README
go env -w GOPROXY=https://goproxy.cn,direct
go mod tidy
Sequence diagram
sequenceDiagram
participant Alice
participant Bob
Alice->>John: Hello John, how are you?
loop Healthcheck
John->>John: Fight against hypochondria
end
Note right of John: Rational thoughts <br/>prevail!
John-->>Alice: Great!
John->>Bob: How about you?
Bob-->>John: Jolly good!
Gantt diagram
gantt
dateFormat YYYY-MM-DD
title Adding GANTT diagram to mermaid
excludes weekdays 2014-01-10
section A section
Completed task :done, des1, 2014-01-06,2014-01-08
Active task :active, des2, 2014-01-09, 3d
Future task : des3, after des2, 5d
Future task2 : des4, after des3, 5d
$...$
āŖ\cup:
$$...$$
matrix
AD
As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem) ,$$ inverts the negative Laplacian, in the sense that .$$
The latter quantity is $$ \int_0^1 x(1-x),dx = \frac 1 2 - \frac 1 3 = \frac 1 6~~.$$
Hence, we have that
$$ \begin{align*} \frac{\pi^2}{6}&=\frac{4}{3}\frac{(\arcsin 1)^2}{2}\ &=\frac{4}{3}\int_0^1\frac{\arcsin x}{\sqrt{1-x^2}},dx\ &=\frac{4}{3}\int_0^1\frac{x+\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}}{\sqrt{1-x^2}},dx\ &=\frac{4}{3}\int_0^1\frac{x}{\sqrt{1-x^2}},dx +\frac{4}{3}\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!(2n+1)}\int_0^1x^{2n}\frac{x}{\sqrt{1-x^2}},dx\ &=\frac{4}{3}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!(2n+1)}\left[\frac{(2n)!!}{(2n+1)!!}\right]\ &=\frac{4}{3}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\ &=\frac{4}{3}\left(\sum_{n=1}^{\infty}\frac{1}{n^2}-\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}\right)\ &=\sum_{n=1}^{\infty}\frac{1}{n^2} \end{align*} $$ B_Scheiner Consider the function
$\pi \cot(\pi z)$ which has poles at$z=\pm n$ where n is an integer. Using the L'hopital rule you can see that the residue at these poles is 1. Now consider the integral$\int_{\gamma_N} \frac{\pi\cot(\pi z)}{z^2} dz$ where the contour$\gamma_N$ is the rectangle with corners given by Ā±(N + 1/2) Ā± i(N + 1/2) so that the contour avoids the poles of$\cot(\pi z)$ . The integral is bouond in the following way:$\int_{\gamma_N} |\frac{\pi\cot(\pi z)}{z^2} |dz\le Max |(\frac{\pi\cot(\pi z)}{z^2}) | Length(\gamma_N)$ . It can easily be shown that on the contour$\gamma_N$ that$\pi \cot(\pi z)< M$ where M is some constant. Then we have$\int_{\gamma_N} |\frac{\pi\cot(\pi z)}{z^2} |dz\le M Max |\frac{1}{z^2} | Length(\gamma_N) = (8N+4) \frac{M}{\sqrt{2(1/2+N)^2}^2}$ where (8N+4) is the lenght of the contour and$\sqrt{2(1/2+N)^2}$ is half the diagonal of$\gamma_N$ . In the limit that N goes to infinity the integral is bound by 0 so we have$\int_{\gamma_N} \frac{\pi\cot(\pi z)}{z^2} dz =0$ by the cauchy residue theorem we have 2ĻiRes(z = 0) + 2Ļi$\sum$Residues(z$\ne$ 0) = 0. At z=0 we have Res(z=0)=$-\frac{\pi^2}{3}$, and$Res (z=n)=\frac{1}{n^2}$ so we have$2\pi iRes(z = 0) + 2\pi i\sum Residues(z\ne 0) = -\frac{\pi^2}{3}+2\sum_{1}^{\infty} \frac{1}{n^2} =0$ Where the 2 in front of the residue at n is because they occur twice at +/- n. We now have the desired result$\sum_{1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$ . Meadara I saw this proof in an extract of the College Mathematics Journal. Consider the Integeral :$I = \int_0^{\pi/2}\ln(2\cos x)dx$ From$2\cos(x) = e^{ix} + e^{-ix}$ , we have:$$\int_0^{\pi/2}\ln\left(e^{ix} + e^{-ix}\right)dx = \int_0^{\pi/2}\ln\left(e^{ix}(1 + e^{-2ix})\right)dx=\int_0^{\pi/2}ixdx + \int_0^{\pi/2}\ln(1 + e^{-2ix})dx$$ The Taylor series expansion of$\ln(1+x)=x -\frac{x^2}{2} +\frac{x^3}{3}-\cdots$ Thus ,$\ln(1+e^{-2ix}) = e^{-2ix}- \frac{e^{-4ix}}{2} + \frac{e^{-6ix}}{3} - \cdots $ , then for$I$ : $$I = \frac{i\pi^2}{8}+\left[-\frac{e^{-2ix}}{2i}+\frac{e^{-4ix}}{2\cdot 4i}-\frac{e^{-6ix}}{3\cdot 6i}-\cdots\right]0^\frac{\pi}{2}$$ $$I = \frac{i\pi^2}{8}-\frac{1}{2i}\left[\frac{e^{-2ix}}{1^2}-\frac{e^{-4ix}}{2^2}+\frac{e^{-6ix}}{3^2}-\cdots\right]0^\frac{\pi}{2}$$ By evaluating we get something like this.. $$I = \frac{i\pi^2}{8}-\frac{1}{2i}\left[\frac{-2}{1^2}-\frac{0}{2^2}+\frac{-2}{3^2}-\cdots\right]0^\frac{\pi}{2}$$ Hence $$\int_0^{\pi/2}\ln(2\cos x)dx=\frac{i\pi^2}{8}-i\sum{k=0}^\infty \frac{1}{(2k+1)^2}$$ So now we have a real integral equal to an imaginary number, thus the value of the integral should be zero. Thus, $\sum{k=0}^\infty \frac{1}{(2k+1)^2}=\frac{\pi^2}{8}$ But let $\sum{k=0}^\infty \frac{1}{k^2}=E$ .We get$\sum_{k=0}^\infty \frac{1}{(2k+1)^2}=\frac{3}{4} E$ And as a result$$\sum_{k=0}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$$ dustin I have another method as well. From skimming the previous solutions, I don't think it is a duplicate of any of them In Complex analysis, we learn that$\sin(\pi z) = \pi z\Pi_{n=1}^{\infty}\Big(1 - \frac{z^2}{n^2}\Big)$ which is an entire function with simple zer0s at the integers. We can differentiate term wise by uniform convergence. So by logarithmic differentiation we obtain a series for$\pi\cot(\pi z)$ . $$ \frac{d}{dz}\ln(\sin(\pi z)) = \pi\cot(\pi z) = \frac{1}{z} - 2z\sum_{n=1}^{\infty}\frac{1}{n^2 - z^2} $$ Therefore, $$ -\sum_{n=1}^{\infty}\frac{1}{n^2 - z^2} = \frac{\pi\cot(\pi z) - \frac{1}{z}}{2z} $$ We can expand$\pi\cot(\pi z)$ as $$ \pi\cot(\pi z) = \frac{1}{z} - \frac{\pi^2}{3}z - \frac{\pi^4}{45}z^3 - \cdots $$ Thus, \begin{align} \frac{\pi\cot(\pi z) - \frac{1}{z}}{2z} &= \frac{- \frac{\pi^2}{3}z - \frac{\pi^4}{45}z^3-\cdots}{2z}\ -\sum_{n=1}^{\infty}\frac{1}{n^2 - z^2}&= -\frac{\pi^2}{6} - \frac{\pi^4}{90}z^2 - \cdots\ -\lim_{z\to 0}\sum_{n=1}^{\infty}\frac{1}{n^2 - z^2}&= \lim_{z\to 0}\Big(-\frac{\pi^2}{6} - \frac{\pi^4}{90}z^2 - \cdots\Big)\ -\sum_{n=1}^{\infty}\frac{1}{n^2}&= -\frac{\pi^2}{6}\ \sum_{n=1}^{\infty}\frac{1}{n^2}&= \frac{\pi^2}{6} \end{align} Elias See evaluations of Riemann Zeta Function$\zeta(2)=\sum_{n=1}^\infty\frac{1}{n^2}$ in mathworld.wolfram.com and a solution by in D. P. Giesy in Mathematics Magazine: D. P. Giesy, Still another elementary proof that$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ , Math. Mag. 45 (1972) 148-149. Unfortunately I did not get a link to this article. But there is a link to a note from Robin Chapman seems to me a variation of proof's Giesy. barto Applying the usual trick 1 transforming a series to an integral, we obtain$$\sum_{n=1}^\infty\frac1{n^2}=\int_0^1\int_0^1\frac{dxdy}{1-xy}$$ where we use the Monotone Convergence Theorem to integrate term-wise. Then there's this ingenious change of variables 2, which I learned from Don Zagier during a lecture, and which he in turn got from a colleague:$$(x,y)=\left(\frac{\cos v}{\cos u},\frac{\sin u}{\sin v}\right),\quad0\leq u\leq v\leq \frac\pi2$$ One verifies that it is bijective between the rectangle$[0,1]^2$ and the triangle$0\leq u\leq v\leq \frac\pi2$ , and that its Jacobian determinant is precisely$1-x^2y^2$ , which means$\frac1{1-x^2y^2}$ would be a neater integrand. For the moment, we have found$$J=\int_0^1\int_0^1\frac{dxdy}{1-x^2y^2}=\frac{\pi^2}8$$ (the area of the triangular domain in the$(u,v)$ plane). There are two ways to transform$\int\frac1{1-xy}$ into something $\int\frac1{1-x^2y^2}$ish: Manipulate$S=\sum_{n=1}^\infty\frac1{n^2}$ : We have$\sum_{n=1}^\infty\frac1{(2n)^2}=\frac14S$ so$\sum_{n=0}^\infty\frac1{(2n+1)^2}=\frac34S$ . Applying the series-integral transformation, we get$\frac34S=J$ so$$S=\frac{\pi^2}6$$ Manipulate$I=\int_0^1\int_0^1\frac{dxdy}{1-xy}$ : Substituting$(x,y)\leftarrow(x^2,y^2)$ we have$I=\int_0^1\int_0^1\frac{4xydxdy}{1-x^2y^2}$ so$$J=\int_0^1\int_0^1\frac{dxdy}{1-x^2y^2}=\int_0^1\int_0^1\frac{(1+xy-xy)dxdy}{1-x^2y^2}=I-\frac14I$$ whence$$I=\frac43J=\frac{\pi^2}6$$ (It may be seen that they are essentially the same methods.) After looking at the comments it seems that this looks a lot like Proof 2 in the article by R. Chapman. See also: Multiple Integral$\int\limits_0^1!!\int\limits_0^1!!\int\limits_0^1!!\int\limits_0^1\frac1{1-xyuv},dx,dy,du,dv$ 1 See e.g. Proof 1 in Chapman's article. 2 It may have been a different one; maybe as in the above article. Either way, the idea to do something trigonometric was not mine. Asier Calbet The sum can be written as the integral:$$\int_0^{\infty} \frac{x}{e^x-1} dx $$ This integral can be evaluated using a rectangular contour from 0 to$\infty$ to$\infty + \pi i$ to $ 0$ . John M. Campbell There is a simple way of proving that$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$ using the following well-known series identity:$$\left(\sin^{-1}(x)\right)^{2} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2 \binom{2n}{n}}.$$ From the above equality, we have that$$x^2 = \frac{1}{2}\sum_{n=1}^{\infty}\frac{(2 \sin(x))^{2n}}{n^2 \binom{2n}{n}},$$ and we thus have that:$$\int_{0}^{\pi} x^2 dx = \frac{\pi^3}{12} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{\int_{0}^{\pi} (2 \sin(x))^{2n} dx}{n^2 \binom{2n}{n}}.$$ Since$$\int_{0}^{\pi} \left(\sin(x)\right)^{2n} dx = \frac{\sqrt{\pi} \ \Gamma\left(n + \frac{1}{2}\right)}{\Gamma(n+1)},$$ we thus have that:$$\frac{\pi^3}{12} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{ 4^{n} \frac{\sqrt{\pi} \ \Gamma\left(n + \frac{1}{2}\right)}{\Gamma(n+1)} }{n^2 \binom{2n}{n}}.$$ Simplifying the summand, we have that$$\frac{\pi^3}{12} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{\pi}{n^2},$$ and we thus have that$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$ as desired.