[Math]Pi(2)
[Math]Pi(2)
接着前一篇，[Math]Pi(1)，下面继续介绍Leonard Euler求Pi的第二个公式。
其实这个公式也是来源一个古老的问题，Basel problem .
证法1.麦克劳伦级数和零点式
sin(x)的 Maclaurin Series为：
$$ ecause sin(x) = x - frac{x^3}{3!} + frac{x^5}{5!} - frac{x^7}{7!} + cdots $$
再把 $frac{sin(x)}{x}$ 表示成零点处的多项式：
egin{align*}
herefore frac{sin(x)}{x} &= 1 - frac{x^2}{3!} + frac{x^4}{5!} - frac{x^6}{7!} + cdots \
&=left(1 - frac{x}{pi}
ight)left(1 + frac{x}{pi}
ight)left(1 - frac{x}{2pi}
ight)left(1 + frac{x}{2pi}
ight)left(1 - frac{x}{3pi}
ight)left(1 + frac{x}{3pi}
ight) cdots \
&= left(1 - frac{x^2}{pi^2}
ight)left(1 - frac{x^2}{4pi^2}
ight)left(1 - frac{x^2}{9pi^2}
ight) cdots
end{align*}
对比前面两式中的x2项的系数有：
$$-left(frac{1}{pi^2} + frac{1}{4pi^2} + frac{1}{9pi^2} + cdots
ight) =-frac{1}{pi^2}sum_{n=1}^{infty}frac{1}{n^2}=-frac{1}{3!}$$
egin{equation}label{E3} sum_{n=1}^{infty}frac{1}{n^2} = frac{pi^2}{6}. end{equation}
证法2.傅里叶级数
1.Fourier series
$$ f(x)=dfrac{a_{0}}{2}+sum_{n=1}^{infty }(a_{n}cos nx+b_{n}sin x),-pi leq xleq pi $$
其中，
egin{align*}
a_{n}&=dfrac{1}{pi }int_{-pi }^{pi }f(x)cos nx;dxqquad n=0,1,2,3,...,\
b_{n}&=dfrac{1}{pi }int_{-pi }^{pi }f(x)sin nx;dxqquad n=1,2,3,... .
end{align*}
f(x) = x2的傅里叶级数
当n = 0，
$$a_{0}=dfrac{1}{pi }int_{-pi }^{pi }x^{2}dx=dfrac{2}{pi }int_{0}^{pi
}x^{2}dx=dfrac{2pi ^{2}}{3}.$$
当n = 1, 2, 3, . . . ，
egin{align*}
a_{n}&=dfrac{1}{pi }int_{-pi }^{pi }x^{2}cos nx;dx=dfrac{2}{pi }int_{0}^{pi }x^{2}cos nx;dx\
&=dfrac{2}{pi } imes dfrac{2pi }{n^{2}}(-1)^{n}=(-1)^{n}dfrac{4}{n^{2}}\
b_{n}&=dfrac{1}{pi }int_{-pi }^{pi }f(x)sin nx;dx=0
end{align*}
$$ ecause int olimits_0^{2pi} x^2cos nx;dx=left[dfrac{2x}{n^{2}}cos nx+left( frac{x^{2}}{n}-dfrac{2}{n^{3}} ight) sin nx ight]|_0^{2pi}=dfrac{2pi }{n^{2}}(-1)^{n}.$$
因此，
egin{equation}
f(x)=dfrac{pi ^{2}}{3}+sum_{n=1}^{infty }left( (-1)^{n}dfrac{4}{n^{2}}cos nx
ight)
end{equation}
将f(π) = π2代入上式有：
egin{align*}
f(pi)&=dfrac{pi ^{2}}{3}+sum_{n=1}^{infty }left( (-1)^{n}dfrac{4}{n^{2}}cos npi
ight)\
&=dfrac{pi ^{2}}{3}+4sum_{n=1}^{infty }left( (-1)^{n}(-1)^{n}dfrac{1}{n^{2}}
ight)\
&=dfrac{pi ^{2}}{3}+4sum_{n=1}^{infty }dfrac{1}{n^{2}}.
end{align*}
最后，我们就可以得到：
egin{equation}label{E5}
sum_{n=1}^{infty }dfrac{1}{n^{2}}=dfrac{pi ^{2}}{4}-dfrac{pi ^{2}}{12}=dfrac{pi ^{2}}{6}
end{equation}
reference