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Evaluate this power series Evaluate the sum $$x+\frac{2}{3}x^3+\frac{2}{3}\cdot\frac{4}{5}x^5+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}x^7+\dots$$ Totally no idea. I think this series may related to the $\sin x$ series because of those missing even powers. Another way of writing this series: $$\sum_{k=0}^{\infty}\frac{(2k)!!}{(2k+1)!!}x^{2k+1}.$$

In this answer, I mention this identity, which can be proven by repeated integration by parts: $$ \int_0^{\pi/2}\sin^{2k+1}(x)\;\mathrm{d}x=\frac{2k}{2k+1}\frac{2k-2}{2k-1}\cdots\frac{2}{3}=\frac{1}{2k+1}\frac{4^k}{\binom{2k}{k}}\tag{1} $$ Your sum can be rewritten as $$ f(x)=\sum_{k=0}^\infty\frac1{(2k+1)}\frac{4^k}{\binom{2k}{k}}x^{2k+1}\tag{2} $$ Combining $(1)$ and $(2)$, we get $$ \begin{align} f(x) &=\int_0^{\pi/2}\sum_{k=0}^\infty\sin^{2k+1}(t)x^{2k+1}\,\mathrm{d}t\\ &=\int_0^{\pi/2}\frac{x\sin(t)\,\mathrm{d}t}{1-x^2\sin^2(t)}\\ &=\int_0^{\pi/2}\frac{-\,\mathrm{d}x\cos(t)}{1-x^2+x^2\cos^2(t)}\\ &=-\frac1{\sqrt{1-x^2}}\left.\tan^{-1}\left(\frac{x\cos(t)}{\sqrt{1-x^2}}\right)\right]_0^{\pi/2}\\ &=\frac1{\sqrt{1-x^2}}\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\\ &=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}\tag{3} \end{align} $$ Radius of Convergence This doesn't appear to be part of the question, but since some other answers have touched on it, I might as well add something regarding it. A corollary of Cauchy's Integral Formula is that the radius of convergence of a complex analytic function is the distance from the center of the power series expansion to the nearest singularity. The nearest singularity of $f(z)$ to $z=0$ is $z=1$. Thus, the radius of convergence is $1$.

Maximum N that will hold this true Find the largest positive integer $N$ such that $$\sqrt{64 + 32^{403} + 4^{N+3}}$$ is an integer Is $N = 1003$?

Note that with $N=2008$ we have $ (2^{N+3}+8)^2=4^{N+3}+2\cdot 8\cdot 2^{N+3}+64=4^{N+3}+2^{2015}+64=64+32^{403}+4^N,$ so we conjecture that the maximal value is $2008$. If $2^{2015}+2^6+2^{2N+6}$ is a perfect square then also $\frac1{64}$ of it, i.e. $2^{2009}+1+2^{2N}=m^2$ for some $m\in\mathbb N$. But if $N> 2008$, we have $$(2^N)^2=2^{2N}<2^{2009}+1+2^{2N}<1+2^{N+1}+2^{2N}=(2^N+1)^2$$ so that $2^N<m<2^N+1$, which is absurd.

absolute value inequalities When answer this kind of inequality $|2x^2-5x+2| < |x+1|$ I am testing the four combinations when both side are +, one is + and the other is - and the opposite and when they are both -. When I check the negative options, I need to flip the inequality sign? Thanks

Another way - no tricks, just systematically looking at all cases, where we can write the inequality without the absolute value sign, to convince you that all possibilities are covered.. Following the definition of the absolute value function, RHS is easy to rewrite as, $$|x+1| = \begin{cases} x+1 & x \ge -1\\ -x-1 &x < -1 \end{cases}$$ Noting $2x^2-5x+2 = (2x-1)(x-2)$, we have the LHS as $$|2x^2-5x+2| = \begin{cases} 2x^2-5x+2 & x \le \frac12 \text{ or } x \ge 2\\ -2x^2+5x-2 &\frac12 < x < 2 \end{cases}$$ Thus we can break the number line into for regions (using break points at $x = -1, \frac12, 2$) and write the inequality in each region without any absolute value symbols. Then we solve the inequality in each region, and combine results in the end. Details follow. Region 1: $x < -1$ Here the inequality is $2x^2-5x+2 < -x - 1$ which is equivalent to $2x^2 - 4x + 3 < 0$ or $2(x-1)^2+1 < 0$, which is not possible. Region 2: $-1 \le x \le \frac12$ Here we have the inequality as $2x^2-5x+2 < x + 1 \iff 2x^2 - 6x + 1 < 0$ which is true when $\frac12(3-\sqrt7) < x < \frac12(3+\sqrt7)$. Taking the part of this solution which is in the region being considered, we have solutions for $x \in (\frac12(3-\sqrt7), \frac12]$ Region 3: $\frac12 < x < 2$ Here the inequality is $-2x^2+5x-2 < x + 1 \iff 2(x- 1)^2 + 2 > 0$ which holds true for all $x$. So the entire region is a solution. Region 4: $2 \le x$ Here the inequality is $2x^2-5x+2 < x + 1$, which is exactly the same form as Region 2, and has the same solutions. Hence $2 \le x < \frac12(3+\sqrt7)$ is the solution from this Region. Combining solutions from all regions, we have $\frac12(3-\sqrt7) < x < \frac12(3+\sqrt7)$ or $|x - \frac32| < \frac{\sqrt7}{2}$ as the complete solution set (which others have already pointed out).

Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$ Question : Is the following true for any $m\in\mathbb N$? $$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$ Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it. Can anyone help? By the way, I've been able to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$ by using $(\star)$. Proof : Let $$f(x)=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{(x-\sin x)(x+\sin x)}{x^2\sin^2 x}.$$ We know that $f(x)\gt0$ if $0\lt x\le {\pi}/{2}$, and that $\lim_{x\to 0}f(x)=1/3$. Hence, letting $f(0)=1/3$, we know that $f(x)$ is continuous and positive at $x=0$. Hence, since $f(x)\ (0\le x\le {\pi}/2)$ is bounded, there exists a constant $C$ such that $0\lt f(x)\lt C$. Hence, substituting $x={(k\pi)}/{(2n+1)}$ for this, we get $$0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.$$ Then, the sum of these from $1$ to $n$ satisfies $$0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.$$ Here, we used $(\star)$. Then, considering $n\to\infty$ leads what we desired.

(First time I write in a math blog, so forgive me if my contribution ends up being useless) I recently bumped into this same identity while working on Fourier transforms. By these means you can show in fact that $$\sum_{k=-\infty}^{+\infty}\frac{1}{(x-k)^2}=\frac{\pi^2}{\sin^2 (\pi x)}.\tag{*}\label{*}$$ Letting $x=\frac13$ yields $$\begin{align}\sum_{k=-\infty}^{+\infty}\frac{1}{(3k-1)^2}&= \sum_{k=0}^{+\infty}\left[\frac{1}{(3k+1)^2}+ \frac{1}{(3k+2)^2} \right]=\\ &=\sum_{k=1}^{+\infty} \frac{1}{k^2} - \sum_{k=1}^{+\infty}\frac{1}{(3k)^2}=\\ &=\frac{8}{9}\sum_{k=1}^{+\infty} \frac{1}{k^2}\\ &\stackrel{\eqref{*}}= \frac{4\pi^2}{27}\end{align},$$ which in the end gives $$\sum_{k=1}^{+\infty} \frac{1}{k^2} = \frac{\pi^2}{6}.\tag{**}\label{**}$$ Now, back to \eqref{*}, we can generalize and take $x=m/n$, with $n$ and $m$ positive integers, obtaining $$\sum_{k=0}^{+\infty}\left[\frac{1}{(nk+m)^2} + \frac{1}{(nk+n-m)^2}\right]=\frac{\pi^2}{n^2 \sin^2\left(\frac{\pi m}{n}\right)}.$$ Finally taking the sum of LHS and RHS of last equation for $m=1,2,\dots,n-1$ yields $$\begin{align}\sum_{m=1}^{n-1}\sum_{k=0}^{+\infty}\left[\frac{1}{(nk+m)^2} + \frac{1}{(nk+n-m)^2}\right]&=\frac{\pi^2}{n^2} \sum_{m=1}^{n-1}\frac{1}{\sin^2\left(\frac{\pi m}{n}\right)}\\ \sum_{k=0}^{+\infty}\sum_{m=1}^{n-1}\left[\frac{1}{(nk+m)^2} + \frac{1}{(nk+n-m)^2}\right]&=\frac{\pi^2}{n^2} \sum_{m=1}^{n-1}\frac{1}{\sin^2\left(\frac{\pi m}{n}\right)}\\ 2\sum_{k=1}^{+\infty}\frac{1}{k^2} - 2\sum_{k=1}^{+\infty}\frac{1}{(nk)^2} &=\frac{\pi^2}{n^2} \sum_{m=1}^{n-1}\frac{1}{\sin^2\left(\frac{\pi m}{n}\right)}\\ \frac{2(n^2-1)}{n^2}\sum_{k=1}^{+\infty}\frac{1}{k^2} &=\frac{\pi^2}{n^2} \sum_{m=1}^{n-1}\frac{1}{\sin^2\left(\frac{\pi m}{n}\right)}. \end{align}$$ Using then \eqref{**} leads to the result $$\sum_{m=1}^{n-1} \frac{1}{\sin^2\left(\frac{\pi m}{n}\right)} = \frac{n^2-1}{3}.$$

Three Variables-Inequality with $a+b+c=abc$ $a$,$b$,$c$ are positive numbers such that $~a+b+c=abc$ Find the maximum value of $~\dfrac{1}{\sqrt{1+a^{2}}}+\dfrac{1}{\sqrt{1+b^{2}}}+\dfrac{1}{\sqrt{1+c^{2}}}$

Trigonometric substitution looks good for this, especially if you know sum of cosines of angles in a triangle are $\le \frac32$. However if you want an alternate way... Let $a = \frac1x, b = \frac1y, c = \frac1z$. Then we need to find the maximum of $$F = \sum \frac{x}{\sqrt{x^2+1}}$$ with the condition now as $xy + yz + zx = 1$. Using this condition to homogenise, we have $$\begin{align} F &= \sum \frac{x}{\sqrt{x^2 + xy + yz + zx}} \\ &= \sum \frac{x}{\sqrt{(x+y)(x+z)}} \\ &= \sum \frac{x\sqrt{(x+y)(x+z)}}{(x+y)(y+z)} \\ &\le \sum \frac{x\left((x+y)+(x+z)\right)}{2\cdot(x+y)(y+z)} \quad \text{by AM-GM}\\ &= \frac12 \sum \left(\frac{x}{x+y}+\frac{x}{x+z} \right) \\ &= \frac12 \sum \left(\frac{x}{x+y}+\frac{y}{y+x} \right) \quad \text{rearranging terms across sums}\\ &= \frac32 \end{align}$$

evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n}$ I am trying to compute the sum $$\sum_{n=1}^\infty \frac{n^2}{3^n}.$$ I would prefer a nice method without differentiation, but if differentiation makes it easier, then that's fine. Can anyone help me? Thanks.

We have \begin{gather*} S=\sum_{n=1}^{\infty}\frac{n^2}{3^n}=\frac{1}{3}\sum_{n=0}^{\infty}\frac{(n+1)^2}{3^{n-1}}=\frac{1}{3}\left(\sum_{n=0}^{\infty}\frac{n^2}{3^{n-1}}+2\sum_{n=0}^{\infty}\frac{n}{3^{n-1}}+\sum_{n=0}^{\infty}\frac{1}{3^{n-1}}\right)=\frac{1}{3}\left(S+2 S_1+S_2\right). \end{gather*} Then $$ S_2=\sum_{n=0}^{\infty}\frac{1}{3^{n-1}}=\frac{3}{2}, $$ $$S_1=\sum_{n=0}^\infty \frac{n}{3^n}=\sum_{n=1}^\infty \frac{n}{3^n}= \frac{1}{3}\sum_{n=1}^\infty \frac{n}{3^{n-1}}= \frac{1}{3}\sum_{n=0}^\infty \frac{n+1}{3^{n}}= \frac{1}{3}\left[\sum_{n=0}^\infty \frac{n}{3^{n}}+\sum_{n=0}^\infty \frac{1}{3^{n}}\right] =\frac{1}{3}\left[S_1+ S_2\right].$$ By solving this equation you get $S_1=\dfrac{3}{4}.$ Thus we have the equation $$ S=\frac{1}{3}(S+2 \cdot \frac{3}{4}+\frac{3}{2}). $$ By solving it you will get $S=\dfrac{3}{2}.$

How prove this inequality $\frac{x}{x^3+y^2+z}+\frac{y}{y^3+z^2+x}+\frac{z}{z^3+x^2+y}\le 1$ for $x+y+z=3$ let $x,y,z$ be positive numbers,and such $x+y+z=3$,show that $$\dfrac{x}{x^3+y^2+z}+\dfrac{y}{y^3+z^2+x}+\dfrac{z}{z^3+x^2+y}\le 1$$ My try:$$(x^3+y^2+z)(\dfrac{1}{x}+1+z)\ge 9$$ so $$\dfrac{x}{x^3+y^2+z}+\dfrac{y}{y^3+z^2+x}+\dfrac{z}{z^3+x^2+y}\le\dfrac{6+xy+yz+xz}{9}\le 1$$ Have other nice methods? Thank you

Note that $2+x^3=x^3+1+1\geqslant 3x$, $y^2+1\geqslant 2y$, thus$$\dfrac{x}{x^3+y^2+z}=\frac{x}{3+x^3+y^2-x-y}\leqslant\frac{x}{3x+2y-x-y}=\frac{x}{2x+y}.$$Similarly, we can get $$\dfrac{y}{y^3+z^2+x}\leqslant\frac{y}{2y+z},\,\,\,\,\,\dfrac{z}{z^3+x^2+y}\leqslant\frac{z}{2z+x}.$$It suffices to show$$\frac{x}{2x+y}+\frac{y}{2y+z}+\frac{z}{2z+x}\leqslant 1\iff \frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}\geqslant 1.$$By Cauchy's inequality, we get$$(y(2x+y)+z(2y+z)+x(2z+x))\left(\frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}\right)\geqslant (x+y+z)^2.$$As $y(2x+y)+z(2y+z)+x(2z+x)=(x+y+z)^2$, so $$\frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}\geqslant 1.$$

Probability task (Find probability that the chosen ball is white.) I have this task in my book: First box contains $10$ balls, from which $8$ are white. Second box contains $20$ from which $4$ are white. From each box one ball is chosen. Then from previously chosen two balls, one is chosen. Find probability that the chosen ball is white. The answer is $0.5$. Again I get the different answer: There are four possible outcomes when two balls are chosen: $w$ -white, $a$ - for another color $(a,a),(w,a),(a,w),(w,w)$. Each outcome has probability: $\frac{2}{10} \cdot \frac{16}{20}; \frac{8}{10} \cdot \frac{16}{20}; \frac{2}{10} \cdot \frac{4}{20}; \frac{8}{10} \cdot \frac{4}{20};$ In my opinion the probability that the one ball chosen at the end is white is equal to the sum of last three probabilities $\frac{8}{10} \cdot \frac{16}{20} + \frac{2}{10} \cdot \frac{4}{20} + \frac{8}{10} \cdot \frac{4}{20}=\frac{21}{25}$. Am I wrong or there is a mistake in the answer in the book?

In case of $(w,a)$ or $(a,w)$ you need to consider that one of these two balls is chosen (randomly as by coin tossing, we should assume). Therefore these cases have to be weighted by a factor of $\frac 12$.

Prove that $ \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$. Prove that $\displaystyle \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$. $\bf{My\; Try}::$ Let $x = \sin \theta$, Then $dx = \cos \theta d\theta$ $\displaystyle = \int _{0}^{1}\frac{\cos \theta }{1-\sin^2 \theta \cdot \sin^2 \alpha}\cdot \cos \theta d\theta = \int_{0}^{1}\frac{\cos ^2 \theta }{1-\sin^2 \theta \cdot \sin ^2 \alpha}d\theta$ $\displaystyle = \int_{0}^{1}\frac{\sec^2 \theta}{\sec^4 \theta -\tan^2 \theta \cdot \sec^2 \theta \cdot \sin^2 \alpha}d\theta = \int_{0}^{1}\frac{\sec^2 \theta }{\left(1+\tan ^2 \theta\right)^2-\tan^2 \theta \cdot \sec^2 \theta\cdot \sin^2 \alpha}d\theta$ Let $\tan \theta = t$ and $\sec^2 \theta d\theta = dt$ $\displaystyle \int_{0}^{1}\frac{1}{(1+t^2)^2-t^2 (1+t^2)\sin^2 \alpha}dt$ Now How can I solve after that Help Required Thanks

Substitute $x=\frac{t}{\sqrt{\cos a+t^2}}$. Then $dx= \frac{\cos a\ dt}{(\cos a+t^2)^{3/2}}$ and \begin{align}\int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 a}dx =& \int_{0}^{\infty}\frac{\sqrt{\cos a}\ \frac1{t^2} }{(t^2+ \frac1{t^2} )\cos a +(1+\cos^2a)}\overset{t\to 1/t}{dt}\\ =& \ \frac{\sqrt{\cos a}}2\int_{0}^{\infty}\frac{d(1-\frac1{t} )}{(t-\frac1{t} )^2\cos a +(1+\cos a)^2}\\ =& \ \frac{\pi}{2(1+\cos a)}\\ \end{align}

How find this sum $\sum_{n=1}^{\infty}n\sum_{k=2^{n-1}}^{2^n-1}\frac{1}{k(2k+1)(2k+2)}$ Find the sum $$\sum_{n=1}^{\infty}n\sum_{k=2^{n - 1}}^{2^{n}\ -\ 1}\dfrac{1}{k(2k+1)(2k+2)}$$ My try: note $$\dfrac{1}{k(2k+1)(2k+2)}=\dfrac{2}{(2k)(2k+1)(2k+2)}=\left(\dfrac{1}{(2k)(2k+1)}-\dfrac{1}{(2k+1)(2k+2)}\right)$$ Then I can't

Are you really sure about the outer $n$? If not, Wolfram Alpha gives a sum that telescopes nicely $$\sum_{k=2^{n-1}}^{2^n-1} \frac{1}{k (2 k+1) (2 k+2)}=-2^{-n-1}-\psi^{(0)}(2^{n-1})+\psi^{(0)}(2^n)-\psi^{(0)}\left(\frac{1}{2}+2^n\right) +\psi^{(0)}\left(\frac{1}{2}+2^{n-1}\right)$$ Update: It is not hard to prove user64494's result manually $$S_2=\sum_{n=1}^{\infty}\sum_{k=2^{n-1}}^{2^n-1}\frac{1}{k(2k+1)(2k+2)}=$$ $$\sum_{n=1}^{\infty}\frac{1}{k(2k+1)(2k+2)}=$$ $$\sum_{n=1}^{\infty}2\left(\frac{1}{2k}-\frac{1}{2k+1}\right)-\left( \frac{1}{2k}-\frac{1}{2k+2}\right)=$$ $$2\left(\sum_{n=1}^{\infty}\frac{1}{2k}-\frac{1}{2k+1}\right)-\frac12\sum_{n=1}^{\infty}\frac1k-\frac{1}{k+1}=$$ Since $\ln{2}=1-\frac12+\frac13-\cdots$, we have $$1-\ln{2}=\sum_{n=1}^{\infty}\frac{1}{2k}-\frac{1}{2k+1}$$ And the second sum is a nicely telescopic one that equals $1$. Therefore $$S_2=2(1-\ln{2})-\frac12=\frac32-2\ln{2}$$ But this is unfortunately not what you have asked for yet. The real sum you are asking for doesn´t get a closed form from either WolframAlpha or Maple, so I don´t think there is an easy way to get it, at least for me.

Limit of $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$ What is the limit of this sequence $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$? Where $a$ is a constant and $n \to \infty$. If answered with proofs, it will be best.

Let $f(x) = 1+x + x^2+x^3+ \cdots$. Then the radius of convergence of $f$ is $1$, and inside this disc we have $f(x) = \frac{1}{1-x}$, and $f'(x) = 1+2x+3x^2+\cdots = \frac{1}{(1-x)^2}$. Suppose $|x|<1$, then we have $xf'(x) = x+2x^2+3x^3+\cdots = \frac{x}{(1-x)^2}$. If we choose $|a| >1$, then letting $x = \frac{1}{a}$ we have $\frac{1}{a} + \frac{2}{a^2}+ \frac{3}{a^3}+ \cdots = \frac{\frac{1}{a}}{(1-\frac{1}{a})^2}= \frac{a}{(a-1)^2}$.

$\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1$ Let $p$ be an odd prime number. Prove that $$\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \left( \frac{3 \cdot 4}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1$$ where $\left( \frac{a}{p}\right)$ is the Legendre symbol. This seems to be a tricky one! I've tried using the property $\left( \frac{ab}{p} \right)=\left( \frac{a}{p}\right) \left( \frac{b}{p} \right)$ and prime factoring all the non-primes but to no avail. I had a quick thought of induction haha, but that was silly. I tried factoring the common Legendre symbols like $\left( \frac{3}{p}\right) \left[ \left( \frac{2}{p} \right) + \left( \frac{4}{p} \right) \right]$ but that didn't bring anything either. And I've been looking for pairwise cancellation with $1$ term leftover, but that does not seem to work. Can you help?

Let $a^\ast$ be the inverse of $a$ modulo $p$. Then $$\left(\frac{a(a+1)}{p}\right)=\left(\frac{a(a+aa^\ast)}{p}\right)=\left(\frac{a^2(1+a^\ast)}{p}\right)=\left(\frac{1+a^\ast}{p}\right).$$ As $a$ ranges from $1$ to $p-2$, the number $1+a^\ast$ ranges, modulo $p$, through the integers from $2$ to $p-1$. But the sum from $1$ to $p-1$ of the Legendre symbols is $0$, so our sum is $-1$.

Show that 2 surfaces are tangent in a given point Show that the surfaces $ \Large\frac{x^2}{a^2} + \Large\frac{y^2}{b^2} = \Large\frac{z^2}{c^2}$ and $ x^2 + y^2+ \left(z - \Large\frac{b^2 + c^2}{c} \right)^2 = \Large\frac{b^2}{c^2} \small(b^2 + c^2)$ are tangent at the point $(0, ±b,c)$ To show that 2 surfaces are tangent, is it necessary and suficient to show that both points are in both surfaces and the tangent plane at those points is the same? Because if we just show that both points are in both surfaces, the surfaces could just intercept each other. And if we just show the tangent plane is the same at 2 points of the surfaces, those points do not need to be the same. Thanks!

The respective gradients of the surfaces are locally perpendicular to them: $$ \nabla f_1 = 2\left(\frac{x}{a^2}, \frac{y}{b^2}, -\frac{z}{c^2} \right) \\ \nabla f_2 = 2\left(x, y, z - \frac{b^2+c^2}{c}\right) $$ At any point of common tangency, the gradients are proportional. Therefore $\nabla f_1 = \lambda \nabla f_2$. This is seen to be true is you substitute $\left(0, \pm b, c\right)$ into the expressions for the gradients, the proportionality constant being $b^2$. QED.

Integral $\int_0^\infty\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}\cdot\frac{\log x}{\sqrt{x^2+1}}\mathrm dx$ I need your assistance with evaluating the integral $$\int_0^\infty\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}\cdot\frac{\log x}{\sqrt{x^2+1}}dx$$ I tried manual integration by parts, but it seemed to only complicate the integrand more. I also tried to evaluate it with a CAS, but it was not able to handle it.

This is not a full answer but how far I got, maybe someone can complete $\frac{1}{x\,\sqrt{2}+\sqrt{2\,x^2+1}}=\frac{1}{\sqrt{2\,x^2}+\sqrt{2\,x^2+1}}=\frac{\sqrt{2\,x^2+1}-\sqrt{2\,x^2}}{\sqrt{2\,x^2+1}^2-\sqrt{2\,x^2}^2}=\sqrt{2\,x^2+1}-\sqrt{2\,x^2}$ So you are looking at $$\int_0^\infty(\sqrt{\frac{2\,x^2+1}{x^2+1}} - \sqrt{\frac{2\,x^2}{x^2+1}}) \log x\,dx$$ I was then looking at $t = \frac{x^2}{x^2+1}$ but then I don't know

The number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ Find the number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ and the equation of type $x+y = x^2 + y^2 - xy$

Added: The approach below is ugly: It would be most comfortable to delete. We look at your second equation. Look first at the case $x\ge 0$, $y\ge 0$. We have $x^2+y^2-xy=(x-y)^2+xy$. Thus $x^2+y^2-xy\ge xy$. So if the equation is to hold, we need $xy\le x+y$. Note that $xy-x-y=(x-1)(y-1)-1$. The only way we can have $xy-x-y\le 0$ is if $(x-1)(y-1)=0$ or $(x-1)(y-1)=1$. In the first case, we have $x=1$ or $y=1$. Suppose that $x=1$. Then we are looking at the equation $1+y=1+y^2-y$, giving $y=0$ or $y=2$. By symmetry we also have the solution $y=1$, $x=0$ or $x=2$. If $(x-1)(y-1)=1$, we have $x=0$, $y=0$ or $x=2$, $y=2$. Now you can do an analysis of the remaining $3$ cases $x\lt 0$, $y\ge 0$; $y\lt 0$, $x\ge 0$; $x\lt 0$, $y\lt 0$. There is less to these than meets the eye. The first two cases are essentially the same. And since $x^2+y^2-xy=\frac{1}{2}((x-y)^2+x^2+y^2)$, we have $x^2+y^2-xy\ge 0$ for all $x,y$, so $x\lt 0$, $y\lt 0$ is impossible.

Understanding an Approximation I am reading the paper A Group-theoretic Approach to Fast Matrix Multiplication and there is an approximation in the paper I don't fully understand. In the proof of Theorem 3.3. it is stated that $$ \frac{\ln (n(n+1)/2)!)}{\ln (1!2!\dots n!)}= 2+ \frac{2-\ln 2}{\ln n}+O\left( \frac{1}{\ln^2 n}\right). $$ This is how far I got: I used Stirling's Approximation to estimate the numerator, leading to $$ \ln \left( \frac{n(n+1)}{2}!\right)= \frac{n(n+1)}{2} \ln \left( \frac{n(n+1)}{2}\right)- \frac{n(n+1)}{2}+O\left(\ln \frac{n(n+1)}{2}\right). $$ For the denominator I also use Stirling's approximation and $\ln (ab) = \ln a + \ln b$, leading to $$ \ln (1!2!\dots n!) = \sum_{k=1}^n k\ln k -k + O(\ln k). $$ I couldn't find a nice closed-form expression for $\sum_{k=1}^n k\ln k$ (is there one?), so I approximated it by $\int x \ln x \, dx = \frac{x^2\ln x} 2 -\frac {x^2} 4$, leading to $$ \ln (1!2!\dots n!) \approx \frac{n^2 \ln n} 2 - \frac{3n^2 } 4 +O(n\ln n). $$ Now, since the leading term of our fraction is, after cancelling, about $\frac{\ln(n^2)}{\ln n}$, I can see where the leading $2$ on the RHS comes from. But what about $\frac{2-\ln 2}{\ln n}$? Can it be calculated by a more precise bound on the denominator, e.g. by avoiding the integral?

The highest order term in $\log \prod\limits_{k=1}^n k!$ is $\frac{n^2}{2}\log n$, so in the desired $$\log \left(\frac{n(n+1)}{2}\right){\Large !} = \left(2 + \frac{2-\log 2}{\log n} + O\left(\frac{1}{\log^2 n}\right)\right)\log \prod_{k=1}^n k!,\tag{1}$$ we get on the right hand side a term $O\left(\frac{n^2}{\log n}\right)$ which we cannot (need not) specify more precisely. Hence in the approximations, we can ignore every term of order $\frac{n^2}{\log n}$ or less. What we need to establish $(1)$ is $$\begin{align} \log \left(\frac{n(n+1)}{2}\right){\Large !} &= \frac{n^2}{2}\log \frac{n(n+1)}{2} - \frac{n^2}{2} + O(n\log n),\tag{2}\\ \log \prod_{k=1}^n k! &= \frac{n^2}{2}\log n - \frac{3n^2}{4} + O(n\log n).\tag{3} \end{align}$$ We obtain $$\begin{align} \log \left(\frac{n(n+1)}{2}\right){\Large !} - 2\log \prod_{k=1}^n k! &= \frac{n^2}{2} \log n(n+1) - \frac{n^2}{2}\log 2 - \frac{n^2}{2}\\ &\quad - \frac{n^2}{2} \log n^2 + \frac{3n^2}{2} + O(n\log n)\\ &= \frac{n^2}{2}\log\frac{n+1}{n} + \frac{n^2}{2}(2-\log 2) + O(n\log n)\\ &= \frac{n^2}{2}(2-\log 2) + O(n\log n)\\ &= \frac{2-\log 2}{\log n}\left(\log \prod_{k=1}^nk! + O(n^2)\right) + O(n\log n) \end{align}$$ and hence $$\log \left(\frac{n(n+1)}{2}\right){\Large !} = \left(2 + \frac{2-\log 2}{\log n}\right)\log \prod_{k=1}^n k! + O\left(\frac{n^2}{\log n}\right).$$ Since $\log \prod k! \in \Theta\left(n^2\log n\right)$, the division yields $$\frac{\log \left(\frac{n(n+1)}{2}\right){\Large !}}{\log \prod\limits_{k=1}^nk!} = 2 + \frac{2-\log 2}{\log n} + O\left(\frac{1}{\log^2 n}\right)$$ as desired. It remains to establish $(2)$ and $(3)$. For that, we use the most significant terms of Stirling's approximation $$\log k! = \left(k+\frac12\right)\log k - k + \frac12\log 2\pi + \frac{1}{12k} + O\left(\frac{1}{k^2}\right).$$ We obtain $(2)$ by ignoring all terms of order less than $n^2$ in Stirling's approximation. For $(3)$, a comparison with the integral $\int_1^n t\log t\,dt$ yields $$\sum_{k=1}^n k\log k = \frac{n^2}{2}\log n - \frac{n^2}{4} + O(n\log n),$$ and hence $$\log \prod_{k=1}^n k! = \sum_{k=1}^n (k\log k - k) + O(n\log n) = \frac{n^2}{2}\log n - \frac{3n^2}{4} + O(n\log n).$$

Radius and Interval of Convergence of $\sum (-1)^n \frac{ (x+2)^n }{n2^n}$ I have found the radius of convergence $R=2$ and the interval of convergence $I =[-2,2)$ for the following infinite series: $\sum_{n=1}^\infty (-1)^n \frac{ (x+2)^n }{n2^n}$ Approach: let $a_n = (-1)^n \frac{ (x+2)^n }{n2^n}$ Take the ratio test $\lim_{x\to -\infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| -\frac{1}{2}(x+2)\right|$ So it converges for $-2 < (x+2) < 2, R = 2$ Test end points if $x = -2$ $\sum_{n=1}^\infty (-1)^n(0) =0 < \infty$ , converges if $x = 2$ $\sum_{n=1}^\infty (-1)^n \frac{4^n}{n2^n} = \sum_{n=1}^\infty (-1)^n \frac{2^{2n}}{n2^n} = \sum_{n=1}^\infty (-1)^n \frac{2^n}{n}$ let $b_n = \frac{2^n}{n}$ $b_{n+1} > b_n$ for every value of $n$ By the Alternating Series Test, $\sum_{n=1}^\infty (-1)^n b_n$ diverges Hence, $R = 2$ and $I = [-2, 2)$ Is my approach correct? Are there any flaws, or better ways to solve?

The radius of convergence is right. The interval of convergence started out OK, you wrote that for sure we have convergence if $-2\lt x+2\lt 2$. But this should become $-4\lt x\lt 0$. The endpoint testing was inevitably wrong, since the incorrect endpoints were being tested. We have convergence at $x=0$ (alternating series) and divergence at $x=-4$ (harmonic series).

In a triangle $\angle A = 2\angle B$ iff $a^2 = b(b+c)$ Prove that in a triangle $ABC$, $\angle A = \angle 2B$, if and only if: $$a^2 = b(b+c)$$ where $a, b, c$ are the sides opposite to $A, B, C$ respectively. I attacked the problem using the Law of Sines, and tried to prove that if $\angle A$ was indeed equal to $2\angle B$ then the above equation would hold true. Then we can prove the converse of this to complete the proof. From the Law of Sines, $$a = 2R\sin A = 2R\sin (2B) = 4R\sin B\cos B$$ $$b = 2R\sin B$$ $$c = 2R\sin C = 2R\sin(180 - 3B) = 2R\sin(3B) = 2R(\sin B\cos(2B) + \sin(2B)\cos B)$$ $$=2R(\sin B(1 - 2\sin^2 B) +2\sin B\cos^2 B) = 2R(\sin B -2\sin^3 B + 2\sin B(1 - \sin^2B))$$ $$=\boxed{2R(3\sin B - 4\sin^3 B)}$$ Now, $$\implies b(b+c) = 2R\sin B[2R\sin B + 2R(3\sin B - 4\sin^3 B)]$$ $$=4R^2\sin^2 B(1 + 3 - 4\sin^2 B)$$ $$=16R^2\sin^2 B\cos^2 B = a^2$$ Now, to prove the converse: $$c = 2R\sin C = 2R\sin (180 - (A + B)) = 2R\sin(A+B) = 2R\sin A\cos B + 2R\sin B\cos A$$ $$a^2 = b(b+c)$$ $$\implies 4R^2\sin^2 A = 2R\sin B(2R\sin B + 2R\sin A\cos B + 2R\sin B\cos) $$ $$ = 4R^2\sin B(\sin B + \sin A\cos B + \sin B\cos A)$$ I have no idea how to proceed from here. I tried replacing $\sin A$ with $\sqrt{1 - \cos^2 B}$, but that doesn't yield any useful results.

$$a^2-b^2=bc\implies \sin^2A-\sin^2B=\sin B\sin C\text{ as }R\ne0$$ Now, $\displaystyle\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)=\sin(\pi-C)\sin(A-B)=\sin C\sin(A-B)\ \ \ \ (1)$ $$\implies \sin B\sin C=\sin C\sin(A-B)$$ $$\implies \sin B=\sin(A-B)\text{ as }\sin C\ne0$$ $$\implies B=n\pi+(-1)^n(A-B)\text{ where }n\text{ is any integer} $$ If $n$ is even, $n=2m$(say) $\implies B=2m\pi+A-B\iff A=2B-2m\pi=2B$ as $0<A,B<\pi$ If $n$ is odd, $n=2m+1$(say) $\implies B=(2m+1)\pi-(A-B)\iff A=(2m+1)\pi$ which is impossible as $0<A<\pi$ Conversely, if $A=2B$ $\displaystyle\implies a^2-b^2=4R^2(\sin^2A-\sin^2B)=4R^2\sin C\sin(A-B)$ (using $(1)$) $\displaystyle=4R^2\sin C\sin(2B-B)=2R\sin B\cdot 2R\sin C=\cdots$

Find the maximum and minimum values of $A \cos t + B \sin t$ Let $A$ and $B$ be constants. Find the maximum and minimum values of $A \cos t + B \sin t$. I differentiated the function and found the solution to it as follows: $f'(x)= B \cos t - A \sin t$ $B \cos t - A \sin t = 0 $ $t = \cot^{-1}(\frac{A}{B})+\pi n$ However, I got stuck here on how to formulate the minimum and maximum points. Any explanation would be appreciated.

$A\cos t+ B \sin t = \sqrt{A^2+B^2} ( \frac{A}{\sqrt{A^2+B^2}} \cos t + \frac{B}{\sqrt{A^2+B^2}} \sin t)$. Choose $\theta$ such that $e^{i \theta} = \frac{A}{\sqrt{A^2+B^2}} + i\frac{B}{\sqrt{A^2+B^2}} $. Then $A\cos t+ B \sin t = \sqrt{A^2+B^2} ( \cos \theta \cos t + \sin \theta \sin t) = \sqrt{A^2+B^2} \cos(\theta-t)$. It follows that the extreme values are $\pm \sqrt{A^2+B^2}$.

Complex numbers problem I have to solve where n is equal to n=80996.

$$\frac{\sqrt3+5i}{4+2\sqrt3i}=\frac{\sqrt3+5i}{4+2\sqrt3i}\cdot\frac{4-2\sqrt3i}{4-2\sqrt3i}=\frac{14\sqrt3+14i}{28}=\frac{\sqrt3+i}2=\cos\frac{\pi}6+i\sin\frac{\pi}6;$$$$\left(\frac{\sqrt3+5i}{4+2\sqrt3i}\right)^n=\left(\cos\frac{\pi}6+i\sin\frac{\pi}6\right)^n.$$

How prove this inequality $\sum\limits_{k=1}^{n}\frac{1}{k!}-\frac{3}{2n}<\left(1+\frac{1}{n}\right)^n<\sum\limits_{k=1}^{n}\frac{1}{k!}$ show that $$\sum_{k=1}^{n}\dfrac{1}{k!}-\dfrac{3}{2n}<\left(1+\dfrac{1}{n}\right)^n<\sum_{k=0}^{n}\dfrac{1}{k!}(n\ge 3)$$ Mu try: I konw $$\sum_{k=0}^{\infty}\dfrac{1}{k!}=e$$ and I can prove Right hand inequality $$\left(1+\dfrac{1}{n}\right)^n=\sum_{k=0}^{n}\binom{n}{k}\dfrac{1}{n^k}$$ and note $$\binom{n}{k}\dfrac{1}{n^k}=\dfrac{1}{k!}\dfrac{(n-k+1)(n-k+2)\cdots(n-1)n}{n^k}\le\dfrac{1}{k!}$$ so $$\left(1+\dfrac{1}{n}\right)^n<\sum_{k=0}^{n}\dfrac{1}{k!}$$ But the left Hand inequality how prove it ? Thank you

I think you mean this inequality$$\sum_{k=0}^{n}\dfrac{1}{k!}-\dfrac{3}{2n}<\left(1+\dfrac{1}{n}\right)^n.$$In fact, we can prove a sharper one $$\left(1+\frac{1}{n}\right)^n+\frac{3}{2n}>e.$$Let $f(x)=\left(1+\dfrac{1}{x}\right)^x+\dfrac{3}{2x}$, then$$f'(x)=\left(1+\dfrac{1}{x}\right)^x\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)-\frac{3}{2x^2}.$$It's suffice to show that $f'(x)<0$ for $x>1$.Actually, notice that $\left(1+\dfrac{1}{x}\right)^x<3$ and $$\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\leqslant \frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}-\frac{1}{1+x}=\frac{1}{x^2(1+x)},$$thus for $x>1$, we have$$\left(1+\dfrac{1}{x}\right)^x\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)-\frac{3}{2x^2}<\frac{3}{x^2(1+x)}-\frac{3}{2x^2}<0,$$therefore $f(x)$ is monotonically decreasing in $[1,+\infty)$. As $\displaystyle\lim_{x\rightarrow+\infty}f(x)=e$, we can get$$\left(1+\dfrac{1}{x}\right)^x+\dfrac{3}{2x}>e$$for all $x>1$, hence$$\left(1+\frac{1}{n}\right)^n+\frac{3}{2n}>e.$$

consecutive convergents Problem: Let $\phi=\frac{1+\sqrt{5}}{2}$ be the golden ratio and let $a$, $b$, $c$, $d$ be positive integers so that $\frac{a}{b}>\phi>\frac{c}{d}$. It is also known that $ad-bc=1$. Prove that $a/b$ and $c/d$ are consecutive convergents of $\phi$. Numerical experimentations point towards the validity of this statement. The converse is well known (and easy to show) but I cannot seem to prove the direct statement. This is not a homework question; I came across it while investigating the geometric discrepancy of a certain lattice point set. Any help would be appreciated.

Note that $\phi=1+\frac{1}{1+\frac{1}{1+\ldots}}$ has convergents $\frac{f_{n+1}}{f_n}$, i.e. ratios of consecutive Fibonacci numbers. Note that I have used lower case for the Fibonacci numbers so as to avoid confusion with the Farey sequence $F_n$. The main idea is to appeal to the properties of Farey sequences. * *Suppose $a, b, c, d$ are positive integers with $a \leq b, c \leq d, \frac{c}{d}<\frac{a}{b}$. Then $\frac{a}{b}$ and $\frac{c}{d}$ are consecutive members of the Farey sequence $F_n$ for some $n$ if and only if $ad-bc=1$. *If $\frac{a}{b}<\frac{c}{d}$ are consecutive members of the Farey sequence $F_n$ for some $n$, then either they are consecutive members in $F_{n+1}$, or $\frac{a}{b}, \frac{a+c}{b+d}, \frac{c}{d}$ are consecutive members in $F_{n+1}$, in which case we have $b+d=n+1$. In other words, as we increase the order of the Farey sequence, $\frac{a+c}{b+d}$ is the first term to appear between $\frac{a}{b}$ and $\frac{c}{d}$. Consider $0<\frac{1}{\phi}<1$. For each $m$, suppose that the Farey sequence $F_m$ is given by $\frac{0}{1}=a_{m, 0}<a_{m, 1}< \ldots <a_{m, |F_m|-1}=\frac{1}{1}$. We partition the interval $[0, 1)$ into $|F_m|-1$ intervals $[a_{m, i}, a_{m, i+1})$. Note that $\frac{1}{\phi}$ must belong to exactly one such interval. Also note that $\frac{1}{\phi}$ is irrational and so cannot be equal to $a_{m, i}$. Thus for each $m$, there is a unique pair of rational numbers $r_m, s_m$ s.t. $r_m, s_m$ are consecutive members of the Farey sequence $F_m$ and $r_m<\frac{1}{\phi}<s_m$. Observe that $\frac{1}{\phi}$ has convergents $\frac{f_{n-1}}{f_n}$. We observe that for $f_n \leq m<f_{n+1}$, we have that $\frac{f_{n-2}}{f_{n-1}}$ and $\frac{f_{n-1}}{f_n}$ are consecutive members of $F_m$. Explanation: This is because we have the identity $f_{n-2}f_n-f_{n-1}^2=(-1)^{n-1}$, so by property $1$ $\frac{f_{n-2}}{f_{n-1}}$ and $\frac{f_{n-1}}{f_n}$ are consecutive members of some Farey sequence $F_k$. Note that we necessarily have $k \geq f_n$, since $\frac{f_{n-1}}{f_n}$ is a member of $F_k$. Therefore $\frac{f_{n-2}}{f_{n-1}}$ and $\frac{f_{n-1}}{f_n}$ will be consecutive members in $F_{f_n}$. (Removing elements doesn't affect the fact that they are consecutive) Now, as we increase the order of the Farey sequence, the first term that appears between them is $\frac{f_{n-2}+f_{n-1}}{f_{n-1}+f_n}=\frac{f_n}{f_{n+1}}$, which cannot appear for $m<f_{n+1}$. Therefore $\frac{f_{n-2}}{f_{n-1}}$ and $\frac{f_{n-1}}{f_n}$ remain as consecutive members in the Farey sequence $F_m$, for $f_n \leq m<f_{n+1}$. Also, as the convergents of $\frac{1}{\phi}$ are alternately greater and smaller than $\frac{1}{\phi}$, we see that $\frac{1}{\phi}$ is strictly between $\frac{f_{n-2}}{f_{n-1}}$ and $\frac{f_{n-1}}{f_n}$. Therefore $\frac{f_{n-2}}{f_{n-1}}$ and $\frac{f_{n-1}}{f_n}$ must be $r_m$ and $s_m$ in some order, i.e. $\{\frac{f_{n-2}}{f_{n-1}},\frac{f_{n-1}}{f_n}\}=\{r_m, s_m\}$. Finally, from the question, $\frac{a}{b}>\phi>\frac{c}{d}$ so $\frac{d}{c}>\frac{1}{\phi}>\frac{b}{a}$. Since $ad-bc=1$, $\frac{b}{a}$ and $\frac{d}{c}$ are consecutive members of some Farey sequence $F_m$. Thus $\{\frac{b}{a}, \frac{d}{c}\}=\{r_m, s_m\}$ Clearly we have $f_n \leq m<f_{n+1}$ for some $n$ so by above $\{r_m, s_m\}=\{\frac{f_{n-2}}{f_{n-1}},\frac{f_{n-1}}{f_n}\}$, so $\{\frac{b}{a}, \frac{d}{c}\}=\{\frac{f_{n-2}}{f_{n-1}},\frac{f_{n-1}}{f_n}\}$. Therefore $\{\frac{a}{b}, \frac{c}{d}\}=\{\frac{f_{n-1}}{f_{n-2}},\frac{f_n}{f_{n-1}}\}$ are consecutive convergents of $\phi$.

$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ ... + \frac{1}{1+2+3+...+n} = ?$ How do I simplify the following series $$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ \frac{1}{1+2+3+4} + \frac{1}{1+2+3+4+5} + \cdot\cdot\cdot + \frac{1}{1+2+3+\cdot\cdot\cdot+n}$$ From what I see, each term is the inverse of the sum of $n$ natural numbers. Assuming there are $N$ terms in the given series, $$a_N = \frac{1}{\sum\limits_{n = 1}^{N} n} = \frac{2}{n(n+1)}$$ $$\Rightarrow \sum\limits_{n=1}^{N} a_N = \sum\limits_{n=1}^{N} \frac{2}{n(n+1)}$$ ... and I'm stuck. I've never actually done this kind of problem before (am new to sequences & series). So, a clear and detailed explanation of how to go about it would be most appreciated. PS- I do not know how to do a telescoping series!!

HINT: $$\frac {2} {n(n+1)} = \frac 2 n - \frac 2 {n+1}$$

Solving for the zero of a multivariate How does one go about solving the roots for the following equation $$x+y+z=xyz$$ There simply to many variables. Anyone have an idea ?

If we fix one of the variables, we get a hyperbola in that plane. So, for example, fixing any $z = z_0,$ this is your relationship: $$ \left(x - \frac{1}{z_0} \right) \left(y - \frac{1}{z_0} \right) = \; 1 + \frac{1}{z_0^2} $$ Makes me think the surface could be connected. Indeed, as $|z| \rightarrow \infty,$ the curve approaches the fixed curve $xy=1.$ Meanwhile, the surface is smooth. The gradient of the function $xyz-x-y-z$ is only the zero vector at $(1,1,1)$ and $(-1,-1,-1)$ but those are not part of the surface. And, in the plane $x+y+z = 0,$ the surface contains the entirety of three lines that meet at $60^\circ$ angles, these being the intersections with the planes $x=0$ or $y=0$ or $z=0.$ EDIT: the surface actually has three connected components, sometimes called sheets in the context of hyperboloids. One is in the positive octant $x,y,z > 0,$ with $xy>1, xz>1, yz > 1$ in addition. The nearest point to the origin is $(\sqrt 3,\sqrt 3,\sqrt 3 ),$ and the thing gets very close to the walls as $x+y+z$ increases. The middle sheet goes through the origin and, near there, is a monkey saddle. The third sheet is in the negative octant. About the middle sheet: you may recall that the real part of $(x+yi)^3$ is $x^3 - 3 x y^2.$ The graph $z =x^3 - 3 x y^2 $ is, near the origin, a monkey saddle, room for two legs and a tail for a monkey sitting on a horse. If we rotate coordinates by $$ x = \frac{u}{\sqrt 3 } - \frac{v}{\sqrt 2 } - \frac{w}{\sqrt 3 }, $$ $$ y = \frac{u}{\sqrt 3 } + \frac{v}{\sqrt 2 } - \frac{w}{\sqrt 3 }, $$ $$ z = \frac{u}{\sqrt 3 } + \frac{2w}{\sqrt 3 }, $$ the surface becomes $$ \color{magenta}{ u \; \left( 1 + \frac{v^2}{6} + \frac{w^2}{6} - \frac{u^2}{9} \right) = \frac{1}{9 \sqrt 2} \; \left( w^3 - 3 w \, v^2 \right).} $$ So, say within the ball $u^2 + v^2 + w^2 < 1,$ the multiplier of $u$ on the left is quite close to $1,$ and we have a monkey saddle. Took a bit of work to confirm: given constants $a+b+c = 0,$ and looking at $$ x=a+t, y = b+t, z = c+t, $$ there are exactly three distinct real values of $t$ that solve $xyz=x+y+z.$

Proving that if these quadratics are equal for some $\alpha$, then their coefficients are equal Let $$P_1(x) = ax^2 -bx - c \tag{1}$$ $$P_2(x) = bx^2 -cx -a \tag{2}$$ $$P_3(x) = cx^2 -ax - b \tag{3}$$ Suppose there exists a real $\alpha$ such that $$P_1(\alpha) = P_2(\alpha) = P_3(\alpha)$$ Prove $$a=b=c$$ Equating $P_1(\alpha)$ to $P_2(\alpha)$ $$\implies a\alpha^2 - b\alpha - c = b\alpha^2 - c\alpha - a$$ $$\implies (a-b)\alpha^2 + (c-b)\alpha + (a-c) = 0$$ Let $$Q_1(x) = (a-b)x^2 + (c-b)x + (a-c)$$ This implies, $\alpha$ is a root of $Q_1(x)$. Similarly, equating $P_2(\alpha)$ to $P_3(\alpha)$ and $P_3(\alpha)$ to $P_1(\alpha)$, and rearranging we obtain quadratics $Q_2(x)$ and $Q_3(x)$ with common root $\alpha$: $$Q_2(x) = (b-c)x^2 + (a-c)x + (b-a)$$ $$Q_3(x) = (c-a)x^2 + (b-a)x + (c-b)$$ $$Q_1(\alpha) = Q_2(\alpha) = Q_3(\alpha) = 0$$ We have to prove that this is not possible for non-constant quadratics $Q_1(x), Q_2(x), Q_3(x)$. EDIT: I also noticed that for distinct $a, b, c \in \{1, 2, 3\}$: $$Q_a(x) + Q_b(x) = -Q_c(x)$$

Denote $$Q_1(x)=P_1(x)-P_2(x)=(a-b)x^2-(b-c)x-(c-a);$$ $$Q_2(x)=P_2(x)-P_3(x)=(b-c)x^2-(c-a)x-(a-b);$$ $$Q_3(x)=P_3(x)-P_1(x)=(c-a)x^2-(a-b)x-(b-c).$$ Then $\alpha$ is a real root of the equations $Q_i(x);$ so that $\Delta_{Q_i(x)}\geq 0 \ \ \forall i=1,2,3;$ where $\Delta_{f(x)}$ denoted the discriminant of a quadratic function $f$ in $x.$ Now, using this we have, $$(b-c)^2+4(a-b)(c-a)\geq 0;$$ $$(c-a)^2+4(a-b)(b-c)\geq 0;$$ $$(a-b)^2+4(b-c)(c-a)\geq 0.$$ Summing these up and using the identity $$\begin{aligned}(a-b)^2+(b-c)^2+(c-a)^2+2\left(\sum_\text{cyc}(a-b)(c-a)\right)\\=[(a-b)+(b-c)+(c-a)]^2=0;\end{aligned}$$ We obtain, $$\begin{aligned}0&\leq 2\left(\sum_\text{cyc}(a-b)(c-a)\right)=2\left(\sum_\text{cyc}(ca+ab-bc-a^2)\right)\\&=-\left[(a-b)^2+(b-c)^2+(c-a)^2\right];\end{aligned}$$ Possible if and only if $(a-b)^2+(b-c)^2+(c-a)^2=0\iff a=b=c.$ We are done

Rudin Example 3.35B Why the $n$th root of $a_n$ is $1/2$? For Baby Rudin Example 3.35(b), I understand how the $\liminf$ and $\limsup$ of the ratio test were found, but I am not clear why $\ \lim \sqrt[n]{a_n } = \frac{1}{2} $. Please help.

The sequence in question is $$\frac{1}{2} + 1 + \frac{1}{8} + \frac{1}{4}+ \frac{1}{32}+ \frac{1}{16}+\frac{1}{128}+\frac{1}{64}+\cdots$$ In case the pattern is not clear, we double the first term, the divide the next by $8$, the double, then divide by $8$, and so on. The general formula for an odd term is $a_{2k-1}=\frac{1}{2^{2k-1}}$. The formula for an even term is $a_{2k}=\frac{1}{2^{2k-2}}$. In the first case, $\sqrt[n]{a_n}=\frac{1}{2}$. In the second, the limit is $\frac{1}{2}$. Since $n$th roots of both the even and odd terms converge to $\frac{1}{2}$, you have your desired result.

Value of sum of telescoping series $$\sum_{n\geqslant1}\frac{1}{\sqrt{n}} -\frac{ 1}{\sqrt {n+2}}$$ In looking at the first five partial sums, I am not convinced the series is telescopic (the middle terms don't cancel out). Thanks in advance!

$\sum_{k=1} ^n \frac{1}{\sqrt{k}}- \frac{1}{\sqrt{k+2}}$ $= \frac{1}{\sqrt{1}}- \frac{1}{\sqrt{3}}+ \frac{1}{\sqrt{2}}- \frac{1}{\sqrt{4}} +\frac{1}{\sqrt{3}}- \frac{1}{\sqrt{5}}+ \frac{1}{\sqrt{4}}- \frac{1}{\sqrt{6}}+ \frac{1}{\sqrt{5}}- \frac{1}{\sqrt{7}}+...+ \frac{1}{\sqrt{n-2}}- \frac{1}{\sqrt{n}}+ \frac{1}{\sqrt{n-1}}- \frac{1}{\sqrt{n+1}}+ \frac{1}{\sqrt{n}}- \frac{1}{\sqrt{n+2}}$ the first five terms dont cancel immediatley, after many terms have been computed they cancel,i will not classify it under telescope because the terms dont entirely cancel

How prove this $a_{n}>1$ let $0<t<1$, and $a_{1}=1+t$, and such $$a_{n}=t+\dfrac{1}{a_{n-1}}$$ show that $a_{n}>1$ My try: since $$a_{1}=1+t>1$$ $$a_{2}=t+\dfrac{1}{a_{1}}=t+1+\dfrac{1}{1+t}-1>2\sqrt{(t+1)\cdot\dfrac{1}{1+t}}-1=2-1=1$$ $$a_{3}=t+\dfrac{1}{a_{2}}=t+\dfrac{1}{t+\dfrac{1}{t+1}}=t+\dfrac{t+1}{t^2+t+1}=1+\dfrac{t^3+t}{t^2+t+1}>1$$ $$\cdots\cdots\cdots$$ But $a_{n}$ is very ugly,so this problem may use other methods.Thank you very much!

Let $\displaystyle \mu = \frac{t + \sqrt{t^2+4}}{2}$, we have $$\mu > 1\quad\text{ and }\quad\mu(t - \mu) = \left(\frac{t + \sqrt{t^2+4}}{2}\right)\left(\frac{t - \sqrt{t^2+4}}{2}\right) = -1$$ From this, we get $$a_{n+1} - \mu = t - \mu + \frac{1}{a_n} = \frac{1}{a_n} - \frac{1}{\mu} = \frac{\mu - \alpha_n}{\mu a_n}$$ This implies if $a_n > 1$, then $$|a_{n+1}-\mu| = \frac{|a_n - \mu|}{\mu a_n} < \frac{|a_n -\mu|}{\mu} < |a_n - \mu|\tag{*1}$$ Notice $$\begin{align} ( 1 - \mu)^2 - (a_1 - \mu)^2 = & (1 - \mu)^2 - (1 + t - \mu)^2 = (1 - \mu)^2 - (1 - \frac{1}{\mu})^2\\ = & (1-\mu)^2(1 - \frac{1}{\mu^2}) > 0 \end{align}$$ We have $a_1 \in (1,2\mu - 1) = (\mu - (\mu - 1),\mu + (\mu - 1))$. Since all $x$ in this interval $> 1$, we can repeatedly apply $(*1)$ to conclude all $a_n$ belongs to this interval and hence $> 1$.

Getting angles for rotating $3$D vector to point in direction of another $3$D vector I've been trying to solve this in Mathematica for $2$ hours, but got the wrong result. I have a vector, in my case $\{0, 0, -1\}$. I want a function that, given a different vector, gives me angles DX and DY, so if I rotate the original vector by an angle of DX around the X axis, and then rotate it by an angle of DY around the Y axis, I'll get a vector with the same direction as the given vector. (I want this so I could input angles to SolidWorks to rotate a part so it will satisfy a constraint I defined in Mathematica.)

So you need a 3×3 rotation matrix $E$ such that $$ E\,\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix} = -\hat{k} $$ This rotation matrix consists of two elementary rotations $$ \begin{aligned} E & = {\rm Rot}(\hat{i},\varphi_x){\rm Rot}(\hat{j},\varphi_y) \\ & = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\varphi_x & -\sin\varphi_x \\ 0 & \sin\varphi_x & \cos\varphi_x \end{pmatrix} \begin{pmatrix} \cos\varphi_y &0 & \sin\varphi_y \\ 0 & 1 & 0 \\ - \sin\varphi_y & 0 & \cos\varphi_y \end{pmatrix} \end{aligned} $$ This is solved with $$ \varphi_x = - \tan^{-1} \left( \frac{y}{\sqrt{x^2+z^2}} \right) \\ \varphi_y = \pi - \tan^{-1} \left( \frac{x}{z} \right) $$ if $\sqrt{x^2+y^2+z^2}=1$ is true. Verification use $(x,y,z) = (\frac{1}{2}, \frac{3}{4}, \frac{\sqrt{3}}{4}) = (0.5, 0.75, 0.4330) $ to get $$ \varphi_x = - \tan^{-1} \left( \frac{\frac{3}{4}}{\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{4}\right)^2}} \right) = -0.8481 \\ \varphi_y = \pi - \tan^{-1} \left( \frac{\frac{1}{2}}{\frac{\sqrt{3}}{4}} \right) = 2.2845$$ With a rotation matrix $$ E = \begin{pmatrix} -0.6547 & 0 & 0.7559 \\ -0.5669 & 0.6614 & -0.4910 \\ -0.5 & -0.75 & -0.4330 \end{pmatrix} $$ and $$ \begin{pmatrix} -0.6547 & 0 & 0.7559 \\ -0.5669 & 0.6614 & -0.4910 \\ -0.5 & -0.75 & -0.4330 \end{pmatrix} \begin{pmatrix} 0.5 \\ 0.75 \\ 0.4330 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -1\end{pmatrix} $$ NOTICE: Some combinations of $(x,y,z)$ will not yield the correct result because it won't be possible with the specific rotation sequence used.

Solution to the limit of a series I'm strugling with the following problem: $$\lim_{n\to \infty}(n(\sqrt{n^2+3}-\sqrt{n^2-1})), n \in \mathbb{N}$$ Wolfram Alpha says the answer is 2, but I don't know to calculate the answer. Any help is appreciated.

For the limit: We take advantage of obtaining a difference of squares. We have a factor of the form $a - b$, so we multiply it by $\dfrac{a+b}{a+b}$ to get $\dfrac{a^2 - b^2}{a+b}.$ Here, we multiply by $$\dfrac{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}$$ $$n(\sqrt{n^2+3}-\sqrt{n^2-1})\cdot\dfrac{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}{\sqrt{n^2+3}+ \sqrt{n^2 - 1}} = \dfrac{n[n^2 + 3 - (n^2 - 1)]}{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}$$ Now simplify and evaluate.

Prove $13|19^n-6^n$ by congruences I am trying to prove $13|19^n-6^n$. With induction its not so bad but by congruences its quite difficult to know how to get started. Any hints?

Because $x^n - y^n$ is divisible by $x-y$ as $$x^n - y^n = (x-y)\sum_{i=0}^{n-1}x^iy^{n-1-i}$$ Substitute $x=19$ and $y = 6$. $$\begin{align*} x^n-y^n =& x^n\left[1-\left(\frac yx\right)^n\right]\\ =& x^n \left[1+\left(\frac yx\right)+\left(\frac yx\right)^2+\cdots+\left(\frac yx\right)^{n-1}\right]\left[1-\left(\frac yx\right)\right]\\ =& \left(x^{n-1}y^0 + x^{n-2}y + x^{n-3}y^2 + \cdots + x^0y^{n-1}\right) (x-y) \end{align*}$$

If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then to prove $x=\sqrt{3}+\sqrt{2}$ If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then we have to prove $x=\sqrt{3}+\sqrt{2}$ The question would have been simple if it asked us to prove the other way round. We can multiply by $x^3$ and solve the quadratic to get $x^3$ but that would be unnecessarily complicated.Also, as $x^3$ has 2 solutions,I can't see how x can have only 1 value. But the problem seems to claim that x can take 1 value only.Nevertheless,is there any way to get the values of x without resorting to unnecessarily complicated means? NOTE: This problem is from a textbook of mine.

$$t+\frac1t=18\sqrt3\iff t^2-(2\cdot9\sqrt3)t+1=0\iff t_{1,2}=\frac{9\sqrt3\pm\sqrt{(9\sqrt3)^2-1\cdot1}}1=$$ $$=9\sqrt3\pm\sqrt{81\cdot3-1}\quad=\quad9\sqrt3\pm\sqrt{243-1}\quad=9\sqrt3\pm\sqrt{242}\quad=\quad9\sqrt3\pm\sqrt{2\cdot121}=$$ $$=9\sqrt3\pm\sqrt{2\cdot11^2}\quad=\quad9\sqrt3\pm11\sqrt2\quad\iff\quad x_{1,2}^3=9\sqrt3\pm11\sqrt2=(a\sqrt3+b\sqrt2)^3=$$ $$=(a\sqrt3)^3+(b\sqrt2)^3+3(a\sqrt3)^2b\sqrt2+3a\sqrt3(b\sqrt2)^2\ =\ 3a^3\sqrt3+2b^3\sqrt2+9a^2b\sqrt2+6ab^2\sqrt3$$ $$\iff3a^3+6ab^2=9=3+6\quad,\quad2b^3+9a^2b=\pm11=\pm2\pm9\iff a=1,\quad b=\pm1.$$

How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola? How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola, knowing the canonical form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=\pm1$ where $a$ and $b$ are constants? Thanks !

Let $$ \frac{y^2-x^2}{x+y+1}=1\\ \Rightarrow y^2-x^2=x+y+1\\ \Rightarrow y^2-x^2-x-y=1 $$ Complete the squares for x and y . You will get rectangular hyperbola. Similar will be the case if $$ \frac{y^2-x^2}{x+y+1}=-1$$

Checking on some convergent series I need some verification on the following 2 problems I attemped: I have to show that the following series is convergent: $$1-\frac{1}{3 \cdot 4}+\frac{1}{ 5 \cdot 4^2 }-\frac{1}{7 \cdot 4^3}+ \ldots$$ . My Attempt: I notice that the general term is given by $$\,\,a_n=(-1)^{n}{1 \over {(2n+1)4^n}} \,\,\text{by ignoring the first term of the given series.}$$ Using the fact that An absolutely convergent series is convergent, $$\sum_{1}^{\infty}|a_n|=\sum_{1}^{\infty} {1 \over {(2n+1)4^n}}\le \sum_{1}^{\infty} {1 \over 4^n}=\sum_{1}^{\infty}{1 \over {2^{2n}}}$$ which is clearly convergent by comparing it with the p-series with $p >1$. I have to show that the following series is convergent:$$1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+ \ldots $$ My Attempt:$$1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+ \ldots \le 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+ \frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\ldots $$. Now, using the fact that $n! >n^2$ for $n \ge 4$ and the fact that omitting first few terms of the series does not affect the characteristics of the series ,we see that $$\frac{1}{4!}-\frac{1}{6!}+\frac{1}{8!}+ \ldots \le \frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+ \frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!}+\ldots =\sum_{4}^{\infty}{1 \over n!} <\sum_{4}^{\infty}{1 \over n^2}$$ and it is clearly convergent by comparing it with the p-series with $p >1$. Now,I am stuck on the third one. I have to show that the following series is convergent:$$\frac{\log 2}{2^2}-\frac{\log 3}{3^2}+\frac{\log 4}{4^2}- \ldots $$ I see that $$\frac{\log 2}{2^2}-\frac{\log 3}{3^2}+\frac{\log 4}{4^2}- \ldots \le \sum_{2}^{\infty} {{\log n} \over {n^2}}= ?? $$ Thanks and regards to all.

Hint: For all sufficiently large $n$ (in fact, $n \ge 1$ suffices for this), we have $\ln{n} \le \sqrt{n}$; thus $$\sum\limits_{n = 2}^{\infty} \frac{\ln n}{n^2} \le \sum\limits_{n = 2}^{\infty} \frac{1}{n^{3/2}}$$ which is a $p$-series.

If $f$ is continuous, why is $f$ with the property $f\left(\frac{x+y}{2}\right)\le \frac{1}{2}f(x)+\frac{1}{2}f(y)$ is convex? If $f$ is continuous, why is $f$ with the property $$f\left(\frac{x+y}{2}\right)\le \frac{1}{2}f(x)+\frac{1}{2}f(y),$$ where $0\le x,y\le 1$ is convex?

By induction we can prove that: if $k,m, l\in\mathbb{N} , k+m=2^l , x,y\in \mbox{domain} f $ then $$f\left( \frac{k}{2^l} \cdot x +\frac{m}{2^l} \cdot y \right)\leq \frac{k}{2^l} \cdot f(x) +\frac{m}{2^l} \cdot f(y). $$ Indeed the asertion is true when $l=1 .$ Suppose that it is true for some $l\geq 1 .$ And let $k+m=2^{l+1} , k=2^l +s , m=2^l-s ,s\in\mathbb{N} . $ We have: $$f\left( \frac{k}{2^{l+1}} \cdot x +\frac{m}{2^{l+1}} \cdot y \right) =f\left( \frac{1}{2} \cdot x +\frac{1}{2}\cdot \left(\frac{s}{2^l} \cdot x+\frac{2^l -s}{2^l} \cdot y\right) \right) \leq \frac{1}{2} \cdot f(x) +\frac{1}{2} \cdot f\left(\frac{s}{2^l} \cdot x+\frac{2^l -s}{2^l} \cdot y\right)\leq \frac{1}{2} \cdot f(x) +\frac{1}{2} \cdot \frac{s}{2^l} \cdot f(x)+\frac{1}{2}\cdot\frac{2^l -s}{2^l} \cdot f(y) = \frac{k}{2^{l+1}} \cdot f(x) +\frac{m}{2^{l+1}} \cdot f(y) .$$ Hence by Induction the asertion holds true for any $l\in\mathbb{N} .$ Now let $1>\alpha >0 , $ and let $\frac{k_l}{2^l} \rightarrow \alpha $ as $l\to \infty .$ Since $f$ continuous we have $$f(\alpha x +(1-\alpha )y ) =\lim_{l\to\infty } f\left(\frac{k_l}{2^l} \cdot x + \left(1-\frac{k_l}{2^l} \right) y\right) \leq \lim_{l\to\infty } \left(\frac{k_l}{2^l} \cdot f(x) + \left( 1-\frac{k_l}{2^l} \right) f(y)\right) =\alpha f(x) +(1-\alpha )f(y) .$$

Different ways of proving a polynomial is irreducible in finite field. Is there a general characterization of irreducible polynomials over a finite field? I was going through a problem in finding whether $p(x):=x^7+x^5+1$ is irreducible over $\mathbb F_2[x]$ or not. If the polynomial is of degree less than or equal to $3$ then we can easily find out if its irreducible or not by finding whether it has a root or not. In this case considering the polynomial $p(x)=f(x)\cdot g(x)$ we may be able to show the irreducibility but this doesn't seem to be a very great idea. Can anyone suggest a better idea ?

The given polynomial is in fact not irreducible. There is at least one decomposition: $$ (x^2+x+1)\cdot(x^5+x^4+x^3+x+1) = x^7+x^5+1 $$ This can be found by resolving the equality for coefficients: $$ (x^2+ax+1)\cdot(x^5+bx^4+cx^3+dx^2+ex+1) = x^7+x^5+1 $$ which, equating term by term, and ignoring the terms of degree $0$ and $7$: $$ (a+b)x^6 + (1+ab+c)x^5 + (b+ac+d)x^4 + (c+ad+e)x^3 + (d+ae+1)x^2 + (e+a)x = x^5 $$ so $$ a+b=0 \\ ab+c=1 \\ b+ac+d = 0\\ c+ad+e=0\\ d+ae=1\\ e+a=0 $$ which are more than enough to find the solution, given that in $\mathbb{Z}_2$, $a^2=a$ and $a+a=0$ for any $a$.

Question regarding 3 x 3 matrices If $A$ is a $3 \times 3$ matrix with real elements and $\det(A)=1$, then are these affirmations equivalent: $$ \det(A^2-A+I_3)=0 \leftrightarrow \det(A+I_3)=6 \text{ and } \det(A-I_3)=0? $$

$\Leftarrow)$ From $\det (A-I)=0$ we know that $1$ is an eigenvalue. Let $x,y$ be the two others (possibly complex, and counting multiplicities). From $\det A=1$ we know that $xy=1$. And $6=\det(A+I)=2(x+1)(y+1)$, so $$ 3=xy+x+y+1=2+x+y, $$ so $x+y=1$. We get a system of two equations on $x,y$, namely $$ x+y=1,\ \ xy=1. $$ That is, $x(1-x)=1$, or $x^2-1x+1=0$. Note that $y$ satisfies the same equation. In any case $A^2-A+I$ has zero as eigenvalue, so $\det(A^2-A+I)=0$. $\Rightarrow)$ We consider the three eigenvalues of $A$, $x,y,z$; they satisfy $xyz=1$ by hypothesis. Since $\det(A^2-A+I)=0$, one of them, say $y$, satisfies $y^2-y+1=0$. This implies that $y$ is non-real, and its conjugate is also an eigenvalue (since $A$ has real entries). So $y=\frac12+i\frac{\sqrt3}2$, $z=\frac12-i\frac{\sqrt3}2$, and $$ x=\frac1{yz}=\frac1{\frac14+\frac34}=1. $$ Now $\det(A-I)=0$ (since $1$ is an eigenvalue) and $$ \det(A+I)=2(y+1)(z+1)=2\left(\frac32+i\frac{\sqrt3}2\right)\left(\frac32-i\frac{\sqrt3}2\right)=2\,\left(\frac94+\frac34\right)=6. $$

How to show that $A_k=(-1)^k\binom nk$? In the identity $$\frac{n!}{x(x+1)(x+2)\cdots(x+n)}=\sum_{k=0}^n\frac{A_k}{x+k},$$prove that $$A_k=(-1)^k\binom nk.$$ My try: The given identity implies $$\frac{1\cdot2\cdots n}{x(x+1)(x+2)\cdots(x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1}+\dots+\frac{A_n}{x+n}.$$ Now putting $A_k=(-1)^k\binom nk,$$$\frac{1\cdot2\cdots n}{x(x+1)(x+2)\cdots(x+n)}=\frac1x-\frac{n}{x+1}+\dots+\frac{(-1)^n}{x+n}.$$ How to proceed further?

HINT: Multiply both side by $x+k$ and then put $x=-k.$

How to solve this reccurence relation? Let a,b,c be real numbers. Find the explicit formula for $f_n=af_{n-1}+b$ for $n \ge 1$ and $f_0 = c$ So I rewrote it as $f_n-af_{n-1}-b=0$ which gives the characteristic equation as $x^2-ax-b=0$. The quadratic formula gives roots $x= \frac{a+\sqrt{a^2+4b}}{-2}, \frac{a-\sqrt{a^2+4b}}{-2}$ Then $f_n=P_1(\frac{a+\sqrt{a^2+4b}}{-2})^n+P_2(\frac{a-\sqrt{a^2+4b}}{-2})^n$ and using the initial condition $t_0=c$ gives $C=P_1+P_2 \Rightarrow P_1=C-P_2$ So $(C-P_2)(\frac{a+\sqrt{a^2+4b}}{-2})^n+P_2(\frac{a-\sqrt{a^2+4b}}{-2})^n$ what next? I tried expanding but that didn't help. I know the answer is something like $cd^n-\frac{b}{a-1}+\frac{bd^n}{a-1}$

Why not consider this? $f_n + m = a(f_{n-1} + m) \Longrightarrow (a-1)m=b$ 1) $a=1$, simple recurrence $f_n = f_{n-1} + b$, $f_n = bn+c$ 2) $a\neq 1$, $m=\frac{b}{a-1}$, $f_n+m = a(f_{n-1}+m)$, geometric sequence $f_n+m=a^n(c+m)$ Hope it is helpful!

How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why \begin{align} \sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6 \end{align} I tried to rewrite it into a geometric series \begin{align} \sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2 \end{align} But I don't know what to do with the $n^2$.

Let me show you a slightly different approach; this approach is very powerful, and can be used to compute values for a large number of series. Let's think of this in terms of power series. You noticed that you can write $$ \sum_{n=0}^{\infty}\frac{n^2}{2^n}=\sum_{n=0}^{\infty}n^2\left(\frac{1}{2}\right)^n; $$ so, let's consider the power series $$ f(x)=\sum_{n=0}^{\infty}n^2x^n. $$ If we can find a simpler expression for the function $f(x)$, and if $\frac{1}{2}$ lies within its interval of convergence, then your series is exactly $f(\frac{1}{2})$. Now, we note that $$ f(x)=\underbrace{0}_{n=0}+\sum_{n=1}^{\infty}n^2x^n=x\sum_{n=1}^{\infty}n^2x^{n-1}. $$ But, notice that $\int n^2x^{n-1}\,dx=nx^n$; so, $$\tag{1} \sum_{n=1}^{\infty}n^2 x^{n-1}=\frac{d}{dx}\left[\sum_{n=0}^{\infty}nx^n\right]. $$ Now, we write $$\tag{2} \sum_{n=0}^{\infty}nx^n=\underbrace{0}_{n=0}+x\sum_{n=1}^{\infty}nx^{n-1}=x\frac{d}{dx}\left[\sum_{n=0}^{\infty}x^n\right]. $$ But, this last is a geometric series; so, as long as $\lvert x\rvert<1$, $$ \sum_{n=0}^{\infty}x^n=\frac{1}{1-x}. $$ Plugging this back in to (2), we find that for $\lvert x\rvert<1$, $$ \sum_{n=0}^{\infty}nx^n=x\frac{d}{dx}\left[\frac{1}{1-x}\right]=x\cdot\frac{1}{(1-x)^2}=\frac{x}{(1-x)^2}. $$ But, plugging this back in to (1), we find that $$ \sum_{n=1}^{\infty}n^2x^{n-1}=\frac{d}{dx}\left[\frac{x}{(1-x)^2}\right]=\frac{x+1}{(1-x)^3} $$ So, finally, $$ f(x)=x\cdot\frac{x+1}{(1-x)^3}=\frac{x(x+1)}{(1-x)^3}. $$ Plugging in $x=\frac{1}{2}$, we find $$ \sum_{n=0}^{\infty}\frac{n^2}{2^n}=f\left(\frac{1}{2}\right)=6. $$

How to put $2x^2 + 4xy + 6y^2 + 6x + 2y = 6$ in canonical form We are given the equation $2x^2 + 4xy + 6y^2 + 6x + 2y = 6$ We did an example of this in class but the equation had less terms. I took a note in class that says : if there are linear terms, I have to rotate... This is what I think I have to do. * *Put the coefficients of this equation in a matrix $A$, then $B = det(A - I \lambda)$ *$det(B)$ must be $ = 0$ *Evaluating this determinant will give me a value for lambda that is the root of the equation. *I can then go on and find eigen lines or eigen vectors. My questions : * *Am I right about the steps to solve it ? Shouldn't it give me a canonical form at the end ? because it my notes, it doesn't. *How do I place the coefficients in the matrix A ? this is my try : $A = \begin{bmatrix}2 & 2\\2 & 6 \\\end{bmatrix}$ the first 2 is for the $2x^2$ term, the second 2 next are for the $4xy$ term which I divided by 2...( I dont really know why btw :\ ) and the 6 is for the $6y^2$ term. Now how do I place the 6x and 2y terms ? $A = \begin{bmatrix}2 & 2\\2 & 6\\3 & 1 \\\end{bmatrix}$ I add another line with those 2 divided by 2 ? edit : Here's my current work I rewrote A as : $ \begin{bmatrix} x & y \\\end{bmatrix} $ $ A = \begin{bmatrix}2 & 2\\2 & 6 \\\end{bmatrix}$ $ \begin{bmatrix} x \\ y \\\end{bmatrix} $ + $ \begin{bmatrix} 6 & 2 \\\end{bmatrix} $ $ \begin{bmatrix} x \\ y \\\end{bmatrix} $ - 6 $B = \begin{bmatrix}2 - \lambda & 2\\2 & 6 - \lambda \\\end{bmatrix}$ $det(B) = (2 - \lambda)(6 - \lambda) - 4$ $det(B) = 12 - 2 \lambda - 6 \lambda + \lambda^2 - 4 $ And I'm stuck here. How do I factor this ? $\lambda^2 - 8 \lambda + 8$ edit : so with the quadration formula I found two roots. $4 + \sqrt{\frac{32}{2}}$ and $4 - \sqrt{\frac{32}{2}}$

Not knowing what exactly "canonical form" is, here is what I get. Translating to get rid of the linear terms: $$ 2(x+2)^2+4(x+2)(y-1/2)+6(y-1/2)^2=\frac{23}{2}\tag{1} $$ With $P=\dfrac{\sqrt{2+\sqrt2}}{2}\begin{bmatrix}1&1-\sqrt2\\-1+\sqrt2&1\end{bmatrix}=\begin{bmatrix}\cos(\pi/8)&-\sin(\pi/8)\\\sin(\pi/8)&\hphantom{+}\cos(\pi/8)\end{bmatrix}$ we have $$ \begin{bmatrix}2&2\\2&6\end{bmatrix} =P^T \begin{bmatrix}4-\sqrt8&0\\0&4+\sqrt8\end{bmatrix} P\tag{2} $$ Therefore, $$ \begin{bmatrix}x+2\\y-1/2\end{bmatrix}^TP^T \begin{bmatrix}4-\sqrt8&0\\0&4+\sqrt8\end{bmatrix} P\begin{bmatrix}x+2\\y-1/2\end{bmatrix} =\frac{23}2 $$ Thus, the curve is an ellipse with its center at $(-2,1/2)$ and its major axis tilted $\pi/8$ clockwise from the $x$-axis. The major and minor axes are $$ \sqrt{\frac{23}8(2\pm\sqrt2)} $$ To get $(1)$, we translated to get rid of the linear terms. So $$ 2(x+h)^2+4(x+h)(y+k)+6(y+k)^2\\ =(2x^2+4hx+2h^2)+(4xy+4kx+4hy+4hk)+(6y^2+12ky+6k^2)\\ =(2x^2+4xy+6y^2)+(4h+4k)x+(4h+12k)y+(2h^2+4hk+6k^2) $$ To match the linear terms, we need $4h+4k=6$ and $4h+12k=2$. Thus, $h=2$ and $k=-1/2$: $$ 2(x+2)^2+4(x+2)(y-1/2)+6(y-1/2)^2\\ =2x^2+4xy+6y^2+6x+2y+\frac{11}2=6+\frac{11}2=\frac{23}2 $$ To get $(2)$, we diagonalized $\begin{bmatrix}2&2\\2&6\end{bmatrix}$. The matrix $P$ rotates counterclockwise by $\pi/8$. After translating by $(2,-1/2)$ and rotating $\pi/8$ counterclockwise, we are left with the ellipse $$ (4-\sqrt8)x^2+(4+\sqrt8)y^2=\frac{23}2 $$

Find the value of $\cos(2\pi /5)$ using radicals This is homework so if there is another example that can illustrate the technique I would happily accept that as guidance. The only thing I have been able to find is a question asking about $\cos(2\pi/7)$, which I think is a much harder problem. I dont have the faintest idea how to solve this and the textbook (Hungerford) doesn't have any examples at all. Ive tried looking for resources online but havent found any that I was able to understand. So can anyone show me how to solve these types of problems? Thanks a bunch.

Note that $2\cdot \dfrac{2\pi}{5} + 3\cdot \dfrac{2\pi}{5} = 2\pi,$ therefore $\cos(2⋅\dfrac{2\pi}{5})=\cos(3⋅\dfrac{2\pi}{5})$. Put $\cos(\dfrac{2\pi}{5})=x$. Using the formulas $\cos2x=2\cos2x−1,\cos 3x=4\cos 3x−3\cos x$, we have $4x^3−2x^2−3x+1=0⇔(x−1)(4x^2+2x−1)=0$. Because $\cos(\dfrac{2\pi}{5})≠1$, we get $4x^2+2x−1=0.$ Another way, $\cos(\dfrac{2\pi}{5})>0$, then $\cos \dfrac{2\pi}{5} = \dfrac{-1 + \sqrt{5}}{2}$.

Finding a 2x2 Matrix raised to the power of 1000 Let $A= \pmatrix{1&4\\ 3&2}$. Find $A^{1000}$. Does this problem have to do with eigenvalues or is there another formula that is specific to 2x2 matrices?

Perform an eigenvalue decomposition of $A$, we then get $$A = \begin{bmatrix} -4/5 & -1/\sqrt2\\ 3/5 & -1/\sqrt2 \end{bmatrix} \begin{bmatrix} -2 & 0\\ 0 & 5 \end{bmatrix} \begin{bmatrix} -4/5 & -1/\sqrt2\\ 3/5 & -1/\sqrt2 \end{bmatrix}^{-1} =VDV^{-1} $$ where $V = \begin{bmatrix} -4/5 & -1/\sqrt2\\ 3/5 & -1/\sqrt2 \end{bmatrix}$ and $D = \begin{bmatrix} -2 & 0\\ 0 & 5 \end{bmatrix}$. Hence, $$A^n = \underbrace{\left(VDV^{-1} \right)\left(VDV^{-1} \right)\cdots \left(VDV^{-1} \right)}_{n \text{ times}} = VD^n V^{-1}$$

Equations and inequalities as parameters: proving that an equation holds. I have $x-y=3$ and $y\le1$ and $x\ge\frac12$. I proved that $\sqrt{(2x-1)^2}+\sqrt{(2y-2)^2}=7$ and that $-\frac52\le y\le 1$ and $\frac12\le x\le4$. How can I prove that $|x+y-5|+|x+y+2|=7$?

ok let us consider following cases : first side is just $-x-y+5$,second case is $x+y+2$,so sum is $5+2=7$ to be more deeply,let us take such situations * *$y=1$ 2.$x=4$ we have $|x+y-5]+|x+y+2|=x+y-5+x+y+2=2*x+2*y-3=2*(x+y)-3 =2*(5)-3=7$ can you continue from this? just consider this situation when $x<0$ then $|x|=-x$,else if $x>=0$ then $|x|=x$.ok let us consider following case,when $y=-5/2$ and $x=1$ ,we have that $-5/2+1-5<0$,that why we will have $-x-y+5$, for second we have $-5/2+1+2>0$ , that why we have second as $x+y+2$,together is is still $7$

Concentric and Tangent Ellipse from 2 Hyperbolas Find the equation of the ellipse that is concentric and tangent to the following hyperbolas: $$\begin{align} -2x^2 + 9y^2 - 20x - 108y + 256 &= 0 \\ x^2 - 4y^2 + 10x + 48y - 219 &= 0 \end{align}$$ I did the math for both equations and the center is the same: $(-5,6)$. I have the equations of each: For the first: $$-2x^2 + 9y^2 - 20x - 108y + 256 = 0$$ $$-\frac{(x+5)^2}{9} + \frac{(y-6)^2}{2} = 1$$ For the 2nd: $$x^2 - 4y^2 + 10x + 48y - 219 = 0$$ $$\frac{(x+5)^2}{100} - \frac{(y-6)^2}{25} = 1$$ I know concentric means the new Ellipse has to have the same center, but I don't know how to make it tangent. Please help. Thanks

Not a complete solution, but this approach will work: An ellipse with centre at the point $(-5,6)$ would be $$\frac{(x+5)^2}{a^2}+\frac{(y-6)^2}{b^2} = 1$$ Now change to a new set of axes ($u, v$) parallel to the $x,y$ axes, but with origin at the point $(-5,6)$. In other words, put $u = x+5$ and $v = y-6$. Referred to the new axes, the equations of the three conics become: $$-\frac{u^2}{9} + \frac{v^2}{2} = 1,$$ $$\frac{u^2}{100} - \frac{v^2}{25} = 1,$$ $$\frac{u^2}{a^2}+\frac{v^2}{b^2} = 1$$ (Note that the values $a$ and $b$ are unchanged by the change in coordinates.) Now, looking at a hyperbola $u^2/a^2-v^2/b^2 = \pm 1$ and an ellipse $u^2/c^2+v^2/d^2 = 1$, you can see (from a sketch) that the ellipse will be a tangent to the hyperbola only when they meet at a point on the $u$ or $v$ axis. Can you finish it from there?

Limit of $\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$ I have to determine the following: $\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$ $\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)=\lim\limits_{x \rightarrow \infty}(\sqrt{x^8(1+\frac{4}{x^8})}-x^4 = \lim\limits_{x \rightarrow \infty}(x^4\sqrt{1+\frac{4}{x^8}}-x^4 = \lim\limits_{x \rightarrow \infty}(x^4(\sqrt{1+\frac{4}{x^8}}-1)= \infty$ Could somebody please check, if my solution is correct?

A short way to (non-rigorously) find the limit is to observe that for large $x$, $$ \sqrt{x^8+4} \approx \sqrt{x^8}=x^4 $$ so that for large $x$ (especially in $\lim_{x \to \infty}$) $$ \sqrt{x^8+4}-x^4 \approx x^4-x^4=0 $$ So the limit must be $0$.

Integral involving a confluent hypergeometric function I have the following integral involving a confluent hypergeometric function: $$\int_{0}^{\infty}x^3e^{-ax^2}{}_1F_1(1+n,1,bx^2)dx$$ where $a>b>0$ are real constants, and $n\geq 0$ is an integer. Wolfram Mathematica returns the following solution: $\frac{a^{n-1}(a+bn)}{(a-b)^{n+2}}$. However, I can't figure out how it arrived at it (I always try to check the solutions "on paper" that Mathematica gives me -- or at least using Gradshteyn and Ryzhik). Can anyone help?

Let's start with the hypergeometric function. We have: \begin{eqnarray} F_{2,1}[1+n,1;b x^2] &=& \sum\limits_{m=0}^\infty \frac{(1+n)^{(m)}}{m!} \cdot \frac{(b x^2)^m}{m!} \\ &=& \sum\limits_{m=0}^\infty \frac{(m+1)^{(n)}}{n!} \cdot \frac{(b x^2)^m}{m!} \\ &=&\left. \frac{1}{n!} \frac{d^n}{d t^n} \left( t^n \cdot e^{b x^2 \cdot t} \right)\right|_{t=1} \\ &=& \frac{1}{n!} \sum\limits_{p=0}^n \binom{n}{p} n_{(p)} (b x^2) ^{n-p} \cdot e^{b x^2} \end{eqnarray} Therefore the integral in question reads: \begin{eqnarray} &&\int\limits_0^\infty x^3 e^{-a x^2} F_{2,1}[1+n,1;b x^2] dx =\\ && \frac{1}{2} \sum\limits_{p=0}^n \frac{n^{(p)}}{p! (n-p)!} b^{n-p} \frac{(n-p+1)!}{(a-b)^{n-p+2}} \\ &&\frac{1}{2} \frac{1}{(a-b)^{n+2}} \sum\limits_{p=0}^n \binom{n}{p} \frac{(n-p+1)^{(p)}}{(n-p+2)^{(p-1)}} \cdot b^{n-p} (a-b)^p \\ && \frac{1}{2} \frac{1}{(a-b)^{n+2}} \sum\limits_{p=0}^n \binom{n}{p} (n-p+1) \cdot b^{n-p} (a-b)^p \\ && \frac{1}{2} \frac{1}{(a-b)^{n+2}} a^{n-1} (a+b n) \end{eqnarray}

Find $f$ if $f(f(x))=\sqrt{1-x^2}$ Find $f$ if $f(f(x))=\sqrt{1-x^2} \land [-1; 1] \subseteq Dom(f)$ $$$$Please give both real and complex functions. Can it be continuous or not (if f is real)

I guessed that $f$ is of the form $\sqrt{ax^2+b}$. Then, $f^2$ is $\sqrt{a^2x^2 + \frac{a^2-1}{a-1}b}$. From here on in, it is algebra: $$ a^2 =-1 \implies a = i ~~~~\text{and}~~~~\frac{a^2-1}{a-1}b = 1 \implies b = \frac{1-i}{2} $$ So we get $f(x) = \sqrt{ix^2 + \frac{1-i}{2}}$. I checked using Wolfram, and $f^2$ appears to be what we want. DISCLAIMER: This is not the only solution.

Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$ How can I prove that $$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$ I think this post can help me, but I'm not sure.

Different approach: $$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\sum_{n=1}^\infty\frac{H_{2n}+\frac{1}{2n+1}}{n^2}$$ $$=\sum_{n=1}^\infty\frac{H_{2n}}{n^2}+\sum_{n=1}^\infty\frac{1}{n^2(2n+1)}$$ where $$\sum_{n=1}^\infty\frac{H_{2n}}{n^2}=4\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}$$ $$=4\sum_{n=1}^\infty\frac{1}{2n}\left(-\int_0^1x^{2n-1}\ln(1-x)dx\right)$$ $$=-2\int_0^1\frac{\ln(1-x)}{x}\sum_{n=1}^\infty\frac{x^{2n}}{n}$$ $$=2\int_0^1\frac{\ln(1-x)\ln(1-x^2)}{x}dx$$ $$=2\int_0^1\frac{\ln^2(1-x)}{x}dx+2\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx$$ $$=2(2\zeta(3))+2(-\frac58\zeta(3))=\frac{11}{4}\zeta(3)$$ and $$\sum_{n=1}^\infty\frac{1}{n^2(2n+1)}=\sum_{n=1}^\infty\frac{1}{n^2}\int_0^1 x^{2n}dx$$ $$=\int_0^1\sum_{n=1}^\infty \frac{x^{2n}}{n^2}=\int_0^1\text{Li}_2(x^2)dx$$ $$=x\text{Li}_2(x^2)|_0^1+2\int_0^1\ln(1-x^2)dx$$ $$=\zeta(2)+2(x-1)\ln(1-x^2)|_0^1-4\int_0^1\frac{x}{1+x}dx$$ $$=\zeta(2)-4(1-\ln(2))$$

solving an expression based on sin $\theta$ If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$, then $x$ must be equal to what? What does the following solution mean? $0 \le \sin^2 \le 1$ This implies $0 \le \frac{x^2 + y^2 + 1 }{ 2x } \le 1$ This implies $\frac{(x - 1)^2 + y^2 }{2x} \le 0 $ This implies $x = 1$. Can you please explain me the solution?

We have $$(x-1)^2=x^2+1-2x\ge0\iff x^2+1\ge 2x$$ and since $\sin^2\theta\ge0$ hence we have $x>0$ and then $$1\ge\sin^2\theta =\frac{x^2+y^2+1}{2x}\ge1\quad (>1\;\text{if}\; y\ne 0\;\text{which gives a contradiction})$$ hence $y=0$ and $$\sin^2\theta=\frac{x^2+1}{2x}=1\iff x=1$$

A fair dice is tossed until a number greater than $4$ appears. The probability that an even number of tosses will be required is A fair dice is tossed until a number greater than $4$ appears. The probability that an even number of tosses will be required is: $A. 1/2$ $B. 3/5$ $C. 1/5$ $D. 2/3$ What I did: The probability should be $(2/3)*(1/3)+(2/3)*(2/3)*(2/3)*(1/3)+(2/3)*(2/3)(2/3)*(2/3)*(2/3)*(1/3)+...$, which should be $2/5$. But this answer doesn't match with the options. Where did I go wrong?

You didn't go wrong. The probability of "success" is 1/3, so the probability of succeding in $2k$ tosses is $$\left(\frac{2}{3}\right)^{2k-1} \frac{1}{3}$$ And $$\sum_{k=1}^\infty\left(\frac{2}{3}\right)^{2k-1} \frac{1}{3}=\frac{1}{3} \frac{3}{2}\sum_{k=1}^\infty\left(\frac{4}{9}\right)^k =\frac{1}{2} \frac{4/9}{1-4/9}=\frac{2}{5}$$

$\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ if........... Help please: If $\sin\alpha+\sin\beta= \sqrt{3} (\cos\beta-\cos\alpha)$ then show that $\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ please tell me how can I approach

Apart from the Prosthaphaeresis Formulas already mentioned with the unmentioned assumption that $\displaystyle \sin\frac{\alpha+\beta}2\ne0$ we can try as follows : Rearranging we have $\displaystyle\sin\alpha+\sqrt3\cos\alpha=\sqrt3\cos\beta-\sin\beta$ As $\displaystyle 1^2+(\sqrt3)^2=4,$ we write $1=2\sin30^\circ,\sqrt3=2\cos30^\circ$ to get $\displaystyle2\sin30^\circ\sin\alpha+2\cos30^\circ\cos\alpha=2\cos30^\circ\cos\beta-2\sin30^\circ\sin\beta$ $\displaystyle\implies \cos(\alpha-30^\circ)=\cos(30^\circ+\beta)$ $\displaystyle\implies \alpha-30^\circ=n360^\circ\pm(30^\circ+\beta)$ Taking the '-' sign, $\displaystyle\implies \alpha-30^\circ=n360^\circ-(30^\circ-\beta)\implies \alpha=n360^\circ-\beta$ This makes both sides equal for all $\alpha,$ so no solution available from here Taking the '+' sign, $\displaystyle\implies \alpha-30^\circ=n360^\circ+(30^\circ-\beta)\implies \alpha-\beta=n360^\circ+60^\circ$ Now, use $\displaystyle2\cos^2A=1+\cos2A$ as $\displaystyle\cos2A=2\cos^2A-1$ for $\displaystyle A=\frac{\alpha-\beta}2$

Doubts in Trigonometrical Inequalities I'm now studying Trigonometrical Inequalities, and I've just got struck when I have modified arguments to my trigonometrical functions, for example: $\sqrt{2} - 2\sin\left(x - \dfrac{\pi}{3} \right) < 0$ when $-\pi < x < \pi$ With some work I've got: $\sin\left(x - \dfrac{\pi}{3} \right) > \dfrac{\sqrt{2}}{2}$ To find bounds: $\sin(x) = \dfrac{\sqrt{2}}{2},\ x = \dfrac{\pi}{4},\ \dfrac{3\pi}{4}$ Resolving to $x + \dfrac{\pi}{3} \implies \dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$ But wolfram gives a way different result, where's my mistake ?

The inequality $$ \sin \left( x - \frac{\pi}{3} \right) > \frac{\sqrt{2}}{2} $$ implies that $$ \frac{\pi}{4} < x - \frac{\pi}{3} < \frac{3\pi}{4}. $$ Adding $\frac{\pi}{3}$ to all three expressions yields $$ \frac{7\pi}{12} < x < \frac{13\pi}{12}. $$ If you impose the initial restriction, then the upper bound is $\pi$.

Help finding the $\lim\limits_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$ I need help finding the $$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$$ I did the following: $$\begin{align*} \lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}} =& \lim_{x \to \infty} \frac{(\sqrt[3]{x} - \sqrt[5]{x})(\sqrt[3]{x} + \sqrt[5]{x})}{(\sqrt[3]{x} + \sqrt[5]{x})(\sqrt[3]{x} + \sqrt[5]{x})}\\ \\ =& \lim_{x \to \infty} \frac{(\sqrt[3]{x})^2 - (\sqrt[5]{x})^2}{(\sqrt[3]{x})^2+2\sqrt[3]{x}\sqrt[5]{x}+(\sqrt[3]{x})^2}\\ \\ =& \lim_{x \to \infty} \frac{x^{2/3}-x^{2/5}}{x^{2/3}+2x^{1/15}+x^{2/5}}\\ \\ =& \lim_{x \to \infty} \frac{x^{4/15}}{2x^{17/15}} \end{align*}$$ Somehow I get stuck. I am sure I did something wrong somewhere.. Can someone please help me out?

$$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}=\lim_{x \to \infty} \frac{x^{1/3}-x^{1/5}}{x^{1/3} + x^{1/5}}=$$ $$=\lim_{x \to \infty} \frac{1-x^{1/5-1/3}}{1 + x^{1/5-1/3}}=\lim_{x \to \infty} \frac{1-x^{-2/15}}{1 + x^{-2/15}}=\lim_{x \to \infty} \frac{1-\frac{1}{x^{2/15}}}{1 + \frac{1}{x^{2/15}}}=1$$

Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers. Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(2,6,3),(3,6,2),(3,2,6),(6,2,3),(6,3,2)}$. Are these the only 10 solutions? First, none of $x$, $y$ or $z$ can be $1$ ($x$, $y$ and $z$ are positive integers) If I let $x=2$, then finding all solutions to $1/y+1/z = 1/2$ leads to $(4,4), (3,6)$ and $(6,3)$ which gives me $(x,y,z)$ as $(2,4,4), (2,3,6), (2,6,3)$ but this also means $(4,4,2), (4,2,4), (3,2,6), (3,6,2), (6,2,3), (6,3,2)$ are all valid triples for this equation. If I let $x=3$, the only different values of $y$ and $z$ are $(3,3)$ How do I prove these are the only ten solutions? (without using any programming) Known result: If we denote $d(n^2)$ as the number of divisors of $n^2$, then the number of solutions of ${\frac {1} {x} }+{\frac {1} {y} } = {\frac {1} {n} }$ = $d(n^2)$ (For positive $x$, $y$) For ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ $z = \frac{xy}{y(x-1)-x}$ where $xy \neq 0$ What happens after that? Question is: how do we sho there are the only ten solutions? I'm not asking for a solution. Assuming $x \le y \le z$ $1 \le y \le \frac{xy}{y(x-1)-x}$ $\Longrightarrow 1 \le x \le y \le \frac{2x}{x-1} $ Got the answer. I'll probably call @mathlove's answer. (Any additional answers I'll view later) Liked @user44197 answer.

We may as well assume $x\le y\le z$ (and then count rearrangements of the variables as appropriate). The smallest variable, $x$, cannot be greater than $3$ (or else $1/x+1/y+1/z\lt1/3+1/3+1/3=1$), nor can it be equal to $1$ (or else $1/x+1/y+1/z=1+1/y+1/z\gt1$). So either $x=2$ or $x=3$. If $x=3$, then $y=z=3$ as well (for the same reason as before), which gives the solution $(x,y,z)=(3,3,3)$. If $x=2$, then $1/2+1/y+1/z=1$ implies $${1\over2}={1\over y}+{1\over z}$$ Applying the inequality $y\le z$ to this equation, we see that $y$ must be greater than $2$ but cannot be greater than $4$, so $y=3$ or $y=4$. Each of these gives a solution, $(x,y,z)=(2,3,6)$ and $(2,4,4)$. Counting rearrangements, we get the OP's $10$ solutions and no others.

Double Euler sum $ \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} $ I proved the following result $$\displaystyle \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} =- \frac{97}{12} \zeta(6)+\frac{7}{4}\zeta(4)\zeta(2) + \frac{5}{2}\zeta(3)^2+\frac{2}{3}\zeta(2)^3$$ After consideration of powers of polylogarithms. You can refer to the following thread . My question is : are there any papers in the literature which dealt with that result? Are my evaluations worth publishing ?

The following new solution is proposed by Cornel Ioan Valean. Based on a few ideas presented in the book, (Almost) Impossible Integrals, Sums, and Series, like the Cauchy product of $(\operatorname{Li}_2(x))^2$, that is $\displaystyle (\operatorname{Li}_2(x))^2=4\sum_{n=1}^{\infty}x^n\frac{H_n}{n^3}+2\sum_{n=1}^{\infty}x^n\frac{H_n^{(2)}}{n^2}-6\sum_{n=1}^{\infty}\frac{x^n}{n^4}$, where if we multiply both sides by $\displaystyle \frac{\log(1-x)}{x}$ and then integrate from $x=0$ to $x=1$, using that $\displaystyle \int_{0}^{1}x^{n-1}\log(1-x)\textrm{d}x=-\frac{H_{n}}{n}$, we get \begin{equation*} \int_0^1 \frac{\log(1-x)}{x}(\operatorname{Li}_2(x))^2 \textrm{d}x=-\frac{1}{3}(\operatorname{Li}_2(x))^3\biggr|_{x=0}^{x=1}=-\frac{35}{24}\zeta(6) \end{equation*} \begin{equation*} =6\sum_{n=1}^{\infty} \frac{H_n}{n^5}-4\sum_{n=1}^{\infty} \frac{H_n^2}{n^4}-2\sum_{n=1}^{\infty} \frac{H_nH_n^{(2)}}{n^3} \end{equation*} \begin{equation*} =5\zeta^2(3)-\frac{17}{3}\zeta(6)-2\sum_{n=1}^{\infty} \frac{H_nH_n^{(2)}}{n^3}, \end{equation*} where the first sum comes from the classical generalization, $ \displaystyle 2\sum_{k=1}^\infty \frac{H_k}{k^n}=(n+2)\zeta(n+1)-\sum_{k=1}^{n-2} \zeta(n-k) \zeta(k+1), \ n\in \mathbb{N},\ n\ge2$, and the second sum, $\displaystyle \sum_{n=1}^{\infty} \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$, is calculated in the mentioned book or in this article. To conclude, we have \begin{equation*} \sum_{n=1}^{\infty}\frac{H_n H_n^{(2)}}{n^3}=\frac{1}{2}\left(5\zeta^2(3)-\frac{101}{24}\zeta(6)\right). \end{equation*} Note the present solution circumvents the necessity of using the value of the series $\displaystyle \sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^3$.

Help with finding tangent to curve at a point Find an equation for the tangent to the curve at $P\left( \dfrac{\pi}{2},3 \right )$ and the horizontal tangent to the curve at $Q.$ $$y=5+\cot x-2\csc x$$ $y\prime=-\csc ^2 x -2(-\csc x \cot x)$ $y\prime= 2\csc x \cot x - \csc ^2 x\implies$ This is the equation of the slope. Now I find the slope at $x=\dfrac{\pi}{2}:$ $$y^{\prime}=2\csc \left( \frac{\pi}{2} \right) \cot \left( \frac{\pi}{2} \right)- \csc ^2 \left( \frac{\pi}{2} \right)$$ $$y^{\prime}= -1$$ Equation for the tangent to the curve at $P$: $$y-3=-1\left( x-\frac{\pi}{2} \right)$$ $$y=-x+\frac{6+\pi}{2}$$ Then I found the horizontal tangent at $Q$, which is where the slope is $0.$ So I set the slope equation equal to $0:$ $$y^{\prime}= 2\csc x \cot x - \csc ^2 x=0$$ $$-\csc ^2 x = -2\csc x \cot x$$ $$-\frac{1}{\sin^2 x}=-2\left( \frac{1}{\sin x} \frac{\cos x}{\sin x} \right )$$ $$\cos x = \frac{1}{2}$$ I don't know how to proceed from here. Since $\cos x = \dfrac{1}{2}$, $\cos^{-1}\left (\dfrac{1}{2}\right)=\dfrac{\pi}{3}$. Plugging this into the equation would give the $y$ value, and I'd be able to find the equation of the tangent line, but I'm confused because there is not just one $x$ to consider, since $x=2n \pm \dfrac{\pi}{3}$. The answer for equation of tangent at $Q$ is $y=5-\sqrt{3}$, and I don't know how to get this. Can you please show how to work this last part in details? Thank you.

The first part looks good. For the second part as you see the solution to the equation gives $$2 \cot x \csc x - \csc^2 x = 0 \Rightarrow x=2n \pi \pm\frac{\pi}{3}.$$ Choose $x=\frac{\pi}{3}$ as it is one such solution. We find the point on $y$ to be $$5+\cot\left( \frac{\pi}{3} \right)-2\csc\left( \frac{\pi}{3} \right)=5-\sqrt{3},$$ And so $y=5-\sqrt{3}$ is indeed a horizontal tangent line, in particular the horizontal tangent line at $x=\frac{\pi}{3}$. You can verify this in WA.

Plane Geometry problem I came across this problem in a mathematics-related facebook group. Could anyone advise on the solution to it(i.e. hints only)? Thank you.

Since triangle ABC is equilateral, we know that angle BAC is 60º. Therefore, angle PAQ is also 60º; we have a circle theorem that tells us that angle POQ is thus twice that measure or 120º. The diameter AD bisects that angle, so angle POD is 60º. Diameter AD is also an altitude of equilateral triangle ABC: thus, if $ \ s \ $ is the length of one side of that triangle, the diameter of the circle is $ \ \frac{\sqrt{3}}{2}s \ $ . The radius OD therefore has length $ \ \frac{\sqrt{3}}{4}s \ $ . Since the altitude AD bisects side BC, then the segment BD has length $ \ \frac{1}{2}s \ $ . Hence, $ \ \tan(\angle BOD ) \ = \ \frac{2}{\sqrt{3}} \ $ . Now, the measure of angle POB, designated as $ \ \phi \ , $ is the difference given by the measure of angle POD (call it $ \ \alpha \ $ ) minus the measure of angle BOD (call that $ \ \beta \ $ ) . We can then use the formula for the tangent of the difference of two angles, with $ \ \tan(\angle POD) \ = \ \tan(\alpha) \ = \ \tan \ 60º \ = \ \sqrt{3} \ $ , to obtain $$ \tan \phi \ = \ \tan( \alpha - \beta ) \ = \ \frac{\tan \alpha \ - \ \tan \beta}{1 \ + \ \tan \alpha \cdot \tan \beta} \ = \ \frac{\sqrt{3} \ - \ \frac{2}{\sqrt{3}}}{1 \ + \ \sqrt{3} \cdot \frac{2}{\sqrt{3}}} \ = \ \frac{ \frac{3 \ - \ 2}{\sqrt{3}}}{1 \ + \ 2 } \ = \ \frac{1}{3 \ \sqrt{3}} \ . $$ We thereby conclude that $ \ \cot \phi \ = \ 3 \ \sqrt{3} \ . $

evaluation of $\lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $ (1) $\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$ (2)$\displaystyle \lim_{x\rightarrow \infty}\left({\sqrt{\lfloor x^2+x \rfloor }-x}\right) = , $where $\lfloor x \rfloor = $ floor function of $x$ $\bf{My\; Try}::$ for (1) one :: We can write as $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln x-\lfloor x \rfloor }{\lfloor x \rfloor}$ and we can say that when $x\rightarrow \infty$, Then $\lfloor x\rfloor \rightarrow x$ So $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln(x)-x}{x} = n\lim_{x\rightarrow \infty}\frac{\ln (x)}{x}-1$ Now Let $\displaystyle L = \lim_{x\rightarrow \infty}\frac{\ln(x)}{x}{\Rightarrow}_{L.H.R} =\lim_{x\rightarrow \infty}\frac{1}{x} = 0$ So $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln x-\lfloor x \rfloor }{\lfloor x \rfloor} = n\cdot 0-1 =-1$ $\bf{My\; Try}::$ for (2)nd one::we can say that when $x\rightarrow \infty$, Then $\lfloor x^2+x\rfloor\rightarrow (x^2+x)$ So $\displaystyle \lim_{x\rightarrow \infty}\left({\sqrt{x^2+x}-x}\right) = \lim_{x\rightarrow \infty}\frac{\left({\sqrt{x^2+x}-x}\right)\cdot \left({\sqrt{x^2+x}+x}\right)}{\left({\sqrt{x^2+x}+x}\right)}$ $\displaystyle \lim_{x\rightarrow \infty}\frac{x}{\left(\sqrt{x^2+x}+x\right)} = \frac{1}{2}$ Now my doubt is can we write when $x\rightarrow \infty$, Then $\lfloor x\rfloor \rightarrow x$ and when $x\rightarrow \infty$, Then $\lfloor x^2+x\rfloor\rightarrow (x^2+x)$ please clear me Thanks

No. You cannot write so. In this case, going back to the definition is the best. So, in order to solove your questions, use the followings : $$x-1\lt \lfloor x\rfloor \le x\Rightarrow \frac{x-1}{x}\le \frac{\lfloor x\rfloor}{x}\lt \frac{x}{x}\Rightarrow \lim_{x\to\infty}\frac{\lfloor x\rfloor}{x}=1$$ where $x\gt0.$

How prove this stronger than Weitzenbock's inequality:$(ab+bc+ac)(a+b+c)^2\ge 12\sqrt{3}\cdot S\cdot(a^2+b^2+c^2)$ In $\Delta ABC$,$$AB=c,BC=a,AC=b,S_{ABC}=S$$ show that $$(ab+bc+ac)(a+b+c)^2\ge 12\sqrt{3}\cdot S\cdot(a^2+b^2+c^2)$$ I know this Weitzenböck's_inequality $$a^2+b^2+c^2\ge 4\sqrt{3}S$$ But my inequality is stronger than this Weitzenbock's inequality. my try: let the semiperimeter, inradius, and circumradius be $s,r,R$ respectively $$a+b+c=2s,ab+bc+ac=s^2+4Rr+r^2,S=rs$$ $$\Longleftrightarrow (s^2+4Rr+r^2)4s^2\ge 12\sqrt{3}\cdot rs[4s^2-2(s^2+4Rr+r^2)]$$ $$\Longleftrightarrow s^3+4Rrs+r^2s\ge 6\sqrt{3}rs^2-24\sqrt{3}Rr^2-6\sqrt{3}r^3$$ and use this Gerretsen inequality: $$r(16R-5r)\le s^2\le 4R^2+4Rr+3r^2$$ and Euler inequality $$R\ge 2r$$ But seems is not usefull, Thank you

Square both sides, put$S^2=\dfrac{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}{16}$ in, we have: $(ab+bc+ac)^2(a+b+c)^4\ge 27\cdot (a+b+c)(a+b-c)(a+c-b)(b+c-a)(a^2+b^2+c^2)^2$ with brutal force method(BW method), WOLG let $a=$Min{$a,b,c$},$b=a+u,c=a+v,u\ge0,v\ge0$ , put in the inequality and rearrange them, we have: $ \iff k_6a^6+k_5a^5+k_4a^4+k_3a^3+k_2a^2+k_1a+k_1a+k_0 \ge0 \iff $ $k_6=+486v^2-486uv+486u^2 \\k_5=1404v^3-648uv^2-648u^2v+1404u^3\\k_4=1935v^4-360uv^3-945u^2v^2-360u^3v+1935u^4\\k_3=1572v^5-60uv^4-588u^2v^3-588u^3v^2-60u^4v+1572u^5\\k_2=760v^6+78uv^5-210u^2v^4-352u^3v^3-210u^4v^2+78u^5v+760u^6\\k_1=216v^7+4uv^6+32u^2v^5-140u^3v^4-140u^4v^3+32u^5v^2+4u^6v+216u^7\\k_0=27v^8+u^2v^6+4u^3v^5-48u^4v^4+4u^5v^3+u^6v^2+27u^8$ with $u^n+v^n\ge u^{n-1}v+uv^{n-1} \ge u^{n-2}v^2+u^2v^{n-2} \ge ...$ , all $k_i \ge 0$ above , and it is trivial that when and only when $u=v=0 \implies k_i=0$ QED.

proving the inequality $\triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$ If $\triangle$ be the area of $\triangle ABC$ with side lengths $a,b,c$. Then show that $\displaystyle \triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$ and also show that equality hold if $a=b=c$. $\bf{My\; Try}::$ Here we have to prove $4\triangle\leq \sqrt{(a+b+c)\cdot abc}$ Using the formula $$\triangle = \sqrt{s(s-a)(s-b)(s-c)},$$ where $$2s=(a+b+c)$$ So $$4\triangle = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{(a+b+c)\cdot(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$ Now using $\bf{A.M\geq G.M}$ for $(b+c-a)\;,(c+a-b)\;,(a+b-c)>0$ $$\displaystyle \frac{(b+c-a)+(c+a-b)+(a+b-c)}{3}\geq \sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$ So we get $\displaystyle (a+b+c)\geq 3\sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$ But I did not understand how can I prove above inequality help Required Thanks

For a triangle $\Delta = \frac{abc}{4R} = rs$ Now in your inequality you can put in the values to get $R \ge 2r$ This is known to be true since the distance between incentre and circumcentre $d^2 = R(R-2r)$ Thus your inequality is proved

if range of $f(x) = \frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4]$. Then $a$ and $b$ are If Range of $\displaystyle f(x) = \frac{x^2+ax+b}{x^2+2x+3}$ is $\left[-5,4\; \right]$ for all $\bf{x\in \mathbb{R}}$. Then values of $a$ and $b$. $\bf{My\; Try}::$ Let $\displaystyle y=f(x) = \frac{x^2+ax+b}{x^2+2x+3} = k$,where $k\in \mathbb{R}$.Then $\displaystyle kx^2+2kx+3k=x^2+ax+b$ $\Rightarrow (k-1)x^2+(2k-a)x+(3k-b) = 0$ Now we will form $2$ cases:: $\bf{\bullet}$ If $(k-1)=0\Rightarrow k=1$, Then equation is $(2-a)x+(3-b)=0$ $\bf{\bullet}$ If $(k-1)\neq 0\Rightarrow k\neq 1$ means either $k>1$ or $k<1$ How can i solve after that Help Required Thanks

To find the places where $f(x)$ is minimal and maximal, differentiate $f$ wrt $x$. Then solve $f'(x)=0$. Call the solution $x_0$ an $x_1$ (and so on if there are move). Now, you know for which $x$ $f(x)$ is minimal/maximal. Calculate $f(x_0)$ and $f(x_1)$. These should be equal to $-5$ and $4$. You only have to know which one is a minimum and which one a maximum, but you should be able to figure that out.

If real number x and y satisfy $(x+5)^2 +(y-12)^2=14^2$ then find the minimum value of $x^2 +y^2$ Problem : If real number x and y satisfy $(x+5)^2 +(y-12)^2=14^2$ then find the minimum value of $x^2 +y^2$ Please suggest how to proceed on this question... I got this problem from [1]: http://www.mathstudy.in/

As mathlove has already identified the curve to be circle $\displaystyle (x+5)^2+(y-12)^2=14^2$ But I'm not sure how to finish from where he has left of without calculus. Here is one of the ways: Using Parametric equation, any point $P(x,y)$ on the circle can be represented as $\displaystyle (14\cos\phi-5, 14\sin\phi+12)$ where $0\le \phi<2\pi$ So, $\displaystyle x^2+y^2=(14\cos\phi-5)^2+(14\sin\phi+12)^2$ $\displaystyle=14^2+5^2+12^2+28(12\sin\phi-5\cos\phi)$ Now can you derive $$-\sqrt{a^2+b^2}\le a\sin\psi-b\cos\psi\le\sqrt{a^2+b^2}?$$ and complete the answer?

Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right. This is the procedure: $$ \sqrt{x-4}-\sqrt{x-5}+1=0\\ \sqrt{x-4}=\sqrt{x-5}-1\\ \text{squaring both sides gives me:}\\ (\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\ x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\ x-4=x-5-\sqrt{x-5}+1\\ x-4=x-4-\sqrt{x-5}\\ \text{substracting x, and adding 4 to both sides}\\ 0=-\sqrt(x-5)\\ \text{switching both sides}\\ \sqrt{x-5}=0\\ \text{sqaring both sides}\\ x-5=0\\ x=5\\ \text{When I place 5 in the equation, I get:}\\ \sqrt{5-4}-\sqrt{5-5}+1=0\\ \sqrt{1}-\sqrt{0}+1=0\\ 1-0+1=0\\ 2=0\\ \text{this means that the equation dosent have any solution, right??}\\ $$ Any advice and suggestion is helpful. Thanks!!!

Easiest way to see it, is take both square roors to the other side and square, to get $$ \begin{split} 1 &= (x-5)-(x-4) - 2\sqrt{(x-5)(x-4)}\\ 2 + 2\sqrt{(x-5)(x-4)} &= 0\\ 1 + \sqrt{(x-5)(x-4)} &= 0\\ \end{split} $$ but $\sqrt{\ldots} \geq 0$ so this is impossible...

Finding the improper integral $\int^\infty_0\frac{1}{x^3+1}\,dx$ $$\int^\infty_0\frac{1}{x^3+1}\,dx$$ The answer is $\frac{2\pi}{3\sqrt{3}}$. How can I evaluate this integral?

$$x^3+1 = (x+1)(x^2-x+1)$$ Logic: Do partial fraction decomposition.Find $A,B,C$. $$\frac{1}{x^3+1} = \frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$$ By comparing corresponding co-efficients of different powers of $x$, you will end up with equations in A,B,C.After solving you get : $$A=\frac{1}{3},B=\frac{-1}{3},C=\frac{2}{3}$$ Then use this: $$\int\frac{1}{x}\,dx=\log x+c$$

Solving an irrational equation Solve for $x$ in: $$\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=\sqrt{5}$$ I used the property of proportions ($a=\sqrt{3+x}$, $b=\sqrt{3-x})$: $$\frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{2a}{2b}=\frac{a}{b}$$ I'm not sure if that's correct. Or maybe the notations $a^3=3+x$, $b^3=3-x$ ? I don't know how to continue. Thank you.

Here is another simple way, exploiting the innate symmetry. Let $\ \bar c = \sqrt{3\!+\!x}+\sqrt{3\!-\!x},\,\ c = \sqrt{3\!+\!x}-\sqrt{3\!-\!x}.\,$ Then $\,\color{#0a0}{\bar c c} = 3\!+\!x-(3\!-\!x) = \color{#0a0}{2x},\ $ so $\,\displaystyle\sqrt{5} = \frac{\bar c}c\, \Rightarrow \color{#c00}{\frac{6}{\sqrt{5}}} = {\frac{1}{\sqrt{5}}\!+\!\sqrt{5}} \,=\, \frac{c}{\bar c}+\frac{\bar c}c \,=\, \frac{(c+\bar c)^2}{\color{#0a0}{c\bar c}} - 2 \,=\,\frac{4(3\!+\!x)}{\color{#0a0}{2x}}-2 \,=\, \color{#c00}{\frac{6}x}\ $ so $\ \color{#c00}x = \ldots$

Solve $7x^3+2=y^3$ over integers I need to solve the following solve $7 x^3 + 2 = y^3$ over integers. How can I do that?

To solve this kind of equations, we have several 'tools' such as using mod, using inequalities, using factorization... In your question, using mod will help you. Since we have $$y^3-2=7x^3,$$ the following has to be satisfied : $$y^3\equiv 2\ \ \ (\text{mod $7$}).$$ However, in mod $7$, $$0^3\equiv 0,$$ $$1^3\equiv 1,$$ $$2^3\equiv 1,$$ $$3^3\equiv 6,$$ $$4^3\equiv 1,$$ $$5^3\equiv 6,$$ $$6^3\equiv 6.$$ So, there is no integer $y$ such that $y^3\equiv 2\ \ \ (\text{mod $7$}).$ Hence, we know that there is no solution.

Algebra 2-Factoring sum of cubes by grouping Factor the sum of cubes: $81x^3+192$ After finding the prime factorization of both numbers I found that $81$ is $3^4$ and $192$ is $2^6 \cdot 3$. The problem is I tried grouping and found $3$ is the LCM so it would outside in parenthesis. The formula for the sum of cubes is $(a+b) (a^2-ab+b^2)=a^3+b^3$ So I tried writing it this way and got:$(3x+2) (9x^2-6x+4)$ and then I realized I was wrong because $b$ from the formula (sum of cubes) was wrong. So how do I find $b$? and Why did I get $2$ for $b$ in the formula?

$$81x^3+192 = 3 (27 x^3 + 64) = 3 ((3x)^3+4^3) \\= 3 (3x + 4) ((3x)^2 - 3x\cdot 4 + 4^2) = 3 (3x+4)(9x^2-12x+16)$$ Since $12^2-4\cdot9\cdot16$ does not have a nice square root further factorization is not possible.

What's the explanation for why n^2+1 is never divisible by 3? What's the explanation for why $n^2+1$ is never divisible by $3$? There are proofs on this site, but they are either wrong or overcomplicated. It can be proved very easily by imagining 3 consecutive numbers, $n-1$, $n$, and $n+1$. We know that exactly one of these numbers must be divisible by 3. $$(n-1)(n)(n+1)=(n)(n-1)(n+1)=(n)(n^2-1)$$ Since one of those first numbers had to have been divisible by $3$, this new product $(n)(n^2-1)$ must also be divisible by $3$. That means that either $n$ (and by extension $n^2$) or $n^2-1$ is divisible by $3$. If one of those has to be divisible by $3$, then $n^2+1$ cannot be. So it is definitely true. My question is why is this true, what is inherent about $1$ more than a square number that makes it not divisible by $3$? Another way of saying this might be to explain it to me as if I don't know algebra.

one of $n-1,n$ or $n+1$ is divisible by $3$. If it is $n$ then so is $n^2$. If it is not $n$, then one of $n-1$ or $n+1$ is divisible by $3$, and hence so is their product $n^2-1$. Thus, either $n^2$ or $n^2-1$ is a multiple of $3$. If$n^2+1$ would be a multiple of three, then one of $2=(n^2+1)-(n^2-1)$ or $1=(n^2+1)-n^2$ would be a multiple of three.

Prove that $\lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $. My attempt: We prove that $$\displaystyle \lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $$ It is sufficient to show that for an arbitrary real number $\epsilon\gt0$, there is a $K$ such that for all $n\gt K$, $$\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| < \epsilon. $$ Note that $$ \displaystyle\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| = \left| \frac{-15}{16n+4} \right| $$ and for $ n > 1 $ $$ \displaystyle \left| \frac{-15}{16n+4} \right| = \frac{15}{16n+4} < \frac{1}{n}. $$ Suppose $ \epsilon \in \textbf{R} $ and $ \epsilon > 0 $. Consider $ K = \displaystyle \frac{1}{\epsilon} $. Allow that $ n > K $. Then $ n > \displaystyle \frac{1}{\epsilon} $. So $ \epsilon >\displaystyle \frac{1}{n} $. Thus $$ \displaystyle\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| = \left| \frac{-15}{16n+4} \right| = \frac{15}{16n+4} < \frac{1}{n} < \epsilon. $$ Thus $$ \displaystyle \lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4}. $$ Is this proof correct? What are some other ways of proving this? Thanks!

Is this proof correct? What are some other ways of proving this? Thanks! Your proof is correct with the caveat that you are a bit more precise about what $\epsilon$ and $K$ mean. Another way to prove this is using l'Hôpital's rule. Let $f(n)=23n+2$, $g(n)=4n+1$, then we can see that $$ \lim_{n\rightarrow\infty} f(n) = \lim_{n\rightarrow\infty} g(n) = \infty $$ In this case, the rule applies because you have an "indeterminant form" of $\infty/\infty$. Then the rule is that $$ \lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)} $$ All that remains is to evaluate $f'(n)=23$, $g'(n)=4$, $$ \lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{23}{4} = \frac{23}{4} $$

How to compute $\lim_{n\to \infty}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{n^2+n-k^2}}$ Find this follow limit $$I=\lim_{n\to \infty}\sum_{k=1}^{n}\dfrac{1}{\sqrt{n^2+n-k^2}}$$ since $$I=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\sqrt{1+\dfrac{1}{n}-\left(\dfrac{k}{n}\right)^2}}$$ I guess we have $$I=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\sqrt{1-(k/n)^2}}=\int_{0}^{1}\dfrac{1}{\sqrt{1-x^2}}dx=\dfrac{\pi}{2}$$ But I can't prove follow is true $$\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\sqrt{1+\dfrac{1}{n}-\left(\dfrac{k}{n}\right)^2}}=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\sqrt{1-(k/n)^2}}$$ I have only prove $$\dfrac{1}{\sqrt{1+\dfrac{1}{n}-\left(\dfrac{k}{n}\right)^2}}<\dfrac{1}{\sqrt{1-\left(\dfrac{k}{n}\right)^2}}$$ Thank you

The difference term-wise is \begin{align} \frac{1}{ \sqrt {1+({\frac{k}{n}})^2}} - \frac{1}{ \sqrt {1 + \frac{1}{n} - {(\frac{k}{n}})^2}}&= \frac{\frac{1}{n}}{ \sqrt {1+({\frac{k}{n}})^2}\cdot \sqrt {1 + \frac{1}{n} - ({\frac{k}{n}})^2} ( \sqrt {1+({\frac{k}{n}})^2}+ \sqrt {1 + \frac{1}{n} - ({\frac{k}{n}})^2}) } \\ &\leq \frac{1}{n}\frac{1}{ \sqrt {1+({\frac{k}{n}})^2}\cdot \sqrt {1- ({\frac{k}{n}})^2} ( \sqrt {1+({\frac{k}{n}})^2}+ \sqrt {1 - ({\frac{k}{n}})^2}) } \end{align} Therefore when we take the difference of the sums we get one extra $\frac{1}{n}$ and the other quantity converges to an integral thus the result is zero which proves your claim. $$ 0 \cdot \int_{0}^{1} \frac{1}{ \sqrt{(1+x^2)(1-x^2)} \sqrt { (1-x^2) }+\sqrt{(1+x^2)}}\rm{d}x=0$$

A second order recurrence relation problem I was asked by a friend with this problem but I can't solve it. Can anyone help? We have a sequence $\left\{a_n\right\}$ that satisfies $a_1=1$, $a_2=2$, $$a_n+\frac{1}{a_n} =\frac{a_{n+1}^2+1}{a_{n+2}}$$ where $n$ is a positive integer. Prove that * *$a_{n+1}=a_n+\frac{1}{a_n}$ *$2n-1\le a_n^2\le3n-2$ *Let $S_n$ be the sum of the sequence $\left\{1\over a_n\right\}$, prove $62<S_{2014}<77$.

(a) By induction, assume $a_{k+1} = a_k + \frac1{a_k}$. Then to get the result for $n=k+1$, consider $$\begin{align*} a_k+\frac1{a_k} =& \frac{a_{k+1}^2+1}{a_{k+2}}\\ a_{k+1}=&\frac{a_{k+1}^2+1}{a_{k+2}}\\ a_{k+2} =& a_{k+1}+\frac1{a_{k+1}} \end{align*}$$ Also prove the base case for $n=1$ holds. (b) Using $a_{n+1}^2 = a_{n}^2+2+\dfrac1{a_n^2}$ and by induction, assume $2k-1\le a_k^2 \le 3k-2$. Then for $n=k+1$, the first inequality is $$a_{k+1}^2 = a_k^2 + 2 +\frac1{a_k^2}\ge2k-1+2 = 2(k+1)-1$$ and the second inequality is similar. Use the fact that $0\le\dfrac1{a_k^2}\le1$. (c) $$\begin{array}{rcl} 2n+1\le&a_{n+1}^2&\le3n+1\\ \sqrt{2n+1}\le& a_{n+1} &\le \sqrt{3n+1}\\ \sqrt{2n+1}\le& a_1+\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_n}&\le \sqrt{3n+1}\\ \sqrt{2\times2014+1}-a_1 \le& \sum_{i=1}^{2014}\frac1{a_i} &\le \sqrt{3\times2014+1}-a_1\\ 62.4\cdots \le& \sum_{i=1}^{2014}\frac1{a_i} &\le 76.7\ldots\\ \end{array}$$

How to prove that $\lim_{(x,y) \to (0,0)} \frac{x^3y}{x^4+y^2} = 0?$ How to prove that $\lim_{(x,y) \to (0,0)} \dfrac{x^3y}{x^4+y^2} = 0?$ First I tried to contradict by using $y = mx$ , but I found that the limit exists. Secondly I tried to use polar coordinates, $x = \cos\theta $ and $y = \sin\theta$, And failed .. How would you prove this limit equals $0$?

Observe that $x^4 + y^2 \geq |x^2y|$ (for instance, because $x^4+y^2+2x^2y = (x^2+y)^2\geq0$ and $x^4+y^2-2x^2y = (x^2-y)^2 \geq0$). Hence $\displaystyle \left|\frac{x^2y}{x^4+y^2}\right| \leq 1$ when $(x,y)\neq (0,0)$ and thus $$\lim_{(x,y)\rightarrow (0,0)} \left|\frac{x^3y}{x^4+y^2}\right| \leq \lim_{(x,y)\rightarrow(0,0)} |x| = 0,$$ so the limit is 0 by the squeeze theorem.

If $n,k\in\mathbb N$, solve $2^8+2^{11}+2^n=k^2$. If $n,k\in\mathbb N$, solve $$2^8+2^{11}+2^n=k^2$$ It's hard for me to find an idea. Some help would be great. Thanks.

HINT: We have $\displaystyle2^8+2^{11}=2304$ Now, $2^8+2^{11}+2^0=2304+1\ne k^2$ $\displaystyle 2^8+2^{11}+2^1=2304+2\equiv2\pmod8$, but $a^2\equiv0,1,4\pmod8$ So, $n\ge2$ let $n=m+2$ where $m\ge0$ $\displaystyle 2^8+2^{11}+2^{m+2}=4(576+2^m)\implies 576+2^m$ must be perfect square Like either method $m\ge2$ let $m=r+2$ where $r\ge0$ $\displaystyle576+2^{r+2}=4(144+2^r)$ Follow this step One Observation : $$(2^4)^2+2\cdot2^4\cdot2^6+(2^6)^2=(2^4+2^6)^2$$

Calculate the pseudo inverse of the matrix The subject is to calculate the pseudo inverse if matrix $\begin{equation*} \mathbf{A} = \left( \begin{array}{ccc} 1 & 0 \\ 2 & 1 \\ 0 & 1 \\ \end{array} \right) \end{equation*}$ My answer is as follows: (SVD decomposition) First, $\begin{equation*} \mathbf{A^TA} = \left( \begin{array}{ccc} 5 & 2 \\ 2 & 2 \\ \end{array} \right) \end{equation*}$, with eigenvalues $\lambda_1 = 6, \lambda_2 = 1$, and eigenvectors $\begin{equation*} \mathbf{x_1} = \frac{1}{\sqrt{5}}\left( \begin{array}{ccc} 2 \\ 1 \\ \end{array} \right) \end{equation*}$, $\begin{equation*} \mathbf{x_2} = \frac{1}{\sqrt{5}}\left( \begin{array}{ccc} -1 \\ 2 \\ \end{array} \right) \end{equation*}$, so the matrix $\begin{equation*} \mathbf{V} = \frac{1}{\sqrt{5}}\left( \begin{array}{ccc} 2 & -1 \\ 1 & 2 \\ \end{array} \right) \end{equation*}$. Second, $\begin{equation*} \mathbf{AA^T} = \left( \begin{array}{ccc} 1 & 2 & 0 \\ 2 & 5 & 1 \\ 0 & 1 & 1 \\ \end{array} \right) \end{equation*}$, with eigenvalues $\lambda_1 = 6, \lambda_2 = 1,\lambda_3 = 0$, and eigenvectors $\begin{equation*} \mathbf{x_1} = \frac{1}{\sqrt{30}}\left( \begin{array}{ccc} 2 \\ 5 \\ 1 \\ \end{array} \right) \end{equation*}$, $\begin{equation*} \mathbf{x_2} = \frac{1}{\sqrt{5}}\left( \begin{array}{ccc} -1 \\ 0 \\ 2 \\ \end{array} \right) \end{equation*}$, $\begin{equation*} \mathbf{x_3} = \frac{1}{\sqrt{6}}\left( \begin{array}{ccc} 2 \\ -1 \\ 1 \\ \end{array} \right) \end{equation*}$ , so the matrix $\begin{equation*} \mathbf{U} = \left( \begin{array}{ccc} \frac{2}{\sqrt{30}} & - \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{6}} \\ \frac{5}{\sqrt{30}} & 0 & - \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{30}} & \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{6}} \\ \end{array} \right) \end{equation*}$, and $\begin{equation*} \mathbf{\Sigma} = \left( \begin{array}{ccc} 6 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{array} \right) \end{equation*}$. Then, the pseudo inverse becomes: $A^+ = V \Sigma^+U^T$. The problem comes to: when I was checking the SVD decomposition, I found $A\ne U\Sigma V^T$. However, I find nothing odds in the calculation. Please help me to point out the error.

Recall, for $\mathbf{\Sigma}$ we take the square roots of the non-zero eigenvalues and populate the diagonal with them, putting the largest in $\mathbf{\Sigma}_{11}$, the next largest in $\mathbf{\Sigma}_{22}$ and so on until the smallest value ends up in $\mathbf{\Sigma}_{mm}$. $$\begin{equation*} \mathbf{\Sigma} = \left( \begin{array}{ccc} \sqrt{6} & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{array} \right) \end{equation*}$$ Everything else is correct (great job), although you can simplify some of the items in $\mathbf{U}$. For example, $\dfrac{2}{\sqrt{6}} = \sqrt{\dfrac{2}{3}}$.

Find the volume of the solid bounded by $z=x^2+y^2+1$ and $z=2-x^2-y^2$. Question: Find the volume of the solid bounded by $z=x^2+y^2+1$ and $z=2-x^2-y^2$. Setting the 2 equations equal w.r.t. $z$, $x^2+y^2+1=2-x^2-y^2 \rightarrow x=\pm\sqrt{\frac 12-y^2}$ Therefore the boundary of $y=\pm\frac {1}{\sqrt2}$. So to find the volume of the solid, take the integration by subtracting the volume above and below the boundaries. $\displaystyle V=\int_{-\frac {1}{\sqrt2}}^{+\frac {1}{\sqrt2}}\int_{-\sqrt{\frac 12-y^2}}^{+\sqrt{\frac 12-y^2}}(2-x^2-y^2)dxdy-\int_{-\frac {1}{\sqrt2}}^{+\frac {1}{\sqrt2}}\int_{-\sqrt{\frac 12-y^2}}^{+\sqrt{\frac 12-y^2}}(x^2+y^2+1)dxdy$ This is what I did. Without solving the equation, can someone tell me if it is correct? Thank you!

Your setup is right. Here is the method you could have done to compute the volume. Assume the density is $f(x,y,z) = 1$, so $$V = \iiint_D \,dx\,dy\,dz$$ We are given that the solid is bounded by $z = x^2 + y^2 + 1$ and $z = 2 - x^2 - y^2$. As I commented under your question, you need to use cylindrical coordinates to evaluate the volume integral. Using the substitutions $x = r\cos(\theta)$, $y = r\sin(\theta)$ and $z = z$, we have $z = r^2 + 1$ and $z = 2 - r^2$. With some knowledge in graphs and functions, we see that the bounds are $$\begin{aligned} r^2 + 1 \leq z \leq 2 - r^2\\ 0 \leq \theta \leq 2\pi\\ 0 \leq r \leq \dfrac{1}{\sqrt{2}} \end{aligned}$$ where $r = \frac{1}{\sqrt{2}}$ is found by solving for $r$ when $r^2 + 1 = 2 - r^2$. So for the volume triple integral, we have $$\begin{aligned} V &= \iiint_D \,dx\,dy\,dz\\ &= \iiint_D r\,dr\,d\theta\,dz\\ &= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}}\int_{r^2 + 1}^{2 - r^2}r\,dz\,dr\,d\theta\\ &= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} r(2 - r^2 - r^2 - 1)\,dr\,d\theta\\ &= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} r(1 - 2r^2)\,dr\,d\theta\\ &= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} (r - 2r^3)\,dr\,d\theta\\ &= \int_{0}^{2\pi}\,d\theta\left.\left(\dfrac{1}{2}r^2 - \dfrac{1}{2}r^4 \right)\right\vert_{r = 0}^{r = \frac{1}{\sqrt{2}}}\\ &= 2\pi \left.\left(\dfrac{1}{2}r^2 - \dfrac{1}{2}r^4 \right)\right\vert_{r = 0}^{r = \frac{1}{\sqrt{2}}}\\ &= 2\pi \cdot \dfrac{1}{8}\\ &= \dfrac{\pi}{4} \end{aligned}$$

Convergence of the integral $\int\limits_{1}^{\infty} \left( \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x+3}} \right) \, dx$ Would someone please help me prove that the integral $$ \int\limits_{1}^{\infty} \left( \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x+3}} \right) \, dx $$ is convergent? Thank you.

Use $$\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+3}}=\frac{\sqrt{x+3}-\sqrt{x}}{\sqrt{x^2+3x}}=\frac{3}{(\sqrt{x+3}+\sqrt{x})\sqrt{x^2+3x}}$$ So, the integrand is positive and $\le \frac{3}{2x\sqrt{x}}$. Here I use $\sqrt{x^2+3x}>x$ and $\sqrt{x+3}>\sqrt{x}$ So the integral converges by the comparison criterium.

how to find $ \lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right)$ How can I find this? $ \lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right)$

Since for any $A>0$ $$\sqrt{A^2 x^2+1}-A|x| = \frac{1}{A|x|+\sqrt{A^2 x^2+1}}<\frac{1}{2A|x|}$$ holds, we have: $$\left|\sqrt{x^2+1}+\sqrt{4x^2+1}-\sqrt{9x^2+1}\right|=\left|\sqrt{x^2+1}-|x|+\sqrt{4x^2+1}-2|x|-\sqrt{9x^2+1}+3|x|\right|\leq \left|\sqrt{x^2+1}-|x|\right|+\left|\sqrt{4x^2+1}-2|x|\right|+\left|\sqrt{9x^2+1}-3|x|\right|<\frac{1}{|x|}\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}\right),$$ hence the limit is $0$.

Is the Series : $\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + ..........+\frac 1 {(n+n)^2} \right) \sin^2 n\theta $ convergent? Is the Series : $$\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) \sin^2 n\theta $$ convergent? Attempt: $$\sum_{n=0}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) \sin^2 n\theta $$ $$= \sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) (1- \cos( 2n\theta)) \frac 1 2$$ $$=\frac 1 2 \left[\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right)\right]- \frac 1 2\left[\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) \cos(2n\theta)\right] $$ The left part by sandwich theorem has limiting value to $0$ The right part by Dirichlets theorem is convergent as $ \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right)$ is a positive, monotonically decreasing sequence and the sequence of partial sum of $\sum_{n=1}^{\infty} \cos 2n\theta$ is bounded. Hene, the given series is convergent. Am i correct?

Notice that $$\sum_{k=1}^n\frac 1{(n+k)^2}=\frac 1{n}\frac 1n\sum_{k=1}^nf\left(\frac kn\right),$$ with $f(x):=\frac 1{1+x^2}$, a continuous positive function. Hence the convergence of the initial series reduces to the convergence of $$\sum_{n\geqslant 1}\frac{\sin^2(n\theta)}n.$$ Write $\sin^2(A)=\frac{1-\cos(2A)}2$. If $\theta$ is not a multiple of $\pi$, the series is divergent because $\sum_{n\geqslant 1}\frac{\sin(2n\theta)}n$ is convergent.

solve equation in positive integers Can anybody help me with this equation? Solve in $\mathbb{N}$: $$3x^2 - 7y^2 +1=0$$ One solution is the pair $(3,2)$, and i think this is the only pair of positive integers that can be a solution. Any idea?

There are infinitely many solutions in positive integers. $7y^2-3x^2=1$ is an example of a "Pell equation", and there are standard methods for finding solutions to Pell equations. For example, the fact that $(x,y)=(2,3)$ is a solution to $7y^2-3x^2=1$ is equivalent to noting that $(2\sqrt7+3\sqrt3)(2\sqrt7-3\sqrt3)=1$. The fundamental unit in $\Bbb Q(\sqrt{21})$ is $55+12\sqrt{21}$; in particular, $(55+12\sqrt{21})(55-12\sqrt{21})=1$. Consequently, if we calculate $(2\sqrt7+3\sqrt3)(55+12\sqrt{21}) = 218 \sqrt{7}+333 \sqrt{3}$, it follows that $(218 \sqrt{7}+333 \sqrt{3})(218 \sqrt{7}-333 \sqrt{3})=1$, or $7\cdot218^2 - 3\cdot333^2=1$. You can get infinitely many solutions $(x_n,y_n)$ to $7y^2 - 3x^2=1$ by expanding $(2\sqrt7+3\sqrt3)(55+12\sqrt{21})^n = y_n \sqrt{7}+x_n \sqrt{3}$. Your solution corresponds to $n=0$, while the previous paragraph is $n=1$; for example, $n=2$ yields $x_2=36627$ and $y_2=23978$.

Prove that $\frac{1}{1*3}+\frac{1}{3*5}+\frac{1}{5*7}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$ Trying to prove that above stated question for $n \geq 1$. A hint given is that you should use $\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$. Using this, I think I reduced it to $\frac{1}{2}(\frac{1}{n^2}-(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}))$. Just not sure if it's correct, and what to do with the second half.

$$\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1}) $$This implies that the sum is $$\frac{1}{2}\big\{\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{2n-1}+\frac{1}{2n-1}-\frac{1}{2n+1}\big\}$$ cancel terms and complete

How to tackle a recurrence that contains the sum of all previous elements? Say I have the following recurrence: $$T(n) = n + T\left(\frac{n}{2}\right) + n + T\left(\frac{n}{4}\right) + n + T\left(\frac{n}{8}\right) + \cdots +n + T\left(\frac{n}{n}\right) $$ where $n = 2^k$, $k \in \mathbb{N} $ and $T(1) = 1$. simplified to: $$T(n) = n \log_2n + \sum_{i=1}^{\log_2n}T\left(\frac{n}{2^i}\right) $$ The Master's theorem is not applicable; neither is the Akra-Bazzi method since $k = \log_2$ is not a constant. What strategy can I use to find a closed form solution? I have a feeling that the closed form is $T(n) = \sum_{i=0}^{\log_2n}\left[j\frac{n}{2^i} \log_2 \left(\frac{n}{2^i} \right)\right] + 1 $ where $j = \max\left(1, 2^{i-1}\right)$ but would like a proof.

I'd just start with $T(1)$ and look for a pattern: $$T(2^1) = 1 \cdot 2^1 + 2^{1-1}T(1)$$ $$T(2^2) = 2\cdot 2^2 + 1\cdot 2^1 + (2^{2-1}) T(1)$$ $$T(2^3) = 3\cdot 2^3 + 2\cdot 2^2 + 2 \cdot 1 \cdot 2^1 + 2^{3-1} T(1)$$ $$T(2^4) = 4\cdot 2^4 + 3\cdot 2^3 + 2 \cdot 2 \cdot 2^2 + 4 \cdot 1 \cdot 2^1 + 2^{4-1} T(1)$$ so that if $T(1) = 1$, $$T(2^n) = 2^{n-1} + \sum_{k=1}^n k \cdot 2^k + \sum_{k=1}^{n-2} (2^{n-k-1}-1)2^k.$$

Center of Mass via integration for ellipsoid I need some help with the following calculation: I have to calculate the coordinates of the center of mass for the ellipsoid $$\left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 + \left( \frac{z}{c} \right)^2 \le 1, \quad z \ge 0$$ with mass-density $\mu(x,y,z)=z^2$. I wanted to use: $$ \begin{align} x & = a r \sin\theta \cos\varphi \\ y & = b r \sin\theta \cos\varphi \\ z & = c r \cos\theta \end{align} $$ whereas $$ \begin{gather} 0 \le r \le 1, \\ 0 \le \theta \le \pi, \\ 0 \le \varphi \le 2\pi \end{gather} $$ and $$\frac{\partial (x,y,z)}{ \partial (r, \theta, \varphi)} = r^2 \sin\theta.$$ Did I choose the right things so far? 1) $$ \begin{align} M & = \int\limits_E µ(x,y,z) d(x,y,z) \\ & = \int_0^1 \hspace{-5pt} \int_0^{\pi} \hspace{-5pt} \int_0^{2\pi} c^2 r^2 \cos^2\theta \cdot r^2 \sin(\theta) d(r, \theta, \varphi) \\ & = c^2 \int_0^1 r^4 dr \int_0^\pi \sin\theta \cdot \cos^2\theta d\theta \int_0^{2\pi} d\varphi \\ & = \frac{4\pi c^2}{15}. \end{align} $$ 2) $$x_s \cdot M = \ldots $$ Here I get $\int_0^{2pi} \cos\varphi \, d \varphi = 0$, so the whole product is zero, so x_s is zero too?? What am I doing wrong?

The mass density is invariant under $x\rightarrow -x$ and $y\rightarrow -y$, so the center of mass must have $x=y=0$. You do still need to find its $z$-coordinate, but since the mass density is only a function of $z$, you can reduce this to a one-dimensional integral. At a given value of $z$, the cross-section is an ellipse with semi-major and semi-minor axes $a\sqrt{1-(z/c)^2}$ and $b\sqrt{1-(z/c)^2}$; this ellipse has area $\pi a b (1-(z/c)^2)$. The mass of a slice of thickness $dz$ at that altitude is therefore $dm=\pi a b (z^2 - z^4/c^2)dz$. The $z$-coordinate of the center of mass is $$ M_z=\frac{\int_{z=0}^{z=c}zdm}{\int_{z=0}^{z=c} dm}=\frac{\int_{0}^{c}(z^3-z^5/c^2)dz}{\int_{0}^{c}(z^2-z^4/c^2)dz}=\frac{\frac{1}{4}c^4-\frac{1}{6c^2}c^6}{\frac{1}{3}c^3-\frac{1}{5c^2}c^5}=\frac{\frac{1}{4}-\frac{1}{6}}{\frac{1}{3}-\frac{1}{5}}c=\frac{5}{8}c. $$

Suppose $f$ is a thrice differentiable function on $\mathbb {R}$ . Showing an identity using taylor's theorem Suppose $f$ is a thrice differentiable function on $\mathbb {R}$ such that $f'''(x) \gt 0$ for all $x \in \mathbb {R}$. Using Taylor's theorem show that $f(x_2)-f(x_1) \gt (x_2-x_1)f'(\frac{x_1+x_2}{2})$ for all $x_1$and $x_2$ in $\mathbb {R}$ with $x_2\gt x_1$. Since $f'''(x) \gt 0$ for all $x \in \mathbb {R}$, $f''(x)$ is an increasing function. And in Taylor's expansion i will be ending at $f''(x)$ but not sure how to bring in $\frac{x_1+x_2}{2}$.

Using the Taylor expansion to third order, for all $y$ there exists $\zeta$ between $(x_1+x_2)/2$ and $y$ such that $$ f(y) = f \left( \frac{x_1+x_2}2 \right) + f'\left( \frac{x_1+x_2}2 \right)\left(y - \frac{x_1 + x_2}2 \right) \\ + \frac{f''(\frac{x_1+x_2}2)}2 \left( y - \frac{x_1 + x_2}2 \right)^2 + \frac{f'''(\zeta)}6 \left( y - \frac{x_1+x_2}2 \right)^3. \\ $$ It follows that by plugging in $x_2$, $$ f(x_2) - f \left( \frac{x_1+x_2}2 \right) > f' \left( \frac{x_1+x_2}2 \right) \frac {x_2-x_1}2 + \frac{f''(\frac{x_1+x_2}2)}2 \left(\frac{x_2-x_1}2 \right)^2 $$ since $f'''(\zeta) > 0$ and $x_2 > \frac{x_1+x_2}2$. Similarly, by plugging in $x_1$, $$ f(x_1) - f \left( \frac{x_1+x_2}2 \right) < f' \left( \frac{x_1+x_2}2 \right) \frac {x_1-x_2}2 + \frac{f''(\frac{x_1+x_2}2)}2 \left(\frac{x_1-x_2}2 \right)^2, $$ (note that the sign that appears in the cubic term is now negative, hence the reversed inequality) which we can re-arrange as $$ f \left( \frac{x_1+x_2}2 \right) - f(x_1) > f' \left( \frac{x_1+x_2}2 \right) \frac {x_2-x_1}2 - \frac{f''(\frac{x_1+x_2}2)}2 \left(\frac{x_1-x_2}2 \right)^2, $$ By adding up, the quadratic terms cancel out and you get your result. Hope that helps,

Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$? Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$? So far I have: $a^2+(a+2)^2+(a+4)^2+1$ $=a^2+a^2+4a+4+a^2+8a+16+1 $ $=3a^2+12a+21$ $=3(a^2+4a+7) $ where do I go from here.. the solution I have is divisible by $3$ not $12$...

If $a$ is odd, then $a = 2b+1$ for some integer $b$. Then $a^2 + 4a + 7 = 4b^2 + 4b + 1 + 8b + 4 + 7 = 4(b^2 + 3b + 3)$, which is evenly divisible by $4$. Combine this with the divisibility by $3$ that you already have, and you're done.

Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ It is known that \begin{align} \arcsin x + \arcsin y =\begin{cases} \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 \le 1 &\text{or} &(x^2+y^2 > 1 &\text{and} &xy< 0);\\ \pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 > 1&\text{and} &0< x,y \le 1;\\ -\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 > 1&\text{and} &-1\le x,y < 0. \end{cases} \end{align} I tried to prove this myself, have no problem in getting the 'crux' $\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2})$ part of the RHS, but face trouble in checking the range of that 'crux' under the given conditions.

Using this, $\displaystyle-\frac\pi2\leq \arcsin z\le\frac\pi2 $ for $-1\le z\le1$ So, $\displaystyle-\pi\le\arcsin x+\arcsin y\le\pi$ Again, $\displaystyle\arcsin x+\arcsin y= \begin{cases} \\-\pi- \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})& \mbox{if } -\pi\le\arcsin x+\arcsin y<-\frac\pi2\\ \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2}) &\mbox{if } -\frac\pi2\le\arcsin x+\arcsin y\le\frac\pi2 \\ \pi- \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})& \mbox{if }\frac\pi2<\arcsin x+\arcsin y\le\pi \end{cases} $ and as like other trigonometric ratios are $\ge0$ for the angles in $\left[0,\frac\pi2\right]$ So, $\displaystyle\arcsin z\begin{cases}\text{lies in } \left[0,\frac\pi2\right] &\mbox{if } z\ge0 \\ \text{lies in } \left[-\frac\pi2,0\right] & \mbox{if } z<0 \end{cases} $ Case $(i):$ Observe that if $\displaystyle x\cdot y<0\ \ \ \ (1)$ i.e., $x,y$ are of opposite sign, $\displaystyle -\frac\pi2\le\arcsin x+\arcsin y\le\frac\pi2$ Case $(ii):$ If $x>0,y>0$ $\displaystyle \arcsin x+\arcsin y$ will be $\displaystyle \le\frac\pi2$ according as $\displaystyle \arcsin x\le\frac\pi2-\arcsin y$ But as $\displaystyle\arcsin y+\arccos y=\frac\pi2,$ we need $\displaystyle \arcsin x\le\arccos y$ Again as the principal value of inverse cosine ratio lies in $\in[0,\pi],$ $\displaystyle\arccos y=\arcsin(+\sqrt{1-y^2})\implies \arcsin x\le\arcsin\sqrt{1-y^2}$ Now as sine ratio is increasing in $\displaystyle \left[0,\frac\pi2\right],$ we need $\displaystyle x\le\sqrt{1-y^2}\iff x^2\le1-y^2$ as $x,y>0$ $\displaystyle\implies x^2+y^2\le1 \ \ \ \ (2)$ So, $(1),(2)$ are the required condition for $\displaystyle \arcsin x+\arcsin y\le\frac\pi2$ Case $(iii):$ Now as $\displaystyle-\frac\pi2\arcsin(-u)\le\frac\pi2 \iff -\frac\pi2\arcsin(u)\le\frac\pi2$ $\arcsin(-u)=-\arcsin u$ Use this fact to find the similar condition when $x<0,y<0$ setting $x=-X,y=-Y$

How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$? How many distinct factors can be made from the number $2^5*3^4*5^3*7^2*11^1$? Hmmm... So I didn't know what to do here so I tested some cases for a rule. If a number had the factors $3^2$ and $2^1$, you can make $5$ distinct factors: $2^1$, $3^1$, $3^2$, $2^1 \cdot 3^1$, $2^1 \cdot 3^2$... I don't see a pattern yet. How does one go about this? I don't think the answer is $5!$....

If $\begin{equation}x = a^p \cdot b^q\cdot c^r+...\end{equation}$ then there are $(p+1)(q+1)(r+1)...$ numbers that divde $x$. Any number that divides $x$ will be of the form $a^\alpha\cdot b^\beta\cdot c^\gamma$ . So we have p p+1 options for $\alpha$ because we need to consider $\alpha = 0$ also. Similarly, we have $q + 1$ options for $\beta$ and $q+1$ options for $\gamma$. Therefore, we multiply these out. To get the intuition of why this is so, let us take the example of the number $24$. $24 = 2^3\cdot3^1$ Any number that divides $24$ will be of the form $2^\alpha\cdot3^\beta$. We have 4 choices for $\alpha:0,1,2,3$ and 2 choices for $\beta:0,1$. So we have $4\cdot2 = 8$ numbers that divide 24. These can be listed out as: $$(2^03^0,2^13^0,2^23^0,2^33^0),(2^03^1,2^13^1,2^23^1,2^33^1)$$ So, $2^5*3^4*5^3*7^2*11^1$ has $6\cdot5\cdot4\cdot3\cdot2 = \boxed{720}$

evaluation of $\int\frac{1}{\sin^3 x-\cos^3 x}dx$ Evaluation of $\displaystyle \int\frac{1}{\sin^3 x-\cos^3 x}dx$ $\bf{My\; Try::}$ Given $\displaystyle \int\frac{1}{\sin^3 x-\cos^3 x}dx = \int\frac{1}{(\sin x-\cos x)\cdot (\sin^2 x-\sin x\cos x+\cos^2 x)}dx$ $\displaystyle = 2\int\frac{(\sin x-\cos x)}{(\sin x-\cos x)^2\cdot (2-\sin 2x)}dx = 2\int \frac{(\sin x-\cos x)}{(1-\sin 2x)\cdot (2-\sin 2x)}dx$ Let $(\sin x+\cos x) = t\;,$ Then $(\cos x -\sin x)dx = dt\Rightarrow (\sin x -\cos x)dx = dt$ and $(1+\sin 2x)=t^2\Rightarrow \sin 2x = (t^2-1)$ So Integral Convert into $\displaystyle 2\int\frac{1}{(2-t^2)\cdot (3-t^2)}dt = 2\int\frac{1}{(t^2-2)\cdot (t^2-3)}dt$ My Question is , is there is any better method other then that Help me Thanks.

You're missing a minus sign at one point, but other than that I think you're OK. Next, use partial fractions: $$ \frac{1}{(2-t^2)(3-t^2)} = \frac{A}{\sqrt{2}-t} + \frac{B}{\sqrt{2}+t} + \frac{C}{\sqrt{3}-t} + \frac{D}{\sqrt{3}+t} $$

Find the value of $\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right)$ I am stuck on this. I would like the algebraic explanation or trick(s) that shows that the equation below has limit of $-2$ (per the book). The wmaxima code of the equation below. $$ \lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right) $$ I've tried factoring out an $x$ using the $\sqrt{x^2} = |x|$ trick. That doesn't seem to work. I get $1 - 1 = 0$ for the other factor meaning the limit is zero...but that's obviously not correct way to go about it :( Thanks.

For $x>0$: For brevity let $A=\sqrt {x^2+2 x}.$ We have $(x+1)^2=A^2 +1>A^2>0$ so $x+1>A>0 . $ ..... So we have $$0<x+1-A=$$ $$=(x+1-A)\frac {x+1+A}{x+1+A}=\frac {(x+1)^2-A^2}{x+1+A}=\frac {1}{x+1+A}<1/x.$$ Therefore $$(i)\quad \lim_{x\to \infty} (x+1-A)=0.$$ For $x>2$: For brevity let $B=\sqrt {x^2-2 x}.$ We have $(x-1)^2=B^2+1>B^2>0$ so $x-1>B>0 . $ ..... So we have $$0<x-1-B=$$ $$=(x-1-B)\frac {x-1+B}{x-1+B}=\frac {(x-1)^2-B^2}{x-1+B}=\frac {1}{x-1+B}<1/(x-1).$$ Therefore $$(ii)\quad\lim_{x\to \infty}(x-1-B)=0.$$ We have $\sqrt {x^2+2 x}-\sqrt {x^2-2 x}=$ $A-B=(A-(x+1))-(B-(x-1))+2. $ From $(i)$ and $(ii)$ we have $$\lim_{x\to \infty} A-B=\lim_{x\to \infty}(A-(x+1))-(B-(x-1))+2=0+0+2=2.$$ The idea is that when $x$ is large, $A$ is close to $x+1$ and $B$ is close to $x-1$ so $A-B$ is close to $(x+1)-(x-1)=2.$

Incongruent solutions to $7x \equiv 3$ (mod $15$) I'm supposed to find all the incongruent solutions to the congruency $7x \equiv 3$ (mod $15$) \begin{align*} 7x &\equiv 3 \mod{15} \\ 7x - 3 &= 15k \hspace{1in} (k \in \mathbb{Z}) \\ 7x &= 15k+3\\ x &= \dfrac{15k+3}{7}\\ \end{align*} Since $x$ must be an integer, we must find a pattern for $k$ that grants this. We know that $\frac{k+3}{7}$ must be equal to some integer, say $m$. Solving for $k$, we have $k=4+7m$. Substituting this into our value for $x$, we get: \begin{align*} x & = \dfrac{15(4+7m) + 3}{7}.\\ &= \dfrac{63}{7} + \frac{105m}{7}.\\ &= 9+15m. \end{align*} So, $x = 9+15m, m\in \mathbb{Z}.$ So, is this what I was looking for? I'm not exactly sure what is meant by incongruent solutions.

We can solve this congruence equation in elementary way also. We shall write $[15]$ to denote the word mod 15. Fine? Note that \begin{align*} &7x\equiv 3[15]\\ -&15x +7x\equiv 3-15[15] ~~\text{because}~~ 15x\equiv 0\equiv 15[15]\\ -&8x\equiv -12[15]\\ &2x\equiv 3\left[\frac{15}{\gcd(15, -4)}\right]~~\text{since}~~ax\equiv ay[m]\Rightarrow x\equiv y\left[\frac{m}{\gcd(a, m)}\right]\\ &2x\equiv 3[15]~~\text{since}~~\gcd(15,-4)=\gcd(15, 4)=1\\ &2x\equiv 3+15[15]~~\text{since} 15\equiv 0[15]\\ &2x\equiv 18[15]\\ &x\equiv 9\left[\frac{15}{\gcd(15, 2)}\right]\\ &x\equiv 9[15] \end{align*} Thus the solution is given by $x\equiv 9[15]$.

Why aren't these two integration methods yielding the same answer? I'm trying to solve this (not homework, if it matters), and both u-substitution and integration by parts are both yielding two different answers. Where am I going wrong? Equation: $$\int \frac{(4x^3)}{(x^4+7)}dx$$ u-substitution answer: $$=\ln\big|(x^4+7)\big|+C$$ integration by parts answer: $$=\int4x^3*(x^4+7)^{-1}dx$$ $$=4x^3*\ln\big|x^4+7\big|-\int 12x^2*(x^4+7)^{-1}dx$$ $$=4x^3*\ln\big|x^4+7\big|-(12x^2*ln\big|x^4+7\big|-\int 24x*(x^4+7)^{-1}dx)$$ $$=4x^3*\ln\big|x^4+7\big|-(12x^2*ln\big|x^4+7\big|-24x*ln\big|x^4+7\big|-\int 24(x^4+7)^{-1}dx)$$ $$= 4x^3*\ln\big|x^4+7\big|-(12x^2*\ln\big|x^4+7\big|-(24x\ln\big|x^4+7\big|-24\ln\big|x^4+7\big|))$$ $$=(4x^3-12x^2+24x-24)(\ln\big|x^4+7\big|)$$

I don't understand it. For $u = x^4+7, du = 4x^3dx$ so $$ \int \frac{4x^3}{x^4+7} dx = \int du/u = \ln |x^4+7| + C. $$ Show work for your by parts results and it will become clear where the error is...

Find value of $r$ and the limit For some $r \in \mathbb Q$, the limit $$\lim_{x \rightarrow \infty}x^r.\frac{1}2.\frac{3}4.\frac{5}6......\frac{2x-1}{2x}$$ exists and is non zero What is that value of $r$ and what is that limit equal to? I rewrote the product $\frac{1}2.\frac{3}4.\frac{5}6......\frac{2x-1}{2x}$ = $\frac{(2x)!}{2^{2x}(x!)^2}$ but it didn't help.

Let $x^r.\frac{1}{2}.\frac{3}{4} \dots \frac{2x-1}{2x} = A \tag{1}$. Clearly $A < x^r. \frac{2}{3} \tag{2}. \frac{4}{5}.\frac{6}{7} \dots \frac{2x}{2x+1}$ and $A > x^r. \frac{1}{2}.\frac{2}{3}.\frac{4}{5} \dots \frac{2x-2}{2x-1} \tag{3}$ Multiplying $(1)$ and $(3)$, $A^2 > x^{2r}.\frac{1}{2(2x)} \text{ or }, A > \frac{x^r}{2 \sqrt{x}}$. The limit does not exist if $r >0.5$. Now we need to check if the limit exists for $r \leq 0.5$ Multiplying $(1)$ and $(2)$, we gwt $A < \frac{x^r}{\sqrt{2x+1}}$ and its clear that $A$ is finite and nonzero for $r=0.5$ and $\lim_{x \to \infty} A \leq \frac{1}{\sqrt{2}}$. I am unable to go further and give the exct value of the limit.

How do I put $\sqrt{x+1}$ into exponential notation? I think $\sqrt{x+1} = x^{1/2} + 1^{1/2}$. Is this incorrect? Why or why not?

Remember the formula for fractional exponents: $$x^\frac{m}{n} = \sqrt[n]{x^m}$$ It can also be written as: $$x^\frac{m}{n} = (\sqrt[n]{x})^m$$ $\sqrt{x+1}$ can be rewritten as $\sqrt[2]{(x+1)^1}$ Using our formula, we can say that: $$\sqrt[2]{(x+1)^1} = (x+1)^\frac{1}{2}$$ Also remember that: $$x^m + y^m \neq (x+y)^m$$ You should be familiar with the Pythagorean triple $3-4-5$. It can be easily seen that: $$3^2 + 4^2 \neq (3+4)^2 \neq 7^2$$ $$3^2 + 4^2 = 5^2$$ I will now go back to the original question. You think that: $$\sqrt{x+1} = x^\frac{1}{2} + 1^\frac{1}{2}$$ Let us expand $x^\frac{1}{2} + 1^\frac{1}{2}$. $$x^\frac{1}{2} + 1^\frac{1}{2}$$ $$=\sqrt[2]{x^1} + \sqrt[2]{1^1}$$ $$=\sqrt{x} + \sqrt{1}$$ $$=\sqrt{x} + 1$$ $$\sqrt{x} + 1 \neq \sqrt{x+1}$$ So, because $\sqrt{x} + 1 \neq \sqrt{x+1}$, then $\sqrt{x+1} \neq x^\frac{1}{2} + 1^\frac{1}{2}$. The correct answer is: $$\sqrt{x+1} = (x+1)^\frac{1}{2}$$ Before I forget, always remember the restrictions on $\sqrt{x+1} = (x+1)^\frac{1}{2}$. If $x < -1$. then the statement is false, because you cannot have negative square roots. TECHNICALLY, it is not false because complex numbers cover negative square roots. But for the purposes of this question, UNLESS you have learned about complex numbers and the number $i$, the statement is false if $x < -1$. RESTRICTIONS ON $\sqrt{x+1} = (x+1)^\frac{1}{2}$: $x ≥ 1$ $\sqrt{x+1} ≥ 0$ $(x+1)^\frac{1}{2} ≥ 0$ I hope I have enlightened you a bit :) EDIT: I do not want you to think that the statement $\sqrt{x}+1=\sqrt{x+1}$ is entirely false. There is a value of $x$ that will make the equation true. Let's find it. $$\sqrt{x}+1=\sqrt{x+1}$$ $$x+2\sqrt{x}+1=x+1$$ $$2\sqrt{x}=0$$ $$4x=0$$ $$x=0$$ Checking for extraneous roots... $$\sqrt{0}+1=\sqrt{0+1}$$ $$1=\sqrt{1}$$ $$1=1$$ So, $\sqrt{x}+1=\sqrt{x+1}$ when $x=0$.

nonlinear diophantine equation $x^2+y^2=z^2$ how to solve a diophantine equation $x^2+y^2=z^2$ for integers $x,y,z$ i strongly believe there is a geometric solution ,since this is a pythagoras theorem form or a circle with radius $z$ $x^2+y^2=z^2$ $(\frac{x}{z})^2+(\frac{y}{z})^2=1\implies x=y=\pm z$ or $0$ so we consider a line passing through points $P_1(- z,0)$ and $P(x,y)$ both on the circle $m=\frac{y}{x+z}$ $x^2+m^2(x+z)^2=z^2$ $(m^2+1)x^2+2xzm^2+(m^2-1)z^2=0$ $((m^2+1)x+(m^2-1)z)(x+z)=0$ $\frac{x}{z}=-\frac{m^2-1}{m^2+1}$ or $-1$ let $m=\frac{a}{b}\implies \frac{x}{z}=\frac{b^2-a^2}{b^2+a^2}$ $\frac{y}{z}=\frac{2a^2}{b^2+a^2}$ how to get explicit $z,x,y$

Euclid's Formula says that in essence, $(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$ for all positive integers $m > n$. This is basically a parametrization of Pythagorean Triplets with two parameters.

Trigonometric Series Proof I am posed with the following question: Prove that for even powers of $\sin$: $$ \int_0^{\pi/2} \sin^{2n}(x) dx = \dfrac{1 \cdot 3 \cdot 5\cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} \times \dfrac{\pi}{2} $$ Here is my work so far: * *Proof by induction $P(1) \Rightarrow n = 2 \Rightarrow $ $$\int_0^{\pi/2} \sin^4(x) dx = \dfrac{1 \cdot 3}{2 \cdot 4} \times \dfrac{\pi}{2} $$ $$ \frac{3\pi}{16} = \frac{3 \pi}{16} $$ Base Case Succeeds. (The left hand side evaluated by MAPLE) $P(n+1) \Rightarrow n = n + 1 \Rightarrow$ $$ \int_0^{\pi/2} \sin^{2n + 1}(x) dx = \int_0^{\pi/2} \sin^{2n} x \sin x dx$$ $$ = \int_0^{\pi/2} (1 - \cos^2x)^n \sin x dx$$ Let $u = \cos x, du = - \sin x dx \Rightarrow$ $$ -\int_0^{\pi/2} (1-u^2)^n du$$ Now I am stuck and unsure what to do with this proof ... Any help would be greatly appreciated (I also tried using the reduction formulas to no avail) EDIT I have completed the proof. I am posting this here for any other people who may have the same question... We will prove by induction that $\forall n \in 2 \mathbb{N}_{>0}$ \begin{align*} \int_0^{\pi/2} \sin^{2n} x dx & = \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{2 \times 4 \times 6 \times \cdots \times 2n} \times \frac{\pi}{2} \tag{1} \end{align*} With $k = 1$ as our base case, we have \begin{align*} \frac{1}{2}x - \frac{1}{4} \sin{2x} \bigg|_{0}^{\pi / 2} & = \frac{\pi}{4} \tag{67} \\ \frac{1}{2} \times \frac{\pi}{2} & = \frac{\pi}{4} \end{align*} Let $n \in 2\mathbb{N}_{>0}$ be given and suppose (1) is true for $k = n$. Then \begin{align*} \int_0^{\pi/2} \sin^{2n+2} x dx & = - \frac{1}{2n+2} \cos^{2n-1}x \sin x \bigg|_{0}^{\pi / 2} + \frac{2n+1}{2n+2} \int_0^{\pi/2} \sin^{2n} x dx \tag{67} \\ & = \frac{2n+1}{2n+2} \int_0^{\pi/2} \sin^{2n} x dx \\ & = \frac{1 \times 3 \times 5 \times \cdots \times (2n + 1)}{2 \times 4 \times 6 \times \cdots \times {(2n+2)}} \times \frac{2n+2}{2n+1} \times \frac{\pi}{2} \\ & = \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{2 \times 4 \times 6 \times \cdots \times 2n} \times \frac{\pi}{2} \\ & = \int_0^{\pi/2} \sin^{2n} x dx \end{align*} Thus, (1) holds for $k = n + 1$, and by the principle of induction, it follows that that (1) holds for all even numbers. $\square$

For $k=1$, it's straightforward to verify$$\int_0^{\pi/2}\sin^2x~dx=\int_0^{\pi/2}\frac{1-\cos 2x}2dx=\frac\pi4$$ Assume $k=n$ we have $$I_n=\int_0^{\pi/2}\sin^{2n}x~dx=\frac{(2n-1)!!}{(2n)!!}\frac\pi2$$ Then for $k=n+1$, $$\begin{align}I_{n+1}&=\int_0^{\pi/2}\sin^{2n}x(1-\cos^2x)dx\\ &=I_n-\int_0^{\pi/2}\sin^{2n}x\cos^2x~dx\\ &=I_n-\left.\frac1{2n+1}\sin^{2n+1}x\cos x\right|_0^{\pi/2}-\frac1{2n+1}\int_0^{\pi/2}\sin^{2n+1}x\sin x~dx\\ &=I_n-\frac1{2n+1}I_{n+1}\end{align}$$ Solve the recurrent relation and obtain $$I_{n+1}=\frac{2n+1}{2n+2}I_n=\frac{(2n+1)!!}{(2n+2)!!}\frac\pi2$$

An inequality for sides of a triangle Let $ a, b, c $ be sides of a triangle and $ ab+bc+ca=1 $. Show $$(a+1)(b+1)(c+1)<4 $$ I tried Ravi substitution and got a close bound, but don't know how to make it all the way to $4 $. I am looking for a non-calculus solution (no Lagrange multipliers). Do you know how to do it?

Solving $ab+bc+ca=1$ for $c$ gives $$ c=\frac{1-ab}{a+b}\tag{1} $$ The triangle inequality says that for non-degenerate triangles $$ |a-b|\lt c\lt(a+b)\tag{2} $$ Multiply $(2)$ by $a+b$ to get $$ |a^2-b^2|\lt1-ab\lt(a+b)^2\tag{3} $$ By $(3)$, we have $(a+b)^2-1+ab\gt0$; therefore, $$ \begin{align} (a+b+1)(a+b+ab-1) &=\left[(a+b)^2-1+ab\right]+(a+b)ab\\ &\gt(a+b)ab\\[6pt] &\gt0\tag{4} \end{align} $$ Furthermore, $\color{#C00000}{(3)}$ implies $$ a\ge1\implies1-b^2\le\color{#C00000}{a^2-b^2\lt1-ab}\implies b\gt a\tag{5a} $$ Similarly, $$ b\ge1\implies1-a^2\le\color{#C00000}{b^2-a^2\lt1-ab}\implies a\gt b\tag{5b} $$ Inequalities $(5)$ imply that if either $a\ge1$ or $b\ge1$, then both $a\gt1$ and $b\gt1$. Consequently, we have both $a\gt b$ and $b\gt a$. Therefore, we must have $$ a\lt1\qquad\text{and}\qquad b\lt1\tag{6} $$ Using $(1)$, $(4)$, and $(6)$, we have $$ \begin{align} abc+a+b+c-2 &=(1+ab)\frac{1-ab}{a+b}+(a+b)-2\\ &=\frac{1-a^2b^2+(a+b)^2-2(a+b)}{a+b}\\ &=\frac{(a+b-1)^2-a^2b^2}{a+b}\\ &=\frac{(a+b+ab-1)(a+b-ab-1)}{a+b}\\ &=-\frac{(a+b+ab-1)(a-1)(b-1)}{a+b}\\ &\lt0\tag{7} \end{align} $$ Therefore, since $ab+bc+ca+1=2$, $(7)$ says $$ \begin{align} (a+1)(b+1)(c+1) &=(abc+a+b+c)+(ab+bc+ca+1)\\[4pt] &\lt2+2\\[4pt] &=4\tag{8} \end{align} $$

If $a$ and $b$ are odd then $a^2+b^2$ is not a perfect square Prove if $a$ and $b$ are odd then $a^2+b^2$ is not a perfect square. We have been learning proof by contradiction and were told to use the Euclidean Algorithm. I have tried it both as written and by contradiction and can't seem to get anywhere.

Let $a=2m+1$ and $b=2n+1$. Assume $a^2+b^2=k^2$. Then: $$(2m+1)^2+(2n+1)^2=k^2 \iff \\ 4(m^2+m+n^2+n)+2=k^2 \iff \\ 4(m^2+m+n^2+n)+2=(2r)^2 \iff \\ 2(m^2+m+n^2+n)+1=2r^2 \iff \\ 2s+1=2r^2,$$ which is a contradiction. Hence, the assumption $a^2+b^2=k^2$ is false.

What steps are taken to make this complex expression equal this? How would you show that $$\sum_{n=1}^{\infty}p^n\cos(nx)=\frac{1}{2}\left(\frac{1-p^2}{1-2p\cos(x)+p^2}-1\right)$$ when $p$ is positive, real, and $p<1$?

Since $0<p<1$, we have \begin{eqnarray} \sum_{n=1}^\infty p^n\cos(nx)&=&\Re\sum_{n=1}^\infty p^ne^{inx}=\Re\sum_{n=1}^\infty(pe^{ix})^n=\Re\frac{pe^{ix}}{1-pe^{ix}}=\Re\frac{pe^{ix}(1-pe^{-ix})}{|1-pe^{ix}|^2}\\ &=&p\Re\frac{-p+\cos x+i\sin x}{|1-p\cos x-ip\sin x|^2}=p\frac{-p+\cos x}{(1-p\cos x)^2+p^2\sin^2x}\\ &=&p\frac{\cos x-p}{1-2p\cos x+p^2\cos^2x+p^2\sin^2x}=\frac{p(\cos x-p)}{1+p^2-2p\cos x}\\ &=&\frac12\left(\frac{1-p^2}{1-2p\cos x+p^2}-1\right). \end{eqnarray}

$f(x) = \arccos {\frac{1-x^2}{1+x^2}}$; f'(0+), f'(0-)? $f(x) = \arccos {\frac{1-x^2}{1+x^2}}$ $f'(x) = 2/(1+x^2)$, but I see graphic, and it is true only for x>=0. For x<=0 => $f'(x) = -2/(1+x^2)$ How can I deduce the second formula or proof that it is.

Method $\#1:$ Let $\displaystyle\arccos\frac{1-x^2}{1+x^2}=y$ $\displaystyle\implies(1) \cos y=\frac{1-x^2}{1+x^2}\ \ \ \ (i)$ Using the definition of Principal values, $\displaystyle \implies(2)0\le y\le\pi \implies 0\le\frac y2\le\frac\pi2\implies \tan\frac y2\ge0$ Applying Componendo and dividendo on $(i),$ $\displaystyle x^2=\frac{1-\cos y}{1+\cos y}=\tan^2\frac y2$ (using $\displaystyle\cos2A=\frac{1-\tan^2A}{1+\tan^2A}$) As $\displaystyle\tan\frac y2\ge0,\implies \tan\frac y2=|x|$ Method $\#2:$ Let $\displaystyle z=\arctan x$ $\displaystyle\implies -\frac\pi2\le z\le\frac\pi2\iff -\pi\le 2z\le\pi$ and $\displaystyle\tan z=x,\frac{1-x^2}{1+x^2}=\cos2z$ $\displaystyle\arccos(2z)=\begin{cases} 2z=2\arctan x &\mbox{if } 0\le 2z\le\pi\iff 0\le z\le\frac\pi2\implies x=\tan z\ge0\ \\-2z=-2\arctan x & \mbox{if } -\pi\le 2z<0\iff -\frac\pi2\le z<0\implies x<0 \end{cases} $

Finding system with infinitely many solutions The question asks to find equation for which the system has infinitely many solutions. The system is: \begin{cases} -cx + 3y + 2z = 8\\ x + z = 2\\ 3x + 3y + az = b \end{cases} How should I approach questions like this? I tried taking it to row reduced echelon form but it got kind of messy. The answer is supposed to be: $$a - c -5 = 0$$ and $$b- 2c +2 = 0$$

You can do row reduction; consider the matrix \begin{align} \left[\begin{array}{ccc|c} -c & 3 & 2 & 8 \\ 1 & 0 & 1 & 2 \\ 3 & 3 & a & b \end{array}\right] &\to \left[\begin{array}{ccc|c} 1 & 0 & 1 & 2 \\ -c & 3 & 2 & 8 \\ 3 & 3 & a & b \end{array}\right]\quad\text{swap 1 and 2}\\ &\to \left[\begin{array}{ccc|c} 1 & 0 & 1 & 2 \\ 0 & 3 & 2+c & 8+2c \\ 0 & 3 & a-3 & b-6 \end{array}\right]\quad\text{reduce first column}\\ &\to \left[\begin{array}{ccc|c} 1 & 0 & 1 & 2 \\ 0 & 3 & 2+c & 8+2c \\ 0 & 0 & a-c-5 & b-2c-14 \end{array}\right]\quad\text{reduce second column}\\ \end{align} The system has infinitely many solutions if and only if the last row is zero, that is \begin{cases} a-c=5\\ b-2c=14 \end{cases}