フーリエ級数展開三角関数、余弦、実数倍角、無限級数の和

学習環境

Pythonで学ぶフーリエ解析と信号処理 (神永正博)(著)、コロナ社)の第2章(フーリエ級数展開)、章末問題2-18の解答を求めてみる。

a_{n} = \frac{2}{2 π} \int_{- π}^{π} \cos a t \cos n t dt

= \frac{1}{π} \int_{- π}^{π} \frac{\cos (a + n) t + \cos (a - n) t}{2} dt

= \frac{1}{π} \int_{0}^{π} (\cos (n + a) t + \cos (n - a) t) dt

= \frac{1}{π} {[\frac{\sin (n + a) t}{n + a} + \frac{\sin (n - a) t}{n - a}]}_{0}^{π}

= \frac{1}{π} (\frac{\sin (n + a) π}{n + a} + \frac{\sin (n - a) π}{n - a})

= \frac{1}{π} (\frac{\cos n π \sin a π}{n + a} - \frac{\cos n π \sin a π}{n - a})

= \frac{1}{π} (\frac{{(- 1)}^{n} \sin a π}{n + a} - \frac{{(- 1)}^{n} \sin a π}{n - a})

= \frac{2}{π} \frac{a {(- 1)}^{n - 1} \sin a π}{n^{2} - a^{2}}

b_{n} = \frac{1}{π} \int_{- π}^{π} \cos a t \sin n t dt

= \frac{1}{π} \int_{- π}^{π} \frac{\sin (n + a) t + \sin (n - a) t}{2} dt

= 0

よって、

\cos a t \sim \frac{\sin a π}{π a} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{a {(- 1)}^{n - 1} \sin a π}{n^{2} - a^{2}} \cos n t

1 = \frac{\sin a π}{π a} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{a {(- 1)}^{n - 1} \sin a π}{n^{2} - a^{2}}

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2} - a^{2}} = \frac{π}{2 a (\sin π a)} (1 - \frac{\sin π a}{π a}) = \frac{π}{2 a} (\frac{1}{\sin π a} - \frac{1}{π a})

\cos a π = \frac{\sin π a}{π a} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{a {(- 1)}^{n - 1} \sin π a}{n^{2} - a^{2}} {(- 1)}^{n}

\cos π a = \frac{\sin π a}{π a} - \frac{2 a (\sin π a)}{π} \sum_{n = 1}^{\infty} \frac{1}{n^{2} - a^{2}}

\sum_{n = n}^{\infty} \frac{1}{n^{2} - a^{2}} = \frac{π}{2 a (\sin π a)} (\frac{\sin π a}{π a} - \cos π a) = \frac{π}{2 a} (\frac{1}{π a} - \frac{1}{\tan π a})

コード(Python)

#!/usr/bin/env python3
import random
import numpy as np
import matplotlib.pyplot as plt

print('2-18.')


def partialsum(t, n, a):
    return np.sin(np.pi * a) / (np.pi * a) + 2 / np.pi * \
        sum(a * (-1) ** (k - 1) * np.sin(a * np.pi) * np.cos(k * t) / (k ** 2 - a ** 2)
            for k in range(1, n + 1))


vectorized = np.vectorize(partialsum)
t = np.linspace(-np.pi, np.pi, 10000)
n = 5
for i in range(5):
    a = random.random() * 5
    print(f'a = {a}')
    fig = plt.figure()
    for m in range(1, n + 1):
        ax = fig.add_subplot(n, 1, m)
        ax.plot(t, np.cos(a * t))
        ax.set_ylim(-2, 2)
        ax.plot(t, vectorized(t, m, a))
    plt.savefig(f'sample8_{i}.png')
    plt.show()

入出力結果

% ./sample8.py
2-18.
a = 2.552785311478355
a = 2.4544664267404053
a = 4.999708433105535
a = 2.037236920387312
a = 2.7287739971046343
%