无穷乘积求值

大白兔奶糖

$\sqrt{2} \sqrt{2+2 \sqrt{2}} \sqrt{2+2 \sqrt{2+2 \sqrt{2}}} \cdots=?$

hbghlyj

Viète's formula$\displaystyle {\frac {2}{\pi }}={\frac {\sqrt {2}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2}}}}{2}}\cdot {\frac {\sqrt {2+{\sqrt {2+{\sqrt {2}}}}}}{2}}\cdots$
\begin{aligned}x_{0}&=2\cos \theta&=\frac{\sin(2\theta)}{\sin\theta}\\
x_{1}&=\sqrt{2+x_{0}}=\sqrt{2+2\cos \theta }=2\cos (\frac{\theta }{2})&=\frac{\sin\theta}{\sin(\frac{\theta}2)}\\
x_{2}&=\sqrt{2+x_{1}}=\sqrt{2+2\cos (\frac{\theta }{2})}=2\cos (\frac{\theta }{4})&=\frac{\sin(\frac{\theta}2)}{\sin(\frac{\theta}4)}\\&\vdots\\
x_{n}&=2\cos (\frac{\theta }{2^{n}})&=\frac{\sin(\frac{\theta}{2^{n-1}})}{\sin(\frac{\theta}{2^n})}\\
x_0x_1\dots x_n&=\frac{\sin(2\theta)}{\sin\theta}\frac{\sin\theta}{\sin(\frac{\theta}2)}\dots\frac{\sin(\frac{\theta}{2^{n-1}})}{\sin(\frac{\theta}{2^n})}&=\frac{\sin(2\theta)}{\sin(\frac{\theta}{2^n})}
\end{aligned}
Cosine Products, Fourier Transforms, and Random Sums
The Fourier transform of $f_n=\prod_{k=1}^n \cos \left(x / 2^k\right)$ is
\[
\hat{f}_n=✱_{k=1}^n \frac{1}{2}\left(\delta_{1 / 2^n}+\delta_{-1 / 2^n}\right)
\]The asterisk indicates a repeated convolution of the factors. Expanding for $n=3$ we see that
\[
\hat{f}_3=\frac{1}{8}\left(\delta_{-7 / 8}+\delta_{-5 / 8}+\cdots \delta_{7 / 8}\right) .
\]
Likewise
\[
\hat{f}_n=\frac{1}{2^n} \sum_{b \in B_n} \delta_b
\]
where $B_n$ is the set of $2^n$ equally spaced numbers from $-1+1 / 2^n$ to $1-1 / 2^n$ with spacing $2 / 2^n=1 / 2^{(n-1)}$.

The sequence of measures $\hat{f}_n$ converges to the uniform density on $[-1,1]$ of total mass 1, which we can write as $\frac12 \chi_{[-1,1]}$. The inverse transform is
\[\mathscr{F}^{-1}\left(\frac12 \chi_{[-1,1]}\right)=\frac{\sin x}{x}\]The spectrum of $(\sin x) / x$ is uniform in the interval $-1 \leq \omega \leq 1$. This means that $\sin x / x$ is a continuous linear combination of the "pure" harmonics $e^{i \omega x}$ with the same weight of $1 / 2$ for each $\omega \in[-1,1]$.

With this proof we have a way to generate a family of similar identities. Let us put point masses at $3^n$ equally spaced points from $-1+1 / 3^n$ to $1-1 / 3^n$ with spacing $2 / 3^n$. Such a measure is the convolution$$✱_{k=1}^n \frac{1}{3}\left(\delta_{-2 / 3^k}+\delta_0+\delta_{2 / 2 k}\right)$$Applying the inverse transform
\[\mathscr{F}^{-1}\left(\frac{1}{3}\left(\delta_{-2 / 3^k}+\delta_0+\delta_{2 / 3^k}\right)\right)=\frac{1}{3}\left(2 \cos \frac{2 x}{3^k}+1\right)\]
and taking limits gives us the infinite product identity
\[\prod_{n=1}^{\infty} \frac{1}{3}\left(1+2 \cos \frac{2 x}{3^n}\right)=\frac{\sin x}{x}\]

hbghlyj

设
$$
t_0=0,\qquad t_{n+1}=\sqrt{2+2t_n}
$$
$f(t)=\sqrt{2+2t}$有唯一的正不动点$x_*=1+\sqrt3$. 如果$0\le t<x_*$, 那么
$$
f(t)-x_*=\frac{(2+2t)-(2+2x_*)}{\,f(t)+x_*\,}
=\frac{2(t-x_*)}{\,f(t)+x_*\,}<0,
$$
所以$f(t)<x_*$. 并且$f$是递增的, 因此从$t_0<t_1$我们得到$f(t_0)<f(t_1)$
$$
t_1< t_2<\cdots< x_*,
$$
所以$(t_n)$是递增的并且有上界$x_*$.
一个递增的有界序列收敛; 记极限为$L$.
在$t_{n+1}=f(t_n)$中取极限得到$L=f(L)$有唯一的正解$x_*$, 所以$L=x_*$.

hbghlyj

在$[\sqrt2,x_*]$上,
$$
f'(t)=\frac1{\sqrt{2+2t}}\le \frac1{\sqrt{2+2\sqrt2}}=:C<1,
$$
所以$f$是一个压缩映射; 根据Banach不动点定理, $t_n\to x_*$.
误差的上界:
\begin{align*}
|t_{n+1}-x_*|&=|f'(\xi_n)|\,|t_n-x_*|&\xi_n\in(t_n,x_*),\\&\le C|t_n-x_*|
\end{align*}
所以$|t_n-x_*| \le C^n|t_0-x_*|$.

设$a_n:=\frac{t_n}{x_*}$. 那么$a_n\to1$并且
$$
|a_n-1|=\frac{|t_n-x_*|}{x_*}\le \frac{|t_0-x_*|}{x_*}C^{-n},
$$
所以$\sum_n |a_n-1|$收敛. 因此无穷乘积$\prod a_n$收敛到一个有限的, 非零的极限$P$.