$∏_{k=1}^n(1+e^{it/2^k})$

hbghlyj

\[\prod _{k=1}^n \left(1+\exp\left(\frac{i t}{2^k}\right)\right)=\sum _{k=0}^{2^n-1} \exp \left(\frac{i k t}{2^n}\right)\]如何证明呢?
来源: Characteristic function (probability theory)
$d_k$为独立 Bernoulli$(\frac12)$ 分布的随机变量, 当$n→∞$时$\prod_{k=1}^nd_k/2^k$趋于$[0,1]$上的均匀分布.
Wolfram reference–CharacteristicFunction 下方展开“Applications”
Verify that the sum $\prod_{k=1}^nd_k/2^k$ where $d_k$ are independent identically distributed BernoulliDistribution[1/2] variates tends in distribution to UniformDistribution[] for large $n$:

1. Subscript[cf, k] =
2. CharacteristicFunction[
3.   TransformedDistribution[Subscript[d, k]/2^k,
4.    Subscript[d, k] \[Distributed] BernoulliDistribution[1/2]], t]

Copy the Code

Use a combinatorial equality for product $\prod_{k=1}^n\text{cf}_k$\[\prod _{k=1}^n \left(\frac{1}{2}+\frac{1}{2} \exp\left(\frac{i t}{2^k}\right)\right)=\frac{\sum _{k=0}^{2^n-1} \exp \left(\frac{i k t}{2^n}\right)}{2^n}\]

hbghlyj

对$n$归纳:
假设$$\prod _{k=1}^n \left(1+\exp\left(\frac{i t}{2^k}\right)\right)=\sum _{k=0}^{2^n-1} \exp \left(\frac{i k t}{2^n}\right)$$则
\begin{align*}\left(1+\exp\left(\frac{i t}{2^{n+1}}\right)\right)\cdot\prod _{k=1}^n \left(1+\exp\left(\frac{i t}{2^k}\right)\right)&=\left(1+\exp\left(\frac{i t}{2^{n+1}}\right)\right)\cdot\sum _{k=0}^{2^n-1} \exp \left(\frac{i k t}{2^n}\right)\\
&=\left(1+\exp\left(\frac{i t}{2^{n+1}}\right)\right)\cdot\sum _{k=0}^{2^n-1} \exp \left(\frac{2ki t}{2^{n+1}}\right)\\
&=\sum _{k=0}^{2^n-1} \exp \left(\frac{2ki t}{2^{n+1}}\right)+\sum _{k=0}^{2^n-1} \exp \left(\frac{(2k+1)i t}{2^{n+1}}\right)\\
&=\sum _{k=0}^{2^{n+1}-1} \exp \left(\frac{ik t}{2^{n+1}}\right)\\
\end{align*}证毕

hbghlyj

反复使用平方差
\[\left(1-\exp \left(\frac{it}{2^n}\right)\right)\prod _{k=1}^n \left(1+\exp\left(\frac{i t}{2^k}\right)\right)=1-\exp(it)\]
即
\[\prod _{k=1}^n \left(1+\exp\left(\frac{i t}{2^k}\right)\right)={1-\exp(it)\over1-\exp \left(\frac{it}{2^n}\right)}\]
右端使用等比数列求和
\[\prod _{k=1}^n \left(1+\exp\left(\frac{i t}{2^k}\right)\right)=\sum _{k=0}^{2^n-1} \exp \left(\frac{i k t}{2^n}\right)\]