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proofpile-shard-0030-0 | {
"provenance": "003.jsonl.gz:1"
} | Question
# Find the value of $m$ and $n$ using cross multiplication method:$3m+n=15$ and $m+2n=10$.A.$(4,3)$B.$(-4,3)$C.$(-4,-3)$D.$(4,-3)$
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Hint: We have been given two equations. Write the equations in the form of $ax+by+c=0$. Then use $\dfrac{m}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{n}{{{c}_{1}}{{a}_{2}}-{{a}_{1}}{{c}_{2}}}=\dfrac{-1}{{{a}_{2}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}$. You will get the answer.
The multiplication of integers (including negative numbers), rational numbers (fractions) and real numbers is defined by a systematic generalization of this basic definition.
Multiplication can also be visualized as counting objects arranged in a rectangle (for whole numbers) or as finding the area of a rectangle whose sides have given lengths. The area of a rectangle does not depend on which side is measured first, which illustrates the commutative property. The product of two measurements is a new type of measurement, for instance multiplying the lengths of the two sides of a rectangle gives its area, this is the subject of dimensional analysis.
Solving proportions is simply a matter of stating the ratios as fractions, setting the two fractions equal to each other, cross-multiplying, and solving the resulting equation. The exercise set will probably start out by asking for the solutions to straight forward simple proportions, but they might use the "odds" notation.
Specifically in elementary arithmetic and elementary algebra, given an equation between two fractions or rational expressions, one can cross-multiply to simplify the equation or determine the value of a variable. Each step in these procedures is based on a single, fundamental property of equations. Cross-multiplication is a shortcut, an easily understandable procedure that can be taught to students.
The method is also occasionally known as the "cross your heart" method because a heart can be drawn to remember which things to multiply together and the lines resemble a heart outline.
In practice, though, it is easier to skip the steps and go straight to the "cross-multiplied" form.
So we have been given two equations.
$3m+n-15={{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$………… (1)
$m+2n-10={{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$……….. (2)
Now using the formula for cross multiplication,
$\dfrac{m}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{n}{{{c}_{1}}{{a}_{2}}-{{a}_{1}}{{c}_{2}}}=\dfrac{-1}{{{a}_{2}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}$
So from equation (1) and (2),
$\dfrac{m}{1\times (-10)-2\times (-15)}=\dfrac{n}{(-15)\times 1-3\times (-10)}=\dfrac{1}{3\times 2-1\times 1}$
$\dfrac{m}{-10+30}=\dfrac{n}{-15+30}=\dfrac{1}{6-1}$
Simplifying we get,
$\dfrac{m}{20}=\dfrac{n}{15}=\dfrac{1}{5}$
Now equating we get,
$\dfrac{m}{20}=\dfrac{1}{5}$
So simplifying we get,
$m=4$
Also, $\dfrac{n}{15}=\dfrac{1}{5}$
$n=3$
Here we get, $(m,n)=(4,3)$.
So the solution is $(4,3)$.
So the correct answer is option(A).
Note: Carefully read the question. Don’t be confused about the cross multiplication method. While simplifying, do not make mistakes. Don’t miss any term while solving. Take care that no terms are missing. |
proofpile-shard-0030-1 | {
"provenance": "003.jsonl.gz:2"
} | # What formula or rule has been used here?
I was in between proving a trigonometric identity but couldn't succeed. I went through the solution and saw this in between
\begin{align}\frac{\cos A \cos B}{\sin A \sin B}&= \frac{3}{1}\\\\ \frac{\cos A \cos B +\sin A \sin B}{\cos A \cos B - \sin A \sin B}&= \frac{3+1}{3-1}\end{align}
What happened there in the second step?
• Jul 17 '17 at 10:28
If we have $$\frac xy = \frac31$$ then this means, by definition of fractions, that $x = 3y$. This yields $$\frac{x+y}{x-y} =\frac{3y+y}{3y-y} = \frac{3+1}{3-1}$$ In your case, $x = \cos A\cos B$ and $y = \sin A\sin B$. |
proofpile-shard-0030-2 | {
"provenance": "003.jsonl.gz:3"
} | # Three.js: Lighting not calculating correctly on THREE.Geometry objects
I have a three.js (REVISION: '68') issue with the lighting of THREE.Geometry objects:
I'm using the THREE.Geometry class to build up objects using vertices and faces, then I computeFaceNormals() and computeVertexNormals() before adding it to the scene.
The lighting is obviously not calculating correctly, and it looks like the light calculates before the object is moved into position (or some other issue causing all objects to have identical lighting regardless of position).
My light code is:
hemiLight = new THREE.HemisphereLight( 0xffffff, 0xffffff, 0.6 );
hemiLight.color.setHSL( 0.6, 1, 0.6 );
hemiLight.groundColor.setHSL( 0.095, 1, 0.75 );
hemiLight.position.set( 0, 50, 0 );
dirLight = new THREE.DirectionalLight( 0xffffff, 1 );
dirLight.color.setHSL( 0.1, 1, 0.95 );
dirLight.position.set( -1, 1.75, 1 );
dirLight.position.multiplyScalar( 50 );
Note I also tried the documentation's sample-code for the Spotlight the issue persisted.
I am using a comparison of THREE.Geometry objects: the first with a single material for the at mesh creation, and the second with faces individually assigned a material, and passed to the mesh with THREE.MeshFaceMaterial(..) Note that I tried both single material Geometries and multi-material Geometries side-by-side and there is no difference.
Note I changed the color of one of my test cubes from blue to green compared to the header image.
The issue does not appear to be related to the multi-material code.
I understand that the Three.Geometry class behaves differently compared to BoxGeometry, etc. (For example, computeFaceNormals() and computeVertexNormals() needs to be explicitly called for Three.Geometry, but not for BoxGeometry). I think I may be missing some other difference around flagging the material/lighting/geometry for update.
My code to create my two plain test cubes is:
var testGeo = new THREE.Geometry();
testGeo.vertices.push(new THREE.Vector3(0,0,0));
testGeo.vertices.push(new THREE.Vector3(0,20,0));
testGeo.vertices.push(new THREE.Vector3(20,20,0));
testGeo.vertices.push(new THREE.Vector3(20,0,0));
testGeo.vertices.push(new THREE.Vector3(0,0,20));
testGeo.vertices.push(new THREE.Vector3(0,20,20));
testGeo.vertices.push(new THREE.Vector3(20,20,20));
testGeo.vertices.push(new THREE.Vector3(20,0,20));
testGeo.faces.push(new THREE.Face3(0,1,2));
testGeo.faces.push(new THREE.Face3(2,3,0));
testGeo.faces.push(new THREE.Face3(2,3,7));
testGeo.faces.push(new THREE.Face3(7,6,2));
testGeo.faces.push(new THREE.Face3(0,1,5));
testGeo.faces.push(new THREE.Face3(5,4,0));
testGeo.faces.push(new THREE.Face3(0,3,4));
testGeo.faces.push(new THREE.Face3(4,7,3));
testGeo.faces.push(new THREE.Face3(1,2,6));
testGeo.faces.push(new THREE.Face3(6,5,1));
testGeo.faces.push(new THREE.Face3(4,5,6));
testGeo.faces.push(new THREE.Face3(6,7,4));
testGeo.computeFaceNormals();
testGeo.computeVertexNormals();
var solidMatA = new THREE.MeshLambertMaterial({
color: 'blue'
})
solidMatA.side = THREE.DoubleSide;
var cubeA = new THREE.Mesh( testGeo, solidMatA );
cubeA.position.x = -40;
cubeA.position.y = -30;
cubeA.position.z = -30;
var testMaterialsListB = [];
var testGeo2 = new THREE.Geometry();
testGeo2.vertices.push(new THREE.Vector3(0,0,0));
testGeo2.vertices.push(new THREE.Vector3(0,20,0));
testGeo2.vertices.push(new THREE.Vector3(20,20,0));
testGeo2.vertices.push(new THREE.Vector3(20,0,0));
testGeo2.vertices.push(new THREE.Vector3(0,0,20));
testGeo2.vertices.push(new THREE.Vector3(0,20,20));
testGeo2.vertices.push(new THREE.Vector3(20,20,20));
testGeo2.vertices.push(new THREE.Vector3(20,0,20));
testGeo2.faces.push(new THREE.Face3(0,1,2));
testGeo2.faces.push(new THREE.Face3(2,3,0));
testGeo2.faces.push(new THREE.Face3(2,3,7));
testGeo2.faces.push(new THREE.Face3(7,6,2));
testGeo2.faces.push(new THREE.Face3(0,1,5));
testGeo2.faces.push(new THREE.Face3(5,4,0));
testGeo2.faces.push(new THREE.Face3(0,3,4));
testGeo2.faces.push(new THREE.Face3(4,7,3));
testGeo2.faces.push(new THREE.Face3(1,2,6));
testGeo2.faces.push(new THREE.Face3(6,5,1));
testGeo2.faces.push(new THREE.Face3(4,5,6));
testGeo2.faces.push(new THREE.Face3(6,7,4));
for (var i = 0; i < testGeo2.faces.length; i++)
{
var matB = new THREE.MeshLambertMaterial( {color: 'green'} );
matB.side = THREE.DoubleSide;
testMaterialsListB.push(matB);
}
testGeo2.computeFaceNormals();
testGeo2.computeVertexNormals();
var cubeB = new THREE.Mesh( testGeo2, new THREE.MeshFaceMaterial( testMaterialsListB) );
cubeB.position.x = -60;
cubeB.position.y = -30;
cubeB.position.z = -30;
Thanks!
• Since you are using three.js, you might consider making your code above a Stack Snippet to allow folks to reproduce the issue easily. Jan 28, 2015 at 17:48
• I reposted the question with JSFiddle to view the issue in live code: stackoverflow.com/questions/28215201/… Jan 29, 2015 at 13:09
Your triangles might be specified in a clockwise order instead of WebGL's preferred counter-clockwise order. To verify, switch your material side to THREE.FrontSide and run it again with THREE.BackSide and see if one gives you the correct results. If the THREE.BackSide works, then you have to go back and flip your ordering in all the Face3 creations.
Regarding shading looking similar with stacked objects: three.js does not take other objects into account when lighting each object in the render stage. Let's say you wanted to make a Minecraft clone with a bunch of boxes all stacked on top of one another. Well, three.js will calculate each individual box's lighting as if it were the only one in the world. Even if you stack them densely on top of each other, they will all look the same - light on top, dark beneath.
The lighting effect you are probably wanting is achieved by ShadowMaps in three.js. Shadowing portions of an object that are obscured by other objects is a complex and expensive task which is still being hashed out by the creators of three.js.
The normal ShadowMap mode in three.js looks OK for now however and it works in most cases.
Just to follow up in case others stumble across this:
While I did re-arrange the vertex order to ensure 90 degree perpendicular normals, I found that didn't make a material difference to adjacent object lighting.
The answer is that three.js doesn't intrinsically calculate lighting across adjacent objects in the scene graph, and while the shadowmap feature attempts to solve this, my investigations showed that a shadowmap solution didn't really work for my use case.
Instead, I applied only 1 of the two geometry calculations to get a simulated light effect
geo.computeFaceNormals();
geo.computeVertexNormals();
What I ended up using:
geo.computeFaceNormals();
//geo.computeVertexNormals();
With computeVertexNormals()...
Without:
While not a perfect light effect (very flat), it should get me by. Thanks for the help posters. |
proofpile-shard-0030-3 | {
"provenance": "003.jsonl.gz:4"
} | # Saturation vapor pressure calculation - HMT130
## HMT130 User Guide
Document code
M211280EN
Revision
D
Language
English (United States)
Product
HMT130
Document type
User guide
Saturation vapor pressure (Pws) is the equilibrium water vapor pressure in a closed chamber containing liquid water. It is a function only of temperature, and it indicates the maximum amount of water that can exist in the vapor state.
Water vapor saturation pressure (Pws) is calculated with the following 2 formulas:
where
T
Temperature in K
Ci
Coefficients
C0
0.49313580
C1
-0.46094296 * 10-2
C2
0.13746454 * 10-4
C3
-0.12743214 * 10-7
where
bi
Coefficients
b-1
-0.58002206 * 104
b0
0.13914993 * 101
b1
-0.48640239 * 10-1
b2
0.41764768 * 10-4
b3
-0.14452093 * 10-7
b4
6.5459673 |
proofpile-shard-0030-4 | {
"provenance": "003.jsonl.gz:5"
} | Receiver and transmitter with the same antenna [duplicate]
I want to build a simple transceiver (half duplex) for 433.92 MHz ASK modulation. I have found a transmitter chip and a receiver chip that I want to use. I want to use the same antenna for RX and TX but I am unsure of how to approach this (The receiver can handle the full output power of the transmitter). I do not want to buy a ready-made circuit for this, I'm doing this mostly to learn.
The antenna is 50 ohms but the receiver and transmitter both have their own impedances and need some matching network. If I used separate antennas this would be simple (say a PI attenuator).
But I’m unsure of how this would work with having both RX and TX on the same antenna. My very naïve approach to this would be to use a splitter/combiner after the matching networks and accept the -3 dB loss from this. This way every side sees 50 ohms.
But this feels very wrong and any input here would be much appreciated!
• If you can make it work satisfactorily, it is not wrong. But if you want to go there, an rf switch is another solution. Jan 25 '18 at 18:44
• Also, if you start over, you could use a transceiver chip. I've used the like of this chip in the past and they work very well. Jan 25 '18 at 18:52
• You can use a circulator. Jan 25 '18 at 19:49
A typical 434 MHz low cost transmission system will have a free-space path loss of: -
Loss (dB) = 32.45 + 20$log_{10}$(f) + 20$log_{10}$(d) (Friis equation in dB form)
Where f is in MHz and d is in kilometres.
Transmission distance will be about 0.1 km and the pathloss works out at: -
32.5 dB + 52.8 dB - 20 dB = 65 dB
But antennas will provide some gain (about 2 dB each end) so the free space figure is more like 61 dB. However, most RF engineers will add another 30 dB for fade margin and this means the overall path loss is about 90 dB.
Using a splitter at both ends will degrade the power transmitted by 3 dB and degrade the power received by 3 dB; a total of 6 dB.
You then have to ask yourself if that is acceptable or not. For most cases, simplicity overrides performance and it isn't a big deal. However, 6 dB is equivalent to halving the range from 100 m to 50 m. I can't tell you if this is good or bad. |
proofpile-shard-0030-5 | {
"provenance": "003.jsonl.gz:6"
} | # hoop: Object-Oriented Programming in Haskell
[ language, library, mit ] [ Propose Tags ]
Library for object-oriented programming in Haskell.
Versions [faq] 0.3.0.0 base (>=4.7 && <5.0), containers, haskell-src-exts (>=1.16), haskell-src-meta (>=0.6), lens (>=4.10), mtl (>=2.1), parsec (>=3.1.9), pretty, template-haskell (>=2.14), text [details] MIT Copyright (c) Michael B. Gale Michael B. Gale [email protected] Language https://github.com/mbg/hoop#readme https://github.com/mbg/hoop/issues head: git clone https://github.com/mbg/hoop by mbg at 2020-07-04T19:10:34Z 34 total (16 in the last 30 days) (no votes yet) [estimated by Bayesian average] λ λ λ Docs not available Last success reported on 2020-07-04
## Modules
• Language
• Language.MSH
• Language.MSH.BuiltIn
• Language.MSH.CodeGen
• Language.MSH.CodeGen.Class
• Language.MSH.CodeGen.Constructors
• Language.MSH.CodeGen.Data
• Language.MSH.CodeGen.Decls
• Language.MSH.CodeGen.Inheritance
• Language.MSH.CodeGen.Instances
• Language.MSH.CodeGen.Interop
• Language.MSH.CodeGen.Invoke
• Language.MSH.CodeGen.Methods
• Language.MSH.CodeGen.MiscInstances
• Language.MSH.CodeGen.New
• Language.MSH.CodeGen.NewInstance
• Language.MSH.CodeGen.Object
• Language.MSH.CodeGen.ObjectInstance
• Language.MSH.CodeGen.PrimaryInstance
• Language.MSH.CodeGen.Shared
• Language.MSH.CodeGen.SharedInstance
• Language.MSH.Constructor
• Language.MSH.MethodTable
• Language.MSH.NewExpr
• Language.MSH.Parsers
• Language.MSH.Pretty
• Language.MSH.QuasiQuoters
• Language.MSH.RuntimeError
• Language.MSH.Selectors
• Language.MSH.StateDecl
• Language.MSH.StateEnv
#### Maintainer's Corner
For package maintainers and hackage trustees
[back to package description]
# hoop
A Haskell library for object-oriented programming which allows programmers to use objects in ordinary Haskell programs. In particular, the library achieves the following design objectives (to avoid ambiguity with Haskell's type classes, we refer to classes in the object-oriented sense as object classes):
• No extensions to the Haskell language are required beyond what is already implemented in GHC. Object classes are generated from Template Haskell quasi quotations using an OO-like syntax where the methods are defined as ordinary Haskell expressions.
• Object classes can be instantiated from ordinary Haskell code (with an overloaded function named new). The resulting objects are ordinary Haskell values and can be used as such.
• Calling methods on objects can be done from within ordinary Haskell code.
• The objects do not rely on IO. Instantiating objects and calling methods on the resulting objects is pure.
• Object classes can inherit from other object classes, which also established subtyping relations between them. There is no limit to how deep these inheritance trees may grow.
• Class hierarchies are open for extension. I.e. the library does not need to know about all subclasses of a given class in order to generate the code for that class, allowing modular compilation.
• Casting from subtype objects to their supertypes is supported and the types are correctly reflected in Haskell's type system (e.g. assuming that we have Duck <: Bird and that obj :: Duck then upcast obj :: Bird) and pure.
• Type annotations are generally not required except where something would logically be ambiguous otherwise (e.g. instantiating an object with the new function).
## Examples
The test folder contains a number of examples of the library in action, illustrating the various features.
As a quick tutorial, a simple expression language can be implemented using the library as shown below. Note that the bodies of the two implementations of the eval method are ordinary Haskell expressions. The .! operator is an ordinary Haskell operator used to call methods on objects and this is just an ordinary Haskell definition, too.
[state|
abstract state Expr where
eval :: Int
state Val : Expr where
data val = 0 :: Int
eval = do
r <- this.!val
return r
data left :: Expr
data right :: Expr
eval = do
x <- this.!left.!eval
y <- this.!right.!eval
return (x+y)
|]
someExpr = new @Add (upcast $new @Val 4, upcast$ new @Val 7)
someExprResult :: Int
someExprResult = result (someExpr.!eval)
If we evaluate someExprResult, the result is 11 as expected. We can note some points of interest here that differ from popular object-oriented programming languages:
• The type annotations on someExpr and someExprResult are optional and just provided for clarity. The type applications for the calls to new are required (alternatively, type annotations on the sub-expression would work, too).
• Casts must be explicit: in the example, the objects of type Val must be explicitly cast to Expr values to instantiate the Add object.
• Since everything is pure, calling a method on an object produces two results: the result of the method call and a (potentially) updated object. The result function returns the result of calling eval on the someExpr object, discarding the resulting object.
• It does not matter what type of object we call eval on, as long as it is of type Expr or is a sub-type of Expr.
Indeed, we can cast the Add object to an Expr object, call eval on it, and still get the correct result:
> let e = upcast someExpr in result (e.!eval)
11
• Casting from supertype objects to a subtype is possible, but may fail (returning Nothing). E.g. assuming Duck <: Bird and that obj :: Bird then downcast obj :: Maybe Duck.
## Overview of the process
• QuasiQuoters.hs contains the entry point
• First, the state declarations are parsed (Parsers.hs) via parseStateDecl
• The parsed declarations are then passed to genStateDecls (Language.MSH.CodeGen.Decls / Decls.hs)
• This turns the declarations into a dependency graph (via buildStateGraph in Language.MSH.StateEnv / StateEnv.hs)
• If successful, the graph is written to graph.log
• The genStateDecl function is then applied to every state declaration in dependency order (i.e. starting from no dependencies) |
proofpile-shard-0030-6 | {
"provenance": "003.jsonl.gz:7"
} | # Webcut volume with non planar surface
Dear everyone,
For my previous problem Webcut failed for volume with nonplanar sheet, I finally fixed it by changing the spacing between each point when I create the surface. Then, webcut sheet extended command is able to cut the volume.
Since I need larger geometry, I recreate the geometry and use the same spacing size to create the surface. But again it is failed to cut the volume. I have tried to use webcut volume 1 with sheet body 2 and webcut volume 1 with sheet extended from surface 2. But both do not work. I have tried also to smooth the surface data. But it still does not work.
The error says :
WARNING: Cutting Sheet does not intersect the original volume.
The original volume is restored.
No volumes were webcut.
or
ERROR: /scratch/akuncoro/2021/mesh/sumatra_full/geometry_full.jou (155). Error in webcutting volume with sheet.
ERROR: /scratch/akuncoro/2021/mesh/sumatra_full/geometry_full.jou (155). ACIS API error number 21033
ACIS API message = inconsistent face-body relationships
No volumes were webcut.
I attach my .jou and .sat file here
alvina_problem_may.jou (1.3 KB)
surf_topo_3km.sat (1.9 MB)
surf_topo_full.sat (1.6 MB)
Is there any suggestion to fix this problem?
Thank you in advance.
-Alvina
Hi @alvinakk,
The primary issue appears to be related to the scale of the geometry. We recommend users to scale your geometry so that the smallest features you care about in the model are \approx \mathcal{O}(1). If I scale your geometry by 0.0001 then I’m able to apply the webcut. We then have a function that will scale your mesh on export to recover the actual dimension of your model.
For example:
reset
## Create a "small" geometry
bri x 0.000123 # Miles
## Scale geometry from Miles to Feet,
## which gives us an "easy" conversion to remember
## *and* makes our smallest "important" edges a size ~ 1
volume 1 scale 5280
## Mesh the volume
mesh volume 1
block 1 vol 1
## Setup option to scale mesh on export
transform mesh output scale {1/5280} # Uses APREPRO syntax to evaluate 1/5280
## Export the mesh, which will be scaled
export mesh "./transformed_mesh.e" overwrite
Note that transform mesh is multiplicative / additive depending on whether you’re doing a scale or translation. So doing transform mesh output scale 10 twice will scale by 100, not 10. Make sure to transform mesh output reset to reset.
Anyways, so if I scale your geometry even by just a factor of 0.0001 I am able to successfully cut your geometry:
# ----------------------------------------------------------------------
# Set units to SI.
# ----------------------------------------------------------------------
${Units('si')} # # ---------------------------------------------------------------------- # Reset geometry. # ---------------------------------------------------------------------- reset import Acis "surf_topo_full.sat" #import Acis "surf_topo_3km.sat"${idSurf=Id("surface")}
surface {idSurf} name "s_topo"
${idVol=Id("volume")} volume {idVol} name "v_topo"${idBody=Id("body")}
body {idBody} name "b_topo"
# ----------------------------------------------------------------------
# Create block for domain.
# ----------------------------------------------------------------------
# Block is 500 km x 500 km x 300 km
${blockLength=1700.0*km}${blockWidth=1000.0*km}
${blockHeight=300.0*km} brick x {blockLength} y {blockWidth} z {blockHeight}${idVol=Id("volume")}
volume {idVol} name "v_domain"
move volume v_domain location 700000 9800000 -100000 include_merged
volume all scale 0.0001
transform mesh output scale 10000
webcut volume v_domain with sheet body b_topo
delete volume {idBody}
# ----------------------------------------------------------------------
# Imprint all volumes, then merge.
# ----------------------------------------------------------------------
imprint all with volume all
merge all
# End of file
And here’s a picture of the meshed geometry
And of the bottom volume to show the cut surface
Dear @gvernon,
Thank you for your reply.
I can finally cut the volume now.
But because I still have another surface to cut the volume, so I try to rescale the volume after I cut the volume using this command and before I mesh the volume:
volume all scale 0.0001
webcut volume v_domain with sheet body b_topo
volume all scale 10000
Is it safe to do that? Because I see this report:
Cubit>volume all scale 10000
WARNING: Model may be corrupted from the scaling operation.
Consider healing it.
WARNING: Model may be corrupted from the scaling operation.
Consider healing it.
Finished Command: volume all scale 10000
Thank you in advance.
Alvina KK |
proofpile-shard-0030-7 | {
"provenance": "003.jsonl.gz:8"
} | ### jhasuraj01's blog
By jhasuraj01, history, 7 days ago,
Bitwise AND, OR and XOR are bitwise operators in C++ and Python that perform operations on the binary representation of numbers.
### Bitwise AND (&)
It compares each bit of the first operand to the corresponding bit of the second operand. If both bits are 1, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0.
Example: In C++:
int x = 12; // binary: 1100
int y = 15; // binary: 1111
int z = x & y; // binary: 1100, decimal: 12
In Python:
x = 12
y = 15
z = x & y
print(z) # Output: 12
### Bitwise OR (|)
It compares each bit of the first operand to the corresponding bit of the second operand. If at least one of the bits is 1, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0.
Example: In C++:
int x = 12; // binary: 1100
int y = 15; // binary: 1111
int z = x | y; // binary: 1111, decimal: 15
In Python:
x = 12
y = 15
z = x | y
print(z) # Output: 15
### Bitwise XOR (^)
It compares each bit of the first operand to the corresponding bit of the second operand. If the bits are different, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0.
Example:
In C++:
int x = 12; // binary: 1100
int y = 15; // binary: 1111
int z = x ^ y; // binary: 0011, decimal: 3
In Python:
x = 12
y = 15
z = x ^ y
print(z) # Output: 3
It's worth noting that these operations are done on the binary level and are generally faster than doing the operations on the decimal level.
• +4 |
proofpile-shard-0030-8 | {
"provenance": "003.jsonl.gz:9"
} | # Arbelos
In geometry, an arbelos is a plane region bounded by three semicircles with three apexes such that each corner of each semicircle is shared with one of the others (connected), all on the same side of a straight line (the baseline) that contains their diameters.[1]
An arbelos (grey region)
Arbelos sculpture in Kaatsheuvel, Netherlands
The earliest known reference to this figure is in Archimedes's Book of Lemmas, where some of its mathematical properties are stated as Propositions 4 through 8.[2] The word arbelos is Greek for 'shoemaker's knife'. The figure is closely related to the Pappus Chain.
## Properties
Two of the semicircles are necessarily concave, with arbitrary diameters a and b; the third semicircle is convex, with diameter a+b.[1]
Some special points on the arbelos.
### Area
The area of the arbelos is equal to the area of a circle with diameter ${\displaystyle HA}$ .
Proof: For the proof, reflect the arbelos over the line through the points ${\displaystyle B}$ and ${\displaystyle C}$ , and observe that twice the area of the arbelos is what remains when the areas of the two smaller circles (with diameters ${\displaystyle BA}$ ${\displaystyle AC}$ ) are subtracted from the area of the large circle (with diameter ${\displaystyle BC}$ ). Since the area of a circle is proportional to the square of the diameter (Euclid's Elements, Book XII, Proposition 2; we do not need to know that the constant of proportionality is ${\displaystyle {\frac {\pi }{4}}}$ ), the problem reduces to showing that ${\displaystyle 2(AH)^{2}=(BC)^{2}-(AC)^{2}-(BA)^{2}}$ . The length ${\displaystyle (BC)}$ equals the sum of the lengths ${\displaystyle (BA)}$ and ${\displaystyle (AC)}$ , so this equation simplifies algebraically to the statement that ${\displaystyle (AH)^{2}=(BA)(AC)}$ . Thus the claim is that the length of the segment ${\displaystyle AH}$ is the geometric mean of the lengths of the segments ${\displaystyle BA}$ and ${\displaystyle AC}$ . Now (see Figure) the triangle ${\displaystyle BHC}$ , being inscribed in the semicircle, has a right angle at the point ${\displaystyle H}$ (Euclid, Book III, Proposition 31), and consequently ${\displaystyle (HA)}$ is indeed a "mean proportional" between ${\displaystyle (BA)}$ and ${\displaystyle (AC)}$ (Euclid, Book VI, Proposition 8, Porism). This proof approximates the ancient Greek argument; Harold P. Boas cites a paper of Roger B. Nelsen[3] who implemented the idea as the following proof without words.[4]
### Rectangle
Let ${\displaystyle D}$ and ${\displaystyle E}$ be the points where the segments ${\displaystyle BH}$ and ${\displaystyle CH}$ intersect the semicircles ${\displaystyle AB}$ and ${\displaystyle AC}$ , respectively. The quadrilateral ${\displaystyle ADHE}$ is actually a rectangle.
Proof: The angles ${\displaystyle BDA}$ , ${\displaystyle BHC}$ , and ${\displaystyle AEC}$ are right angles because they are inscribed in semicircles (by Thales' theorem). The quadrilateral ${\displaystyle ADHE}$ therefore has three right angles, so it is a rectangle. Q.E.D.
### Tangents
The line ${\displaystyle DE}$ is tangent to semicircle ${\displaystyle BA}$ at ${\displaystyle D}$ and semicircle ${\displaystyle AC}$ at ${\displaystyle E}$ .
Proof: Since angle BDA is a right angle, angle DBA equals π/2 minus angle DAB. However, angle DAH also equals π/2 minus angle DAB (since angle HAB is a right angle). Therefore triangles DBA and DAH are similar. Therefore angle DIA equals angle DOH, where I is the midpoint of BA and O is the midpoint of AH. But AOH is a straight line, so angle DOH and DOA are supplementary angles. Therefore the sum of angles DIA and DOA is π. Angle IAO is a right angle. The sum of the angles in any quadrilateral is 2π, so in quadrilateral IDOA, angle IDO must be a right angle. But ADHE is a rectangle, so the midpoint O of AH (the rectangle's diagonal) is also the midpoint of DE (the rectangle's other diagonal). As I (defined as the midpoint of BA) is the center of semicircle BA, and angle IDE is a right angle, then DE is tangent to semicircle BA at D. By analogous reasoning DE is tangent to semicircle AC at E. Q.E.D.
### Archimedes' circles
The altitude ${\displaystyle AH}$ divides the arbelos into two regions, each bounded by a semicircle, a straight line segment, and an arc of the outer semicircle. The circles inscribed in each of these regions, known as the Archimedes' circles of the arbelos, have the same size.
## Variations and generalisations
example of an f-belos
The parbelos is a figure similar to the arbelos, that uses parabola segments instead of half circles. A generalisation comprising both arbelos and parbelos is the f-belos, which uses a certain type of similar differentiable functions.[5]
In the Poincaré half-plane model of the hyperbolic plane, an arbelos models an ideal triangle.
## Etymology
The type of shoemaker's knife that gave its name to the figure
The name arbelos comes from Greek ἡ ἄρβηλος he árbēlos or ἄρβυλος árbylos, meaning "shoemaker's knife", a knife used by cobblers from antiquity to the current day, whose blade is said to resemble the geometric figure. |
proofpile-shard-0030-9 | {
"provenance": "003.jsonl.gz:10"
} | # zbMATH — the first resource for mathematics
Attractors for second order lattice dynamical systems. (English) Zbl 1002.37040
The second order lattice system $\ddot{u}_i+h(\dot u_i)-(u_{i-1}-2u_{i}+u_{i+1})+\lambda u_i+f(u_i)=g_i,\quad i\in \mathbb{Z},$ is considered, where $$\lambda>0$$, $$(g_i)_i\in\ell^2$$, and the nonlinearities $$f$$ and $$g$$ satisfy some regularity and monotonicity assumtpions. The existence of global attractor in a suitable state space ($$\ell^2\times\ell^2$$) is established and its semicontinuity properties are studied.
##### MSC:
37L60 Lattice dynamics and infinite-dimensional dissipative dynamical systems 37L25 Inertial manifolds and other invariant attracting sets of infinite-dimensional dissipative dynamical systems
Full Text:
##### References:
[1] P. W. Bates, K. Lu, and, B. Wang, Attractors for lattice dynamical systems, preprint, 1999. [2] Afraimovich, V.S.; Chow, S.-N.; Hale, J.K., Synchronization in lattices of coupled oscillations, Phys. D, 103, 442-451, (1997) · Zbl 1194.34056 [3] Hale, J.K., Asymptotic behavior of dissipative systems, (1988), Amer. Math. Soc Providence · Zbl 0642.58013 [4] Temam, R., Infinite-dimensional dynamical systems in mechanics and physics, Appl. math. sciences, 68, (1988), Springer-Verlag New York · Zbl 0662.35001 [5] Feireisl, E., Global attractors for semilinear damped wave equations with supercritical exponent, J. differential equations, 116, 431-447, (1995) · Zbl 0819.35097 [6] Zhou, S., Dimension of the global attractor for damped nonlinear wave equations, Proc. amer. math. soc., 127, 3623-3631, (1999) · Zbl 0940.35038 [7] Ghidaglia, J.M.; Temam, R., Attractors for damped nonlinear wave equations, J. math. pure appl., 66, 273-319, (1987) · Zbl 0572.35071 [8] Karachalios, N.I.; Starrakakis, N.M., Existence of a global attractor for semilinear dissipative wave equations on Rn, J. differential equations, 157, 183-205, (1999) · Zbl 0932.35030 [9] Feireisl, E., Long time behavior and convergence for semilinear wave equations on Rn, J. dynam. differential equations, 9, 133-155, (1997) · Zbl 0879.35109
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. |
proofpile-shard-0030-10 | {
"provenance": "003.jsonl.gz:11"
} | ## Differential and Integral Equations
### Sub- and supersolutions for semilinear elliptic equations on all of $\mathbb{R}^n$
#### Article information
Source
Differential Integral Equations, Volume 7, Number 5-6 (1994), 1215-1225.
Dates
First available in Project Euclid: 23 May 2013
Brown, K. J.; Stavrakakis, N. Sub- and supersolutions for semilinear elliptic equations on all of $\mathbb{R}^n$. Differential Integral Equations 7 (1994), no. 5-6, 1215--1225. https://projecteuclid.org/euclid.die/1369329512 |
proofpile-shard-0030-11 | {
"provenance": "003.jsonl.gz:12"
} | # Measurement of cross sections and polarisation observables in η photoproduction from neutrons and protons bound in light nuclei
Witthauer, L.. Measurement of cross sections and polarisation observables in η photoproduction from neutrons and protons bound in light nuclei. 2015, Doctoral Thesis, University of Basel, Faculty of Science.
Preview
25Mb
Official URL: http://edoc.unibas.ch/diss/DissB_11534 |
proofpile-shard-0030-12 | {
"provenance": "003.jsonl.gz:13"
} | ### Ioannis Tsokanos (The University of Manchester)
Thursday, May 12, 2022, 11:10 – 12:00, -101
Abstract:
In this talk, we study the density properties in the real line of oscillating sequences of the form $( g(k) \cdot F(kα) )_{k \in \mathbb{N}}$, where $g$ is a positive increasing function and $F$ a real continuous $1$-periodic function. This extends work by Berend, Boshernitzan and Kolesnik who established differential properties on the function F ensuring that the oscillating sequence is dense modulo 1.
More precisely, when $F$ has finitely many roots in $[0,1)$, we provide necessary and sufficient conditions for the oscillating sequence under consideration to be dense in $\mathbb{R}$. All the related results are stated in terms of the Diophantine properties of $α$, with the help of the theory of continued fractions. |
proofpile-shard-0030-13 | {
"provenance": "003.jsonl.gz:14"
} | 81 views
Instead of walking along two adjacent sides of a rectangular field, a boy took a short cut along the diagonal and saved a distance equal to half the longer side. Then, the ratio of the shorter side to the longer side is $:$
1. $1/2$
2. $2/3$
3. $1/4$
4. $3/4$
1 Answer
option D correct ans
let length (l) > breath (b)
l+ b = root ( l^2 + b^2 ) + l/2
l/2 +b = root (l^2 + b ^2 )
square both side
we get
b/l= 3/4
by
0 votes
1 answer
1
134 views
0 votes
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2
94 views
0 votes
1 answer
3
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4
81 views |
proofpile-shard-0030-14 | {
"provenance": "003.jsonl.gz:15"
} | ## 1. Add image to the slide page (bottom)
The following minimal working example shows how on can include an image in the title page using \titlegraphics command:
% Add image to the title page
\documentclass{beamer}
\usepackage{tikz}
\title{A new presentation}
\author{The author}
\institute{The Institute that pays him}
% Add the image inside titlegraphics macro
\titlegraphic{
\includegraphics[width=\textwidth]{Sample Image}
}
\begin{document}
\begin{frame}
\titlepage
\end{frame}
\end{document}
Compiling this code yields:
It should be noted that titlegraphics content will move the title page details (title, author, institute, etc) to the above. So sometimes we need to fix also the height of the image in the \includegraphics command using height=<value> (e.g. height=0.5\textwidth).
## 2. Add image to the slide page (top)
As the title of a presentation is positioned at the top of a title slide, we can include an image just before the title text inside \title{} command:
% Add image to the title page (top position)
\documentclass{beamer}
% Add the image inside title macro
\title{
\centering\includegraphics[width=\textwidth]{Sample Image}\\
A new presentation
}
\author{The author}
\institute{The Institute that pays him}
\begin{document}
\begin{frame}
\titlepage
\end{frame}
\end{document}
Compiling this code yields:
We reached the end of this tutorial, If you would like to add a background image to the title slide, I invite you to read this tutorial: “How do you add a background image in LaTeX Beamer? |
proofpile-shard-0030-15 | {
"provenance": "003.jsonl.gz:16"
} | ?
Free Version
Difficult
# Titration Curve for a Weak Acid
APCHEM-UDVYEN
A weak acid of unknown molarity is titrated with $NaOH$. The graph above was obtained.
At which of the following points on the titration curve is $[A^-]$ closest to twice that of $[HA]$?
A
A
B
B
C
C
D
D |
proofpile-shard-0030-16 | {
"provenance": "003.jsonl.gz:17"
} | # Diffie-Hellman exchange
So for a Diffie-Hellman problem, I am given the prime $p$ for the Diffie-Hellman exchange. I am also given $g$, secret number for machine as $A$, secret number for station as $B$, a Diffie-Hellman shielded Login Name $V$, and Diffie-Hellman shielded password $W$.
For this problem, I am given three users and see which one accessed files to which they had no clearance to. So I computed $x=g^A\pmod p$ and $y=g^B\pmod p$, then $x^A\pmod p$ and $y^B\pmod p$ which gave me my secret common key. Where I am confused is as to how to now unshield the DHS key and how I can use $V$ and $W$ to do so.
What do I do next?
According to a textbook I am reading, it says that the equation $DHS*u = 1\pmod p$ has a solution in $\mathbb N_p$ and this is the solution for $UDHS$, or the unshielding of DHS. But I am confused by what this means. Help is appreciated.
-
Actually, you should be computing $x^B \bmod p$ and $y^A \bmod p$ to derive the shared secret. – poncho Oct 22 '13 at 19:26
Within the DH protocol, there's no standard way to do "Diffie-Hellman shielded Logic name" and "password". It is certainly possible to design a protocol that uses DH which does it, however the details of that protocol are outside the Diffie-Hellman protocol. – poncho Nov 4 '13 at 17:14 |
proofpile-shard-0030-17 | {
"provenance": "003.jsonl.gz:18"
} | #### VT Markets APP
Trade CFDs on FX, Gold and more
### US Producer Prices Index fell unexpectedly, Reflecting a Drop in Energy Costs.
###### August 12, 2022
US stocks slid on Thursday and erased gains on speculation the rally that followed softer inflation data went too far, with Federal Reserve still setting monetary policy tight. A key measure of US producer prices unexpectedly fell for the first time in more than two years, mainly reflecting a drop in energy costs. A similar result to the consumer prices report on Wednesday, both the overall and core figures were softer than forecast. However, inflation remains stubbornly high and will likely keep the Fed on a hawkish path to curb it. Meanwhile, equities have been bolstered by a better-than-expected earnings season, and those companies that have trailed analysts’ estimates were rewarded with the biggest gains in at least five years.
The benchmarks, S&P500 and Dow Jones Average Industrial were both little changed down on Thursday after the market consumed CPI numbers. Five out of eleven sectors stayed in positive territory, as Energy and Financial sectors performed best among all groups, rising 3.19% and 1.02% respectively. It’s worth noting that big Tech underperformed as Nasdaq 100 more than 20% above its June lows, and the index slid 0.6% on daily basis for the day.
Main Pairs Movement
US dollar was slightly lower on Thursday, following a dramatic 1% loss the previous day when data showed U.S. inflation was not as hot as anticipated in July. The DXY index edged lower since the Asia trading session and touched a daily-low level below 104.6, and then rebounded to a level above 105.2.
The GBP/USD slid with a 0.11 % loss on daily basis, as the market amid a risk-off impulse while the greenback weakened. The cables witnessed fresh upbeat transactions during the Asian trading session and then lost bullish momentum and fell to a level below 1.220. Apart from that, investors needed to keep an eye out for the critical GDP report on Friday, to confirm the slowdown of economic growth across the UK. Meantime, EURUSD has turned sideways around 1.032, and the pairs advanced with a 0.2% gain for the day.
Gold declined with a 0.15% loss on daily basis, as Federal officials keep their hawkish stances. XAUUSD oscillate in a range from $1,783 to$1,799 marks. WTI and Brent oil both surged on Thursday, rising 2.62% and 2.13% respectively.
Technical Analysis
EURUSD (4-Hour Chart)
The EUR/USD pair advanced on Thursday, preserving its upside traction and extending its previous rebound toward the 1.036 mark after the release of softer-than-expected US PPI data. The pair is now trading at 1.03281, posting a 0.28% gain daily. EUR/USD stays in the positive territory amid a weaker US dollar across the board, as the easing US inflation figures lend support to market sentiment and kept the safe-haven greenback to remain on the back foot. The US Producer Price Index (PPI) declined to 9.8% every year in July, which came in lower than the market’s expectations and pushed the EUR/USD pair higher. For the Euro, European indexes struggle to post advances and the EUR/USD pair up for the fifth consecutive day.
For the technical aspect, the RSI indicator is 66 as of writing, suggesting that the upside is more favoured as the RSI stays above the mid-line. As for the Bollinger Bands, the price failed to climb higher but hovered around the upper band, therefore some upside traction can be expected. In conclusion, we think the market will be slightly bullish as the pair is testing the 1.0325 resistance line. A sustained strength above that level might open the road to additional gains.
Resistance: 1.0325, 1.0438, 1.0484
Support: 1.0282, 1.0158, 1.0111
GBPUSD (4-Hour Chart)
The GBP/USD pair edged higher on Thursday, failing to gather bullish momentum and remaining under pressure below the 1.225 mark during the US session amid risk-off market sentiment. At the time of writing, the cable stays in positive territory with a 0.11% gain for the day. The cooler-than-expected US inflation report and the upbeat US Initial Jobless Claims figure both exerted bearish pressure on the safe-haven greenback and underpinned the GBP/USD pair. The economic data showed that supply-chain conditions are improving and inflationary pressures on the wholesale side have also begun to ease. For the British pound, the Bank of England Chief Economist Huw Pill said on Thursday that higher rates in the short term could also mean some slowing in the UK economy.
For the technical aspect, the RSI indicator is 61 as of writing, suggesting that sellers remain on the sidelines as the RSI on the four-hour chart stays near 60. For the Bollinger Bands, the price failed to preserve upside traction and started to retreat, therefore a continuation of the downside trend can be expected. In conclusion, we think the market will be bearish as long as the 1.2248 resistance line holds. On the upside, if the pair climbs above that level and starts using it as support, bulls could show interest and lift the pair higher.
Resistance: 1.2248, 1.2317, 1.2381
Support: 1.2154, 1.2068, 1.2027
XAUUSD (4-Hour Chart)
Despite the renewed weakness witnessed in the US dollar amid the softer-than-expected US PPI report on Thursday, the pair XAU/USD struggled to gather bullish momentum and retreated to the $1,787 area to erase most of its daily gains during the US trading session. XAU/USD is trading at 1,789.87 at the time of writing, losing 0.13% daily. Signs that inflation might have peaked already continue to support speculations for a less aggressive policy tightening by the Fed, as the softer-than-expected US PPI data have also reinforced market expectations. For the time being, a 50 bps rate hike by the Fed seems likely in the September meeting. Moreover, the risk-on market mood might keep a lid on any further gains for the safe-haven metal. For the technical aspect, the RSI indicator is 52 as of writing, suggesting the pair’s indecisiveness in the near term as the RSI indicator stays near the mid-line. For the Bollinger Bands, the price witnessed fresh selling and dropped below the moving average, therefore the downside traction should persist. In conclusion, we think the market will be bearish as the pair is heading to test the 1785 support line. A break below that level could favour the bear skewed the risk to the downside. Resistance:$1,811, $1,822,$1,831
Support: $1,785,$1,769, \$1,756
Economic Data |
proofpile-shard-0030-18 | {
"provenance": "003.jsonl.gz:19"
} | Question
# The displacement of a particle executing SHM is given by$$y\,=\,5\,\sin \, 4t\,+\,\displaystyle \frac{\pi}{3}$$If $$T$$ is the time period and the mass of the particle is $$2$$ g, the kinetic energy of the particle when $$t\,=\,\displaystyle \frac{T}{4}$$ is given by
A
0.4 J
B
0.5 J
C
3 J
D
0.3 J
Solution
## The correct option is C 0.3 JThe displacement of particle, executing SHM is$$y\, = \, 5 \sin \, 4 t \, + \, \displaystyle \frac {\pi}{3}$$......(i)Velocity of particle$$\displaystyle \frac{dy}{dt} \, = \, \displaystyle \frac{5d}{dt} \, \sin \, 4t \, + \, \displaystyle \frac{\pi}{3}$$$$= \, 5 \, cos \, 4 \, t \, + \, \displaystyle \frac{\pi}{3}$$$$= \, 20 \, cos \, 4 \, t \, + \,\displaystyle \frac{\pi}{3}$$Velocity at $$t \,= \, \displaystyle \frac{T}{4}$$$$\displaystyle \frac{dy}{dt}_{t \, = \, \displaystyle \frac{T}{4}} \,= \,20 \, cos \, 4 \, \times \, \displaystyle \frac{T}{4} \, + \,\displaystyle \frac{\pi}{3}$$Or $$u \,= \, 20\, cos \, T \, + \,\displaystyle \frac{\pi}{3}$$.......(ii)Now, putting value of T in eQ. (II), WE GET$$U \, = \, 20 \cos \, \displaystyle \frac{\pi}{2} \, + \, \displaystyle \frac{\pi}{3}$$$$= \, - \, 20 \sin \, \displaystyle \frac{\pi}{3}$$$$= \, - \, 20 \, \times \, \displaystyle \frac{\overline{3}}{2}$$$$= \, - \, 10 \, \times \, \overline{3}$$The kinetic energy of particle,$$KE \, = \, \displaystyle \frac{1}{2} \, mu^2$$$$\because \, m \, = \, 2g \, = \, 2 \, \times \, 10^{-3} \, kg$$$$= \, \displaystyle \frac{1}{2} \, \times \, 2 \, \times \, 10^{-3} \, \times \, -10 \, \overline{3}^2$$$$= \, 10^{-3} \, \times \, 100 \, \times \, 3$$$$3 \, \times \, 10^{-1}$$$$KE \, = \, 0.3 \, J$$Physics
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proofpile-shard-0030-19 | {
"provenance": "003.jsonl.gz:20"
} | Theorem 8.6 The diagonals of a parallelogram bisect each other Given : ABCD is a Parallelogram with AC and BD diagonals & O is the point of intersection of AC and BD To Prove : OA = OC & OB = OD Proof : Since, opposite sides of Parallelogram are parallel. The Equation 2 gives. Verify your number to create your account, Sign up with different email address/mobile number, NEWSLETTER : Get latest updates in your inbox, Need assistance? Hence diagonals of a parallelogram bisect each other [Proved]. Thus the two diagonals meet at their midpoints. This is exactly what we did in the general case, and it's the simplest way to show that two line segments are equal. Thus the two diagonals meet at their midpoints. google_ad_client = "pub-9360736568487010"; . In AOD and BOC OAD = OCB AD = CB ODA = OBC AOD BOC So, OA = OC & OB = OD Hence Proved. Prove that the diagonals of a parallelogram bisect each other. That is, each diagonal cuts the other into two equal parts. Then the two diagonals are c = a + b (Eq 1) d = b - a (Eq 2) Now, they intersect at point 'Q'. We show that these two midpoints are equal. Start studying Geometry. In a quadrangle, the line connecting two opposite corners is called a diagonal. The position vectors of the midpoints of the diagonals AC and BD are (bar"a" + bar"c")/2 and (bar"b" + bar"d")/2. If possible I would just like a push in the right direction. The angles of a quadrilateral are in the ratio 3: 5: 9: 13. In this video, we learn that the diagonals of a parallelogram bisect each other. To prove that AC and BD bisect each other, you have to prove that AE = EC = BE = ED. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. Google Classroom Facebook Twitter Angles EDC and EAB are equal in measure for the same reason. 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Your mobile number below, for any content/service related issues please contact on this.! Quadrangle, the diagonals of a parallelogram bisect each other Greycells18 Media Limited its. We are given a parallelogram ABCD, shown in figure 10.2.13 lesson, we will use triangles! Figure 10.2.13 of parallelogram bisect each other their midpoints other are the diagonals diagonals of parallelogram bisect other. Intersects another line segment and separates it into two equal parts two diagonals meet their. To itself 0 and Answer separates it into two equal parts are the diagonals ( lines linking opposite prove that the diagonals of a parallelogram bisect each other bisect. On below numbers, Kindly Sign up for a concave quadrilateral ️ by... That all four angles at point E are congruent and a E is congruent itself! Personalized experience intersection point E '' attempt to prove that AE = EC = BE = ED of! 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( please explain briefly and if possible with proof and example ) Thank you linking opposite corners called... ) bisect each other, then it is a parallelogram bisect each.. Push in the given figure, LMNQ is a parallelogram bisect each other, then it is a bisect! Be = ED a call from us give your mobile number below, any! Can divide it into two equal parts is called a bisector the same reason for class-9 ( Class-IX or... The right direction 5: 9: 13 attempt to prove that the diagonals of the rectangle perpendicular that four... Example ) Thank you can divide it into two triangles, please refer the. A trapezium in which PQ rectangle bisect each other into two equal parts is called a bisector EAB equal... Self this is so in measure for the same reason the indicated coordinates show... ) Thank you, show the diagonals AC and BD bisect each other to the,! Is a trapezium in which, in the ratio 3: 5: 9: 13 or grade-9.... With AC and BD have the same Twitter ∴ diagonals bisect each other, B E and E... This video is suited for class-9 ( Class-IX ) or grade-9 kids this shows that the diagonals and call intersection... Ac } $bisect Answer to your question ️ prove by vector method that diagonals... 5: 9: 13 a call from us give your mobile number below for. Angles which the meet point E '' prove that AC and BD bisect each other then. Quadrilateral are in the ratio 3: 5: 9: 13 in a quadrangle, the diagonals bisects other... The parallelogram and convince your self this is so ( please explain briefly if! Point E are congruent and a E is congruent to itself opposite corners ) bisect each,! Figure above drag any vertex to reshape the parallelogram and convince your self is., we will use congruent triangles parallelogram bisect each other figure 10.2.13 point E congruent... Of the rectangle bisect each other same mid-point ∴ diagonals AC and BD have the same reason for... Diagonals meet at their midpoints and convince your self this is so prove that the diagonals of parallelogram! Vocabulary, terms, and more with flashcards, games, and other study tools point are. Intersection point E '' the angles of a parallelogram bisect each...... Abc D is an quadrilateral with AC and BD bisect each other - Mathematics TopperLearning.com... \Overline { AC }$ bisect proof and example ) Thank you like a push in the direction. The ratio 3: 5: 9: 13 reshape the parallelogram and convince your self this is so all!, show the diagonals of a parallelogram bisect each other of the rectangle bisect other...... Q.E.D number below, for any content/service related issues please contact this. 9 0 0 and Answer meet at their midpoints given above is quadrilateral ABCD and we to! Given a parallelogram bisect each other point E are 9 0 0 and Answer I just! To itself congruent to itself connecting two opposite corners is called a bisector = ED for personalized... The same reason hereto get an Answer to your question ️ prove vector. Any parallelogram, the diagonals of a square bisect each other in the ratio 3 5!
D.K. Metcalf Womens Jersey |
proofpile-shard-0030-20 | {
"provenance": "003.jsonl.gz:21"
} | 1. ## Statistics proof
Given that random variable $X$, its mean value $\mu$, its variance $\sigma ^2$, and the mean value $X^2$, which is $\mu _{X^2}$, prove that $\sigma ^2=\mu _{X^2} - \mu _X^2$.
Letting X be the values $x_1,x_2,...,x_n$ and $P(X = x_i) =p_i$, $(i=1,2,...,n)$
$\therefore \sigma ^2 = \sum_{i=1}^n (x_i-\mu _x)^2 \cdot p_i$
$= \sum_{i=1}^n x_i^2 \cdot p_i - 2\mu _X \cdot \bigg (\sum_{i=1}^n x_i \cdot p_i \bigg ) + \mu _X^2 \bigg ( \sum_{i=1}^n p_i \bigg )$ I'm pretty lost how to get this step and the rest.
$=\mu_{X^2}-2\mu _X \cdot \mu _X + \mu _X^2$
$=\mu _{X^2} - \mu _X^2$
I should have taken a picture of the book. That was a ton of code to write, even though it doesn't look like much at all.
2. Originally Posted by chengbin
Given that random variable $X$, its mean value $\mu$, its variance $\sigma ^2$, and the mean value $X^2$, which is $\mu _{X^2}$, prove that $\sigma ^2=\mu _{X^2} - \mu _X^2$.
Letting X be the values $x_1,x_2,...,x_n$ and $P(X = x_i) =p_i$, $(i=1,2,...,n)$
$\therefore \sigma ^2 = \sum_{i=1}^n (x_i-\mu _x)^2 \cdot p_i$
$= \sum_{i=1}^n x_i^2 \cdot p_i - 2\mu _X \cdot \bigg (\sum_{i=1}^n x_i \cdot p_i \bigg ) + \mu _X^2 \bigg ( \sum_{i=1}^n p_i \bigg )$ I'm pretty lost how to get this step Mr F says: Just expand!
and the rest. Mr F says: By definition: ${\color{red}\sum_{i=1}^n p_i = 1}$, ${\color{red}\sum_{i=1}^n x_i p_i = \mu_X}$ and ${\color{red}\sum_{i=1}^n x^2_i p_i = \mu_{X^2}}$.
$=\mu_{X^2}-2\mu _X \cdot \mu _X + \mu _X^2$
$=\mu _{X^2} - \mu _X^2$
I should have taken a picture of the book. That was a ton of code to write, even though it doesn't look like much at all.
.. |
proofpile-shard-0030-21 | {
"provenance": "003.jsonl.gz:22"
} | class resistics.time.reader_ats.TimeReaderATS(dataPath: str)[source]
Data reader for ATS formatted data
For ATS files, header information is XML formatted. The end time in ATS header files is actually one sample past the time of the last sample. The dataReader handles this and gives an end time corresponding to the actual time of the last sample.
Notes
The raw data units for ATS data are in counts. To get data in field units, ATS data is first multipled by the least significat bit (lsb) defined in the header files,
data = data * leastSignificantBit,
giving data in mV. The lsb includes the gain removal, so no separate gain removal needs to be performed.
For electrical channels, there is additional step of dividing by the electrode spacing, which is provided in metres. The extra factor of a 1000 is to convert this to km to give mV/km for electric channels
data = (1000 * data)/electrodeSpacing
Finally, to get magnetic channels in nT, the magnetic channels need to be calibrated.
Methods
setParameters() Set data format parameters dataHeaders() Headers to read in readHeaders() Specific function for reading the headers for internal format lineToKeyAndValue(line) Separate a line into key and value with = as a delimiter
dataHeaders(self)[source]
Return the data headers in the internal file format
Returns
Common headers with information about the recording
readHeader(self)[source]
setParameters(self) → None[source] |
proofpile-shard-0030-22 | {
"provenance": "003.jsonl.gz:23"
} | Cours de Jean-Pierre Serre, no. 1 (1981) , 204 p.
### Sommaire
History p. 1 Siegel formula p. 3 Tamagawa p. 8 I - Integration on $p$-adic manifolds page p. 9 Notation p. 9 Measure attached to a form $\omega$ p. 10 Rational summation p. 15 Igusa p. 16 Vector bundles p. 17 Smooth schemes p. 18 Number of points of classical and exceptional groups p. 20 Decomposition of a measure p. 24 Connection with densities p. 28 Digression : Liftable solutions p. 31 Smooth case p. 32 Real case p. 37 Use of resolution of singularities p. 40 Oesterlé : hypersurfaces p. 41 Łojasiewicz inequality p. 48 Oesterlé : general case p. 49 II - Adeles p. 55 History p. 55 Definition p. 55 $\mathbb{A}_K/K$ compact p. 58 Haar measure on $\mathbb{A}_K$ p. 61 Characters and duality p. 64 Haar measure for dual groups p. 66 Compatible measures with respect to duality p. 67 Adelic integration and heuristic formulas (Goldbach,...) p. 69 Adelic points of algebraic varieties p. 75 Properties of the functor $V\mapsto V(\mathbb{A}_K)$ p. 78 Restriction of scalars p. 84 Algebraic groups and adelic points p. 88 Abelian varieties (S. Bloch) p. 88 Weak approximation p. 90 Strong approximation p. 91 Adeles, classes and genera p. 99 Tensors page p. 102 Vector bundles p. 104 Adelic measures p. 106 Convergent case p. 108 Algebraic groups p. 111 Tori p. 114 Convergence p. 117 Tamagawa number (semi-simple case) p. 122 Theorem of Ono – Weil conjecture p. 123 Tamagawa number (reductive case) p. 126 $\tau(\mathbb{G}_m)=1$ p. 131 Tori (Ono) p. 133 $\tau(PGL_n)=n \tau(SL_n)$ p. 135 Tamagawa $\leftrightarrow$ Siegel (mass formula) p. 136 Tamagawa $\leftrightarrow$ Siegel (the two-groups game) p. 145 Correction to Ono's theory p. 146 Positive definite quadratic forms over $\mathbb{Q}$ p. 147 Proof that $\tau(SO_n)=2$ for $n=2,3,4$ p. 152 Proof of Siegel's formula p. 156 Proof ($m\geq 5$) p. 163 Proof ($m=4$) p. 165 Proof ($m=2$) p. 166 Proof ($m=3$) p. 169 Remarks on Siegel's proof p. 172 Application to modular forms p. 176 III - $SL_n$ p. 180 The Minkowski-Hlawka theorem p. 180 Proof p. 184
@book{CJPS_1981__1_,
author = {Serre, Jean-Pierre},
title = {Adeles and {Tamagawa} numbers},
series = {Cours de Jean-Pierre Serre},
publisher = {Harvard},
number = {1},
year = {1981},
language = {en},
url = {http://www.numdam.org/item/CJPS_1981__1_/}
}
TY - BOOK
AU - Serre, Jean-Pierre
TI - Adeles and Tamagawa numbers
T3 - Cours de Jean-Pierre Serre
PY - 1981
DA - 1981///
IS - 1
PB - Harvard
UR - http://www.numdam.org/item/CJPS_1981__1_/
LA - en
ID - CJPS_1981__1_
ER -
Serre, Jean-Pierre. Adeles and Tamagawa numbers. Cours de Jean-Pierre Serre, no. 1 (1981), 204 p. http://numdam.org/item/CJPS_1981__1_/ |
proofpile-shard-0030-23 | {
"provenance": "003.jsonl.gz:24"
} | # Fun with statistics – transformations of random variables part 2
I recently posted on how to find the distribution of functions of random variables, i.e., the distribution of $Y=g(X)$, where $X$ is a random variable with known distribution and $y=g(x)$ is some function.
As a natural extension of this concept we may ask ourselves what happens if we have two random variables involved. Let us start with one function of two random variables, i.e., given $X$ and $Y$ and knowing their joint PDF $f_{X,Y}(x,y)$ or their joint CDF $F_{X,Y}(x,y) = {\rm Pr}[X \leq x, Y \leq y]$ we would like to calculate the distribution of $Z = g(X,Y)$ where $z=g(x,y)$ is a function with two arguments, e.g., $z=x+y$.
Again, there are multiple ways of addressing this problem. A natural way would be to calculate the CDF of $Z$ directly, i.e., $F_Z(z) = {\rm Pr}[Z \leq z] = {\rm Pr}[g(X,Y) \leq z]$. In other words, we need to compute the probability of the event that relates to all realization of $X$ and $Y$ which satisfy $g(X,Y) \leq z$. This is easily done by integrating the joint PDF $f_{X,Y}(x,y)$ over all points in the set ${\mathcal D}_z$ which contains all points $(x,y)$ for which $g(x,y) \leq z$. Written out, we have
$$F_Z(z) = {\rm Pr}[Z \leq z] = {\rm Pr}[g(X,Y) \leq z] = \iint_{{\mathcal D}_z} f_{X,Y}(x,y) {\rm d}x {\rm d} y$$
Whether or not this approach is easy to follow depends on two things: (1) how easy it is to parametrize the set ${\mathcal D}_z$ and (2) how easy it is to integrate the joint PDF over ${\mathcal D}_z$.
Let us make an example considering the simple function $g(x,y) = x+y$. Then ${\mathcal D}_z$ contains all points $(x,y)$ for which $x+y \leq z$, i.e., $y\leq z-x$ or $x \leq z-y$. Geometrically, this is the set of points that are on the lower-left of a line with slope -1 and offset $z$, i.e., a line passing through $(z,0)$ and $(0,z)$. The integral over this set is relatively simple, as we can directly write it as
$$\displaystyle F_Z(z) = \int_{-\infty}^{+\infty} \int_{-\infty}^{z-y} f_{X,Y}(x,y) {\rm d}x {\rm d} y = \int_{-\infty}^{+\infty} \int_{-\infty}^{z-x} f_{X,Y}(x,y) {\rm d}y {\rm d} x$$.
Another example is $g(x,y) = \max(x,y)$. Since $\max(x,y) \leq z \Leftrightarrow ((x \leq z) \;\mbox{and} \; (y \leq z))$ we can argue
$$F_Z(z) = {\rm Pr}[\max(X,Y) \leq z] = \int_{-\infty}^z \int_{-\infty}^z f_{X,Y}(x,y) {\rm d}x {\rm d} y$$.
Geometrically, ${\mathcal D}_z$ contains all points on the “lower left” of the point $(z,z)$, i.e., the intersection of the half-planes below $y=z$ and left of $x=z$.
The second extension is to consider two functions of two random variables. Say we are given the distribution of $X$ and $Y$ via their joint PDF, we would like to find the joint PDF of $Z=g(X,Y)$ and $W=h(X,Y)$. There is a closed-form expresion for it as a direct extension of the closed-form expression for the PDF of one function of one random variable. It reads as
$$f_{Z,W}(z,w) = \sum_{i=1}^N \frac{1}{|{\rm det} \ma{J}(x_i,y_i)|} f_{X,Y}(x_i,y_i)$$,
where $(x_i,y_i)$ are all solutions to the system of equations $z=g(x,y)$, $w=h(x,y)$ in $x$ and $y$. Here, $\ma{J}$ is the Jacobian matrix given by
$$\ma{J} = \left[ \begin{array}{cc} \frac{\partial g}{x} & \frac{\partial g}{y} \\ \frac{\partial h}{x} & \frac{\partial h}{y} \end{array}\right]$$.
Moreover, the term ${\rm det} \ma{J}(x_i,y_i)$ means that we first compute the determinant of the Jacobian matrix (in terms of $x$ and $y$) and then insert $x_i(z,w)$ and $y_i(z,w)$.
Example? How about the joint distribution of $X+Y$ and $X-Y$? In this case, solving for $z=x+y$ and $w=x-y$ for $x$ and $y$ is simple, we have one solution given by $x_1 = (z+w)/2$ and $y_1 = (z-w)/2$. The Jacobian matrix is given by
$$\ma{J} = \left[ \begin{array}{cc} 1& 1 \\ 1 & -1 \end{array}\right]$$
and hence its determinant is $-2$ everywhere. This gives the solution for $f_{Z,W}(z,w)$ in the form
$f_{Z,W}(z,w) = \frac{1}{2} f_{X,Y}((z+w)/2,(z-w)/2)$.
As in the 1-D case, this direct solution depends heavily on our ability to solve the given functions for $x$ and $y$, which may be tedious for complicated functions.
Interestingly, the first case where we considered one function of one random variable can be solved also via this approach, simply by creating another “auxiliary” variable, and then marginalizing over it. So once we have $Z=g(X,Y)$ we make up another $W=h(X,Y)$, choosing it such that the remaining calculations are simple. For instance, for $g(x,y) = x+y$ we may choose $h(x,y) = y$. Then, the Jacobian matrix becomes
$$\ma{J} = \left[ \begin{array}{cc} 1& 1 \\ 0 & 1 \end{array}\right]$$
with determinant one. Moreover, we have $x_1 = z-w$ and $y_1 = w$. Therefore, we get
$$f_{Z,W}(z,w) = f_{X,Y}(z-w,w)$$.
The final step is marginalizing out the auxiliary $W$ which gives
$$f_Z(z) = \int_{-\infty}^{+\infty} f_{X,Y}(z-w,w) {\rm d}w.$$
Looks much like a convolution integral, doesn’t it? In fact, if $X$ and $Y$ are statistically independent, we can write $f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y)$ and hence we obtain
$$f_Z(z) = \int_{-\infty}^{+\infty} f_{X}(z-w)\cdot f_Y(w) {\rm d}w = f_X(x) * f_Y(y),$$
where $*$ denotes convolution. This shows very easily that the PDF of the sum of two random variables is the convolution of their PDFs, if they are statistically independent. |
proofpile-shard-0030-24 | {
"provenance": "003.jsonl.gz:25"
} | # cdlib.algorithms.r_spectral_clustering¶
r_spectral_clustering(g_original: object, n_clusters: int = 2, method: str = 'vanilla', percentile: int = None) → cdlib.classes.node_clustering.NodeClustering
Spectral clustering partitions the nodes of a graph into groups based upon the eigenvectors of the graph Laplacian. Despite the claims of spectral clustering being “popular”, in applied research using graph data, spectral clustering (without regularization) often returns a partition of the nodes that is uninteresting, typically finding a large cluster that contains most of the data and many smaller clusters, each with only a few nodes. This method allows to compute spectral clustering with/withouth different regualarization functions designed to address such a limitation.
Supported Graph Types
Undirected Directed Weighted
Yes No No
Parameters: g_original – a networkx/igraph object n_clusters – How many clusters to look at method – one among “vanilla”, “regularized”, “regularized_with_kmeans”, “sklearn_spectral_embedding”, “sklearn_kmeans”, “percentile”. percentile – percentile of the degree distribution to perform regularization. Value in [0, 100]. Mandatory if method=”percentile” or “regularized”, otherwise None NodeClustering object
>>> from cdlib import algorithms
>>> import networkx as nx
>>> G = nx.karate_club_graph()
>>> coms = algorithms.r_spectral_clustering(G, n_clusters=2, method="regularized", percentile=20)
Zhang, Yilin, and Karl Rohe. “Understanding Regularized Spectral Clustering via Graph Conductance.” arXiv preprint arXiv:1806.01468 (2018). |
proofpile-shard-0030-25 | {
"provenance": "003.jsonl.gz:26"
} | # Prove $\varphi(x)$ to be primitive recursive
Let $\varphi(x)=2x$ if $x$ is a perfect square, $\varphi(x) = 2x+1$ otherwise. Show $\varphi$ is primitive recursive.
In proving $\varphi$ to be a p.r. function I think it could come in handy the following theorem:
Let $\mathcal C$ be a PRC class. Let the functions $g$, $h$ and the predicate $P$ belong to $\mathcal C$, let
$$f(x_1,\ldots, x_n) = \begin{cases} g(x_1, \ldots, x_n) \;\;\;\;\;\text{ if } P(x_1, \ldots, x_n)\\ h(x_1,\ldots,x_n) \;\;\;\;\;\text{ otherwise} \end{cases}$$ Then $f$ belongs to $\mathcal C$ because $$f(x_1, \ldots, x_n) = g(x_1, \ldots, x_n) \cdot P(x_1, \ldots, x_n) + g(x_1, \ldots, x_n) \cdot \alpha(P(x_1, \ldots, x_n))$$ where
$$\alpha(x) = \begin{cases} 1 \;\;\;\;\;\text{ if } x = 0\\ 0 \;\;\;\;\;\text{ if } x \neq 0 \end{cases}$$
and $\alpha(x)$ is p.r.
So similarly I would say that $\varphi(x)$ is p.r. as
$$\varphi(x) = \begin{cases} 2x \;\;\;\;\;\;\;\;\;\;\;\text{ if } x = t \cdot t \\ 2x+1 \;\;\;\;\;\text{ otherwise} \end{cases}$$ hence $$\varphi(x) = 2x \cdot P(x_1, \ldots, x_n) + (2x+1) \cdot \alpha(P(x_1, \ldots, x_n))$$ and $P$ is a primitive recursive predicate as $x \cdot y$ is p.r. and also $x = y$.
Does everything hold? Is there anything wrong? If so, since I am tackling this kind of exercise for the fist time, will you please tell me what's the proper way to solve this?
-
You need to show that you can write an "if,then,else"-function, i.e. $f(x,y,z) = y$ if $x$ is true and $z$ otherwise. Then you need to show that you can test whether or not $x$ is a perfect square. This you can do by iterating $i$ from 1 to $x$ testing if $i^2 = x$. – Pål GD Jan 17 '13 at 18:54
To see if $x$ is a perfect square is easy (for example, by adding 1 + 3 + 5 ... you get the succesive squares); once that is settled your problem is solved. Think of such problems primarily as programming asignments (in a rather cruel programming language). |
proofpile-shard-0030-26 | {
"provenance": "003.jsonl.gz:27"
} | Dolphins of the open ocean are classified as Type II Odontocetes(toothed whales). These animals use ultrasonic "clicks" with afrequency of about 54.8 kHz to navigateand find prey.
(a) Suppose a dolphin sends out a series ofhigh-pitched clicks that are reflected back from the bottom of theocean 61 m below. How much time elapsesbefore the dolphin hears the echoes of the clicks? (The speed ofsound in seawater is approximately 1530 m/s.)
1 s
(b) What is the wavelength of 54.8 kHzsound in the ocean?
2 mm |
proofpile-shard-0030-27 | {
"provenance": "003.jsonl.gz:28"
} | Math Help - general solution to differential equation
1. general solution to differential equation
For our practice final one of the questions is
Find the general solution to the following differential equation: y'' - 4y' + 4y =x2 - 3x + 2
I understand that one would add the particular solution and general solution. For the particular solution I got 1/4x2-1/4x+1/8 and for the general solution I got Ae2x+Be2x. However he gave us the solution and it was Ae-2x+Bxe-2x+1/4x2-1/4x+1/8. Does anyone see where I went wrong? I can show more work if thats helpful.
2. Re: general solution to differential equation
Originally Posted by JML2618
For our practice final one of the questions is
Find the general solution to the following differential equation: y'' - 4y' + 4y =x2 - 3x + 2
I understand that one would add the particular solution and general solution. For the particular solution I got 1/4x2-1/4x+1/8 and for the general solution I got Ae2x+Be2x. However he gave us the solution and it was Ae-2x+Bxe-2x+1/4x2-1/4x+1/8. Does anyone see where I went wrong? I can show more work if thats helpful.
Your mistake was in not dealing with the fact that your characteristic polynomial has a single root of order 2.
That makes your homogeneous solution of the form
$y[x]=c_1e^{2x}+c_2xe^{2x}$ (note the factor x in the 2nd term)
go use this form for your homogeneous solution and resolve your particular solution.
3. Re: general solution to differential equation
oh okay i understand now thanks for the reply! Do you use this whenever it's in the form y''+y'+y and ce^ax +ce^ax when it's y''+y'?
4. Re: general solution to differential equation
or is it that the roots are equal?
5. Re: general solution to differential equation
Originally Posted by JML2618
or is it that the roots are equal?
Pauls Online Notes : Differential Equations - Repeated Roots
6. Re: general solution to differential equation
Notice that $Ae^{2x}+ Be^{2x}= Ce^{2x}$ with C= A+ B. Those are not two independent solutions and cannot give the general solution to a second order equation. |
proofpile-shard-0030-28 | {
"provenance": "003.jsonl.gz:29"
} | ## Intermediate Algebra (6th Edition)
Published by Pearson
# Chapter 11 - Section 11.2 - Arithmetic and Geometric Sequences - Exercise Set: 14
#### Answer
$-5120$
#### Work Step by Step
To find the $n$th term of a geometric sequence, we use the formula $a_n = a_1 \cdot r^{n-1}$ where $a_1$ is the first term and $r$ is the common ratio. $a_6 = 5\cdot(-4)^{6-1} = 5\cdot(-4)^5 = 5\cdot(-1024) = -5120$
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. |
proofpile-shard-0030-29 | {
"provenance": "003.jsonl.gz:30"
} | # Principle of superposition in uniform compression of bar
With the following quantities defined as follows:
Normal stress along x, $\sigma_x = \frac{F_{nx}}{S}$
Strain along x, $\epsilon_x= \frac{\Delta L_x}{L_x}$
and Poisson's Law: $\epsilon_y=\epsilon_z=-\nu \epsilon_x= -\nu\frac{\sigma_x}{E}$, as well as Hooke's Law: $\epsilon_x=\frac{1}{E}\sigma_x$,
with $F_{nx}, L_x,S, E, \nu$ being the normal force applied, length along x, area, Young's modulus and Poisson's coefficient respectively, we are trying to find the change of volume of a uniformly compressed parallelepiped $(\sigma_x=\sigma_y=\sigma_z=\sigma=-p)$.
I do not understand the passage: $\epsilon_x=\frac{\Delta L_x}{L_x}=\frac{\sigma_x}{E}-\frac{\nu}{E}(\sigma_y+\sigma_z)$ which is supposedly validated by the superposition principle. This already makes little sense to me because of Hooke's law implying $\frac{\nu}{E}(\sigma_y+\sigma_z)=0$. In this context, I interpret the superposition principle as the sum of the inputs(stresses) equalling the sum of the outputs (extensions). What's happening here? As you might guess, this is a new topic for me.
• I think I finally understand what you are asking. See the ADDENDUM to my answer below. Aug 30, 2017 at 12:27
The general equations for the three strains are: $$\epsilon_x=\frac{\sigma_x-\nu(\sigma_y+\sigma_z)}{E}$$ $$\epsilon_y=\frac{\sigma_y-\nu(\sigma_x+\sigma_z)}{E}$$ $$\epsilon_z=\frac{\sigma_z-\nu(\sigma_x+\sigma_y)}{E}$$ For the case of uniaxial loading $\sigma_x=\sigma$ in the x-direction, while the stresses in the y and z directions are zero ($\sigma_y=\sigma_z=0$),we get from the above three equations:$$\epsilon_x=\frac{\sigma}{E}$$ $$\epsilon_y=-\nu\frac{\sigma}{E}=-\nu\epsilon_x$$ $$\epsilon_z=-\nu\frac{\sigma}{E}=-\nu\epsilon_x$$
Now let's consider a different kind of loading where, instead of the load being only in the x direction, there are also equal stresses in the y and z directions, such that $$\sigma_x=\sigma_y=\sigma_z=\sigma$$If we substitute these into the three general equations for the strains in terms of the stresses, we obtain:$$\epsilon_x=\epsilon_y=\epsilon_z=(1-2\nu)\frac{\sigma}{E}$$This equation can also be obtained by starting with a uniaxial stress in the x-direction, then superimposing an equal stress in the y-direction, and then superimposing an equal stress in the z direction.
If $\sigma=-p$, the linear strains in the three directions are $$\epsilon=-(1-2\nu)\frac{p}{E}$$The volumetric strain $\epsilon_v$ is three times the linear strain, so,$$\epsilon_v=-3(1-2\nu)\frac{p}{E}$$ Since the original volume is $L_xL_yL_z$, the change in volume is $$\Delta V=-3(1-2\nu)\frac{p}{E}L_xL_yL_z$$ ADDENDUM
Here's another way of looking at it. Suppose you start out with a uniaxial stress of $\sigma_x$ on the body. Then, the strains in the three directions are $$\epsilon_x=\frac{\sigma_x}{E}$$ $$\epsilon_y=-\nu\epsilon_x=-\nu\frac{\sigma_x}{E}$$and$$\epsilon_z=-\nu\epsilon_x=-\nu\frac{\sigma_x}{E}$$ If, instead, you start out with a unixial stress of $\sigma_y$ on the body, then the strains in the three directions are $$\epsilon_y=\frac{\sigma_y}{E}$$ $$\epsilon_x=-\nu\epsilon_y=-\nu\frac{\sigma_y}{E}$$and$$\epsilon_z=-\nu\epsilon_y=-\nu\frac{\sigma_y}{E}$$
If, instead, you start out with a unixial stress of $\sigma_z$ on the body, then the strains in the three directions are $$\epsilon_z=\frac{\sigma_z}{E}$$ $$\epsilon_x=-\nu\epsilon_z=-\nu\frac{\sigma_z}{E}$$and$$\epsilon_y=-\nu\epsilon_z=-\nu\frac{\sigma_z}{E}$$
If, instead, you impose all three of these stresses simultaneously on the body, the strains you get are obtained by linearly superimposing (i.e., adding together) the three strains from the uniaxial stress cases: $$\epsilon_x=\frac{\sigma_x-\nu(\sigma_y+\sigma_z)}{E}$$ $$\epsilon_y=\frac{\sigma_y-\nu(\sigma_x+\sigma_z)}{E}$$ $$\epsilon_z=\frac{\sigma_z-\nu(\sigma_x+\sigma_y)}{E}$$
• Yes, the addendum clarified everything! Aug 30, 2017 at 19:17 |
proofpile-shard-0030-30 | {
"provenance": "003.jsonl.gz:31"
} | ## Reblog: Calculus Tidbits
[Feature photo above by Olga Lednichenko via Flickr (CC BY 2.0).]
This week I have a series of quotes about calculus from my first two years of blogging. The posts were so short that I won’t bother to link you back to them, but math humor keeps well over the years, and W. W. Sawyer is (as always) insightful.
I hope you enjoy this “Throw-back Thursday” blast from the Let’s Play Math! blog archives:
## Finding the Limit
Eldest daughter had her first calculus lesson last night: finding the limit as delta-t approached zero. The teacher found the speed of a car at a given point by using the distance function, calculating the average speed over shorter and shorter time intervals. Dd summarized the lesson for me:
“If you want to divide by zero, you have to sneak up on it from behind.”
## Harmonic Series Quotation
This kicked off my week with a laugh:
Today I said to the calculus students, “I know, you’re looking at this series and you don’t see what I’m warning you about. You look and it and you think, ‘I trust this series. I would take candy from this series. I would get in a car with this series.’ But I’m going to warn you, this series is out to get you. Always remember: The harmonic series diverges. Never forget it.”
—Rudbeckia Hirta
Learning Curves Blog: The Harmonic Series
quoting Alexandre Borovik
## So You Think You Know Calculus?
Rudbeckia Hirta has a great idea for a new TV blockbuster:
## Common Sense and Calculus
And here’s a quick quote from W. W. Sawyer’s Mathematician’s Delight:
If you cannot see what the exact speed is, begin to ask questions. Silly ones are the best to begin with. Is the speed a million miles an hour? Or one inch a century? Somewhere between these limits. Good. We now know something about the speed. Begin to bring the limits in, and see how close together they can be brought.
Study your own methods of thought. How do you know that the speed is less than a million miles an hour? What method, in fact, are you unconsciously using to estimate speed? Can this method be applied to get closer estimates?
You know what speed is. You would not believe a man who claimed to walk at 5 miles an hour, but took 3 hours to walk 6 miles. You have only to apply the same common sense to stones rolling down hillsides, and the calculus is at your command.
## Reblog: Patty Paper Trisection
[Feature photo above by Michael Cory via Flickr (CC BY 2.0).]
I hear so many people say they hated geometry because of the proofs, but I’ve always loved a challenging puzzle. I found the following puzzle at a blog carnival during my first year of blogging. Don’t worry about the arbitrary two-column format you learned in high school — just think about what is true and how you know it must be so.
I hope you enjoy this “Throw-back Thursday” blast from the Let’s Play Math! blog archives:
One of the great unsolved problems of antiquity was to trisect any angle using only the basic tools of Euclidean geometry: an unmarked straight-edge and a compass. Like the alchemist’s dream of turning lead into gold, this proved to be an impossible task. If you want to trisect an angle, you have to “cheat.” A straight-edge and compass can’t do it. You have to use some sort of crutch, just as an alchemist would have to use a particle accelerator or something.
One “cheat” that works is to fold your paper. I will show you how it works, and your job is to show why …
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## Reblog: Solving Complex Story Problems
[Dragon photo above by monkeywingand treasure chest by Tom Praison via flickr.]
Over the years, some of my favorite blog posts have been the Word Problems from Literature, where I make up a story problem set in the world of one of our family’s favorite books and then show how to solve it with bar model diagrams. The following was my first bar diagram post, and I spent an inordinate amount of time trying to decide whether “one fourth was” or “one fourth were.” I’m still not sure I chose right.
I hope you enjoy this “Throw-back Thursday” blast from the Let’s Play Math! blog archives:
Cimorene spent an afternoon cleaning and organizing the dragon’s treasure. One fourth of the items she sorted was jewelry. 60% of the remainder were potions, and the rest were magic swords. If there were 48 magic swords, how many pieces of treasure did she sort in all?
[Problem set in the world of Patricia Wrede’s Enchanted Forest Chronicles. Modified from a story problem in Singapore Primary Math 6B. Think about how you would solve it before reading further.]
How can we teach our students to solve complex, multi-step story problems? Depending on how one counts, the above problem would take four or five steps to solve, and it is relatively easy for a Singapore math word problem. One might approach it with algebra, writing an equation like:
$x - \left[\frac{1}{4}x + 0.6\left(\frac{3}{4} \right)x \right] = 48$
… or something of that sort. But this problem is for students who have not learned algebra yet. Instead, Singapore math teaches students to draw pictures (called bar models or math models or bar diagrams) that make the solution appear almost like magic. It is a trick well worth learning, no matter what math program you use …
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## Reblog: Putting Bill Gates in Proportion
[Feature photo above by Baluart.net.]
Seven years ago, one of my math club students was preparing for a speech contest. His mother emailed me to check some figures, which led to a couple of blog posts on solving proportion problems.
I hope you enjoy this “Throw-back Thursday” blast from the Let’s Play Math! blog archives:
## Putting Bill Gates in Proportion
A friend gave me permission to turn our email discussion into an article…
Can you help us figure out how to figure out this problem? I think we have all the information we need, but I’m not sure:
The average household income in the United States is $60,000/year. And a man’s annual income is$56 billion. Is there a way to figure out what this man’s value of $1mil is, compared to the person who earns$60,000/year? In other words, I would like to say — $1,000,000 to us is like 10 cents to Bill Gates. ### Let the Reader Beware When I looked up Bill Gates at Wikipedia, I found out that$56 billion is his net worth, not his income. His salary is $966,667. Even assuming he has significant investment income, as he surely does, that is still a difference of several orders of magnitude. But I didn’t research the details before answering my email — and besides, it is a lot more fun to play with the really big numbers. Therefore, the following discussion will assume my friend’s data are accurate… [Click here to go read Putting Bill Gates in Proportion.] ## Bill Gates Proportions II Another look at the Bill Gates proportion… Even though I couldn’t find any data on his real income, I did discover that the median American family’s net worth was$93,100 in 2004 (most of that is home equity) and that the figure has gone up a bit since then. This gives me another chance to play around with proportions.
So I wrote a sample problem for my Advanced Math Monsters workshop at the APACHE homeschool conference:
The median American family has a net worth of about $100 thousand. Bill Gates has a net worth of$56 billion. If Average Jane Homeschooler spends \$100 in the vendor hall, what would be the equivalent expense for Gates?
## Reblog: The Handshake Problem
[Feature photo above by Tobias Wolter (CC-BY-SA-3.0) via Wikimedia Commons.]
Seven years ago, our homeschool co-op held an end-of-semester assembly. Each class was supposed to demonstrate something they had learned. I threatened to hand out a ten question pop quiz on integer arithmetic, but instead my pre-algebra students voted to perform a skit.
I hope you enjoy this “Throw-back Thursday” blast from the Let’s Play Math! blog archives:
If seven people meet at a party, and each person shakes the hand of everyone else exactly once, how many handshakes are there in all?
In general, if n people meet and shake hands all around, how many handshakes will there be?
1-3 narrators
### Props
Each friend will need a sheet of paper with a number written on it big and bold enough to be read by the audience. The numbers needed are 0, 1, 2, 3, … up to one less than the number of friends. Each friend keeps his paper in a pocket until needed.
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## Reblog: In Honor of the Standardized Testing Season
[Feature photo above by Alberto G. Photo right by Renato Ganoza. Both (CC-BY-SA-2.0) via flickr.]
Quotations and comments about the perils of standardized testing, now part of my book Let’s Play Math.
I hope you enjoy this “Throw-back Thursday” blast from the Let’s Play Math! blog archives:
The school experience makes a tremendous difference in a child’s learning. Which of the following students would you rather be?
I continued to do arithmetic with my father, passing proudly through fractions to decimals. I eventually arrived at the point where so many cows ate so much grass, and tanks filled with water in so many hours. I found it quite enthralling.
— Agatha Christie
An Autobiography
…or…
“Can you do Addition?” the White Queen asked. “What’s one and one and one and one and one and one and one and one and one and one?”
“I don’t know,” said Alice. “I lost count.”
“She can’t do Addition,” the Red Queen interrupted. “Can you do Subtraction? Take nine from eight.”
“Nine from eight I can’t, you know,” Alice replied very readily: “but—”
“She can’t do Subtraction,” said the White Queen. “Can you do Division? Divide a loaf by a knife — what’s the answer to that?”
Claim your two free learning guide booklets, and be one of the first to hear about new books, revisions, and sales or other promotions.
## Reblog: The Case of the Mysterious Story Problem
[Feature photo above by Carla216 via flickr (CC BY 2.0).]
Seven years ago, I blogged a revision of the first article I ever wrote about homeschooling math. I can’t even remember when the original article was published — years before the original (out of print) editions of my math books.
I hope you enjoy this “Throw-back Thursday” blast from the Let’s Play Math! blog archives:
I love story problems. Like a detective, I enjoy sifting out clues and solving the mystery. But what do you do when you come across a real stumper? Acting out story problems could make a one-page assignment take all week.
You don’t have to bake a pie to study fractions or jump off a cliff to learn gravity. Use your imagination instead. The following suggestions will help you find the clues you need to solve the case… |
proofpile-shard-0030-31 | {
"provenance": "003.jsonl.gz:32"
} | # How do you simplify (x^2-x-6)/(4x^3)*(x+1)/(x^2+5x+5)?
Jul 18, 2015
Try factoring and find:
$\frac{{x}^{2} - x - 6}{4 {x}^{3}} \cdot \frac{x + 1}{{x}^{2} + 5 x + 5}$
$= \frac{\left(x - 3\right) \left(x + 2\right) \left(x + 1\right)}{4 {x}^{3} \left(x + \frac{5 + \sqrt{5}}{2}\right) \left(x + \frac{5 - \sqrt{5}}{2}\right)}$ (factoring)
$= \frac{{x}^{3} - 7 x - 6}{4 {x}^{5} + 20 {x}^{4} + 20 {x}^{3}}$ (multiplying)
#### Explanation:
Going in one direction, multiply up to get:
$\frac{{x}^{2} - x - 6}{4 {x}^{3}} \cdot \frac{x + 1}{{x}^{2} + 5 x + 5}$
$= \frac{\left({x}^{2} - x - 6\right) \left(x + 1\right)}{4 {x}^{3} \left({x}^{2} + 5 x + 5\right)}$
$= \frac{{x}^{3} - 7 x - 6}{4 {x}^{5} + 20 {x}^{4} + 20 {x}^{3}}$
Going in the other direction, factor to get:
$\frac{{x}^{2} - x - 6}{4 {x}^{3}} \cdot \frac{x + 1}{{x}^{2} + 5 x + 5}$
$\frac{\left(x - 3\right) \left(x + 2\right)}{4 {x}^{3}} \cdot \frac{x + 1}{{x}^{2} + 5 x + 5}$
$= \frac{\left(x - 3\right) \left(x + 2\right) \left(x + 1\right)}{4 {x}^{3} \left(x + \frac{5 + \sqrt{5}}{2}\right) \left(x + \frac{5 - \sqrt{5}}{2}\right)}$
No common factors to cancel, so this cannot be simplified. |
proofpile-shard-0030-32 | {
"provenance": "003.jsonl.gz:33"
} | # Generic properties of $p$-groups
I have the impression that most people in algebra believe the following statement to be true, but I have no reference for it.
Fix a natural number $n$. Consider for each prime $p$ the set of all groups of order $p^n$. Then is the following true?
• There is a $p'$ such that for all $p\geq p'$ the number of isomorphism classes of groups of order $p^n$ does only depend on n.
• We can write down generic presentation for all the groups of order $p^n$ with fixed $n$ and $p\geq p'$. That means that we can give a presentation where each word in the relation subgroup has the same form. It depends only on the chosen prime $p$.
• Furthermore many group theoretic properties are shared for groups with the same generic presentation (but for different primes). Is it true that these groups have the same nilpotency degree? Is it true that the sizes of the conjugacy classes depend polynomially on $p$? Is it true that the number of cojugacy classes of a certain subgroup also depends polynomially on $p$? What can be said about the ($G$-)poset of subgroups?
I'd like to know what is already known and maybe given a reference.
-
I doubt the first bullit point because $n\equiv 1\pmod p$ allows a very "$p$-specific" construction of a semidirect product – Hagen von Eitzen Jan 30 '13 at 10:18
But there are only finitely many such $p$. Now choose $p'$ bigger than any of these $p$. – Curufin Jan 30 '13 at 10:36
Do you mean "depends only on $n$" in the second point? – Martin Brandenburg Jan 30 '13 at 10:48
No. I fix $n$. Generic means for me: I have a presentation where I just replace $p$ with a certain prime. $\langle a\mid a^{p^n}\rangle$ is the generic presentation of a cyclic group with $p^n$ elements. – Curufin Jan 30 '13 at 10:58
This is a duplicate of math.stackexchange.com/questions/263876. Your first statement is false for $n=5$. – Derek Holt Jan 30 '13 at 12:18
Your conjectures are sort of true, but the reality is much more complicated than you have phrased it.
## Summary
Counting $p$-groups for large $p$ (compared to the nilpotency class) is the same as counting certain finite dimensional (restricted) Lie algebras. Such counts are organized into a tree, with smaller groups being the parents of larger groups. Such counts involve linear algebra and an orbit calculation. The orbit calculation involves counting points in a variety. In all cases known before 2011 or so, the orbit calculations were “PORC”, polynomial on residue classes, so that while no single number, and no single polynomial work, there are finitely many polynomials that work, and the choice of which polynomial is based solely on the residue of $p$ mod some fixed number $n$. Rather than organize $p$-groups by order, it may be better to organize them by coclass: the number of times a child is much larger than the parent. In 2012, a parent group was found so that the orbit calculations involved in a non-PORC way counting the points on a variety. Further investigation found that the conjugacy classes of the group and its descendants were also not PORC.
However, to my mind the calculations for $p^5$ (which are PORC) are extremely similar to these, so to me the essence of the conjecture still holds, but its specific expression is now known to be flawed. I found Vaughan-Lee's recent papers to be a very readable introduction to these ideas, though I learned them from the books mentioned below.
## Lazard correspondence
Given a $p$-group $G$, define the subgroups $G_{n+1} = [G,G_n] (G_n)^p$ with $G_1 = G$, so that $G_2 = \Phi(G)$, and $G_n/G_{n+1}$ is an elementary abelian $p$-group centralized by $G$. This called the lower exponent $p$ series. We define a restricted $p$-Lie algebra on $L(G) = \oplus_n G_n/G_{n+1}$ with $[a_i,b_j]_L = v$ and $(a_i)^p_L = w$ where $$v_{\ell} = \begin{cases} [a_i,b_j]_G & \ell = i+j \\ 0 & \text{otherwise} \end{cases} \qquad w_{\ell} = \begin{cases} (a_i)^p_G & \ell = i + 1 \\ 0 & \text{otherwise} \end{cases}$$ In other words, the commutator and $p$th power map are the same in $L(G)$ as in $G$, except that we have to be careful which quotient group everything happens in. If $G_p = 1$, then one can recover the group $G$ from $L(G)$ using the Baker-Campbell-Hausdorff formula for the exponential, so that counting $G$s is the same as counting $L(G)$s.
In Higman's case, this correspondence is fairly clear: $L(G) = G/\Phi(G) \oplus \Phi(G)$ and so every element of $L(G)$ has the form $(\bar g, h)$ and $[ (\bar a, b), (\bar c, d) ] = ( \bar 1, [a,c] )$ and $(\bar a,b)^p = (\bar 1, a^p)$. Any basis of $L(G)$ is a minimal generating set of $G$ (after taking any arbitrary choice of pre-images), and the restricted Lie algebra relations give the relations of the group. Higman showed that if $G_3=1$, then enumerating these Lie algebras was PORC.
## $p$-group generation algorithm
To organize the calculation, we view $G/G_n$ as the parent of $G/G_{n+1}$. Given a parent $G/G_n$ that we've already constructed, we try to find all descendants $G/G_{n+1}$. This calculation is described in O'Brien's 1990 article. Again, the gist is just linear algebra and orbit calculations, so one tends to get PORC behavior.
These techniques were used to correct earlier calculations of $p^6$, and to enumerate the groups of order $p^7$. In all cases the answers turn out to be PORC. Each presentation depends on $p$, occasionally requiring elements to be chosen from the orbit space of a variety over some characteristic $p$-field. The properties of each presentation are (more or less by definition of the $p$-group generating algorithm) the same: in particular the nilpotency class and $p$-class is constant on these varieties, indeed the entire structure of $G/G_n$ is constant.
## Coclass
Organizing $p$-groups by their order is in many ways a bad idea. Philip Hall suggested using iso-clinism as a better equivalence relation than order, and this method was used in much of the earlier work. However, the coclass proved to be a very nice unifying method. In many cases there is a single (parameterized) expression for an infinite sequence of groups, whose properties (nilpotency class and conjugacy classes) are parameterized in a very simple way. The coclass of $G$ is $n-c$ where $|G|=p^n$ and $c$ is the nilpotency class. A group of coclass 0 is cyclic of order $p$, and groups of coclass 1 are called maximal class. For $p=2$, these are the dihedral, the semidihedral, and quaternion groups; each of these have nice parameterized expressions, and most of the time one has an easy time dealing with the entire family. The coclass conjectures give nice information on the groups in each family using a $p$-adic space group (a single group with a simple presentation whose finite quotients are the mainline groups in that family). The non-mainline groups are topic of current study. du Sautoy's zeta functions, and Eick's computational research group have made significant progress on these groups.
## Non-PORC behavior
Recently a group of order $p^9$ whose descendants of order $p^{10}$ are not PORC was discovered by du Sautoy and Vaughan-Lee (2012). In a followup (fairly expository) paper they also show the number of conjugacy classes and the size of the automorphism group are not PORC. In two followup expository papers Vaughan-Lee revisits and simplifies Higman's original PORC calculations.
## Bibliography
Books
• Dixon, J. D.; du Sautoy, M. P. F.; Mann, A.; Segal, D. Analytic pro-p groups. Cambridge University Press, Cambridge, 1999. xviii+368 pp. ISBN: 0-521-65011-9 MR1720368 DOI:10.1017/CBO9780511470882
• Leedham-Green, C. R.; McKay, S. The structure of groups of prime power order. Oxford University Press, Oxford, 2002. xii+334 pp. ISBN: 0-19-853548-1 MR1918951
• Holt, Derek F.; Eick, Bettina; O'Brien, Eamonn A. “Handbook of computational group theory.” Chapman & Hall/CRC, Boca Raton, FL, 2005. xvi+514 pp. ISBN: 1-58488-372-3 MR2129747 DOI:10.1201/9781420035216
Articles
- |
proofpile-shard-0030-33 | {
"provenance": "003.jsonl.gz:34"
} | ## anonymous 3 years ago $\int _a^b \frac{dx}{y}$,where $y^2=ax^2+bx+c$
1. experimentX
reduce y in to this from (px + q)^2 + r
2. anonymous
where $y^2=ax^2+bx+c$
3. anonymous
sry i meant y^2
4. experimentX
do the same .. it wouldn't make any difference
5. anonymous
$1/a(x+b/2a)^2-b^2/4a+c/a 6. anonymous \[1/a(x+b/2a)^2-b^2/4a+c/a$
7. anonymous
i tried a euclidian sub of t $t=y+x\sqrt a$
8. experimentX
yeah .. then substitute u = Sqrt(a)(x + b/x) should be of the form |dw:1362742001381:dw|
9. anonymous
yeah to the first post or euclidean?
10. experimentX
??
11. anonymous
do you agree with the sub of$t=y+x\sqrt{a}$
12. experimentX
where did you get that y ... there is no such y.
13. experimentX
check your equation again ... don't put that y there ... there is no y involved. I think you are confused.
14. anonymous
i have to go now but i 'll try it |
proofpile-shard-0030-34 | {
"provenance": "003.jsonl.gz:35"
} | # Tag Info
29
It is an access hatch used during construction and maintenance. Credit: NASA-KSC Credit: NASA This part got at least some media coverage during the scrubbing of STS-121, when a Engine Cutoff (ECO) sensor, a fuel gauge, mounted behind that cover, inside the Liquid Hydrogen (LH2) tank, malfunctioned, causing that launch to be delayed, while the sensors were ...
27
I did a crude spreadsheet sim using the Rogers Commission report to get throttle times, to wit: Throttle down to 94% at 24 seconds Throttle down to 65% at 42 seconds Throttle up to 104% at 65 seconds I neglected startup propellant consumption and assumed step function throttling. I took liftoff O2 load to be 1,387,457 lb and H2 load to be 234,265 lb. I ...
25
That's the intertank - the cylinder that connected the bottom of the LO2 tank to the top of the LH2 tank. It didn't contain propellant, but did contain the forward interface with the Solid Rocket Boosters, and was built for lightness and strength, with skin-stringer construction. The ribs you see were the stringers. The intertank is a steel / aluminum ...
24
The breakup of Challenger occurred about 73 seconds into flight. Main engine cutoff normally occurs about 510 seconds into flight, implying that about 86% of the fuel would be remaining. (Many sources give 480 seconds, but I suspect that's a simple division of the tankage mass by the full-throttle consumption rate; looking at actual mission reports supports ...
22
The shuttle external tank held the propellants for the shuttle main engines. It was filled from spherical tanks positioned at the perimeter of the launch pad. Insulated lines ran from the spheres, through the Mobile Launcher, and into the Orbiter through two tombstone-shaped Tail Service Masts. Then through the Orbiter Main Propulsion System plumbing into ...
21
No, the previously used External Tanks (ETs) disintegrated in the atmosphere before they fell into the sea. Notably, Buzz Aldrin and others proposed different ideas for reuse of the tank in orbit, and allegedly NASA said that they would be willing to take external tanks to orbit if a private company would use them. No private effort ever stepped up to the ...
20
This part of the External Tank is called the "LH2 Tank aft dome". There are really two large circular penetrations on it. They are the ones offset from the center of the tank. One is the access hatch/manhole. (this description is from the LO2 tank part of the linked document. Further down it says the LH2 "manhole fitting was similar to those on both the ...
17
Based on this description of the Space Shuttle flight profile, no external tank would ever have completed so much as a single orbit. An external tank would achieve essentially the same orbital apogee as the orbiter itself, but that is all. The shuttle fired its OMS engines to achieve an actual orbit AFTER tank separation. This means that the tank remained on ...
13
I don't have a great reference for this, but it was to reduce cost on the throw-away External Tank. By using the same interface into the Orbiter used to supply propellants to the main engines, the cost and complexity of adding a dedicated loading interface to the tank was avoided. It was not a tremendous complexity hit to the Orbiter Main Propulsion system ...
13
No reuse, but... NASA did not at any point actually reuse an external tank in any way. However... They made plans to allow reuse in orbit. NASA did have tentative plans for utilization of the tanks in orbit. These plans were scrapped, however. The primary factors being (1) decreased payload capacity to stable orbit †, (2) risk of insulating foam falling ...
12
tl;dr - the parts at the rear of ET-94 where the foam was removed were painted orange for display. The foam was not dyed but started out a light cream color. It slowly turned orange when exposed to light. Here is a picture of foam that was trimmed off during the stringer crack problem on STS-133. You can see the internal foam is lighter, and the metal is ...
10
This was when planning safe disposal of the External Tank on certain types of launch abort. Source: Space Shuttle Abort Evolution page 9-10 A major design activity was conducted preparing for Shuttle launches on the West coast from Vandenberg AEB. Though not flown due to new program directions following the Challenger accident, a lot of design work and ...
8
As a design decision, if you throw away the External Tank and it has engines attached, you are throwing away the engines. Since the Orbiter was returning for sure anyway, the decision was to leave the engines on the orbiter. Once the ET is done its job, the engines are not needed to make orbit. The OMS pods provide enough punch for the needed orbital ...
8
Setting aside the political/managerial issues this is about mass. The final version of the tank had a "dry" (empty tanks) mass of about 26.5 tons and a fuelled mass of about 760 tons. The surface area is a roughly 2600 $m^2$ and the total mass of the thermal protection system is just over 2 tons (all masses and dimensions from wikipedia). So any replacement ...
7
The original External Tank nose cone design was indeed blunt - almost like a fireplug, as seen in this 1975 concept art. [Image source - lost in the mists of time to me, but NASA somewhere] However, wind tunnel testing at the Arnold Engineering Development Center revealed that this configuration caused unsteady aerodynamic buffeting at some conditions. A ...
7
A few facts: SRB Burn Time is 127 seconds Start of Challenger Incident- 64 s Vehicle breakup- 72 s Nominal time to orbit- 510 s. So the SRBs were about half-way done with their burn time before the vehicle started to break up. The Space Shuttle Main Engine actually produced less than 1 g thrust until about the time of SRB separation. If the SRBs could have ...
7
A big issue with boosting the tank to orbit, would be the foam insulation. It was believed it would come off in chunks, like popcorn, causing immense amounts of orbital debris, potentially in the orbit you wished to store it at. Which could be really bad news.
6
Reusability. The whole idea of the shuttle was to discard all the parts that are simple, cheap and easy to replace and recover everything expensive, complex and hard to replace. Of course the reality, involving meddling by parties other than NASA, never mind failures in the process the shuttle was designed (not so much the design itself as the process of ...
5
Safety distances are decided by modelling the worst-case scenario (an explosion of the rocket right on the launch pad). An explosion results in an overpressure which drops off as distance increases, the safe distance is one where the overpressure is limited to a survivable level. The same goes for structures: you can calculate how much overpressure a ...
5
There were three attach points. The forward bipod that you show in your answer, and two aft attach points. At each attach point a large explosive bolt held the tank and Orbiter together. Large umbilical door openings in the aft of the Orbiter let the aft bolts pass through and also had all the fluid and electrical connections. After separation tile-covered ...
5
There are generally speaking 2 conops to get into orbit with the Space Shuttle. The first one, and most commonly used, requires two thrusts post Main Engine CutOff (MECO) to achieve orbit. This one drops the ET in the Indian Ocean. The "Direct Insert" requires only one OMS maneuver, and the tank landed in the Pacific Ocean. So the External tank might ...
4
What the equations I used completely ignore is initial thrust to gross launch mass which'd surely affect gravity drag? ... What is the assumption behind the 9.7 km/s delta-v on the Wikipedia page as to gravity drag fraction and initial launch acceleration? 9.7km/s is towards the high end of delta-v to orbit requirements. It varies with both the aerodynamics ...
3
Launch vehicles don't stage to get the earlier stages out of the way, they stage to get rid of excess mass so they can actually reach orbit with a useful payload. The Shuttle ET was a bit of an exception due to various compromises in its design, and a poor design in that it hauled 27 metric tons of mass up to just short of a circular orbit that could ...
3
Your question is based on a misunderstanding- the External Tank propellant tanks did in fact have relief valves. The 02 valve relieved at 24 psid and the H2 valve relieved at 36 psid. From the 1982 Press Manual, pages 92-95 GremlinWrangler's comment about the inadvisability of venting hydrogen is well founded - see Flight Rule A5-154 whose rationale ...
2
I believe you are right, it is the External Tank. Why does it have problems? The external tanks were deposited in either the Indian or Pacific Oceans, clear across the world. So, what would that do for a low southward inclination launch from Vandenburg? It would cut across South America, then over Africa, the Middle East, and Russia, with only a brief stay ...
Only top voted, non community-wiki answers of a minimum length are eligible |
proofpile-shard-0030-35 | {
"provenance": "003.jsonl.gz:36"
} | Limited access
Two identical stars of mass $M$ and radius $R$ are separated by a distance $d$ ($d>>R$). The two stars are in circular orbit around their combined center of mass. They are observed to be moving at a speed ${v}_{1}$.
Two other identical stars each with a mass of $2M$ (but with the same radius $R$) are found to be in the same configuration; specifically separated by a distance $d$ and orbiting their combined center of mass. They are moving at a speed ${v}_{2}$.
What is the ratio of $\cfrac {{v}_{2}}{{v}_{1}}$?
A
$\cfrac{1}{4}$
B
$1$
C
$\sqrt {2}$
D
$2$
Select an assignment template |
proofpile-shard-0030-36 | {
"provenance": "003.jsonl.gz:37"
} | Using normal approximation to estimate the probability of winning more than you lose in 100 plays
You play a game where if the two cards you pick from a deck of cards (without replacement) have consecutive rank (i.e. 2 and 3, A and K, A and 2, etc.), you win. The game pays 81 USD to 5 USD, meaning if you win, you have a net gain of 81 USD and if you lose, you lose 5 USD. You have two scenarios: play 10 times or play 100 times.
I calculated the chance of winning to be $\binom {2}{1} \frac {52}{52} *\frac {4}{51} =\frac {8}{51}$
I then calculated the winning E(X)=$100* \frac{8}{52} =15.6$ SE(X)= $\sqrt {(\frac{8}{51}*\frac{43}{51})} *\sqrt{100}=3.64$
The solution given for the P(win more than lose) is 0.997
How can the probability of winning more than losing be the case when we only win 15.6 games out of 100 plays?
What am I missing here?
Thank you!
• In the $15.6$ games you win you win $\$81\cdot15.6=\$1263.6$ and in the $84.4$ games you lose, you only lose $\$5\cdot84.4=\$422.$ Where can I go to play this game? – saulspatz Jul 9 '18 at 4:56
• lol only in the land of arbitrary math problems. How did they arrive the 0.997 approximation though? If we use the formula $\frac{Observed \, Value - 15.6}{3.64}$ what would we plug in for the Observed value though? – pino231 Jul 9 '18 at 6:36 |
proofpile-shard-0030-37 | {
"provenance": "003.jsonl.gz:38"
} | # Atomes froids dans des réseaux optiques - Quelques facettes surprenantes d'un système modèle
Abstract : This thesis is devoted to the experimental study of atoms trapped and cooled in several types of optical structures. In order to characterise these media, we used different techniques such as time-of-flight techniques, direct imaging of the atomic
cloud and pump-probe spectroscopy. We thus obtained information about kinetic temperature, spatial diffusion of atoms and atomic motion in the optical potential wells.
We first studied the dynamics of cesium atoms in three-dimensional bright optical lattices when a magnetic field is applied. In particular, we showed that optical lattices operating in the jumping regime do provide good trapping and cooling efficiencies and that a motionnal narrowing effect gives birth to narrow
vibrational sidebands on pump-probe transmission spectra. Still with cesium atoms, we created and characterised a three-dimensional bright optical lattice obtained with only two laser beams through the Talbot effect, and also a random medium
generated by a speckle field.
We endly studied a brownian motor'' for $^(87)$Rb atoms in a grey asymmetric potential. The results of the experimental study are in good qualitative agreement with semi-classical Monte-Carlo numerical simulations.
Keywords :
Document type :
Theses
https://tel.archives-ouvertes.fr/tel-00006734
Contributor : Cécile Robilliard <>
Submitted on : Tuesday, August 24, 2004 - 12:26:07 PM
Last modification on : Thursday, December 10, 2020 - 12:37:00 PM
Long-term archiving on: : Friday, April 2, 2010 - 8:27:16 PM
### Identifiers
• HAL Id : tel-00006734, version 1
### Citation
Cécile Mennerat-Robilliard. Atomes froids dans des réseaux optiques - Quelques facettes surprenantes d'un système modèle. Physique Atomique [physics.atom-ph]. Université Pierre et Marie Curie - Paris VI, 1999. Français. ⟨tel-00006734⟩
Record views |
proofpile-shard-0030-38 | {
"provenance": "003.jsonl.gz:39"
} | • Create Account
### #ActualBaneTrapper
Posted 03 October 2012 - 08:24 AM
I' once had the exact same problem. The chance is that you are linking against release library when you are actually debugging.
http://en.sfml-dev.o...g60228#msg60228
I believe this would do it, but am going to work atm, Thx for help.
EDIT:: SFML WORKS!
Name is fixed, my error was:
while at Project properties, Linker, Input, Aditional dependecies, i did it like this
sfml-window.lib
while debug needs to be
sfml-window-d.lib
Thanks for the help!
### #2BaneTrapper
Posted 03 October 2012 - 08:23 AM
I' once had the exact same problem. The chance is that you are linking against release library when you are actually debugging.
http://en.sfml-dev.o...g60228#msg60228
I believe this would do it, but am going to work atm, Thx for help.
EDIT:: SFML WORKS!
Name is fixed, my error was:
while at Project properties, Linker, Input, Aditional dependecies, i did it like this
sfml-window.lib
while debug needs to be
sfml-window-d.lib
### #1BaneTrapper
Posted 02 October 2012 - 11:22 PM
I' once had the exact same problem. The chance is that you are linking against release library when you are actually debugging. |
proofpile-shard-0030-39 | {
"provenance": "003.jsonl.gz:40"
} | Openlumify
I worked for a “data discovery and analytics platform” company1 for a couple of years, and got to see first-hand the power of graph-based datastores for analysis, hypothesis-testing and pattern discovery. It truly was an example of intelligence amplification and allowed the analysts I worked with to work more effectively than ever.
Since that day I've had a desire to map things out, to build a kind of mega-mindmap2 to really understand some of the more complex and complicated situations — ancient and modern history, the middle east, US politics, the powerful people behind every conspiracy theory imaginable…
So I was quite happy to see that someone somewhere had started an opensource graph-based analysis suite of a similar sort. Unfortunately by the time I found it, it had become abandonware, moving from open- to (assumed) closed-source by the IP owner, who was then bought by another company, then again, renamed, ad absurdum infinitum. I grabbed the last opensource I could find and have started on the long journey to revive it.
So far I've updated the build (Maven! Eek) to work with Java 9+ and refactored everything to theorg.openlumify package. It builds, it runs … but I can't login.
If only there were more hours in the night or day!
Some plans
In no particular order …
• get it running
• build containers
• make it work in Kubernetes
• make sure it has a timeline and a map (at least!)
• pick a decent backend datastore and implement it
• setup a project website
• host it somewhere online e.g. for investigative journos
• karma!
1. I don't want to say their name, even after all these years, since I once tweeted the name — quite innocuously — and it eventually came back through the ‘peer network’ (no hierarchies, man). Probably not a huge deal, but I was super creeped out that they were actively watching for public mentions and performing sentiment analysis and someone with fewer than a hundred followers would actually catch any attention at all. ↩︎
2. Hey, I've loved mindmaps since the day I discovered them. Feels like I've been drawing and redrawing the same few maps — with alterations — for most of my life, anytime I was feeling lost or overwhelmed. Most recently I've recreated them in Simplemind, but there are notebooks upon notebooks exploring my plans, desires, perceived strengths and weaknesses. ↩︎ |
proofpile-shard-0030-40 | {
"provenance": "003.jsonl.gz:41"
} | Apunts de Matemàtiques per a l'accés a la UIB per a majors de 25 anys
You can not select more than 25 topics Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
#### 58 lines 2.1KB Raw Permalink Blame History
``````%% qrcode.ins %% Copyright 2014 by Anders O.F. Hendrickson %% %% This work may be distributed and/or modified under the %% conditions of the LaTeX Project Public License, either version 1.3 %% of this license or (at your option) any later version. %% The latest version of this license is in %% http://www.latex-project.org/lppl.txt %% and version 1.3 or later is part of all distributions of LaTeX %% version 2005/12/01 or later. %% %% This work has the LPPL maintenance status `maintained'. %% %% The Current Maintainer of this work is Anders O.F. Hendrickson. %% %% This work consists of the files qrcode.dtx and qrcode.ins %% and the derived file qrcode.sty. \input docstrip.tex \keepsilent \usedir{tex/latex/qrcode} \preamble This is a generated file. Copyright (C) 2014 by Anders Hendrickson This work may be distributed and/or modified under the conditions of the LaTeX Project Public License, either version 1.3 of this license or (at your option) any later version. The latest version of this license is in http://www.latex-project.org/lppl.txt and version 1.3 or later is part of all distributions of LaTeX version 2005/12/01 or later. \endpreamble \generate{\file{qrcode.sty}{\from{qrcode.dtx}{package}}} \obeyspaces \Msg{*************************************************************} \Msg{* *} \Msg{* To finish the installation you have to move the following *} \Msg{* file into a directory searched by TeX: *} \Msg{* *} \Msg{* qrcode.sty *} \Msg{* *} \Msg{* To produce the documentation run the file qrcode.dtx *} \Msg{* through LaTeX. *} \Msg{* *} \Msg{* Happy TeXing! *} \Msg{* *} \Msg{*************************************************************} \endbatchfile`````` |
proofpile-shard-0030-41 | {
"provenance": "003.jsonl.gz:42"
} | # Find an equation of the tangent line to the curve at the given point. y =...
###### Question:
Find an equation of the tangent line to the curve at the given point. y = x4 + 5x2 - x, (1,5) y = Show My Work (Requiredi 2
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proofpile-shard-0030-42 | {
"provenance": "003.jsonl.gz:43"
} | How to Detect if Motor is not Working?
sorry for my english, I am from Indonesia.
I try to make a (car) relay active - switch on, if the motor/fan is working normal, so when the fan is out, not working or broken, automaticly the relay is off.
But it is not working.
Edited Transistor NPN 2N3055
Motor 12 Vdc / 3 A
Car Relay Hela 12V
• Please explain how you believe your current circuit should work. – Ignacio Vazquez-Abrams Jan 7 '16 at 9:23
• What fault conditions on the motor are you trying to detect? – Icy Jan 7 '16 at 9:25
• As long as the motor works fine, the relay should still on. – Herbrata Moeljo Jan 7 '16 at 9:29
• 'not working' could be seized - high current; open circuit - zero current; disconnected load - low current + possibly lots of other conditions. – Icy Jan 7 '16 at 9:33
• so you require base current of 3.75mA (150/40) - and you have sized 270R resistor to give 10mA at 3A motor current. If this is rated current typical operating current will be much lower perhaps less than 1A - try using 1 or 2 Ohm resistor instead of 270R - the BD139 can take 1.5A of base current - and this will also give a reduced voltage drop on the motor at full torque. – Icy Jan 7 '16 at 10:01
Start by assuming the motor is drawing 3 A, and the relay coil is drawing 150 mA. Then the transistor collector current is also drawing 150 mA. Because the transistor should be acting as a switch, its base current should be in the range of 1/10 to 1/20 the collector current, or about 10 mA. Then the voltage across RSENSE should be $$V_{RSENSE} = 0.7 + I_b \times RBASE$$ For a 47 ohm RBASE, this works out to $$V_{RSENSE} = 0.7 + .01 \times 47 = 0.7 + .47 = 1.17\text{ volts}$$ and the power dissipated in RSENSE $$P = i VSENSE= 3\times1.17 = 3.5\text{ watts}$$ so you'll need RSENSE to be at least 3.5 watts, and a 5-watt resistor would be a reasonable choice. |
proofpile-shard-0030-43 | {
"provenance": "003.jsonl.gz:44"
} | # Find out the rules from examples, then solve it
Fill the blank rectangles with numbers 6 to 16
Find out the rules from examples
Hint:
No common divisors except 1
• In examples 2 and 3, do you realize that 9 is not a prime number? – Cordfield Aug 29 '17 at 9:15
• @Cordfield : Yes I realize it. – Jamal Senjaya Aug 29 '17 at 9:31
• Your “Hint” is pretty heavy; you might as well have come out and said explicitly what rule we were missing. You might want to write hints that are lightweight; pointing us in a direction without shoving the answer into our face. – Peregrine Rook Aug 30 '17 at 4:52
First, the board look like this.
AA 5 BB CC DD
EE FF GG
HH II JJ 17 KK
And I found three rules.
1. If there is a dot between two cells, two numbers should be consecutive. If there is no dot, two numbers should not be consecutive.
2. Greatest common divisor of two adjacent cells should be 1. In other words, two numbers should be coprime.
3. By rule 2, difference of two adjacent cells should be an odd number.
By rule 1,
BB, II, and JJ are obvious.
e1 5 6 o1 e2
o2 o3 o4
e3 15 16 17 e4
Now we have numbers from 7 to 14.
By rule 3,
e1 ~ e4 should be an even number, and o1 ~ o4 should be an odd number.
And the candidates are:
e1 = 8, 10, 12, 14
e2 = 8, 10, 12, 14
e3 = 8, 10, 12, 14
e4 = 8, 10, 12, 14
o1 = 7, 9, 11, 13
o2 = 7, 9, 11, 13
o3 = 7, 9, 11, 13
o4 = 7, 9, 11, 13
By rule 1, some numbers can be removed from the candidates.
e1 = 8, 10, 12, 14
e2 = 8, 10, 12, 14
e3 = 8, 10, 12
e4 = 8, 10, 12, 14
o1 = 9, 11, 13
o2 = 7, 9, 11, 13
o3 = 9, 11, 13
o4 = 7, 9, 11, 13
By rule 2, more numbers can be removed.
e1 cannot be 10, e3 cannot be 10 and 12, o1 cannot be 9, o3 cannot be 9
e1 = 8, 12, 14
e2 = 8, 10, 12, 14
e3 = 8
e4 = 8, 10, 12, 14
o1 = 11, 13
o2 = 7, 9, 11, 13
o3 = 11, 13
o4 = 7, 9, 11, 13
Since e3 has one candidate,
we can fill in e3. Then o2 should be 7 or 9.
e1 5 6 o1 e2
o2 o3 o4
8 15 16 17 e4
e1 = 12, 14
e2 = 10, 12, 14
e4 = 10, 12, 14
o1 = 11, 13
o2 = 7, 9
o3 = 11, 13
o4 = 7, 9, 11, 13
Let's assume:
o2 is 9.
e1 5 6 o1 e2
9 o3 o4
8 15 16 17 e4
Since 7 can only be filled in o4, it should be 7, and e4 should be 8, which is impossible.
Therefore,
o2 should be 7.
e1 5 6 o1 e2
7 o3 o4
8 15 16 17 e4
e1 = 12, 14
e2 = 10, 12, 14
e4 = 10, 12, 14
o1 = 11, 13
o3 = 11, 13
o4 = 9, 11, 13
In order,
e1 should be 12, o4 should be 9, e4 should be 10, e2 should be 14, o1 should be 11, and o3 should be 13.
12 5 6 11 14
7 13 9
8 15 16 17 10
And this is my final answer.
I think my explanation is messy and those formattings are horrible. Feel free to point errors or improve formatting of my answer.
• Rule 3 implies rule 2 – boboquack Aug 29 '17 at 9:24
• @boboquack Oh, you are right. For God's sake. – Otami Arimura Aug 29 '17 at 9:27
• @boboquack no, there are lots of coprime pairs with an even difference. – Kruga Aug 29 '17 at 14:42
• @Kruga however no even numbers can be next to each other, meaning that they have to be at the places they are, meaning the odd numbers must be in the remaining places and as a result all the differences are odd – boboquack Aug 29 '17 at 21:30
• @boboquack ah, I see what you mean. But it still depends on the shape of the grid and the preexisting numbers. – Kruga Aug 30 '17 at 7:25
the rule is simple:
dot between squares means those numbers should be consecutive. and if there is no dot, they cannot be consecutive and the difference has to be odd number.
and the numbers that you are supposed to fill is given as
$5$ to $17$
so firstly you can easily see that
5 to 6 and 17 to 16 then 16 to 15.
the rest becomes:
14 5 6 9 12
11 x 13 x 7
10 15 16 17 8
• 14 breaks your configuration – boboquack Aug 28 '17 at 9:12
• @boboquack forgot 14 :)) it is ojay now. – Oray Aug 28 '17 at 9:14
• Nop wrong answer. – Jamal Senjaya Aug 28 '17 at 9:15
• @JamalSenjaya it is valid for all given examples. :/ – Oray Aug 28 '17 at 9:16
• @JamalSenjaya check it again, change the rule after examining the examples a little deeper. – Oray Aug 28 '17 at 9:22
I found the same rules that Oray found, and I found the same solution:
$\array{14&~\mathbf5&~\mathit6&~9&12\\11&\textbf X&13&\textbf X&~7\\10&\mathit{15}&\mathit{16}&\mathbf{17}&~8}$
where $\mathbf{5}$ and $\mathbf{17}$ (bold) are the given numbers, and $\mathit{6}$, $\mathit{15}$, and $\mathit{16}$ (italic) are the trivially forced ones.
Unfortunately, I found seven other solutions following the same rules:
$\array{10&~\mathbf5&~\mathit6&~9&14\\~7&\textbf X&13&\textbf X&11\\~8&\mathit{15}&\mathit{16}&\mathbf{17}&12}$ $\quad\qquad\array{12&~\mathbf5&~\mathit6&11&14\\~7&\textbf X&13&\textbf X&~9\\~8&\mathit{15}&\mathit{16}&\mathbf{17}&10}$ $\quad\qquad\array{12&~\mathbf5&~\mathit6&~9&14\\~7&\textbf X&13&\textbf X&11\\~8&\mathit{15}&\mathit{16}&\mathbf{17}&10}$
$\array{12&~\mathbf5&~\mathit6&11&14\\~9&\textbf X&13&\textbf X&~7\\10&\mathit{15}&\mathit{16}&\mathbf{17}&~8}$ $\quad\qquad\array{14&~\mathbf5&~\mathit6&13&10\\11&\textbf X&~9&\textbf X&~7\\12&\mathit{15}&\mathit{16}&\mathbf{17}&~8}$
$\array{10&~\mathbf5&~\mathit6&~9&14\\13&\textbf X&11&\textbf X&~7\\12&\mathit{15}&\mathit{16}&\mathbf{17}&~8}$ $~~~\text{and}\,~~\array{10&~\mathbf5&~\mathit6&11&14\\13&\textbf X&~9&\textbf X&~7\\12&\mathit{15}&\mathit{16}&\mathbf{17}&~8}$
A bit more discussion: we know from the problem statement that we are required to use the integers 6 through 16.
The fact that a dot implies that the adjoining numbers are consecutive (differ by 1) forces the 6, 15, and 16, as previously stated, leaving 7, 8, 9, 10, 11, 12, 13, and 14 to be filled in. The fact that all pairs of adjacent numbers differ by an odd number (> 1) gives us this template:
$$\array{E_1&~\mathbf5&~\mathit6&O_1&E_2\\O_2&\textbf X&O_3&\textbf X&O_4\\E_3&\mathit{15}&\mathit{16}&\mathbf{17}&E_4}$$ where the $E$s are even numbers and the $O$s are odd numbers.
$O_1$ and $O_3$ can’t be 7, because they are adjacent to 6 (without a dot, so not consecutive), so $O_2$ or $O_4$ must be 7. The number below the 7 must be 8, because $O_2$ and $E_3$ are separated by a dot, and so are $O_4$ and $E_4$, and the number below 7 can’t be 6, because we already know where that is. That narrows it down to these possibilities:
$\array{E_1&~\mathbf5&~\mathit6&O_1&E_2\\~7&\textbf X&O_3&\textbf X&O_4\\~8&\mathit{15}&\mathit{16}&\mathbf{17}&E_4}$ $\quad\text{or}\quad\array{E_1&~\mathbf5&~\mathit6&O_1&E_2\\O_2&\textbf X&O_3&\textbf X&~7\\E_3&\mathit{15}&\mathit{16}&\mathbf{17}&~8}$
For my next step, I tried to solve for the other bottom pair of edge numbers i.e., $O_2~\&~E_3$ or $O_4~\&~E_4$, whichever one wasn’t 7 & 8. I got the following pairs: {9,10}, {11,10}, {11,12}, {13,12}, and {13,14}. Each of those, except for {13,14}, led to at least one solution, as shown above.
So I guess there’s a pattern that we haven’t spotted. :-(
Rules
Rule 1 - Even Odd Even Odd and so on (matrix style)
Rule 2 - Dots mean subtracted by one
Rule 3 - No dots, no common divisor except 1
Solution - Same as @Otami Arimura
12 5 6 11 14
7 x 13 x 9
8 15 16 17 10 |
proofpile-shard-0030-44 | {
"provenance": "003.jsonl.gz:45"
} | Chrysler will pay another $1 million to California as part of a parallel administrative settlement agreement with the California Air Resources Board (CARB), and will provide similar remedies for California-certified vehicles with the catalyst or OBD defects. The lawsuit is the result of a joint EPA-CARB investigation of Chrysler’s 1996 through 2001 Cherokees, Grand Cherokees, Wranglers, Dakota trucks, and Ram vans, wagons, and pickup trucks. The investigation disclosed that a significant percentage of the vehicles experience excessive deterioration or failure of the catalytic converter. The deterioration of the catalytic converters in the named models results from a design defect in the original converter installed on each of the vehicles. As a result of this design defect—in some of the identified Chrysler vehicles—the internal components of the converter move around excessively, causing the device’s ceramic core to break up. ### Comments My 1998 Jeep Grand Cherokee Laredo has a strange rattle in the catalytic converter. My dealer says there is no recall action. How can I find out if my car is part of the proposed Chrysler action? Thanks. Norita J. Halvorsen. My catalytic converter is really rattling, making a terrible noise. I was at the Daimler-Chrysler office in Hendersonville NC today and asked them about the warranty on my 1997 Jeep Grand Cherokee and they told me they hadn't received anything from the Company about the warranty on this vehicle. When will you start sending out these warranty notices. I want to get mine fixed if it is under warranty. Thanks, Doris Grant when my cat. converter went bad at a little over 78,000 miles the dealer fixed it.the dealer said that an 02sensor had to be removed to get to the cat.converter.it cost me 200.00 bucks.i think i was taken for a ride.also,my transmission blew up at the same time the converter did.my mechanic said the converter probably killed my transmission.should i seek compensation from the dealer for charging me for the sensor to fix the converter and what should i do about the transmission cost? I took my Grand Cherokee in for the recall. The converter was already rattling so they replaced the sensor and had to wait 2 weeks for them to get the converter in. Returned to have converter installed when i returned to pick up Jeep was told they had to carry the Jeep to a different place to have the converter cut off and weld the new one back one. The one they replaced it with is round doesn't fit in the same spot so they rerouted the pipes. After repairs were done engine is not running good real rough idel then noticed transmission fluid under the jeep when checked the transfer case is leaking. I checked online didn't find a converter that is round they all are shaped like a muffler with a flat top & bottom. Never received anything from anyone wanting to know how my service went. Is it possible this not a converter they put on my Jeep? My catalytic converter on my 1997 Grand Cherokee Laredo has been causing the "Check Engine" to light up. I cannot get the local Jeep dealership to check the problem; moreover, they've told me I do not have any claim, but they won't even check the vehicle. What do I do from here? I now have this issue: Bad catalytic converter which lead to the manifold cracking, found by dealership ($1800.00+ to repair). Took it to Dan Fast Muffler for an estimate. They have three more sitting waiting to be fixed with the same issues! Where is the JD and EPA with getting this settlement and recall issued? If I get this repaired will I be able to get a refund if the recall does finially appear?
My cat was doing that for a while and I just removed it because I did not have the money to fix it. Will I be able to take my jeep in for the recall anyway?
My '96 Grand CherokeeReceived the letter from mfgr to take car in for cat recall. Made appointment 2 weeks in advance-left car all day-dealer says part is on nation-wide back order & may take weeks or months to get new part. Meanwhile car started stalling at idle & running rough but showing no codes(classic symptoms of clogged up cat) I removed cat & sure enough, all the honeycomb material was dislodged and loose inside. I removed the material and re-installed cat-all is fine now, EXCEPT car is due for bi-annual Emission test soon.. Obviously it wont pass with cat empty,so I'm in Government Limbo 'til dealer gets new part ...
My '96 had same problems. Cat. started rattling so I had it removed. Due to "light wallet" at the time I had a straight pipe installed and ever since engine stalls and and back fires. Sometimes good for3-5 days, other times can't leave my driveway for 15 mins.! Today my muffler blew open from a huge backfire. Getting new one put on tomorrow and wondering now if I am part of this recall and if these "mysto" problems are from the cat. converter. As weel, will i blow another muffler without the converter installed. Mechanics I have spoken to say the pipe has nothing to do with any of this but I'm thinking otherwise.
Who do I contact for warranty?
The comments to this entry are closed. |
proofpile-shard-0030-45 | {
"provenance": "003.jsonl.gz:46"
} | Björn smedman in Probability 30 minutes
# Variational Coin Toss
Variational inference is all the rage these days, with new interesting papers coming out almost daily. But diving straight into Huszár (2017) or Chen et al (2017) can be a challenge, especially if you’re not familiar with the basic concepts and underlying math. Since it’s often easier to approach a new method by first applying it to a known problem I thought I’d walk you through variational inference applied to the classic “unfair coin” problem.
If you like thinking in code you can follow along in this Jupyter notebook.
## The Usual Bayesian Treatment of an Unfair Coin
Okay, so we got the usual problem statement: You’ve found a coin in a magician’s pocket. Since you found it in a magician’s pocket you’re not sure it’s a “fair coin”, i.e. one that has a 50% chance of landing heads up when you toss it; it could be a “magic” coin, one that comes up heads with probability $z$ instead.
So you toss the coin a few times and it comes up tails, heads, tails, tails and tails. What does that tell you about $z$? Let’s bring out the (hopefully) familiar Bayesian toolbox and find out.
### Choosing a Prior
First we need to place a prior probability distribution over $z$. Since we found the coin in a magician’s pocket we think that $z$ may very well be quite far from $0.5$. At the same time; there’s nothing obviously strange about the coin, so it seems improbable that $z$ would be very high or very low. We therefore choose $p(z) = \text{Beta}(\alpha = 3, \beta = 3)$ as our prior.
Figure 1. Prior probability distribution over $z$.
### Classic Bayesian Inference
Now that we have a prior over $z$ and we’ve tossed the coin a few times we’re ready to infer the posterior. Call the outcome of the coin tosses $\vect{x}$. Then according to Bayes theorem the posterior $p(z \given \vect{x})$ is
$$$$p(z \given \vect{x}) = \frac{p(\vect{x} \given z) p(z)}{p(\vect{x})},$$$$
where $p(\vect{x} \given z)$ is usually referred to as the “likelihood of $z$” and $p(\vect{x})$ is called the model “evidence”. $p(z)$ is of course just the prior.
With our choice of prior over $z$, and conditional probability of $\vect{x}$ given $z$, we could quite easily derive a closed form expression for the posterior. But since the subject here is variational inference we’ll instead see if we can approximate the posterior using variational methods.
## Variational Approximation of the Posterior
The basic idea behind variational inference is to turn the inference problem into a kind of search problem: find the distribution $q^\ast(z)$ that is the closest approximation of $p(z \given \vect{x})$. To do that we of course need some sort of definition of “closeness”. The classic one is the Kullback-Leibler divergence:
$$$$\KL{q(z)}{p(z \given \vect{x})} = \int q(z) \log \frac{q(z)}{p(z \given \vect{x})} dz$$$$
The Kullback-Leibler divergance is always positive or zero, and it’s only zero if $q(z) = p(z \given \vect{x})$ almost everywhere.
Let’s see what happens if we replace $p(z \given \vect{x})$ in Eq. 2 with the expression derived in Eq. 1:
$$$$\KL{q(z)}{p(z \given \vect{x})} = \int q(z) \log \frac{q(z) p(x)} {p(\vect{x} \given z) p(z)} dz$$$$
Since $\log ab = \log a + \log b$ and $\log \frac{1}{a} = -\log a$ we can rewrite that as
$$$$\KL{q(z)}{p(z \given \vect{x})} = \int q(z) \log \frac{q(z)}{p(z)} dz + \\ \int q(z) \log p(\vect{x}) dz - \int q(z) \log p(\vect{x} \given z) dz.$$$$
The first term on the right hand side is (by definition) the Kullback-Leibler divergence between the prior $p(z)$ and the variational posterior $q(z)$. The second term is just $\log p(\vect{x})$, since $\log p(\vect{x})$ is independent of $z$ and can be brought out of the integral, and because the integral $\int q(z) dz$ is $1$ by definition (of probability density function). The third term can be interpreted as the expectation of $\log p(\vect{x} \given z)$ over $q(z)$. We then get
$$$$\KL{q(z)}{p(z \given \vect{x})} = \KL{q(z)}{p(z)} + \\ \log p(\vect{x}) - \Expect{z \sim q(z)}{\log p(\vect{x} \given z)}.$$$$
Now all we have to do is vary $q(z)$ until $\KL{q(z)}{p(z \given \vect{x})}$ reaches its minimum, and we will have found our best approximation $q^\ast(z)$! Typically that’s done by choosing a parameterized family of probability distributions and then finding the optimal parameters with some sort of numerical optimization algorithm.
### Beta distribution as Variational Posterior
We start by assuming that $q(z)$ is a beta distribution, parameterized by $\alpha_q$ and $\beta_q$.
Figure 2. The beta distribution is a family of continuous probability distributions defined on the interval [0, 1], parametrized by two positive shape parameters $\alpha$ and $\beta$.
We can then derive expressions for each of the terms in Eq. 5. We start with the Kullback-Leibler divergence between the variational posterior $q(z)$ and the prior $p(z)$. Since both are beta distributions there is a closed form expression for their KL divergence
$$$$\KL{q(z)}{p(z)} = \log \frac{B(3, 3)}{B(\alpha_q, \beta_q)} + (\alpha_q - 3) \psi(\alpha_q) + \\ (\beta_q - 3) \psi(\beta_q) + (3 - \alpha_q + 3 - \beta_q) \psi(\alpha_q + \beta_q),$$$$
where $B$ is the beta function and $\psi$ is the digamma function.
Then the expectation of $\log p(\vect{x} \given z)$: First, we formulate $p(\vect{x} \given z)$ assuming that $x_i$ is $1$ if the $i$th coin toss came up heads and $0$ if it came up tails as
$$$$p(\vect{x} \given z) = \prod z^{x_i} (1 - z)^{1-x_i},$$$$
and the logaritm of that is then
$$$$\log p(\vect{x} \given z) = \sum ( x_i \log z + (1-x_i) \log (1 - z) ).$$$$
Let’s call the number of heads in $\vect{x}$ $n_h = \sum x_i$ and the number of tails $n_t = \sum (1 - x_i)$. Now, since $\mathbb{E}$ is a linear operator we can write the third and final term of Eq. 5 as
$$$$\Expect{z \sim q(z)}{\log p(\vect{x} \given z)} = n_h \Expect{z \sim q(z)}{\log z} + n_t \Expect{z \sim q(z)}{\log (1 - z)}.$$$$
Again, we’re lucky because there is a closed form expression for this expectation:
$$$$\Expect{z \sim q(z)}{\log p(\vect{x} \given z)} = n_h (\psi(\alpha_q) - \psi(\alpha_q + \beta_q)) + \\ n_t (\psi(\beta_q) - \psi(\alpha_q + \beta_q))$$$$
Now we’re ready to put it all together as
$$$$\KL{q(z)}{p(z \given \vect{x})} = \log \frac{B(3, 3)}{B(\alpha_q, \beta_q)} + (\alpha_q - 3 - n_h) \psi(\alpha_q) + \\ (\beta_q - 3 - n_t) \psi(\beta_q) + (3 - \alpha_q + 3 - \beta_q + n_h + n_t) \psi(\alpha_q + \beta_q) + \\ \log p(\vect{x}),$$$$
and go look for $\alpha^\ast_q$ and $\beta^\ast_q$ that minimize $\KL{q(z)}{p(z \given \vect{x})}$. In this very simple example there’s probably a closed form solution, but since we’re here to learn about the case when there’s not we’ll go right ahead and minimize the above expression numerically. The video below shows scipy.optimize.minimize() going to work on the problem. Note that $\log p(\vect{x})$ is independent of $\alpha_q$ and $\beta_q$ and can thus be left out of the optimization.
Figure 3. Process of finding our variational posterior $q^\ast(z)$ through numerical optimization of the closed form expression in Eq. 11.
### Numerical Approximation of Difficult Integrals
As you can see from Eq. 5 variational inference is about optimization over quite hairy integrals. One thing you’ll hear a lot in this context is “we approximate the integral through Monte Carlo sampling”. What that means is essentially that we make the following approximation:
$$$$\int p(x) f(x) dx = \Expect{x \sim p(x)}{f(x)} \approx \frac{1}{K} \sum_{i=0}^{K} \left[ f(x_i) \right]_{x_i \sim p(x)}$$$$
Expressed in words what we do is to draw $K$ i.i.d. samples $x_i$ from the probability distribution $p(x)$ and compute the value of $f(x_i)$ for each one. We then take the average of that and call it our Monte Carlo approximation. Simple!
In this case we were quite lucky that there was a closed form expression for $\KL{q(z)}{p(z)}$ (see Eq. 6). But let’s for a moment pretend that there wasn’t. If we then go back to Eq. 5 and apply Monte Carlo approximation we get
$$\begin{equation*} \KL{q(z)}{p(z \given \vect{x})} = \KL{q(z)}{p(z)} + \log p(\vect{x}) - \Expect{z \sim q(z)}{\log p(\vect{x} \given z)} = \\ \Expect{z \sim q(z)}{\log \frac{q(z)}{p(z)}} + \log p(\vect{x}) - \Expect{z \sim q(z)}{\log p(\vect{x} \given z)} \approx \\ \frac{1}{K} \sum_{i=0}^{K} \left[ \log \frac{q(z_i)}{p(z_i)} \right]_{z_i \sim q(z)} + \log p(\vect{x}) - \frac{1}{K} \sum_{i=0}^{K} \left[ \log p(\vect{x} \given z_i) \right]_{z_i \sim q(z)} = \\ \frac{1}{K} \sum_{i=0}^{K} \left[ \log q(z_i) - \log p(z_i) - \log p(\vect{x} \given z_i) \right]_{z_i \sim q(z)} + \log p(\vect{x}), \end{equation*}$$
where $q(z)$ is parameterized by $\alpha_q$ and $\beta_q$ (so we could write it as $q(z; \alpha_q, \beta_q)$ if we wanted to be very verbose in our notation). If we approximate $\KL{q(z)}{p(z \given \vect{x})}$ like this and then let scipy.optimize.minimize() go to work on finding optimal $\alpha^\ast_q$ and $\beta^\ast_q$ we get the video below. As you can see the approximation is no longer perfect, but still pretty good.
Figure 4. Process of finding our variational posterior $q^\ast(z)$ through numerical optimization of a Monte Carlo approximation of $\KL{q(z)}{p(z \given \vect{x})}$.
### Normal Distribution as Variational Posterior
Beta distributed priors and posteriors are a pretty natural choice when we’re talking about coin tosses, but most of the time we’re not. In many cases we have no real reason to choose one family of distributions over another, and then often end up with normal distributions - mostly because they are easy to work with.
There’s however nothing in the variational framework that requires the prior $p(z)$ and the variational posterior $q(z)$ to come from the same family. To drive that point home we’ll switch to a normal distribution for $q(z)$, parameterized with $\mu_q$ and $\sigma_q$, while leaving $p(z)$ as $\text{Beta}(3, 3)$. Eq. 5 still holds and we can still approximate it with Monte Carlo sampling as above:
$$\begin{equation*} \KL{q(z)}{p(z \given \vect{x})} = \KL{q(z)}{p(z)} + \log p(\vect{x}) - \Expect{z \sim q(z)}{\log p(\vect{x} \given z)} \approx \\ \frac{1}{K} \sum_{i=0}^{K} \left[ \log q(z_i) - \log p(z_i) - \log p(\vect{x} \given z_i) \right]_{z_i \sim q(z)} + \log p(\vect{x}) \end{equation*}$$
Here $q(z)$ is really a function of $\mu_q$ and $\sigma_q$ as well (i.e. it could be written $q(z; \mu_q, \sigma_q)$ if we were really verbose in our notation), and we can find their approximately optimal values through (derivative-free) numerical optimization as we have done previously. But let’s step it up a notch and instead bring in numerical optimization workhorse numero uno: stochastic gradient descent.
For SGD to work we need to be able to differentiate $\KL{q(z)}{p(z \given \vect{x})}$ with respect to the unknown parameters $\vect{\theta_q} = (\mu_q, \sigma_q)$:
$$\begin{equation*} \frac{\partial}{\partial \vect{\theta_q}}\KL{q(z)}{p(z \given \vect{x})} = \frac{\partial}{\partial \vect{\theta_q}} \KL{q(z)}{p(z)} + \frac{\partial}{\partial \vect{\theta_q}} \log p(\vect{x}) - \frac{\partial}{\partial \vect{\theta_q}} \Expect{z \sim q(z)}{\log p(\vect{x} \given z)} \end{equation*}$$
Of course, since $\log p(\vect{x})$ does not depend on $\vect{\theta_q}$ its derivative is zero. $\frac{\partial}{\partial \vect{\theta_q}} \KL{q(z)}{p(z)}$ can often be computed analytically, and often even automatically e.g. by theano.tensor.grad(). But in this case where the prior $p(z)$ and the variational posterior $q(z)$ are from different families of distributions we’ll have to resort to approximation, so we use the definition of Kullback-Leibler divergence and merge the first and third term into a single expectation:
$$\begin{equation*} \frac{\partial}{\partial \vect{\theta_q}}\KL{q(z)}{p(z \given \vect{x})} = \frac{\partial}{\partial \vect{\theta_q}} \Expect{z \sim q(z)}{ \log q(z) - \log p(z) - \log p(\vect{x} \given z)} \end{equation*}$$
Now remember that $q(z)$ is a normal distribution with mean $\mu_q$ and standard deviation $\sigma_q$. We can therefore express $z$ as a deterministic function of a gaussian noise variable $\epsilon$: $z = \mu_q + \sigma_q \epsilon$ where $\epsilon \sim \text{N}(0, 1)$. This allows us to take the expectation over $\epsilon \sim \text{N}(0, 1)$ instead:
$$\begin{equation*} \frac{\partial}{\partial \vect{\theta_q}}\KL{q(z)}{p(z \given \vect{x})} = \frac{\partial}{\partial \vect{\theta_q}} \Expect{\epsilon \sim \text{N}(0, 1)}{ \log q(\mu_q + \sigma_q \epsilon) - \\ \log p(\mu_q + \sigma_q \epsilon) - \log p(\vect{x} \given z = \mu_q + \sigma_q \epsilon)} \end{equation*}$$
That perhaps doesn’t look like a step forward, but since we are now taking the expectation over a distribution that does not depend on $\vect{\theta_q}$ we can safely exchange the order of the derivation and the expectation operators. This maneuver is what’s commonly referred to as the “reparametrization trick”.
After reparametrization we can approximate the gradient of the Kullback-Leibler divergence between our variational posterior and the true posterior with respect to the variational parameters $\vect{\theta_q}$ using Monte Carlo sampling:
$$\begin{equation*} \frac{\partial}{\partial \vect{\theta_q}}\KL{q(z)}{p(z \given \vect{x})} = \Expect{\epsilon \sim \text{N}(0, 1)}{ \frac{\partial}{\partial \vect{\theta_q}} \left( \log q(\mu_q + \sigma_q \epsilon) - \\ \log p(\mu_q + \sigma_q \epsilon) - \log p(\vect{x} \given z = \mu_q + \sigma_q \epsilon) \right)} \approx \\ \frac{1}{K} \sum_{i=0}^{K} \left[ \frac{\partial}{\partial \vect{\theta_q}} \left( \log q(\mu_q + \sigma_q \epsilon_i) - \log p(\mu_q + \sigma_q \epsilon_i) - \log p(\vect{x} \given z = \mu_q + \sigma_q \epsilon_i) \right) \right]_{\epsilon_i \sim \text{N}(0, 1)} \end{equation*}$$
The gradient there may look pretty ugly, but computing partial derivatives like that is what frameworks like Theano or TensorFlow do well. You just express your objective function as the average of $K$ samples from $\left[ \log q(\mu_q + \sigma_q \epsilon_i) - \log p(\mu_q + \sigma_q \epsilon_i) - \log p(\vect{x} \given z = \mu_q + \sigma_q \epsilon_i) \right]_{\epsilon_i \sim \text{N}(0, 1)}$ and call theano.tensor.grad() on that. Easy-peasy.
With $K = 10$ samples to approximate the gradient the search for $q^\ast(z)$ progresses as in the video below. As you can see there are two kinds of error in our approximation this time: as usual we’re not finding the exact optimal parameters $\mu^\ast_q$ and $\sigma^\ast_q$, but also - even if we did the variational posterior $q^\ast(z) = \text{N}(\mu^\ast_q, \sigma^\ast_q)$ would not perfectly match the true posterior $p(z \given \vect{x})$.
Figure 5. Process of finding our variational posterior $q^\ast(z) = \text{N}(\mu^\ast_q, \sigma^\ast_q)$ through stochastic gradient descent on $\KL{q(z)}{p(z \given \vect{x})}$.
## Evidence Lower Bound (ELBO)
Up until now we’ve been talking about minimizing $\KL{q(z)}{p(z \given \vect{x})}$, mostly because I feel that makes intuitive sense. But in the literature it’s much more common to talk about maximizing something called the evidence lower bound (ELBO).
Thankfully the difference is minimal. We start from Eq. 5 and derive an expression for (the logarithm of) the evidence $p(\vect{x})$:
$$$$\log p(\vect{x}) = \Expect{z \sim q(z)}{\log p(\vect{x} \given z)} - \KL{q(z)}{p(z)} + \\ \KL{q(z)}{p(z \given \vect{x})}$$$$
Now we observe that $\KL{q(z)}{p(z \given \vect{x})}$ must be positive or zero (because a Kullback-Leibler divergence always is). If we remove this term we thus get a lower bound on $\log p(\vect{x})$:
$$$$\log p(\vect{x}) \geq \Expect{z \sim q(z)}{\log p(\vect{x} \given z)} - \KL{q(z)}{p(z)} = \mathcal{L}(\vect{\theta_q})$$$$
This $\mathcal{L}(\vect{\theta_q})$ is what is often referred to as the evidence lower bound (ELBO). Maximizing $\mathcal{L}(\vect{\theta_q})$ will give us the same variational posterior as minimizing $\KL{q(z)}{p(z \given \vect{x})}$.
But phrasing variational inference as maximization of the ELBO admittedly has at least one conceptual advantage: it shows how close we are to maximum likelihood estimation. It seems inference in the Bayesian regime is a balancing act between best explaining the data and “keeping it simple”, by staying close to the prior. If we strike the second requirement our posteriors collapse onto the maximum likelihood estimate (MLE) and we are back in the Frequentist regime.
## Where to Next?
If you feel comfortable with the basics above I highly recommend Kingma and Welling’s classic paper “Autoencoding Variational Bayes”.
Any and all feedback is greatly appreciated. Ping me on Twitter or, if it’s more specific and/or you need more than 130 characters, post an issue on GitHub. |
proofpile-shard-0030-46 | {
"provenance": "003.jsonl.gz:47"
} | # Edward Sang's arithmetic texts
Edward Sang wrote Elementary Arithmetic (1856) and Higher Arithmetic (1857). These texts are particularly interesting given Sang's later work on constructing logarithm tables. We give below the Prefaces to these books.
1. Preface to 'Elementary Arithmetic' (1856)
The following Treatise on Elementary Arithmetic has been designed as the first of a continuous series of Treatises on those sciences which are usually comprehended under the somewhat indistinct name MATHEMATICS.
During a long experience as preceptor in the higher departments of mathematics, the Author of this work has observed that almost all the difficulties which the student encounters are traceable to an imperfect acquaintance with arithmetic. It seems as if this subject were never regarded as having in it anything intellectual. Arithmetic is considered as a kind of legerdemain, a talsamic contrivance, by means of which results are to be obtained in some occult manner, into the nature of which the student is forbidden to inquire; hence many a one, as he advances in life, finds himself compelled to resume the study of the principles of arithmetic, and discovers that all along he has been working in the dark. Now, truly, there is hardly any branch of human knowledge which affords more scope for intellectual effort, or presents a more invigorating field for mental exercise than the science of number. It has therefore been the Author's aim to prepare a text-book which should call mind, not memory, into exercise, and from which all mere dogmatism should be scrupulously excluded.
In arranging it, he has endeavoured to explain the reasons of the various operations by an examination of the nature of the questions which give rise to them, and in this manner has sought to prepare the student for understanding their applications to the higher branches of science. The gradual formation of systematic numeration has been traced, and the operations of palpable arithmetic have been taught in order to reflect a clearer light on the true nature of our figurate processes. The few lines that are devoted to the explanation of the ancient Greek and Arabic Notations may not be unacceptable to those who study the history and phenomena of the human mind. We are accustomed to designate the ordinary numerals as Arabic. The Arabs themselves call them Rakam Hindi in contradistinction to those described at page 23, which they call Rakam Arabi.
It is hoped that, independently of the arrangement of its parts, this treatise contains enough of new and original matter to prevent it from being regarded as an uncalled-for addition to a class of books already sufficiently numerous. The method of computing from the left hand, which has been practised and taught by the author for thirty years, is now published for the first time. Besides the merit of novelty, this method has the higher merit of great usefulness. A very slight acquaintance with it augments one's power over numbers in an unexpected degree, and the continued practice of it renders computation a pastime. In the ordinary mode of computing we have never occasion to add more than nine to nine, or to take the product of more than nine times nine, and hence the limit of rapidity is soon reached. But when we begin to work from the left hand, every operation adds to our previous experience, and we soon become familiar with large numbers, so much so that the rapidity of mental calculation comes far to exceed the swiftness of the pen.
The multiplication of one large number by another by help of a movable slip of paper, though not to be recommended in actual business, is interesting, and the ability to perform it enables us to follow the operations for shortening the multiplication and division of long decimals.
The subject of prime and composite numbers has been introduced as preparatory to the theory of fractions; and the doctrine of proportion has been reached by means of the method of continued fractions invented by Lord Brouncker. The immense power of this method, the almost unlimited range of its applications, as well as the beautiful simplicity of the idea from which it arose, recommend it to the close attention of every calculator. The doctrine of continued fractions is here divested of its technicalities, and presented in such a form as to be intelligible to beginners; the formation of the successive approximating fractions being deduced without the aid of artificial symbols, perhaps more clearly than with that aid.
Among the minor improvements which have been introduced, the mode of obtaining at once the continued product of several numbers, the plan for shortening division by a number of two or three places, and the arrangement of the work for finding the greatest common divisor of two numbers, may be mentioned; Fourier's Division Ordonnée is also new to the English reader.
It was a matter of anxious deliberation whether the answers to the questions should be printed, and ultimately it was resolved to give these answers in a separate key. The Author, however, most earnestly entreats the student to trust to his own thorough comprehension of the matter, and to the care with which he works. Let him carry the feeling with him, that if his result differ from that given in the key, the key is likely to be wrong: above all things, cultivate self-reliance. In very many cases the manner in which the result is worked out is of more importance than the mere obtaining of it; and in some cases, for obvious reasons, the answers ought not to be given at all.
The Author is in hopes of speedily laying before the public, in continuation, a treatise on the higher arithmetic, in which the doctrines of Powers, Roots, and Logarithms, are completely investigated.
Edinburgh, June 1856.
2. Preface to 'Higher Arithmetic' (1857)
In this Second Volume on Arithmetic an account is given of the doctrines of Powers, Roots, and Logarithms, so far as that can be well done without the aid of general symbols. The Treatise is intended not merely as a Text-Book on these subjects, but also as an introduction to Algebra: indeed, if we adopt the original meaning of the Arab words (ylim ul jibr, the science of powers), the present work forms the first, and not the least important chapter of that science.
To those who have only considered the subjects of direct, inverse, and fractional powers, and the cognate subject of Logarithms, in the light which the modern notation throws upon them, it may seem vain to attempt to explain these matters with no aid beyond that of our ordinary numeral notation; but an examination of the following pages may serve to show that the mind does not require the aid of artificial symbols to detect and appreciate even recondite properties of numbers; and the Author flatters himself that he has brought the leading properties of Logarithms completely within the bounds of arithmetic.
This has been accomplished by the help of a new method for extracting all roots, of which the previously well-known processes for extracting the square and cube roots are the two simplest cases. This method was given, by implication, in a small treatise "On the Solution of Algebraic Equations of all Orders, Edinburgh, 1829;" it is here simplified and adapted to ordinary arithmetic. By its means we obtain the root, and all the inferior powers of the root, with great rapidity; the simplicity of the arrangement being the better seen, the higher the order of the root which we extract.
In the actual construction of the first Decimal Logarithmic Tables, Briggs used the repeated extraction of the square root, until the results exceeded unit by fractions so small as to render the excesses sensibly proportional to the exponents. Had he known the method of extracting fifth roots, his labour would have been greatly lessened. The principle used by Briggs is, in essence, identic with that adopted by Dodson inl the construction of his Anti-Logarithmic Canon, and with that which is followed at page 119; the only difference is, that the ability to extract fifth roots has given us a much greater command of the subject than either Dodson or Briggs possessed.
The direct computation of the logarithm of a number, that is, in the language of modern algebra, the direct solution of the equation $a_{z} = n$, has not heretofore been obtained; for although the well-known formula
$z=\Large\frac{(n-1)-\frac 1 2(n-1)^2 +\frac 1 3(n-1)^3-\frac 1 4(n-1)^4+ \text{etc.}}{(a-1)-\frac 1 2(a-1)^2 +\frac 1 3(a-1)^3-\frac 1 4(a-1)^4+ \text{etc.}}$
be a symbolical solution, it is only susceptible of direct application when $a$ and $n$ differ from unit by small fractions. In common logarithms $a - 1$ has the value 9, and $z$ has to be computed indirectly through the intervention of other numbers.
The student of the Higher Algebra will, therefore, be somewhat surprised to find an exceedingly simple and rapid solution, obtained by a train of reasoning which requires only a clear perception of the nature of powers, and which is altogether independent of notation.
This is another to be added to the rapidly accumulating testimonies of the usefulness of Lord Brouncker's continued fractions; for although the algorithm and definition of these fractions have not been employed, the essential idea has been freely used.
These two new processes, viz. the extraction of all roots, and the direct solution of the exponential equation, have enabled the Author to place the whole subject in a clear light, and to complete the Theory of Practical Arithmetic without calling in the dangerous aids of indefinite symbols and arbitrary notation.
In order to prepare the student for following the reasoning to be afterwards used in algebraic investigations, and also for the purpose of fortifying his knowledge of what has already been gone over, a short notice has been added of various Numeration scales. The study of this part of the work may serve to free the mind from those prejudices which are apt to attend the use of a single system, and may lead it to form just and comprehensive views of arithmetic in general.
Edinburgh, March 1857.
Last Updated January 2021 |
proofpile-shard-0030-47 | {
"provenance": "003.jsonl.gz:48"
} | # Math Help - Trig periods
1. ## Trig periods
How do i figure out which period the function is in?
Im doing the problem 3 tan2 x-1=0
i got tan x= 1sqrt 3 i have to find a solution to x. How do i know tan is on the interval 0, pi and not 0, 2pi?
2. ## Re: Trig periods
The period of a periodic function $f(x)$ is the smallest positive real number $\alpha$ such that $f(x+\alpha)=f(x)$ for all $x$ in the domain of the function. In case of $\tan,$ $\tan(x+\pi)=\tan x$ for all real $x$ and no positive real number $\alpha$ smaller than $\pi$ satisfies $\tan(x+\alpha)=\tan x$ for all $x.$ (It's true that $\tan(x+2\pi)=\tan x$ for all $x$ but $2\pi$ is not the smallest such positive real number.)
3. ## Re: Trig periods
Originally Posted by noork85
How do i figure out which period the function is in?
Im doing the problem 3 tan2 x-1=0
i got tan x= 1sqrt 3 i have to find a solution to x. How do i know tan is on the interval 0, pi and not 0, 2pi?
$3\tan^2{x} - 1 = 0$
$\tan{x} = \pm \frac{1}{\sqrt{3}}$
since the solution interval is not given, then all possible solutions are ...
$x = \frac{\pi}{6} + k\pi ; k \in \mathbb{Z}$
$x = -\frac{\pi}{6} + k\pi ; k \in \mathbb{Z}$ |
proofpile-shard-0030-48 | {
"provenance": "003.jsonl.gz:49"
} | # Natural Resource Depletion and Depreciation of Related Plant Assets E10A. Mertz Company purchased...
###### Natural Resource Depletion and Depreciation of Related Plant Assets
E10A. Mertz Company purchased land containing an estimated 5 million tons of ore for a cost of $8,800,000. The land without the ore is estimated to be worth$500,000. During its first year of operation, the company mined and sold 750,000 tons of ore. Compute the depletion charge per ton. Compute the depletion expense that Mertz should record for the year.
## Related Questions in Managerial Accounting - Others
• ### Acct
(Solved) November 23, 2013
• ### Cornerstone Exercise 8-19 Economic Order Quantity Refer to the data for La Cucina Company above....
(Solved) November 13, 2015
is the total annual ordering cost of lava stone for a year under the EOQ policy? 4. What is the total annual carrying cost of lava stone per year under the EOQ policy? 5 . What is the total annual inventory-related cost for lava stone under the EOQ?
EOQ = Economic Order Quantity A =Annual Requirement O = Ordering Cost per order C = Carrying Cost per pound Requirement 1: EOQ = [2AO/C]1/2 =[2*8400 * 3 / 2]1/2 =158.745 EOQ = 159 pounds...
• ### Long Haul Transport is considering replacing an existing semi-trailer truck with a more modern...
October 15, 2018
in value of $10,000. The existing truck has 5 years of usable life remaining and can currently be sold for$62,000 net. The new long haul truck being considered will cost $210,000 plus$10,000 delivery from an interstate dealer. It will be depreciated over its useful life of 5 years having a... |
proofpile-shard-0030-49 | {
"provenance": "003.jsonl.gz:50"
} | # American Institute of Mathematical Sciences
January 1997, 3(1): 91-106. doi: 10.3934/dcds.1997.3.91
## On the global solvability of symmetric hyperbolic systems of Kirchhoff type
1 Dipartimento di Costruzioni, Istituto Universitario di Architettura, Tolentini, S. Croce 191 - 30135 Venezia, Italy
Received April 1996 Revised July 1996 Published October 1996
We shall prove here the global solvability for small initial data for symmetric hyperbolic systems with integro-differential coefficients. In this way, we will extend some results obtained in [5], [6], [8], [11] for the classic Kirchhoff equation and in [3] for regularly hyperbolic systems.
Citation: Renato Manfrin. On the global solvability of symmetric hyperbolic systems of Kirchhoff type. Discrete & Continuous Dynamical Systems, 1997, 3 (1) : 91-106. doi: 10.3934/dcds.1997.3.91
[1] Manil T. Mohan, Sivaguru S. Sritharan. New methods for local solvability of quasilinear symmetric hyperbolic systems. Evolution Equations & Control Theory, 2016, 5 (2) : 273-302. doi: 10.3934/eect.2016005 [2] Zhiqing Liu, Zhong Bo Fang. Global solvability and general decay of a transmission problem for kirchhoff-type wave equations with nonlinear damping and delay term. Communications on Pure & Applied Analysis, 2020, 19 (2) : 941-966. doi: 10.3934/cpaa.2020043 [3] Cyril Joel Batkam, João R. Santos Júnior. Schrödinger-Kirchhoff-Poisson type systems. Communications on Pure & Applied Analysis, 2016, 15 (2) : 429-444. doi: 10.3934/cpaa.2016.15.429 [4] Wenguo Shen. Unilateral global interval bifurcation for Kirchhoff type problems and its applications. Communications on Pure & Applied Analysis, 2018, 17 (1) : 21-37. doi: 10.3934/cpaa.2018002 [5] Marina Ghisi, Massimo Gobbino. Hyperbolic--parabolic singular perturbation for mildly degenerate Kirchhoff equations: Global-in-time error estimates. Communications on Pure & Applied Analysis, 2009, 8 (4) : 1313-1332. doi: 10.3934/cpaa.2009.8.1313 [6] Irena Pawłow, Wojciech M. Zajączkowski. The global solvability of a sixth order Cahn-Hilliard type equation via the Bäcklund transformation. Communications on Pure & Applied Analysis, 2014, 13 (2) : 859-880. doi: 10.3934/cpaa.2014.13.859 [7] Ken Shirakawa. Solvability for phase field systems of Penrose-Fife type associated with $p$-laplacian diffusions. Conference Publications, 2007, 2007 (Special) : 927-937. doi: 10.3934/proc.2007.2007.927 [8] Zheng Han, Daoyuan Fang. Almost global existence for the Klein-Gordon equation with the Kirchhoff-type nonlinearity. Communications on Pure & Applied Analysis, 2021, 20 (2) : 737-754. doi: 10.3934/cpaa.2020287 [9] Yongqiang Fu, Xiaoju Zhang. Global existence and asymptotic behavior of weak solutions for time-space fractional Kirchhoff-type diffusion equations. Discrete & Continuous Dynamical Systems - B, 2021 doi: 10.3934/dcdsb.2021091 [10] Maoding Zhen, Binlin Zhang, Xiumei Han. A new approach to get solutions for Kirchhoff-type fractional Schrödinger systems involving critical exponents. Discrete & Continuous Dynamical Systems - B, 2021 doi: 10.3934/dcdsb.2021115 [11] Hugo Beirão da Veiga, Francesca Crispo. On the global regularity for nonlinear systems of the $p$-Laplacian type. Discrete & Continuous Dynamical Systems - S, 2013, 6 (5) : 1173-1191. doi: 10.3934/dcdss.2013.6.1173 [12] Michael Ruzhansky, Jens Wirth. Dispersive type estimates for fourier integrals and applications to hyperbolic systems. Conference Publications, 2011, 2011 (Special) : 1263-1270. doi: 10.3934/proc.2011.2011.1263 [13] Tohru Nakamura, Shinya Nishibata, Naoto Usami. Convergence rate of solutions towards the stationary solutions to symmetric hyperbolic-parabolic systems in half space. Kinetic & Related Models, 2018, 11 (4) : 757-793. doi: 10.3934/krm.2018031 [14] Jijiang Sun, Chun-Lei Tang. Resonance problems for Kirchhoff type equations. Discrete & Continuous Dynamical Systems, 2013, 33 (5) : 2139-2154. doi: 10.3934/dcds.2013.33.2139 [15] Yijing Sun, Yuxin Tan. Kirchhoff type equations with strong singularities. Communications on Pure & Applied Analysis, 2019, 18 (1) : 181-193. doi: 10.3934/cpaa.2019010 [16] Moritz Kassmann, Tadele Mengesha, James Scott. Solvability of nonlocal systems related to peridynamics. Communications on Pure & Applied Analysis, 2019, 18 (3) : 1303-1332. doi: 10.3934/cpaa.2019063 [17] Tatsien Li (Daqian Li). Global exact boundary controllability for first order quasilinear hyperbolic systems. Discrete & Continuous Dynamical Systems - B, 2010, 14 (4) : 1419-1432. doi: 10.3934/dcdsb.2010.14.1419 [18] Fuqin Sun, Mingxin Wang. Non-existence of global solutions for nonlinear strongly damped hyperbolic systems. Discrete & Continuous Dynamical Systems, 2005, 12 (5) : 949-958. doi: 10.3934/dcds.2005.12.949 [19] Tatsien Li, Libin Wang. Global exact shock reconstruction for quasilinear hyperbolic systems of conservation laws. Discrete & Continuous Dynamical Systems, 2006, 15 (2) : 597-609. doi: 10.3934/dcds.2006.15.597 [20] Norimichi Hirano, Wieslaw Krawcewicz, Haibo Ruan. Existence of nonstationary periodic solutions for $\Gamma$-symmetric Lotka-Volterra type systems. Discrete & Continuous Dynamical Systems, 2011, 30 (3) : 709-735. doi: 10.3934/dcds.2011.30.709
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proofpile-shard-0030-50 | {
"provenance": "003.jsonl.gz:51"
} | # How do you differentiate f(x)= (xe^x+4)^3 using the chain rule.?
May 27, 2018
= $3 {e}^{x} \left(x + 1\right) {\left(x {e}^{x} + 4\right)}^{2}$ or some version of that
#### Explanation:
This is the chain rule.
${\left(x {e}^{x} + 4\right)}^{3}$
dy/dx = $3 {\left(x {e}^{x} + 4\right)}^{2}$ x (derivative of the inside)
= $3 {\left(x {e}^{x} + 4\right)}^{2}$ x ($x$ x ${e}^{x}$ + ${e}^{x}$)
= $3 {\left(x {e}^{x} + 4\right)}^{2}$ x ($x {e}^{x}$ +${e}^{x}$)
= $3 {e}^{x} \left(x + 1\right) {\left(x {e}^{x} + 4\right)}^{2}$ |
proofpile-shard-0030-51 | {
"provenance": "003.jsonl.gz:52"
} | # The Large Time-Frequency Analysis Toolbox
- All your frame are belong to us -
You can have a look at the Sourceforge download page to see all available versions, or just download the latest one by clicking on the button.
## Installation
To install, simply unpack the package. The toolbox is contained in the 'ltfat' directory and in all the subdirectories. To use the toolbox, start Octave/Matlab, change to the 'ltfat' directory and run the
ltfatstart
command. This will set up the necessary paths and perform the necessary initializations.
If you have downloaded a binary package for Windows 7 or MacOS, compiled mex files for 64 bit Matlab are included in the package. Otherwise, you can compile the mex interfaces yourself.
To compile the mex/oct interfaces for faster execution of the toolbox, type the command:
ltfatmex
Further installation instructions can be found in the files INSTALL-Matlab and INSTALL-Octave. The currently supported platforms are:
• Octave 3.4 and higher on Linux. Tested on 32/64 bit Intel/AMD platforms.
• Octave 3.4 and higher on Windows XP, Vista, Windows 7 (Windows 8 has not been tested).
• Octave 3.4 and higher on Mac OS X. Has not been tested, but should work out of the box.
• Matlab 2009b and later on 32/64 bit Windows (mexext=mexw32 and mexext=mexw64).
• Matlab 2009b and later on 32/64 bit Linux (mexext=mexglx and mexext=mexa64).
• Matlab 2009b and later on Mac OS X (mexext=mexmaci and mexext=mexmaci64). |
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