5+20怎么算？幼儿园难题第一弹来了！

· · 算法·理论

我们都知道，像 5+20 这样的难题已经不能是幼儿园学 前班的小朋友能做的了，但是他又很具有思维价值，所以 各大幼儿园学后班都会放置这样一道压轴题，难倒了一众 小朋友，所以才诞生了这一篇超详细，一定可以把你教会 （fei）教程！

设\ 5+20=A\sum_{n=1}^{\infty} \frac{1}{n^{2}}+(\oiint P dy dz+Q dzdx+Rdxdy)+\oint_{|w-i|=2}\frac{dw}{(w-i)^{2}} 其中A为待定常数\ ,\ P=x^{2}+y,Q=\cos^{z}x^{2},R=(x+y)z, \sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{1}{1^{2} }+\frac{1}{2^{2} }+\frac{1}{3^{2} }+\frac{1}{4^{2} }+……，w\ 为复变量 令\ f(x)=x^{2}\ ,\ 其可展开为\ S(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}(a_{n}\cos nx+b_{n}\sin nx) 其中\ ，a_{n}=\frac{1}{\pi }\int_{-\pi}^{\pi}f(x)\cos nx dx,b_{n}=\frac{1}{\pi }\int_{-\pi}^{\pi}\sin nxdx 则\ b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}x^{2}\sin nx dx=0,a_{0}=\frac{1}{\pi}\int_{-\pi}^{\pi}x^{2}dx=\frac{2}{3}\pi^{2} 又\ a_{n}=\frac{1}{\pi }\int_{-\pi}^{\pi}x^{2}\cos n\pi dx =\frac{1}{\pi}(\frac{1}{n}x^{2} \sin nxdx+\frac{2}{n^{2}}\cos nx-\frac{2}{n^{3}}\sin nx)|_{-\pi }^{\pi} =\frac{4}{n^{2}}\cos nxdx=(-1)^{n}\frac{4}{n^{2}} 则\ f(x)\overset{展开}{\sim}S(x)=\frac{\pi^{2}}{3}+\sum_{n=1}^{\infty}(-1)^{n}\frac{4}{n^{2}}\cos nx =>S(\pi)=\frac{1}{2}(\lim_{x \to -\pi-}f(x)+\lim_{x \to -\pi+}f(x))=\pi^{2} 又\ S(\pi)=\frac{\pi^{2}}{3}+\sum_{n=1}^{\infty}(-1)^{n}\frac{4}{n^{2}}\cos n\pi=\frac{\pi^{2}}{3}=>\sum_{n=1}^{\infty}\frac{4}{n^{2}} 则\ \pi^{2}=\frac{\pi^{2}}{3} +\sum_{n=1}^{\infty}\frac{4}{n^{2}}=>\sum_{n=1}^{\infty}\frac{4}{n^{2}}=\frac{2}{3}\pi^{2}=>\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2} }{6} 令\ w=i+2e^{i\theta},有dw=2ie^{i\theta}d\theta\ ,则 \ \ \ \ \ 5+20 =A\sum_{n=1}^{\infty} \frac{1}{n^{2}}+(\oiint P dy dz+Q dzdx+Rdxdy)+\oint_{|w-i|=2}\frac{dw}{(w-i)^{2}} =A\frac{\pi^{2} }{6}+\underset{\Omega:x^{2}+y^{2}+z^{2}\le 1}{\int \int \int }(\frac{aP}{ax}+\frac{aQ}{ay}+\frac{aR}{aZ})dV+\int_{0}^{2\pi}\frac{2ie^{i\theta }}{4e^{2i\theta}}d\theta =A\frac{\pi^{2} }{6}+\underset{\Omega }{\int\int\int}(2x+0+x+y)dV+\frac{i}{2} \int_{0}^{2\pi}e^{-i\theta}d\theta =A\frac{\pi^{2} }{6}+\underset{\Omega }{\int\int\int}(3x+y)dV+\frac{-1}{2}e^{-i\theta}d\theta =A\frac{\pi^{2}}{6}+\int_{0}^{2\pi}d\theta\int_{0}^{\pi}d\varphi\int_{0}^{1}(3r\cos \theta +r\sin \theta)r^{2}\sin^{2}\varphi-\frac{e^{-2\pi i}-e^{0}}{2} =A\frac{\pi^{2}}{6}+\int_{0}^{2\pi}d\theta\int_{0}^{\pi}(\frac{3}{4} \cos \theta +\frac{1}{4}\sin \theta)\sin^{2}\varphi d\varphi -\frac{\left | - \right | }{2} =A\frac{\pi^{2}}{6}+\int_{0}^{2\pi}(\frac{3}{4} \cos \theta +\frac{1}{4}\sin \theta)\cdot 2\cdot \frac{1}{2}\cdot \pi d\theta =A\frac{\pi^{2}}{6}+0+0=A\frac{\pi^{2}}{6}=>5+20=A\frac{\pi^{2}}{6} 则\ A=\frac{6}{\pi^{2}}(5+20)= \frac{30}{\pi^{2}}(1+4)=5\cdot \frac{30}{\pi^{2}} 则\ 5+20=A\frac{\pi^{2}}{6}=5\cdot \frac{30}{\pi^{2}}\cdot \frac{\pi^{2}}{6}=5\cdot \frac{30}{6}=5\cdot 5=5^{2}=25

综上所述， 5+20=25

好啦，这道幼儿园难题就这样被我们秒了，我是老L，
每日一个斜眼法，让我们在下一次幼儿园难题中见面！