let's just put this aside, and let pairing be part of our system. Indeed, then in that case you need the axiom of replacement for infinite collections.
The classic example is $V_{\omega+\omega}$. We start with $V_0=\varnothing$, and for $n$, $V_{n+1}=\mathcal P(V_n)$. When we reach $\omega$, $V_\omega=\bigcup\{V_n\mid n<\omega\}$, and so we continue again with power sets and unions.
It is not hard to check that $V_{\omega+\omega}$ actually satisfies all the axioms of $\sf ZF$ except Replacement. Including pairing, just to be clear. But now consider the function given by $f(0)=V_\omega$ and $f(n+1)=\mathcal P(f(n))=V_{\omega+n+1}$. The range of $f$ is exactly $\{V_{\omega+n}\mid n<\omega\}$, and it is easy to see that this is not an element of $V_{\omega+\omega}$.
So indeed Replacement fails, and since pairing and union hold, it fails for infinite sets.
• Firstly, thank you Asaf! I try to recap: (1) yes, we need the replacement for putting togheter an infinite number of sets; (2) since replacement implies pairing, in a minimal setting we need it even for putting togheter a finite number of sets. – Michele Cirillo Dec 5 '17 at 11:19
• Yeah. While I can't quite sign off on that (2) point, I don't see any naive way to prove pairing without a whole bunch of axioms. So let's say that it seems like Replacement might even be needed for finite sets in minimal settings. – Asaf Karagila Dec 5 '17 at 11:27
• Ok, my mistake was that I haven't in mind the actual set of axioms that I used in the reasoning. We could formulate the point (2) in this way: (2') the possibility of build a set of two elements can be implied by different other axioms of my theory other than replacement (e.g. infinite+power set); but replacement might be needed in some cases, whenever other axioms are not assumed – Michele Cirillo Dec 5 '17 at 11:35 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14166 | # a solid right cylinder is 18 cm high and the radius of the base is 7 cm .two equal right cones are drilled on the plane faces of the cylinder , height of each cone being1/3rd the height of the cylinder and the radius of base of each cone being equal to base of the cylinder . find the total surface area of the remaining solid
Dear Student,
The following figure can be drawn based on the statements given.
Total surface area of the cone=${\mathrm{\pi r}}^{2}+\mathrm{\pi rl}$
Total surface area of solid left
=1100-713.68
=386.32
Regards,
• -3
fbdxf
• 0
What are you looking for? |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14167 | # Leonhard Euler
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Leonhard Paul Euler (15 April 1707 - 18 September 1783) was a pioneering Swiss mathematician and physicist and widely considered to be one of the greatest mathematicians of all time. His numerous works (over 900 publications) in many areas had a decisive influence on the development of mathematics.
## Biography
Leonhard Euler's father was pastor Paul Euler, who had studied theology at the University of Basel and had attended Jacob Bernoulli's lectures there. He also lived in Jacob's Bernoulli's house with Johann Bernoulli. Later he became a Protestant minister and married Margaret Brucker, the daughter of another Protestant minister.
Their son Leonhard Euler was born on 15 April 1707 in the town of Basel, Switzerland, but the family moved to Riehen when he was one year old. His father gave him the basic knowledge of mathematics.
At the age of 7, Euler returned to Basel and lived with his grandmother on his mother's side. During that time he went to school. However this school was known as a rather poor one, and Euler learnt no mathematics at all. Thus he read mathematics texts on his own and took some private lessons. His father wanted him to go to church and sent him to the University of Basel to prepare for the ministry. Euler entered the University in 1720, at the age of 14, first to obtain a general education. Johann Bernoulli soon discovered Euler's great potential for mathematics. Euler was writing that time:
... I soon found an opportunity to be introduced to a famous professor Johann Bernoulli. ... True, he was very busy and so refused flatly to give me private lessons; but he gave me much more valuable advice to start reading more difficult mathematical books on my own and to study them as diligently as I could; if I came across some obstacle or difficulty, I was given permission to visit him freely every Sunday afternoon and he kindly explained to me everything I could not understand ...
In 1723 he gained his Master's degree with a dissertation comparing the natural philosophy systems of Newton and Descartes. On his father's wishes, Euler he began to study theology, but spent all his spare time studying mathematics. He wrote two articles on reverse trajectory which were highly valued by his teacher Bernoulli. In 1727 Euler applied for a position as physics professor at Basel university, but was turned down.
At this time a new centre of science had appeared in Europe - the Petersburg Academy of Sciences. As Russia had few scientists of its own, many foreigners were invited to work at this centre - Euler among them. On the 24 May 1727 Euler arrived in Petersburg. His great talents were soon recognised. Among the areas he worked in include his theory of the production of the human voice, the theory of sound and music, the mechanics of vision, and his work on telescopic and microscopic perception. On the basis of this last work, not published until 1779, the construction of telescopes and microscopes was made possible.
In his study of colour effects, Euler hoped to make use of the observation of the conjunction of Venus and Moon, due to take place on the 8th of September 1729. However, no such effects were observed during this conjunction, and Euler was forced to wait for the eclipse of the sun which would take place in 1748. He observed this eclipse in Berlin, where he moved in 1741. Here he worked in the Prussian Academy of Sciences and was appointed as head of the Berlin Observatory, and was also tutor to the nieces of King Frederich II of Prussia.
Euler's works were not devoted solely to the natural sciences. A true renaissance man, he also involved himself in the philosophical debates of the day, and triumphantly declared himself a firm believer in the freedom of the will. Such views won him few friends in Germany, and the book in which he thus expressed himself was published for the first time in Russia, where Euler returned in 1766. Here he found many who agreed with his views, among them enemies of the views of Leibnitz and Voltaire.
In 1763 Catherine II came to the throne. She carried out reforms in the Academy of Sciences and aimed to make it a more prestigious institution. When Euler returned to Petersburg with his two elder sons they were given a two-storey house on the banks of the Neva and Euler given a position at the head of the Academy of Sciences.
When Euler renurned to Petersburg, he had already reconsidered his views on the atmosphere of planets. The work of Lomonosov and Bernoulli in this field led him to conclude that the atmosphere on the Earth and on other planets must be considerably more transparent than he had thought. Euler took a very active role in the observation of the movement of Venus across the face of the sun, despite the fact that at this time he was nearly blind. He had already lost one eye in the course of an experiment on light diffraction in 1738, and an eye disease and botched operation in 1771 led to an almost total loss of vision.
In contrast to most intellectuals of his time he was conservative and a convinced Christian. There is a story, which is often told in books, saying that once at the court of Catherine the Great he met the French philosopher Denis Diderot, who was a convinced atheist and tried to convince the Russians of atheism, much to the annoyance of Catherine. Therefore she asked Euler to stop him. Euler thought about it and when Catherine invited Diderot to have a theological discussion with Euler, Euler said: “${\displaystyle {\frac {a+b^{n}}{n}}=x}$, therefore God exists, answer!” Diderot, who knew almost nothing about algebra knew not what to answer and therefore returned to Paris. This story however is almost certainly an urban myth and Diderot knew enough algebra to answer Euler. However it is said that Euler published some not really serious proofs of the existence of God, which may well be, since at that time people were wondering about the possibility to give an algebraic proof of the existence of God.
This did not, however, stop Euler's creative output. Until his death in 18 September 1783, the Academy was presented with over 500 of his works. The Academy continued to publish them for another half century after the death of the great scientist. When Euler died the mathematician and philosopher Marquis de Condorcet said “...et il cessa de calculer et de vivre” ("and he stopped calculating and living"). |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14168 | # Distance covered by point on the rim
Given a circular wheel that is rotating at the rate of 25 revolutions per minute. If the radius of the wheel is 50 cms, what could be the distance covered by a point on the rim in one second (Given the π = 3.1416)
Any takes? Thanks.
-
Just use conversion factors. One revolution is $2\pi\times50\text{ cm}$ (why?), and there are sixty seconds in a minute... – Ｊ. Ｍ. Dec 14 '11 at 10:14
Is the wheel rolling along the ground? Or just rotating about a fixed axle? In the first case, the distance covered depends on the initial position of the point. – TonyK Dec 14 '11 at 10:18
The number of revolutions per second is $$\def\textstyle{} 25{\textstyle {\text{ rev} \over \text{ min}}}\cdot\textstyle{1\over 60}\textstyle{\text{min}\over \text{sec} } ={25\over 60}{\text{rev}\over\text{sec}} ={5\over12}{\text{rev}\over\text{sec}}.$$ The point travels $\pi\cdot2\cdot50=100\pi {\text{ cm}\over\text{rev}}$ . So, in one second, the distance would be $$\underbrace{100\pi \textstyle {\text{ cm}\over\text{rev}}}_{\text{ dist per rev}}\cdot\underbrace{\textstyle {5\over12}\textstyle{\text{rev}\over\text{sec}}\cdot 1\text{sec}}_{ \text{number of revs.}}={125\pi\over3} \text{ cm}.$$
Yes, that's correct. The point travels $125\pi/3$ (which is the same as you have) centimeters per second. (In the above, the wheel is rotating at $5/12$ revs per second, which gives the angle as $\theta={5\over 12}\cdot 2\pi= {5\pi\over6}$.) – David Mitra Dec 14 '11 at 10:55 |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14169 | # If 1 2 3 12 x A é ù = ê ú ë û – and 1 y B x é ù ê ú = ê ú ê ú ë û be such that 6 AB 8 é ù = ê ú ë û , then:
Question:
If $A=\left[\begin{array}{ccc}1 & 2 & x \\ 3 & -1 & 2\end{array}\right]$ and $B=\left[\begin{array}{l}y \\ x \\ 1\end{array}\right]$ be such that $\mathrm{AB}=\left[\begin{array}{l}6 \\ 8\end{array}\right]$, then:
1. $y=2 x$
2. $y=-2 x$
3. $y=x$
4. $y=-x$
Correct Option: 1
JEE Main Previous Year 1 Question of JEE Main from Mathematics Inverse Trigonometric Functions chapter.
JEE Main Previous Year Online April 12, 2014
Solution:
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MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14170 | +0
0
281
3
The vertex of the right isosceles triangle is the center of the square. What is the area of the overlapping region?
May 27, 2021
#1
0
The area of the overlapping region is 20.
May 27, 2021
#2
0
Could you explain your reasoning a bit to help me understand?
Guest May 27, 2021
#3
+26321
+1
$$\begin{array}{|rcll|} \hline A &=& \dfrac{5x}{2} + 5*\left(\dfrac{(5-x)+5}{2}\right) \\ A &=& \dfrac{5x}{2} + 5*\left(\dfrac{10-x}{2}\right) \\ A &=& \dfrac{5}{2}\left( x+10-x \right) \\ A &=& \dfrac{5}{2}*10 \\ \mathbf{A} &=& \mathbf{25} \\ \hline \end{array}$$ |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14171 | ## BdMO National Higher Secondary 2014/2
Discussion on Bangladesh Mathematical Olympiad (BdMO) National
samiul_samin
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### BdMO National Higher Secondary 2014/2
Closing his eyes Tawsif begins to plca knights on a Chess board of $21\times 23$.After plcing how many knights Tawsif will be sure that in next move at least one knight will attack another one?
samiul_samin
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### Re: BdMO National Higher Secondary 2014/2
Hint
Answer
Last edited by samiul_samin on Sun Feb 18, 2018 7:03 pm, edited 1 time in total.
samiul_samin
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### Re: BdMO National Higher Secondary 2014/2
Simillar problem
BdMO National Secondary 2014/3
Last edited by samiul_samin on Sun Feb 18, 2018 7:46 pm, edited 1 time in total.
samiul_samin
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### Re: BdMO National Higher Secondary 2014/2
Chess
Knowing basic Chess rules is better to solve this problem.It is really an interesting problem which is also important for programmers!
ahmedittihad
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### Re: BdMO National Higher Secondary 2014/2
Do not spam in the forum.
Frankly, my dear, I don't give a damn.
samiul_samin
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### Re: BdMO National Higher Secondary 2014/2
ahmedittihad wrote:
Sun Feb 18, 2018 7:43 pm
Do not spam in the forum.
What do you mean?
ahmedittihad
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Joined: Mon Mar 28, 2016 6:21 pm
### Re: BdMO National Higher Secondary 2014/2
What I mean is, you have posted 4 posts which you could have done in 1. That's spamming.
Frankly, my dear, I don't give a damn.
samiul_samin
Posts: 1007
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### Re: BdMO National Higher Secondary 2014/2
ahmedittihad wrote:
Sun Feb 18, 2018 7:49 pm
What I mean is, you have posted 4 posts which you could have done in 1. That's spamming.
Oh,Sorry.I don't know it.I am really sorry.I will not do it later.
samiul_samin
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### Re: BdMO National Higher Secondary 2014/2
Here is the detailed Solution |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14172 | Address 4615 70th Ave, Kenosha, WI 53144 (262) 652-2626 http://www.tg3electronics.com/contact-us
# linear fit standard error Somers, Wisconsin
Previous company name is ISIS, how to list on CV? There is no contradiction, nor could there be. It is sometimes useful to calculate rxy from the data independently using this equation: r x y = x y ¯ − x ¯ y ¯ ( x 2 ¯ − Notice that it is inversely proportional to the square root of the sample size, so it tends to go down as the sample size goes up.
Confidence intervals were devised to give a plausible set of values the estimates might have if one repeated the experiment a very large number of times. In this analysis, the confidence level is defined for us in the problem. Approximately 95% of the observations should fall within plus/minus 2*standard error of the regression from the regression line, which is also a quick approximation of a 95% prediction interval. So now I need to find the confidance interval of a.
S is known both as the standard error of the regression and as the standard error of the estimate. Error t value Pr(>|t|) (Intercept) 5.000e+00 2.458e-16 2.035e+16 <2e-16 *** xdata 1.000e+00 3.961e-17 2.525e+16 <2e-16 *** --- Signif. Other regression methods besides the simple ordinary least squares (OLS) also exist. Today, I’ll highlight a sorely underappreciated regression statistic: S, or the standard error of the regression.
And the uncertainty is denoted by the confidence level. Jim Name: Jim Frost • Tuesday, July 8, 2014 Hi Himanshu, Thanks so much for your kind comments! The latter case is justified by the central limit theorem. In light of that, can you provide a proof that it should be $\hat{\mathbf{\beta}} = (\mathbf{X}^{\prime} \mathbf{X})^{-1} \mathbf{X}^{\prime} \mathbf{y} - (\mathbf{X}^{\prime} \mathbf{X})^{-1} \mathbf{X}^{\prime} \mathbf{\epsilon}$ instead? –gung Apr 6 at 3:40 1
What is the probability that they were born on different days? Sign Me Up > You Might Also Like: How to Predict with Minitab: Using BMI to Predict the Body Fat Percentage, Part 2 How High Should R-squared Be in Regression That is, we are 99% confident that the true slope of the regression line is in the range defined by 0.55 + 0.63. This would be quite a bit longer without the matrix algebra.
Retrieved 2016-10-17. ^ Seltman, Howard J. (2008-09-08). Discover... The standard error of the forecast for Y at a given value of X is the square root of the sum of squares of the standard error of the regression and We are working with a 99% confidence level.
Sieve of Eratosthenes, Step by Step Public huts to stay overnight around UK 2002 research: speed of light slowing down? However, more data will not systematically reduce the standard error of the regression. Estimation Requirements The approach described in this lesson is valid whenever the standard requirements for simple linear regression are met. Note, however, that the critical value is based on a t score with n - 2 degrees of freedom.
Is it correct to write "teoremo X statas, ke" in the sense of "theorem X states that"? In a simple regression model, the percentage of variance "explained" by the model, which is called R-squared, is the square of the correlation between Y and X. What's the bottom line? Return to top of page.
Confidence intervals The formulas given in the previous section allow one to calculate the point estimates of α and β — that is, the coefficients of the regression line for the Is there a way to view total rocket mass in KSP? In other words, α (the y-intercept) and β (the slope) solve the following minimization problem: Find min α , β Q ( α , β ) , for Q ( α For example in the following output: lm(formula = y ~ x1 + x2, data = sub.pyth) coef.est coef.se (Intercept) 1.32 0.39 x1 0.51 0.05 x2 0.81 0.02 n = 40, k
But if it is assumed that everything is OK, what information can you obtain from that table? The coefficients, standard errors, and forecasts for this model are obtained as follows. Is a food chain without plants plausible? The standard error of the forecast is not quite as sensitive to X in relative terms as is the standard error of the mean, because of the presence of the noise
When n is large such a change does not alter the results appreciably. It is also possible to evaluate the properties under other assumptions, such as inhomogeneity, but this is discussed elsewhere.[clarification needed] Unbiasedness The estimators α ^ {\displaystyle {\hat {\alpha }}} and β Confidence intervals for the mean and for the forecast are equal to the point estimate plus-or-minus the appropriate standard error multiplied by the appropriate 2-tailed critical value of the t distribution. Although the OLS article argues that it would be more appropriate to run a quadratic regression for this data, the simple linear regression model is applied here instead.
The standard error of the mean is usually a lot smaller than the standard error of the regression except when the sample size is very small and/or you are trying to Thanks! The critical value is a factor used to compute the margin of error. AP Statistics Tutorial Exploring Data ▸ The basics ▾ Variables ▾ Population vs sample ▾ Central tendency ▾ Variability ▾ Position ▸ Charts and graphs ▾ Patterns in data ▾ Dotplots
The standard error of the estimate is a measure of the accuracy of predictions. share|improve this answer edited Apr 7 at 22:55 whuber♦ 145k17284544 answered Apr 6 at 3:06 Linzhe Nie 12 1 The derivation of the OLS estimator for the beta vector, $\hat{\boldsymbol The original inches can be recovered by Round(x/0.0254) and then re-converted to metric: if this is done, the results become β ^ = 61.6746 , α ^ = − 39.7468. {\displaystyle To illustrate this, let’s go back to the BMI example. Fitting so many terms to so few data points will artificially inflate the R-squared. Is there a way to view total rocket mass in KSP? To find the critical value, we take these steps. That is, R-squared = rXY2, and that′s why it′s called R-squared. Can 「持ち込んだ食品を飲食するのは禁止である。」be simplified for a notification board? So basically for the second question the SD indicates horizontal dispersion and the R^2 indicates the overall fit or vertical dispersion? –Dbr Nov 11 '11 at 8:42 4 @Dbr, glad Is there a succinct way of performing that specific line with just basic operators? –ako Dec 1 '12 at 18:57 1 @AkselO There is the well-known closed form expression for What is the probability that they were born on different days? The standard error of the forecast gets smaller as the sample size is increased, but only up to a point. In the multivariate case, you have to use the general formula given above. –ocram Dec 2 '12 at 7:21 2 +1, a quick question, how does$Var(\hat\beta)\$ come? –loganecolss Feb A model does not always improve when more variables are added: adjusted R-squared can go down (even go negative) if irrelevant variables are added. 8. Suppose our requirement is that the predictions must be within +/- 5% of the actual value. |
MathCode-Pile_decontaminated_orig_math-related_devided_processed_train-00004-of-00114-3158c787ea8296d3_doc_14173 | # Derivation of nonlinear Gibbs measures from many-body quantum mechanics
```@article{Lewin2014DerivationON,
title={Derivation of nonlinear Gibbs measures from many-body quantum mechanics},
author={Mathieu Lewin and Phan Th{\`a}nh Nam and Nicolas Rougerie},
journal={arXiv: Mathematical Physics},
year={2014}
}```
• Published 1 October 2014
• Physics
• arXiv: Mathematical Physics
We prove that nonlinear Gibbs measures can be obtained from the corresponding many-body, grand-canonical, quantum Gibbs states, in a mean-field limit where the temperature T diverges and the interaction behaves as 1/T. We proceed by characterizing the interacting Gibbs state as minimizing a functional counting the free-energy relatively to the non-interacting case. We then perform an infinite-dimensional analogue of phase-space semiclassical analysis, using fine properties of the quantum…
Classical field theory limit of 2D many-body quantum Gibbs states
• Physics
• 2018
Nonlinear Gibbs measures play an important role in many areas of mathematics, including nonlinear dispersive equations with random initial data and stochastic partial differential equations. In
Classical field theory limit of many-body quantum Gibbs states in 2D and 3D
• Physics
• 2020
We provide a rigorous derivation of nonlinear Gibbs measures in two and three space dimensions, starting from many-body quantum systems in thermal equilibrium. More precisely, we prove that the
Gibbs Measures of Nonlinear Schrödinger Equations as Limits of Quantum Many-Body States in Dimension d ≤ 3
• Physics
• 2016
We prove that Gibbs measures of nonlinear Schr\"odinger equations arise as high-temperature limits of thermal states in many-body quantum mechanics. Our results hold for defocusing interactions in
From Bosonic Grand-Canonical Ensembles to Nonlinear Gibbs Measures
In a recent paper, in collaboration with Mathieu Lewin and Phan Th{\`a}nh Nam, we showed that nonlinear Gibbs measures based on Gross-Pitaevskii like functionals could be derived from many-body
Gibbs Measures of Nonlinear Schrödinger Equations as Limits of Many-Body Quantum States in Dimensions \$\${d \leqslant 3}\$\$d⩽3
• Physics
• 2016
We prove that Gibbs measures of nonlinear Schrödinger equations arise as high-temperature limits of thermal states in many-body quantum mechanics. Our results hold for defocusing interactions in
De Finetti theorems, mean-field limits and Bose-Einstein condensation
These notes deal with the mean-field approximation for equilibrium states of N-body systems in classical and quantum statistical mechanics. A general strategy for the justification of effective
Gibbs measures as unique KMS equilibrium states of nonlinear Hamiltonian PDEs.
• Mathematics
• 2021
The classical Kubo-Martin-Schwinger (KMS) condition is a fundamental property of statistical mechanics characterizing the equilibrium of infinite classical mechanical systems. It was introduced in
A Microscopic Derivation of Gibbs Measures for Nonlinear Schrödinger Equations with Unbounded Interaction Potentials
We study the derivation of the Gibbs measure for the nonlinear Schrodinger equation (NLS) from many-body quantum thermal states in the high-temperature limit. In this paper, we consider the nonlocal
Cylindrical Wigner measures
In this paper we study the semiclassical behavior of quantum states acting on the C*-algebra of canonical commutation relations, from a general perspective. The aim is to provide a unified and
## References
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Exponential Relaxation to Equilibrium for a One-Dimensional Focusing Non-Linear Schrödinger Equation with Noise
• Mathematics, Physics
• 2014
We construct generalized grand-canonical- and canonical Gibbs measures for a Hamiltonian system described in terms of a complex scalar field that is defined on a circle and satisfies a nonlinear
The mean-field approximation and the non-linear Schrödinger functional for trapped Bose gases
• Physics, Mathematics
• 2014
We study the ground state of a trapped Bose gas, starting from the full many-body Schrodinger Hamiltonian, and derive the nonlinear Schrodinger energy functional in the limit of large particle
Mean field limit for bosons and propagation of Wigner measures
• Physics
• 2009
We consider the N-body Schrodinger dynamics of bosons in the mean field limit with a bounded pair-interaction potential. According to the previous work [Ammari, Z. and Nier, F., “Mean field limit for
REMARKS ON THE QUANTUM DE FINETTI THEOREM FOR BOSONIC SYSTEMS
• Mathematics
• 2013
The quantum de Finetti theorem asserts that the k-body density matrices of a N-body bosonic state approach a convex combination of Hartree states (pure tensor powers) when N is large and k fixed. In
Théorèmes de de Finetti, limites de champ moyen et condensation de Bose-Einstein
These lecture notes treat the mean-field approximation for equilibrium states of N body systems in classical and quantum statistical mechanics. A general strategy to justify effective models based on
Long time dynamics for the one dimensional non linear Schr\"odinger equation
• Mathematics
• 2010
In this article, we first present the construction of Gibbs measures associated to nonlinear Schr\"odinger equations with harmonic potential. Then we show that the corresponding Cauchy problem is
Examples of Bosonic de Finetti States over Finite Dimensional Hilbert Spaces
According to the Quantum de Finetti Theorem, locally normal infinite particle states with Bose–Einstein symmetry can be represented as mixtures of infinite tensor powers of vector states. This note
Statistical mechanics of the nonlinear Schrödinger equation
• Mathematics
• 1988
AbstractWe investigate the statistical mechanics of a complex fieldø whose dynamics is governed by the nonlinear Schrödinger equation. Such fields describe, in suitable idealizations, Langmuir waves |
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# Plasmonic nanostar photocathodes for optically-controlled directional currents
## Abstract
Plasmonic nanocathodes offer unique opportunities for optically driving, switching, and steering femtosecond photocurrents in nanoelectronic devices and pulsed electron sources. However, angular photocurrent distributions in nanoplasmonic systems remain poorly understood and are therefore difficult to anticipate and control. Here, we provide a direct momentum-space characterization of multiphoton photoemission from plasmonic gold nanostars and demonstrate all-optical control over these currents. Versatile angular control is achieved by selectively exciting different tips on single nanostars via laser frequency or linear polarization, thereby rotating the tip-aligned directional photoemission as observed with angle-resolved 2D velocity mapping and 3D reconstruction. Classical plasmonic field simulations combined with quantum photoemission theory elucidate the role of surface-mediated nonlinear excitation for plasmonic field enhancements highly concentrated at the sharp tips (Rtip = 3.4 nm). We thus establish a simple mechanism for femtosecond spatiotemporal current control in designer nanosystems.
## Introduction
Femtosecond optical control over nanoscale currents is essential for ultrafast electron diffraction and microscopy1,2,3,4,5,6, x-ray free-electron lasers7,8,9, and terahertz optoelectronic circuits10,12,12. Among recent advances, plasmonic metal nanostructures have shown considerable versatility and promise as bright photocathodes8,9,13, femtosecond photodiodes10,11, and carrier-envelope-phase-sensitive photodetectors14,15 that can be integrated into nanoscale, chip-based devices. In plasmonic systems, the mapping from optical field parameters onto near-field electron dynamics is primarily governed by the particle geometry and corresponding field enhancements, which can be crafted with high precision by synthetic or lithographic methods11,13,16,17,18. Particle geometry defines the surface field-enhanced hot-spot regions where conduction electrons build up during collective oscillations, become excited, and escape as photoemission or photovoltaic currents. Geometry and particle array patterning also govern the frequency response of plasmonic systems, which have been tailored for broadband photodetection19, photocurrent polarity control20, and selective multi-mode lasing with narrow spectral linewidths21. Spectral characteristics can even be controlled at the single-particle level for asymmetric particles that support multiple resonances, such as gold nanostars16,22,23,24.
With nanostars and other multi-resonant particles, important opportunities for spatiotemporal photocurrent control emerge via frequency- and polarization-selective excitation of different plasmonic hot spots, which are often spatially separated and oriented in different directions16,17,22,24,25. Electric near-field hot spots have been extensively investigated in nanoplasmonic systems, with photoemission electron microscopy (PEEM) studies establishing the correlation between photoemission and plasmonic hot spots with ~20 nm spatial resolution16,17. Furthermore, these techniques have been combined with optical pulse shaping26 to achieve coherent control over spatial photoemission distributions on femtosecond timescales18. However, direct observation of the corresponding photoelectron momentum-space distributions has remained a challenge, requiring angle-resolved photoelectron mapping from single, resonantly excited nanoparticles. Such capabilities have only appeared recently25,27,28,29 as a comprehensive understanding of photocurrent distributions is becoming crucial for the design and implementation of nanocathodes in nascent ultrafast nanoelectronics and electron imaging applications. Full photoelectron momentum and energy characterization has been achieved by Lehr et al.28,29 on individual gold nanorods and bow-tie nanoantennas using time-of-flight momentum PEEM28,29, which serves to clarify nanoplasmonic angular photoemission distributions and phenomena such as the transition into the optical field emission regime. Angular photoelectron mapping and steering have also been demonstrated for gold30 and tungsten31,32 nanotips, primarily in the field emission regime. Despite these advances, many important aspects of the nanoplasmonic photoelectron emission mechanism and opportunities for angular control remain to be elucidated, particularly in the multiphoton regime.
Here, we demonstrate optically-controlled directional multiphoton photoemission (MPPE) from single gold nanostars, using two-dimensional (2D) photoelectron velocity mapping and 3D reconstruction for detailed characterization of the angular and energy distributions. We begin with an examination of the plasmonic properties and directional photoemission from single-nanostar tips resonantly excited using a pulsed femtosecond laser. Individual tips behave as locally bright, point-like electron sources with a high degree of spatial coherence. We then demonstrate the selective excitation of different tips on a single nanostar via optical frequency and polarization control, yielding wide angular switching/steering of the tip-aligned photoemission currents. Correlated scanning electron microscopy (SEM) and 3D photoemission modeling clarify the effects of nanostar tip geometry and plasmonic field enhancements. As the overall directional effects are not contingent on laser intensity, the present method for optically controlling photocurrents can be extended from the weak-field (multiphoton) regime into the strong-field (optical field emission) regime. However, weak-field MPPE processes are emphasized here due to the minimal nanostar heating, sub-single-electron femtosecond pulses (10−5 photoelectrons from each pulse on average) that preclude space-charge effects, and <1 eV photoelectron kinetic energy spreads for high temporal coherence. Photocurrent control timescales approaching the attosecond range may be achievable, fundamentally limited only by the nonlinear photoemission decay associated with plasmon dephasing.
## Results
### Single-tip excitation and photoemission properties
Gold nanostars are synthesized with sharp 3.4 ± 0.4 nm radii tips and sorted by size (see Methods) to select for simple geometries with an average number of three tips lying in the surface plane (Supplementary Fig. 1). Electron micrographs for a selection of representative nanostars are shown in Fig. 1a. Plasmonic properties are readily characterized using normal-incidence laser light for these primarily in-plane geometries, with the nanostars supporting localized surface plasmon resonances across the near-infrared tuning range of the Ti:sapphire oscillator (700–1000 nm, 50 fs, 75 MHz). A scanning sample stage enables diffraction-limited photoemission mapping for locating single nanostars dispersed on a transparent, conductive indium tin oxide (ITO)-coated glass substrate27,33 (see Methods). We then collect photoelectron velocity distributions as a function of laser frequency, polarization, and intensity with the velocity map imaging (VMI) electrostatic lens configuration depicted in Fig. 1b34.
We characterized the resonance properties for three sample nanostars via spectral (Fig. 1c) and polarization dependences (Fig. 1d), with the peak polarization indicated in the correlated SEM insets aligned along a particular tip for each nanostar. In each case, the n-photon photoemission (nPPE) rate varies as $${{\it{\Gamma }}}_{n{\mathrm{PPE}}}\sim {\mathrm{cos}}^{2n}( {\theta - \theta _{{\mathrm{tip}}}} )$$ as the polarization angle θ is rotated away from the resonant tip, consistent with the field projection onto a single, well-defined hot-spot axis. The essential role of the plasmon resonance in driving these behaviors has been established in previous photoemission studies correlated with dark field microscopy for gold nanorods33, confirming the direct relation between surface plasmon (Mie) resonances in linear scattering spectra and the peaks observed in MPPE spectra. Due to the E2n In electric near-field sensitivity of the n-photon process, nanostars with multiple similar tips may have only one dominant hot spot (e.g., Star 2). Note, for instance, that a mere 10% difference in field enhancement results in a factor of two difference in the 4PPE rate.
Electrons must absorb multiple visible or near-infrared photons (~1.5 eV) to overcome typical metal work functions (~5 eV) and escape into the vacuum. The total MPPE rate is described by a sum over process orders,
$${{{\it{\Gamma} }}_{{\mathrm{MPPE}}} = \mathop {\sum}\limits_{\it{n}} {{{\it{\Gamma} }}_{{\it{n}}{\mathrm{PPE}}}} = \mathop {\sum}\limits_{\it{n}} {\sigma _{\it{n}}\left( {\omega ,\theta } \right){\it{I}}_0^{\it{n}}} } ,$$
(1)
in which I0 is the input laser intensity and σn(ω, θ) is the nPPE cross-section as a function of laser frequency and polarization. The plasmonic field enhancement effect is therefore included within σn(ω, θ). We determined multiphoton process orders for a random selection of resonantly excited nanostars by measuring photoemission rate as a function of peak laser pulse intensity. The process order summary in Fig. 1e reveals a sigmoidal transition from 3PPE to 4PPE centered around 830 nm (1.5 eV), indicating a nanostar tip work function of $$\phi \approx 3\hbar \omega _{830} =\;$$4.5 eV at which three-photon-excited electrons from the Fermi level begin to overcome the surface potential barrier. Intensity-dependence plots and power-law fits are shown for the three sample nanostars, with the non-integer Star 2 process order (n = 3.6) representing weighted contributions from both 3PPE and 4PPE processes. From Eq. (1), the log–log slope is
$${\frac{{{\mathrm{dlog}}_{10}\left( {{{\it{\Gamma} }}_{{\mathrm{MPPE}}}} \right)}}{{{\mathrm{dlog}}_{10}\left( {I_0} \right)}} = \frac{{\sum\nolimits_n {\sigma _n\left( {\omega ,\theta } \right)I_0^nn} }}{{ \sum\nolimits_n {\sigma _n\left( {\omega ,\theta } \right)I_0^n} }} = \sum\limits_n {w_nn} } ,$$
(2)
which is the weighted average over process orders with weight factors $$w_n = \sigma _n\left( {\omega ,\theta } \right)I_0^n/\mathop {\sum }\nolimits_{\tilde n} \sigma _{\tilde n}\left( {\omega ,\theta } \right)I_0^{\tilde n}$$ representing the relative contributions of each process order, such that $$\mathop {\sum }\nolimits_n w_n = 1$$. With only 3PPE and 4PPE processes contributing, Eq. (2) yields an effective process order of 3.5 when $$\sigma _3\left( {\omega ,\theta } \right)I_0^3 = \sigma _4\left( {\omega ,\theta } \right)I_0^4$$, as is the case at 830 nm in Fig. 1e. It should be noted that peak photocurrents of 108 s−1 during the peak laser pulse intensity correspond to an average of only 10−5 photoelectrons from each pulse when multiplied by the laser pulse duration (10−13 s). With a laser repetition rate of 75 MHz, we can thus achieve photoemission rates >100 s−1 with negligible probability of two electron emission events occurring within a single laser pulse, thereby precluding any space-charge effects.
These measurements verify that all signal is predominantly due to MPPE rather than optical field emission or thermionic emission for the range of intensities utilized in these studies. For further verification, the Keldysh parameter, $$\gamma = \sqrt {\phi /2U_{\mathrm{p}}}$$, is commonly used to characterize the transition from weak- to strong-field emission35, where ϕ is the work function and $$U_{\mathrm{p}} = e^2E^2/\left( {4m_e\omega ^2} \right)$$ is the ponderomotive energy. Perturbative MPPE is dominant for γ > 2 and optical field emission is dominant for γ < 1, with the transition occurring in the 1 < γ < 2 range9. For peak input pulse intensities I0 < 3 × 108 W cm−2 and simulated field enhancements $$\left| {E/E_0} \right|\, <\, 100$$ described below, Keldysh parameters γ > 3 fall within the MPPE regime and corroborate the measured intensity dependences.
### Single-tip photoelectron velocity distributions
Selecting Star 1 (Figs. 1c and 2a) as a representative single-tip emitter, we measured and modeled the characteristic photoelectron velocity distribution from a resonantly-excited nanostar tip. First, we performed finite element simulations using the nanostar geometry measured via correlated SEM for insight into the plasmonic field enhancements. After we account for a 3 nm Pt coating applied prior to SEM imaging for improved contrast (see Methods), the nanostar arm angles, lengths, and widths in our models are determined from the correlated SEM micrographs, with the arm shapes and tip radii (3.4 nm) obtained by statistical transmission electron microscopy (TEM) characterization (Supplementary Fig. 1). Both the influence of the ITO-coated substrate and the HEPES (4-(2-hydroxyethyl)-1-piperazineethanesulfonic acid) ligand layer on the nanostar surface were accounted for in the simulations. Further details of the finite element modeling are shown in Supplementary Fig. 2 and discussed in Supplementary Note 1. The simulated surface-normal field distribution in Fig. 2b highlights all of the tips in the linear case, with a particularly strong peak enhancement of $$\left| {E_ \bot /E_0} \right| \approx 50$$ at the resonantly-excited tip. This tip becomes clearly dominant when the field enhancement is raised to the eighth power for the relevant 4PPE process, with the nonlinear enhancement strongly confined to its apex. Photoemission distributions measured for this hemispherical hot spot geometry may also guide expectations for convex hot spots in other systems, including etched nanotip photocathodes4,6.
The measured 2D (vz-projected) photoelectron velocity map in Fig. 2c shows clear directionality along the resonant tip axis, as would be predicted for electrons emitted outward from the field-enhanced surface region. The 3D velocity distribution (Fig. 2e) was reconstructed using the basis set expansion (BASEX) algorithm36, which relies on the approximation of cylindrical symmetry around the resonant x-axis tip to compensate for information lost in the vz projection (Supplementary Note 2). Tip-aligned directional emission is also observed for Stars 2 and 3 (Supplementary Fig. 3), which are resonantly excited in the transition regime and the 3PPE regime, respectively. While both 3PPE and 4PPE processes lead to similarly tip-aligned photoemission distributions, photon energies just above the 3PPE onset may be preferred to reduce the kinetic energy spread (Supplementary Fig. 3) for optimal temporal coherence of the photoelectron waveform1.
Photoemission generally proceeds by a combination of volume and surface mechanisms for momentum conservation37,38. The directional distributions observed here indicate the dominant role of surface scattering, in which case the multiphoton excitation rate depends on the surface-normal electric field37,39,40,40, $$E_ \bot ^{2n}$$. In contrast, volume excitation mechanisms depend on internal fields as E2n (Supplementary Fig. 2) and produce hot electrons throughout the nanostar that would escape from various surfaces around the tips and body, leading to largely isotropic rather than directional velocity distributions. We therefore restricted our present theoretical investigations to surface-mediated MPPE, which is further supported by the tip-localized photoemission observed in nanostar PEEM studies16.
Calculations based on the coherent surface MPPE theory developed by Yalunin et al.40 were carried out to determine photoemission rates and velocity distributions from each nanostar surface area element, using the full SEM-correlated nanostar geometry and simulated plasmonic fields shown in Fig. 2b, with a fine surface mesh shown in Supplementary Fig. 2. The full 3D photoelectron velocity distribution was then determined by summing all photoemission contributions over the entire nanostar surface (see Methods). An initial Fermi–Dirac electron distribution within the gold was considered at a pulse-averaged temperature of 2500 K, calculated via the two-temperature model (Supplementary Fig. 6 and Supplementary Note 4) with a linear nanostar absorption cross-section of 104 nm2 determined via the finite element simulations. The emission process at each nanostar surface area element consists of a heated Fermi gas impinging on the surface barrier and becoming excited via coherent multiphoton absorption into external Volkov (field-dressed) states. The outgoing photocurrent includes direct and surface-rescattered quantum amplitudes40. Optical field penetration into the gold is neglected and all excitation therefore occurs from the external evanescent component of the initial free-electron wavefunctions, which accounts for momentum non-conservation at the surface barrier. Due to the short decay length of the external electron wavefunction (~2 Å) and the small quiver amplitude of the emitted electrons (~1.5 Å) relative to the plasmonic field decay length (~1.5 nm), the external field is approximated to be uniform in calculating the emission amplitude. The evanescent plasmonic field then ponderomotively accelerates the emitted electrons outward from the surface, uniformly shifting the kinetic energy distribution by Up as described further in the Methods. The ponderomotive energy $$U_{\mathrm{p}} = E^2e^2/\left( {4m_e\omega ^2} \right) = I_0\left| {E/E_0} \right|^2e^2/\left( {2c\varepsilon _0m_e\omega ^2} \right)$$ for the present input intensity (I0 = 2 × 108 W cm−2) and calculated tip field enhancement (|E/E0| = 50) is only 0.03 eV. Although only a minor contribution in the present studies, the ponderomotive energy scales linearly with the input intensity and quadratically with the field enhancement and thus may become important in other similar systems.
Theoretical 2D (Fig. 2d) and 3D (Fig. 2f) distributions reiterate and confirm the tip-aligned directionality, with good agreement in the photoelectron angular (Fig. 2g) and kinetic energy (Fig. 2h) distributions. The slight downward directionality in the theoretical 3D distribution is due to the ITO image charge field, which skews the tip field enhancement toward the substrate (also see Supplementary Fig. 4). However, the otherwise strong agreement between experimental and theoretical distributions indicates that such substrate symmetry-breaking effects are minor. With regard to substrate effects, we also mention the possibility of gap resonances occurring in nanoscale junctions between nanoparticles and conducting substrates separated by an organic molecular layer such as the ~1 nm HEPES ligand layer on the nanostars, as observed previously with Au nanospheres separated from an Au film substrate by an organic spacer layer41. While no direct evidence of a gap resonance effect is observed in the present experiments nor in the finite element simulations (Supplementary Fig. 2), due primarily to the in-plane resonance excitation with in-plane polarized light, such effects can be difficult to accurately account for in the simulations as they depend sensitively on the details of the nanostar–substrate interface, which we do not directly observe.
Although we have investigated the near-field contributions to photoemission dynamics and distributions in detail in order to gain general insights for arbitrary emitter geometries, the nanostar tips studied here are among the sharpest nanoplasmonic geometries available and for most purposes can be treated as point-like electron sources. Electrons may therefore be emitted into a broad angular distribution (Fig. 2g) but still retain a high degree of spatial coherence, determined by the small source radius (3.4 nm) and photoelectron angular uncertainty (~70°). Nanostar tip spatial coherence characteristics are comparable to those of state-of-the-art femtosecond electron sources operating in the linear photoemission regime42 (Supplementary Note 3), which are only an order of magnitude from the fundamental limit imposed by the uncertainty principle. Furthermore, single-nanostar tips remain stable emission sources in the present intensity range during hours of measurements, including at least half an hour of continuous exposure while collecting velocity maps, as demonstrated in Supplementary Fig. 5. We calculate only modest lattice temperature increase of 200 K during a laser pulse (Supplementary Figs. 6, 7 and Supplementary Note 4) with 13 ns between pulses sufficient for equilibration back to room temperature. The high degree of stability and spatial coherence enables ultrafast point projection3,4,6 and diffractive2,3,5 imaging of nearby nanoscale objects following possible electron beam manipulation such as acceleration or collimation via nanoengineered electron optics7. For higher optical intensities, a variety of new physical behaviors emerge in the strong fields and gradients at point-like nanotips43, including the onset of tunneling emission and subsequent quiver or sub-cycle dynamics44. Such effects are negligible in the present MPPE studies, but the theory readily extends into intermediate- and strong-field regimes40 and the influence of strongly varying plasmonic fields on the photoelectron trajectories can be included in the manner demonstrated by Dombi et al.13 when necessary.
### Selective excitation of multiple nanostar hot spots
Building on previous investigations of selective nanostar tip excitation16, the excitation of in-plane tips with in-plane polarization control in the present studies provides a particularly clear mapping between optical parameters and tip hot spots. While the presence of many tips and the effect of near-field tip–tip coupling22 can lead to complicated optical parameter mappings in some cases, we emphasize the simple, typical nanostar behaviors here and proceed to demonstrate selective tip excitation by independently tuning frequency and polarization. Multiple tips are involved in both plasmon resonance modes for the four-tip nanostar in Fig. 3a, but simulations in Fig. 3b reveal that only one tip hot spot is dominant for each mode. Spectra and polarization plots show two distinct plasmon peaks at nearly orthogonal peak polarization angles, with the photoemission rate at either peak showing minimal contributions from the other. Entirely frequency-controlled tip selectivity can thus be achieved for an isotropic polarization state (circular or unpolarized) or for a linear polarization state oriented between the two resonance modes. The frequency sensitivity depends on the spectral peak widths, relative amplitudes, and separation.
An intermediate laser frequency setting exists between the spectral peaks at which both plasmon modes are equally excited at their respective linear polarization angles. This is demonstrated by the four-lobed polarization dependence measured between the resonances at 835 nm in Fig. 3a and reiterated by the calculations in Fig. 3b for the correlated nanostar geometry. At the intermediate frequency, the linear polarization angle may thus be tuned to select between the two resonance modes and corresponding tip hot spots, demonstrating polarization-controlled selective tip excitation. The relative mode strength (i.e., relative lobe amplitude in the polarization plots) can be continuously adjusted via frequency tuning. Overall, the 2D optical parameter space sampled in Fig. 3 is described by the nPPE photoemission cross-section, $$\sigma_{n} (\omega ,\theta ) = A_{1}^{( n)}(\omega ){\mathrm{cos}}^{2n}(\theta - \theta_{1} ) + A_{2}^{( n )}( \omega ){\mathrm{cos}}^{2n}( \theta - \theta_{2})$$, in which $$A_1^{\left( n \right)}\left( \omega \right)$$ and $$A_2^{\left( n \right)}\left( \omega \right)$$ are the n-photon spectra of the two plasmon resonance modes. In addition to the strong theoretical agreement with the observed spectral and polarization behaviors (Fig. 3b), MPPE rates are calculated to within an order of magnitude of the experimental measurements by integrating the theoretical current density, JMPPE, over the nanostar surface, accounting for both 3PPE and 4PPE contributions. Reserving other aspects of the theory, this level of quantitative agreement corresponds to calculated fields within 50% of the experimental values for a 3PPE process ( E6), which is good considering the sensitivity of the field enhancements to the precise tip radius, the charge distribution during plasmon oscillation (i.e., the overall nanostar geometry), and the surface dielectric environment due to the HEPES surface ligands (see Methods and Supplementary Note 1). Note that while the spectra of the nanostar in Fig. 3 happen to coincide with the 3PPE-to-4PPE cross-over around 830 nm (Fig. 1e), the observed behaviors are a general feature of multi-resonance geometries and are also demonstrated in the next section with a nanostar studied entirely in the 3PPE regime.
### Frequency- and polarization-controlled photoemission
The nanostar in Fig. 4a displays simple multi-resonance behavior, with a higher-energy (blue) dipolar resonance mode aligned with the shorter tip and a lower-energy (red) dipolar resonance mode aligned with the longer tip. These two tips are approximately orthogonal and can be individually addressed, as established in the polarization dependence at different excitation frequencies (Fig. 4b). Instead of maintaining the circular polarization, tip selectivity can be achieved with frequency tuning alone, as discussed in the previous section and further demonstrated here via MPPE simulations (Fig. 4c). Velocity distributions measured (Fig. 4d) and calculated (Fig. 4e) at each frequency are directionally aligned with the corresponding resonant tip axis and photoemission directionality is rotated by a full 90° upon frequency-controlled switching between tips. The average electron kinetic energy decreases with excitation frequency by conservation of energy. When both tip modes are excited at intermediate frequencies (e.g., Fig. 4c, 775 nm), the resulting velocity distribution is simply a linear combination of the individual tip angular distributions. Thus, this linear combination allows for a continuous steering of the average emission angle, although the total angular distribution is broadened by arising from two separate point-like sources.
Polarization-controlled directional emission is presented in Fig. 5 for the same nanostar as in Fig. 4, but exclusively at the intermediate frequency setting at which both tips are equally resonantly enhanced. Simulations demonstrate switching of photoemissive regions between the two tip hot spots as the linear polarization is rotated out of alignment with one mode and into alignment with the other (Fig. 5a). The photoemission directionality is rotated by 90° (Fig. 5b, c) in the same manner observed via frequency control, due to the same underlying process of selective hot-spot excitation. Intermediate polarizations again result in a linear combination of the two tip angular distributions, with the relative weights determined by the polarization dependence (Fig. 5a). Although full polarization contrast is demonstrated by complete alignment along either tip mode, the polarization plot indicates that much less angular tuning is necessary to strongly favor one tip over the other due to the cos6(θ − θtip) polarization sensitivity for each plasmon mode. Strong tip selectivity can be achieved with a 90:10 emission ratio by only ±10° tuning away from the intermediate polarization, at which the emission ratio is 50:50. Therefore, as a benefit of the MPPE nonlinearity, photoemission directionality can be rotated by 90° with only 20° polarization rotation. This fine degree of tip discrimination also indicates possibilities for utilizing higher tip densities for a more continuous control of angular photoemission.
## Discussion
Versatile photoemission switching/steering has been demonstrated by independently tuning one of two optical degrees of freedom (frequency or linear polarization), leaving the other available for modifying the control characteristics. For example, polarization can be utilized for tip selection and corresponding manipulation of angular currents, while frequency can be simultaneously utilized to control the electron kinetic energy distributions and relative tip photoemission enhancement. Such possibilities illustrate how techniques developed for coherent control over nanoplasmonic hot spots using femtosecond optical amplitude, frequency (phase), and polarization shaping18,45 can be directly applied to photocurrent degrees of freedom. The photoemission switching timescale is fundamentally limited by the plasmon dephasing (T2) time and by the optical cycle of the excitation laser field, which defines the fastest timescale on which polarization and frequency can be manipulated. Due to the E2n process nonlinearity, the nPPE current decays 2n-times faster than the plasmonic field, which typically dephases in 10 fs or less33,46. Thus, the 3PPE and 4PPE decay times are comparable to the 1–3 fs optical cycles for visible and near-infrared frequencies. This suggests that spatiotemporal control over hot-spot excitation and directional current generation may be achieved on timescales approaching the attosecond range, even in the weak-field MPPE intensity regime.
Gold nanostars behave as prototypical nanoplasmonic cathodes, with the multi-tip geometries shown to provide direct maps between laser parameters and excitation of different hot spots. Individual tip hot spots have been extensively characterized for sample nanostars via polarization and spectral studies, correlated SEM imaging, and finite element simulations. Angle-resolved photoelectron velocity measurements demonstrate corresponding frequency and polarization control over photoemission current direction, with all experiments corroborated by 3D surface-mediated photoemission calculations that can be carried out for arbitrary nanoplasmonic geometries. Although volume-mediated excitation processes must also be considered in general, the observed directionality and agreement between experiment and theory strongly underscores the dominant role of surface-mediated MPPE at the sharp nanostar tips. The results presented here highlight opportunities for implementing designer plasmonic nanoparticles and nanostructures as all-optical photocurrent control elements in a variety of applications, including femtosecond electron imaging and diffraction, polarization-sensitive photodetection, site-selective photocatalysis, and terahertz nanoelectronics.
## Methods
### Nanostar synthesis and sorting
Nanostars are synthesized via a seedless growth method, in which HEPES buffer functions both as a nucleation and a shape-directing agent. A 1 M stock HEPES solution is made by dissolving the buffer salt in Millipore water (18.2 MΩcm). The pH of the HEPES solution is measured and adjusted to 7.38 using concentrated NaOH solution. Nanostars are synthesized by adding 0.2 mM (final concentration) gold (III) chloride trihydrate (HAuCl4) to 110 mM HEPES buffer and vortexing for 1 min after HAuCl4 addition. The growth solution is left in the dark for 24 h to allow for growth and stabilization. To improve size homogeneity and to achieve the desired resonances within the tuning range of Ti:sapphire laser, density gradient centrifugation is carried out to sort the nanostars. A continuous linear sucrose gradient is created by using a custom mixing program of the gradient maker (BioComp Instruments) to mix 50% and 60% weight-to-volume sucrose solutions at an angle. At this point, 500 μL of 8–10 nM concentrated nanostar solutions are layered on the top of the prepared gradient and centrifuged at 4400 × g for 4 h. The samples are fractionated at intervals of 4 mm from the meniscus (BioComp Instruments) and centrifuged at 15,000 r.p.m. for 25 min to remove excess sucrose. Each fraction is then dialyzed in Thermo Fisher 20K Slide-A-Lyzer Dialysis Cassettes for 24 h to remove remaining sucrose from the solution. TEM images are taken of each fraction to choose the population with desired size and aspect ratio of the branches. Fraction 7 out of 23 is selected for the present studies.
### Sample preparation
Nanostar samples are prepared on glass coverslips sputter coated with a 10 nm ITO film, which provides ohmic conductivity for charge neutralization along with visible transparency for back-side illumination and excitation of the nanostars. A 50 nm alphanumeric gold grid is patterned onto the ITO via negative photomask lithography using an uncoated copper TEM grid (LF-400) as the mask, enabling particle locating in correlated SEM-SPIM (scanning photoelectron imaging microscopy) studies. The ITO substrates are ultraviolet ozone cleaned to remove hydrocarbons and to permit surface wetting, after which a 50 μL aliquot of aqueous nanostar solution is spin coated for 5 min on the substrate at 1500 r.p.m. The dilution is optimized for a final particle coverage of ~0.05 nanostars μm−2, such that ~20 nanostars can be studied in a 20 × 20 μm2 SPIM scan, with negligible probability of two nanostars overlapping within the 500 nm diffraction-limited laser spot.
### Scanning photoelectron imaging microscopy
The SPIM technique combines scanning photoemission microscopy/spectroscopy for single-particle characterization with a VMI electrostatic lens for transverse (vx, vy) photoelectron velocity mapping. Femtosecond pulses are generated via a 75 MHz Ti:sapphire oscillator (KMLabs Swift, 700–1000 nm), with second harmonic generation (350–500 nm) and an optical parametric oscillator (KMLabs, 510–780 nm signal output tuning range) providing broad tunability throughout the visible and near-IR. For optimal frequency tunability without spatial walk-off, no external prism dispersion compensation is utilized for the majority of the present studies. Group velocity dispersion in the system thus results in 100–200 fs pulse durations at the sample across the laser tuning range. As a check, several measurements were also performed with dispersion-compensated (~50 fs) pulses and found to yield indistinguishable photoelectron velocity distributions. A high-vacuum-compatible (<5 × 10−7 Torr) reflective microscope objective (NA = 0.65) focuses the pulsed laser beams to a diffraction-limited spot (~500 nm) on the sample in a normal-incidence, back-illuminated configuration (Fig. 1b).
The ITO-coated sample coverslip is situated on a scanning copper stage, both of which are held at −4500 V and together serve as the repeller electrode in the VMI stack (see Fig. 1b). The VMI lens provides a linear velocity-to-position mapping onto a spatially resolved chevron microchannel plate electron multiplier, in which a single electron is multiplied up to 107 electrons. The amplified electron signal is then accelerated onto a P47 phosphor screen, the fluorescence from which is imaged on a 1.3 megapixel CCD camera running at 20 frames per second. Single event (x, y) coordinates are determined via centroiding and converted to (vx, vy) velocity with a calibration factor of 4150 m s−1 px−1 measured previously27. Three quartered piezoelectric posts with capacitive sensor feedback provide fine scanning over a 30 × 30 μm2 area, with xyz piezo motors providing extended positioning over larger millimeter distances. Single plasmonic nanoparticles can thus be identified and studied for hours or days with minimal drift correction. The nanostars are cleaned prior to all studies via brief (~1 s) exposure to ~1 GW cm−2 of second harmonic light at 400 nm, which removes adlayers (i.e., water) that develop during exposure to ambient air25.
### Transmission and scanning electron microscopy
Nanostars are examined via TEM (FEI Tecnai T12 SpiritBT, 100 kV, LaB6) after drop casting 15 μL of the aqueous nanostar solution onto a carbon-coated TEM grid for 10 min, removing the excess solution, and drying in air. Correlated SEM studies (FEI Nova NanoSEM 630, 10 kV, through lens detector, field immersion mode) are performed on the glass/ITO sample substrates following SPIM studies using the Au grid reference to locate the same nanostars in both systems. The sample is coated with a 3 nm Pt film prior to imaging to enhance conductivity, thereby improving contrast and reducing degradation due to charging and carbon buildup during electron beam exposure. Refer to Supplementary Fig. 1 for TEM and SEM statistical characterization of the nanostar dimensions.
### Surface MPPE modeling
We implement the theory developed by Yalunin et al.40 to model surface-mediated MPPE velocity distributions for coherent excitation at a metal–vacuum interface, modified into a two-step process to account for the high-gradient evanescent surface plasmonic fields. In the first step, free-electron initial states are excited into field-dressed final states (Volkov states) with the ponderomotive quiver energy appearing in the energy conservation equation,
$$\frac{{{{\hbar}^{2}} k^{2}}}{{2m_e}} + n{\hbar} \omega = \frac{{p^2}}{{2m_e}} + U_{\mathrm{p}} + E_{\mathrm{F}} + \phi ,$$
(3)
in which $$\hbar k$$ is the initial state momentum, p is the final state drift momentum, Up is the ponderomotive energy, EF is the Fermi energy, and ϕ is the work function. In the second step, as the electron leaves the evanescent surface plasmon field, the ponderomotive energy is fully converted into kinetic energy corresponding to the surface-normal momentum. The differential multiphoton photocurrent is given by
$${\frac{{{\mathrm{d}}J_{{\mathrm{MPPE}}}}}{{{\mathrm{d}}^{3}k}} = \frac{{2{\hbar}}}{{\left( {2{\uppi}} \right)^{3} m_{e}}}\mathop {\sum }\limits_{n {\ge} n_{{\mathrm{min}}}} \frac{{k_z}}{{e^{\left( {{{\hbar}^{2}} k^{2}/2m_e - E_{\mathrm{F}}} \right)/kT} \, + \, 1}}P_n\left( {k_z} \right)} ,$$
(4)
which is the collision rate of Fermi sea electrons on the surface potential barrier, scaled by the dimensionless excitation and emission probability, Pn(kz), and summed over all allowed multiphoton process orders as determined by energy conservation.
The photoemission rate can be written in terms of external momentum via coordinate transformation
$$\begin{array}{*{20}{c}} {\hbar k_x = p_x} \end{array},$$
(5a)
$$\begin{array}{*{20}{c}} {\hbar k_y = p_y} \end{array},$$
(5b)
$$\begin{array}{*{20}{c}} {\frac{{{\hbar}^{2} k_z^2}}{{2m_e}} + n{\hbar} \omega = \frac{{p_z^2}}{{2m_e}} + U_{\mathrm{p}} + E_{\mathrm{F}} + \phi .} \end{array}$$
(5c)
The Jacobian determinant is $$\left| {{\mathrm{d}}^3k/{\mathrm{d}}^3p} \right| = p_z/\left( {{\hbar}^{4} k_z} \right)$$ and the photoemission probability is thus given in external momentum coordinates by
$${\frac{{{\mathrm{d}}J_{{\mathrm{MPPE}}}}}{{{\mathrm{d}}^3p}} = \frac{2}{{h^3m_e}}\mathop {\sum }\limits_n \frac{{p_z}}{{e^{\left( {p^2/2m_e + U_{\mathrm{p}} + \phi - n\hbar {\upomega}} \right)/kT} \, + \, 1}}P_n\left( {k_z\left( {p_z} \right)} \right)} ,$$
(6)
in which all process orders may contribute to a final momentum state at finite temperature due to the exponential Fermi–Dirac tail, although most processes are negligible at room temperature except the dominant multiphoton order dictated by energy conservation. The photoemission probability derived by Yalunin et al.40 via Green’s function solution to the time-dependent Schrödinger equation is given here in terms of external coordinates as
$${P_n\left( {k_z\left({p_z} \right)} \right) = \frac{{{\hbar}^{2}}}{{\sqrt{2m_e}(E_{\mathrm{F}} + \phi) }}\frac{{\sqrt {\frac{{p_z^2}}{{2m_e}} \, + \, U_{\mathrm{p}} \, + \, \phi \, - \, n\hbar \omega } }}{{p_z}}\left| {I_n\left( {p_z} \right) + R_n^{\left( 1 \right)}R_n^{\left( 2 \right)}I_n\left( { - p_z} \right)} \right|^2} ,$$
(7a)
$${I_n\left( {p_z} \right) = \frac{{\sqrt {2m_e} }}{h}\int_{0}^{2\pi } \left( {i\sqrt {n\hbar \omega - U_{\mathrm{p}} - \frac{{p_z^2}}{{2m_e}}} + \frac{{p_z}}{{\sqrt {2m_e} }} - \sqrt {2U_{\mathrm{p}}} } \right)e^{iS\left( q \right)}{\mathrm{d}}q},$$
(7b)
$${S\left( q \right) = nq + 2\frac{{p_z}}{{\hbar \omega }}\sqrt {\frac{{U_{\mathrm{p}}}}{{m_e}}} {\mathrm{cos}}\left( q \right) - \frac{{U_{\mathrm{p}}}}{{2\hbar \omega }}{\mathrm{sin}}\left( {2q} \right)} ,$$
(7c)
$${R_n^{\left( 1 \right)} = - \frac{{\sqrt {\frac{{p_z^2}}{{2m_e}} + U_{\mathrm{p}} + E_{\mathrm{F}} + \phi } - \frac{{p_z}}{{\sqrt {2m_e} }}}}{{\sqrt {\frac{{p_z^2}}{{2m_e}} + U_{\mathrm{p}} + E_{\mathrm{F}} + \phi } + \frac{p_z}{\sqrt {2m_e} }}}} ,$$
(7d)
$${R_n^{\left( 2 \right)} = {\mathrm{J}}_0\left( { - 4\frac{{p_z}}{{\hbar \omega }}\sqrt {\frac{{U_{\mathrm{p}}}}{{m_e}}} } \right)} .$$
(7e)
The integral terms $$I_n\left(\! { \pm p_z} \right)$$ represent the outward- and backward-moving excited waves outside the medium. The reflection coefficients on the backward-moving wave account for rescattering on the surface potential barrier, where $$R_n^{(1)}$$ is the reflection coefficient for a step-down potential (height $$E_{\mathrm{F}} + \phi$$) and $$R_n^{(2)}$$ accounts for the effect of the oscillating triangular barrier in the applied optical field. Interference between the direct and rescattered waves can have significant effects on the final emission amplitude, as discussed in detail by Yalunin et al.40. Finally, we include the ponderomotive energy transfer in the present case of an evanescent plasmonic field via the coordinate transformation $$p_z^2/2m_e \to p_z^2/2m_e - U_{\mathrm{p}}$$.
To calculate the full 3D photoelectron velocity distribution for a given nanostar, the photoemission contributions from each surface area element are calculated using the surface field enhancements determined via finite element simulation. The photoemission distribution is calculated with respect to each surface normal ($$\hat p_{z,{\mathrm{rel}}}$$) and rotated into the global frame ($$\hat p_z$$) via Cartesian rotation matrices. Only a single rotation about axis $$\hat p_z \times \hat p_{z,{\mathrm{rel}}}$$ is required for azimuthally-isotropic distributions. A sample uniform surface mesh and nonlinear surface-normal field enhancement distribution is shown in Fig. 6, with additional details of the finite element domain presented in Supplementary Fig. 2. Empirical values are utilized for all model inputs, with the exception of the Fermi–Dirac electron temperature, which is calculated at 2500 K via a two-temperature model (Supplementary Fig. 6 and Supplementary Note 4). It should finally be noted that geometries with concave surface regions (including nanostars) may allow for the intersection of emitted electrons with other surfaces of the emitter geometry. These effects are not presently accounted for, although they should be negligible for the tip-like emission described in the present work.
## Data availability
All data required to reproduce the findings in this study are available from the corresponding authors upon reasonable request.
## Code availability
The MATLAB script for surface-mediated multiphoton photoemission calculations is available from the corresponding authors upon reasonable request.
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## Acknowledgements
We would like to acknowledge Dr. K. Culver’s contributions to the nanostar synthesis process along with her role in stimulating these investigations. J.P., F.M., and D.J.N would also like to thank Prof. M. Müller and Prof. M. Raschke for illuminating conversations. Photoemission studies (D.J.N.) were supported by the Air Force Office of Scientific Research (FA9550-15-1-0090) with additional funds for laser development and apparatus construction provided by the National Science Foundation (CHE 1665271, PHY 1734006). Nanostar synthesis (T.W.O.) was supported by the National Science Foundation (CHE 1808502).
## Author information
Authors
### Contributions
J.P., F.M., and D.J.N designed the experiments. P.C. and T.W.O. developed the synthetic methods, synthesized the gold nanostars, and characterized the samples. J.P. and F.M. performed the experiments, analysis, and calculations. All authors contributed to the writing of the manuscript and provided approval of the final version.
### Corresponding authors
Correspondence to Teri W. Odom or David J. Nesbitt.
## Ethics declarations
### Competing interests
The authors declare no competing interests.
Peer review information Nature Communications thanks Christoph Lienau, Gerd Schönhense and the other, anonymous, reviewer(s) for their contribution to the peer review of this work. Peer reviewer reports are available.
Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
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Pettine, J., Choo, P., Medeghini, F. et al. Plasmonic nanostar photocathodes for optically-controlled directional currents. Nat Commun 11, 1367 (2020). https://doi.org/10.1038/s41467-020-15115-0
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