計算メモ．とても楽しい本です．
以下の書籍で知りました．
【目次】
第 1 章 ヴィエトから統計的独立性の概念へ
第 2 章 ボレルとその後
第 3 章 正規法則
第 4 章 素数は賽を振る
第 5 章 気体分子運動論から連分数へ
CHAPTER1 - FROM VIETA TO THE NOTION OF STATISTICAL INDEPENDENCE
A formula of Vieta
$\cos (2\theta) = 2\cos^{2}\theta - 1$より$\cos(\pi/2^{n+1})= \dfrac{\sqrt{2 + 2\cos(\pi/2^{n})}}{2}$だから\begin{aligned}
& \cos(\pi / 2) = 0 \\
& \cos(\pi / 2^{2}) = \frac{\sqrt{2}}{2} \\
& \cos(\pi / 2^{3}) = \frac{\sqrt{2 + \sqrt{2}}}{2} \\
& \cos(\pi / 2^{4}) = \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} \\
& \quad \vdots
\end{aligned}
& \cos(\pi / 2) = 0 \\
& \cos(\pi / 2^{2}) = \frac{\sqrt{2}}{2} \\
& \cos(\pi / 2^{3}) = \frac{\sqrt{2 + \sqrt{2}}}{2} \\
& \cos(\pi / 2^{4}) = \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} \\
& \quad \vdots
\end{aligned}
Another look at Vieta's formula
$m=0,1,2,...,2^{k - 1} - 1$に対して\begin{aligned}
\epsilon_{k} (t)
&=
\begin{cases}
\, 0 & (2m / 2^{k} \leq t < (2m + 1) / 2^{k}) \\
\, 1 & ((2m + 1) / 2^{k} \leq t < (2m + 2) / 2^{k})
\end{cases} \\
\end{aligned}
\epsilon_{k} (t)
&=
\begin{cases}
\, 0 & (2m / 2^{k} \leq t < (2m + 1) / 2^{k}) \\
\, 1 & ((2m + 1) / 2^{k} \leq t < (2m + 2) / 2^{k})
\end{cases} \\
\end{aligned}
\begin{aligned}
r_{k}(t)
&= 1 - 2\epsilon_{k} (t) \\
&=
\begin{cases}
\, 1 & (2m / 2^{k} \leq t < (2m + 1) / 2^{k}) \\
\, -1 & ((2m + 1) / 2^{k} \leq t < (2m + 2) / 2^{k})
\end{cases}
\end{aligned}
r_{k}(t)
&= 1 - 2\epsilon_{k} (t) \\
&=
\begin{cases}
\, 1 & (2m / 2^{k} \leq t < (2m + 1) / 2^{k}) \\
\, -1 & ((2m + 1) / 2^{k} \leq t < (2m + 2) / 2^{k})
\end{cases}
\end{aligned}
\begin{aligned}
t
& = \sum_{k = 1}^{\infty} \frac{\epsilon_{k}(t)}{2^{k}} \\
1 - 2t
& = \sum_{k = 1}^{\infty} \frac{1}{2^{k}} - 2 \sum_{k = 1}^{\infty} \frac{\epsilon_{k}(t)}{2^{k}} \\
& = \sum_{k = 1}^{\infty} \frac{r_{k}(t) }{2^{k}}
\end{aligned}
t
& = \sum_{k = 1}^{\infty} \frac{\epsilon_{k}(t)}{2^{k}} \\
1 - 2t
& = \sum_{k = 1}^{\infty} \frac{1}{2^{k}} - 2 \sum_{k = 1}^{\infty} \frac{\epsilon_{k}(t)}{2^{k}} \\
& = \sum_{k = 1}^{\infty} \frac{r_{k}(t) }{2^{k}}
\end{aligned}
\begin{aligned}
\int_{0}^{1} e^{ix(1 - 2t)} \,\mathrm{d}t
&=e^{ix} \biggl[-\frac{e^{-2ixt}}{2ix}\biggr]_{0}^{1} \\
&=\frac{e^{ix} - e^{-ix}}{2ix} \\
&=\frac{\sin x}{x}
\end{aligned}
\int_{0}^{1} e^{ix(1 - 2t)} \,\mathrm{d}t
&=e^{ix} \biggl[-\frac{e^{-2ixt}}{2ix}\biggr]_{0}^{1} \\
&=\frac{e^{ix} - e^{-ix}}{2ix} \\
&=\frac{\sin x}{x}
\end{aligned}
\begin{aligned}
& \int_{0}^{1} \exp\biggl[ix\frac{r_{k}(t)}{2^{k}}\biggr] \,\mathrm{d}t \\
&= \sum_{m=0}^{2^{k - 1} - 1} (e^{ix/2^{k}} + e^{- ix/2^{k}}) \cdot \frac{1}{2^{k}} \\
&= \cos(x/2^{k})
\end{aligned}
& \int_{0}^{1} \exp\biggl[ix\frac{r_{k}(t)}{2^{k}}\biggr] \,\mathrm{d}t \\
&= \sum_{m=0}^{2^{k - 1} - 1} (e^{ix/2^{k}} + e^{- ix/2^{k}}) \cdot \frac{1}{2^{k}} \\
&= \cos(x/2^{k})
\end{aligned}
An accident or a beginning of something deeper?
$r_{k}(t)$が$\pm 1$の値をとる区間を合計すると，それぞれ$1/2$の長さになるから\begin{aligned}
\int_{0}^{1} e^{ic_{k} r_{k}(t)} \,\mathrm{d}t
&= \frac{e^{ic_{k}} + e^{-ic_{k}}}{2}
= \cos c_{k}
\end{aligned}
\int_{0}^{1} e^{ic_{k} r_{k}(t)} \,\mathrm{d}t
&= \frac{e^{ic_{k}} + e^{-ic_{k}}}{2}
= \cos c_{k}
\end{aligned}
$t \in (0, 1)$に対し
\begin{aligned}
& \biggl|\sum_{k=1}^{n} \frac{r_{k}(t)}{2^{k}} - (1 - 2t) \biggr| \\
& = \biggl|\sum_{k=n + 1}^{\infty} \frac{r_{k}(t)}{2^{k}} \biggr| \\
& \leq \sum_{k=n + 1}^{\infty} \frac{1}{2^{k}} = \frac{1/2^{n + 1}}{1 - 1/2} \to 0 \quad(n \to \infty)
\end{aligned}
より，$(0, 1)$上一様収束．& \biggl|\sum_{k=1}^{n} \frac{r_{k}(t)}{2^{k}} - (1 - 2t) \biggr| \\
& = \biggl|\sum_{k=n + 1}^{\infty} \frac{r_{k}(t)}{2^{k}} \biggr| \\
& \leq \sum_{k=n + 1}^{\infty} \frac{1}{2^{k}} = \frac{1/2^{n + 1}}{1 - 1/2} \to 0 \quad(n \to \infty)
\end{aligned}
$(\frac{1}{2})^{n} = \frac{1}{2}\cdots \frac{1}{2}$ ($n$ times)
同じ$k$同士が隣り合うように並び替えれば\begin{aligned}
& \prod_{k=1}^{n} e^{ic_{k}\delta_{k}} \cdot \prod_{k=1}^{n} \mu\{r_{k} (t) = \delta_{k}\} \\
&=\prod_{k=1}^{n} e^{ic_{k}\delta_{k}} \mu\{r_{k} (t) = \delta_{k}\}
\end{aligned}
& \prod_{k=1}^{n} e^{ic_{k}\delta_{k}} \cdot \prod_{k=1}^{n} \mu\{r_{k} (t) = \delta_{k}\} \\
&=\prod_{k=1}^{n} e^{ic_{k}\delta_{k}} \mu\{r_{k} (t) = \delta_{k}\}
\end{aligned}
次もよくやる式変形．因数分解したものと展開したものの関係．
\begin{aligned}
& \sum_{x(1)\in X_{1}, ..., x(n) \in X_{n}} \prod_{k=1}^{n} A_{x(k)} \\
& =\sum_{x(1)\in X_{1}, ..., x(n) \in X_{n}} A_{x(1)} A_{x(2)} \cdots A_{x(n)} \\
&= \biggl( \sum_{x(1)\in X_{1}} A_{x(1)} \biggr) \cdots \biggl( \sum_{x(n)\in X_{n}} A_{x(n)} \biggr) \\
&=\prod_{k=1}^{n} \biggl(\sum_{x(k)\in X_{k}} A_{x(k)} \biggr)
\end{aligned}
& \sum_{x(1)\in X_{1}, ..., x(n) \in X_{n}} \prod_{k=1}^{n} A_{x(k)} \\
& =\sum_{x(1)\in X_{1}, ..., x(n) \in X_{n}} A_{x(1)} A_{x(2)} \cdots A_{x(n)} \\
&= \biggl( \sum_{x(1)\in X_{1}} A_{x(1)} \biggr) \cdots \biggl( \sum_{x(n)\in X_{n}} A_{x(n)} \biggr) \\
&=\prod_{k=1}^{n} \biggl(\sum_{x(k)\in X_{k}} A_{x(k)} \biggr)
\end{aligned}
Heads or tails
$r_{1}(t), ..., r_{n}(t)$のうち$l$個が$+1$をとるとき，残りの$(n - l)$個は$-1$をとるから，\begin{aligned}
& r_{1}(t)+ \cdots + r_{n}(t) \\
& = (+1) \cdot l + (- 1) \cdot (n - l) \\
& = 2l - n.
\end{aligned}
& r_{1}(t)+ \cdots + r_{n}(t) \\
& = (+1) \cdot l + (- 1) \cdot (n - l) \\
& = 2l - n.
\end{aligned}
$\cos x = \dfrac{e^{ix} + e^{-ix}}{2}$より
\begin{aligned}
\cos^{n} x
&= \frac{1}{2^{n}} \sum_{k = 0}^{n} \binom{n}{k} e^{ikx} e^{-i(n - k)x} \\
&= \frac{1}{2^{n}} \sum_{k = 0}^{n} \binom{n}{k} e^{i(2k - n)x}
\end{aligned}
だから，\cos^{n} x
&= \frac{1}{2^{n}} \sum_{k = 0}^{n} \binom{n}{k} e^{ikx} e^{-i(n - k)x} \\
&= \frac{1}{2^{n}} \sum_{k = 0}^{n} \binom{n}{k} e^{i(2k - n)x}
\end{aligned}
\begin{aligned}
& \frac{1}{2\pi} \int_{0}^{2\pi} e^{-i(2l-n)x} \cos^{n} x \,\mathrm{d}x \\
&= \frac{1}{2^{n}} \sum_{k = 0}^{n} \binom{n}{k}
\cdot \underbrace{\frac{1}{2\pi} \int_{0}^{2\pi} e^{i2(k - l)x} \,\mathrm{d}x}_{=\delta_{kl}} \\
&= \frac{1}{2^{n}} \binom{n}{l}
\end{aligned}
& \frac{1}{2\pi} \int_{0}^{2\pi} e^{-i(2l-n)x} \cos^{n} x \,\mathrm{d}x \\
&= \frac{1}{2^{n}} \sum_{k = 0}^{n} \binom{n}{k}
\cdot \underbrace{\frac{1}{2\pi} \int_{0}^{2\pi} e^{i2(k - l)x} \,\mathrm{d}x}_{=\delta_{kl}} \\
&= \frac{1}{2^{n}} \binom{n}{l}
\end{aligned}
Independence and "Independence" Problems
Problems
1
$m=0,1,2,...,3^{k - 1} - 1$に対して\begin{aligned}
\eta_{k} (t)
&=
\begin{cases}
\, 0 & (3m / 3^{k} \leq t < (3m + 1) / 3^{k}) \\
\, 1 & ((3m + 1) / 3^{k} \leq t < (3m + 2) / 3^{k}) \\
\, 2 & ((3m + 2) / 3^{k} \leq t < (3m + 3) / 3^{k})
\end{cases} \\
\end{aligned}
\eta_{k} (t)
&=
\begin{cases}
\, 0 & (3m / 3^{k} \leq t < (3m + 1) / 3^{k}) \\
\, 1 & ((3m + 1) / 3^{k} \leq t < (3m + 2) / 3^{k}) \\
\, 2 & ((3m + 2) / 3^{k} \leq t < (3m + 3) / 3^{k})
\end{cases} \\
\end{aligned}
2
\begin{aligned}
\frac{\sin x}{x}
= \prod_{k = 1}^{\infty} \frac{1 + 2\cos\frac{2x}{3^{k}}}{3}
\end{aligned}
を示す．\frac{\sin x}{x}
= \prod_{k = 1}^{\infty} \frac{1 + 2\cos\frac{2x}{3^{k}}}{3}
\end{aligned}
\begin{aligned}
s_{k}(t)
& = 1 - \eta_{k} \\
&=
\begin{cases}
\, 1 & (3m / 3^{k} \leq t < (3m + 1) / 3^{k}) \\
\, 0 & ((3m + 1) / 3^{k} \leq t < (3m + 2) / 3^{k}) \\
\, -1 & ((3m + 2) / 3^{k} \leq t < (3m + 3) / 3^{k})
\end{cases}
\end{aligned}
s_{k}(t)
& = 1 - \eta_{k} \\
&=
\begin{cases}
\, 1 & (3m / 3^{k} \leq t < (3m + 1) / 3^{k}) \\
\, 0 & ((3m + 1) / 3^{k} \leq t < (3m + 2) / 3^{k}) \\
\, -1 & ((3m + 2) / 3^{k} \leq t < (3m + 3) / 3^{k})
\end{cases}
\end{aligned}
\begin{aligned}
t
& = \sum_{k = 1}^{\infty} \frac{\eta_{k}(t)}{3^{k}} \\
1 - 2t
& = \sum_{k = 1}^{\infty} \frac{2}{3^{k}} - 2\sum_{k = 1}^{\infty} \frac{\eta_{k}(t)}{3^{k}} \\
& = 2\sum_{k = 1}^{\infty} \frac{s_{k}(t) }{3^{k}}
\end{aligned}
t
& = \sum_{k = 1}^{\infty} \frac{\eta_{k}(t)}{3^{k}} \\
1 - 2t
& = \sum_{k = 1}^{\infty} \frac{2}{3^{k}} - 2\sum_{k = 1}^{\infty} \frac{\eta_{k}(t)}{3^{k}} \\
& = 2\sum_{k = 1}^{\infty} \frac{s_{k}(t) }{3^{k}}
\end{aligned}
\begin{aligned}
\frac{\sin x}{x}
&= \int_{0}^{1} e^{ix(1 - 2t)} \,\mathrm{d}t \\
&= \int_{0}^{1} \exp\biggl(i 2x \sum_{k = 1}^{\infty} \frac{s_{k}(t) }{3^{k}} \biggr) \,\mathrm{d}t \\
&=\prod_{k = 1}^{\infty}
\underbrace{\int_{0}^{1} \exp\biggl(i 2x \frac{s_{k}(t) }{3^{k}} \biggr) \,\mathrm{d}t}_{=\frac{1}{3} (e^{-i2x/3^{k}} + 1 + e^{i2x/3^{k}} )} \\
&= \prod_{k = 1}^{\infty} \frac{1 + 2\cos(2x/3^{k})}{3}
\end{aligned}
\frac{\sin x}{x}
&= \int_{0}^{1} e^{ix(1 - 2t)} \,\mathrm{d}t \\
&= \int_{0}^{1} \exp\biggl(i 2x \sum_{k = 1}^{\infty} \frac{s_{k}(t) }{3^{k}} \biggr) \,\mathrm{d}t \\
&=\prod_{k = 1}^{\infty}
\underbrace{\int_{0}^{1} \exp\biggl(i 2x \frac{s_{k}(t) }{3^{k}} \biggr) \,\mathrm{d}t}_{=\frac{1}{3} (e^{-i2x/3^{k}} + 1 + e^{i2x/3^{k}} )} \\
&= \prod_{k = 1}^{\infty} \frac{1 + 2\cos(2x/3^{k})}{3}
\end{aligned}
一般化する．
$m=0,1,2,...,p^{k - 1} - 1$に対して
\begin{aligned}
&\xi_{k} (t) \\
&=
\begin{cases}
\, 0 & (pm / p^{k} \leq t < (pm + 1) / p^{k}) \\
\, 1 & ((pm + 1) / p^{k} \leq t < (pm + 2) / p^{k}) \\
\, \vdots \\
\, p - 1 & ((pm + p - 1) / p^{k} \leq t < (pm + p) / p^{k})
\end{cases} \\
\end{aligned}
&\xi_{k} (t) \\
&=
\begin{cases}
\, 0 & (pm / p^{k} \leq t < (pm + 1) / p^{k}) \\
\, 1 & ((pm + 1) / p^{k} \leq t < (pm + 2) / p^{k}) \\
\, \vdots \\
\, p - 1 & ((pm + p - 1) / p^{k} \leq t < (pm + p) / p^{k})
\end{cases} \\
\end{aligned}
$\sin x/ x$を出すために$1 - 2t $は変えられないから，先に$1 - 2t $の計算をして$u_{k}$を決めている．
\begin{aligned}
&u_{k}(t) = p - 1 - 2\xi_{k} (t) \\
&=
\begin{cases}
\, p - 1 & (pm / p^{k} \leq t < (pm + 1) / p^{k}) \\
\, p - 3 & ((pm + 1) / p^{k} \leq t < (pm + 2) / p^{k}) \\
\, \vdots \\
\, -(p - 1) & ((pm + p - 1) / p^{k} \leq t < (pm + p) / p^{k})
\end{cases} \\
\end{aligned}
&u_{k}(t) = p - 1 - 2\xi_{k} (t) \\
&=
\begin{cases}
\, p - 1 & (pm / p^{k} \leq t < (pm + 1) / p^{k}) \\
\, p - 3 & ((pm + 1) / p^{k} \leq t < (pm + 2) / p^{k}) \\
\, \vdots \\
\, -(p - 1) & ((pm + p - 1) / p^{k} \leq t < (pm + p) / p^{k})
\end{cases} \\
\end{aligned}
\begin{aligned}
t
& = \sum_{k = 1}^{\infty} \frac{\xi_{k}(t)}{p^{k}} \\
1 - 2t
& = \sum_{k = 1}^{\infty} \frac{p-1}{p^{k}} - 2\sum_{k = 1}^{\infty} \frac{\xi_{k}(t)}{p^{k}} \\
& = \sum_{k = 1}^{\infty} \frac{u_{k}(t) }{p^{k}}
\end{aligned}
t
& = \sum_{k = 1}^{\infty} \frac{\xi_{k}(t)}{p^{k}} \\
1 - 2t
& = \sum_{k = 1}^{\infty} \frac{p-1}{p^{k}} - 2\sum_{k = 1}^{\infty} \frac{\xi_{k}(t)}{p^{k}} \\
& = \sum_{k = 1}^{\infty} \frac{u_{k}(t) }{p^{k}}
\end{aligned}
\begin{aligned}
\frac{\sin x}{x}
&= \int_{0}^{1} e^{ix(1 - 2t)} \,\mathrm{d}t \\
&= \int_{0}^{1} \exp\biggl(ix \sum_{k = 1}^{\infty} \frac{u_{k}(t) }{p^{k}} \biggr) \,\mathrm{d}t \\
&=\prod_{k = 1}^{\infty}
\int_{0}^{1} \exp\biggl(i x \frac{u_{k}(t) }{p^{k}} \biggr) \,\mathrm{d}t
\end{aligned}
\frac{\sin x}{x}
&= \int_{0}^{1} e^{ix(1 - 2t)} \,\mathrm{d}t \\
&= \int_{0}^{1} \exp\biggl(ix \sum_{k = 1}^{\infty} \frac{u_{k}(t) }{p^{k}} \biggr) \,\mathrm{d}t \\
&=\prod_{k = 1}^{\infty}
\int_{0}^{1} \exp\biggl(i x \frac{u_{k}(t) }{p^{k}} \biggr) \,\mathrm{d}t
\end{aligned}
\begin{aligned}
&\int_{0}^{1} \exp\biggl(i x \frac{u_{k}(t) }{p^{k}} \biggr) \,\mathrm{d}t \\
&=
\begin{cases}
\, \displaystyle
\frac{2}{p} \sum_{l = 0}^{p/2 - 1} \cos [x (2l + 1) / p^{k}]& (p: \text{even}) \\
\, \displaystyle
\frac{1}{p} \biggl\{ 1 + 2 \sum_{l = 1}^{(p - 1)/2} \cos [x \cdot 2l / p^{k}] \biggr\} & (p: \text{odd})
\end{cases}
\end{aligned}
&\int_{0}^{1} \exp\biggl(i x \frac{u_{k}(t) }{p^{k}} \biggr) \,\mathrm{d}t \\
&=
\begin{cases}
\, \displaystyle
\frac{2}{p} \sum_{l = 0}^{p/2 - 1} \cos [x (2l + 1) / p^{k}]& (p: \text{even}) \\
\, \displaystyle
\frac{1}{p} \biggl\{ 1 + 2 \sum_{l = 1}^{(p - 1)/2} \cos [x \cdot 2l / p^{k}] \biggr\} & (p: \text{odd})
\end{cases}
\end{aligned}
3
$\delta_{i} \in \{-1, 1\}$とする．\begin{aligned}
&\int_{0}^{1} r_{k_{1}} r_{k_{2}} \cdots r_{k_{s}} \,\mathrm{d}t \\
&= \sum_{\delta_{1},..., \delta_{s}}
\biggl[
\prod_{i = 1}^{s} \delta_{i}
\cdot \mu\{r_{k_{1}}(t) = \delta_{1}, \cdots , r_{k_{s}}(t) = \delta_{s} \} \biggr] \\
&= \sum_{\delta_{1},..., \delta_{s}}
\biggl[
\prod_{i = 1}^{s} \delta_{i}
\cdot \prod_{j = 1}^{s} \mu\{r_{k_{j}}(t) = \delta_{j}\} \biggr] \\
&= \sum_{\delta_{1},..., \delta_{s}}
\prod_{i = 1}^{s} \delta_{i} \cdot \mu\{r_{k_{i}}(t) = \delta_{i}\} \\
&= \prod_{i = 1}^{s}
\underbrace{\sum_{\delta_{i}} \delta_{i} \cdot \mu\{r_{k_{i}}(t) = \delta_{i}\}}_{=0} \\
&=0
\end{aligned}
&\int_{0}^{1} r_{k_{1}} r_{k_{2}} \cdots r_{k_{s}} \,\mathrm{d}t \\
&= \sum_{\delta_{1},..., \delta_{s}}
\biggl[
\prod_{i = 1}^{s} \delta_{i}
\cdot \mu\{r_{k_{1}}(t) = \delta_{1}, \cdots , r_{k_{s}}(t) = \delta_{s} \} \biggr] \\
&= \sum_{\delta_{1},..., \delta_{s}}
\biggl[
\prod_{i = 1}^{s} \delta_{i}
\cdot \prod_{j = 1}^{s} \mu\{r_{k_{j}}(t) = \delta_{j}\} \biggr] \\
&= \sum_{\delta_{1},..., \delta_{s}}
\prod_{i = 1}^{s} \delta_{i} \cdot \mu\{r_{k_{i}}(t) = \delta_{i}\} \\
&= \prod_{i = 1}^{s}
\underbrace{\sum_{\delta_{i}} \delta_{i} \cdot \mu\{r_{k_{i}}(t) = \delta_{i}\}}_{=0} \\
&=0
\end{aligned}
4
\begin{aligned}
\int_{0}^{1} w_{0}(t) w_{0}(t) \,\mathrm{d}t
& =1.
\end{aligned}
$n \geq 1$のとき，$r_{k}^{2}(t) \equiv 1$より\int_{0}^{1} w_{0}(t) w_{0}(t) \,\mathrm{d}t
& =1.
\end{aligned}
\begin{aligned}
\int_{0}^{1} w_{n}(t) w_{n}(t) \,\mathrm{d}t
&=1.
\end{aligned}
$m \neq n$なら，$w_{m}(t) w_{n}(t) $を$r_{k}$の積で表したとき，各$r_{k}$は高々2回現れる．また，$m \neq n$だから，1回しか現れないものが必ず存在する．2回現れるものについては$r_{k}^{2}(t) \equiv 1$より\int_{0}^{1} w_{n}(t) w_{n}(t) \,\mathrm{d}t
&=1.
\end{aligned}
\begin{aligned}
& w_{m}(t) w_{n}(t)
= r_{k_{1}}(t) \cdots r_{k_{s}}(t) \\
& (k_{1} < \cdots < k_{s})
\end{aligned}
の形になる．よって，問題3. より& w_{m}(t) w_{n}(t)
= r_{k_{1}}(t) \cdots r_{k_{s}}(t) \\
& (k_{1} < \cdots < k_{s})
\end{aligned}
\begin{aligned}
\int_{0}^{1} w_{m}(t) w_{n}(t) \,\mathrm{d}t
&=0.
\end{aligned}
\int_{0}^{1} w_{m}(t) w_{n}(t) \,\mathrm{d}t
&=0.
\end{aligned}
\begin{aligned}
& \int_{0}^{1} f(t) w_{n}(t) \,\mathrm{d}t = 0 \\
& (n = 0, 1, 2, ...)
\end{aligned}
とする．& \int_{0}^{1} f(t) w_{n}(t) \,\mathrm{d}t = 0 \\
& (n = 0, 1, 2, ...)
\end{aligned}
5
CHAPTER2 - BOREL AND AFTER
"Laws of large numbers"
\begin{aligned}
& \lim_{n \to \infty}
\sum_{|2l - n| > \epsilon n} \frac{1}{2} \binom{n}{l} \\
&=
\end{aligned}
& \lim_{n \to \infty}
\sum_{|2l - n| > \epsilon n} \frac{1}{2} \binom{n}{l} \\
&=
\end{aligned}
$r_{i}^{2}(t) \equiv 1$より
\begin{aligned}
& \int_{0}^{1} \biggl[ \sum_{i = 1}^{n} r_{i}(t) \biggr]^{2} \,\mathrm{d}t \\
&= \sum_{i = 1}^{n} \int_{0}^{1} r_{i}^{2} (t) \,\mathrm{d}t
+ 2 \sum_{i < j} \underbrace{\int_{0}^{1} r_{i} (t) r_{j}(t) \,\mathrm{d}t}_{=0} \\
&=n
\end{aligned}
& \int_{0}^{1} \biggl[ \sum_{i = 1}^{n} r_{i}(t) \biggr]^{2} \,\mathrm{d}t \\
&= \sum_{i = 1}^{n} \int_{0}^{1} r_{i}^{2} (t) \,\mathrm{d}t
+ 2 \sum_{i < j} \underbrace{\int_{0}^{1} r_{i} (t) r_{j}(t) \,\mathrm{d}t}_{=0} \\
&=n
\end{aligned}
Borel and "normal numbers"
ゼロにならないのは$r_{i}^{4}$と$r_{i}^{2}r_{j}^{2}$の項だけだから，\begin{aligned}
& \int_{0}^{1} \biggl[ \sum_{i = 1}^{n} r_{i}(t) \biggr]^{4} \,\mathrm{d}t \\
&= \sum_{i = 1}^{n} \int_{0}^{1} r_{i}^{4} (t) \,\mathrm{d}t
+ \sum_{i \neq j} \int_{0}^{1} r_{i}^{2} (t) r_{j}^{2} (t) \,\mathrm{d}t .
\end{aligned}
2項目の選び方は，（$n$から2コの添字を選ぶ方法）×（4つの積から2コ選ぶ方法）だから& \int_{0}^{1} \biggl[ \sum_{i = 1}^{n} r_{i}(t) \biggr]^{4} \,\mathrm{d}t \\
&= \sum_{i = 1}^{n} \int_{0}^{1} r_{i}^{4} (t) \,\mathrm{d}t
+ \sum_{i \neq j} \int_{0}^{1} r_{i}^{2} (t) r_{j}^{2} (t) \,\mathrm{d}t .
\end{aligned}
\begin{aligned}
\binom{n}{2} \binom{4}{2}
&= \frac{n(n-1)}{2} \frac{4\cdot 3}{2}
\end{aligned}
通り．\binom{n}{2} \binom{4}{2}
&= \frac{n(n-1)}{2} \frac{4\cdot 3}{2}
\end{aligned}
4乗ではなく2乗だと，$1/n$が出てきて収束が言えない．3乗だと$r_{i}$の奇数次しか現れないので積分がゼロになってしまう．
almost every number $t$ has (asymptotically!) the same number of zeros and ones in its binary expansion!
Mark Kac, "Statistical Independence in Probability, Analysis and Number Theory"
