フーリエ級数展開三角関数、正弦、実数倍角

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2-17の解答を求めてみる。

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

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

= 0

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

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

= \frac{2}{2 π} \int_{0}^{π} (\cos (a - n) t - \cos (a + n) t) dt

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

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

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

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

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

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

よって求めるフーリエ展開は、

\begin{array}{l} a \notin ℤ \\ \sin a t \sim \frac{2}{π} \sum_{k = 1}^{\infty} \frac{n {(- 1)}^{n} \sin a π}{a^{2} - n^{2}} \sin n t \end{array}

コード(Python)

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

print('2-17.')


def partialsum(t, n, a):
    return 2 / np.pi * \
        sum(k * (-1) ** k * np.sin(a * np.pi) * np.sin(k * t) / (a ** 2 - k ** 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(a)
    fig = plt.figure()
    for m in range(1, n + 1):
        ax = fig.add_subplot(n, 1, m)
        ax.plot(t, np.sin(a * t))
        ax.set_ylim(-2, 2)
        ax.plot(t, vectorized(t, m, a))
    plt.savefig(f'sample7_{i}.png')
    plt.show()

入出力結果

% ./sample7.py
2-17.
4.11686011187972
1.264113176239896
4.464191894422588
0.6797740784195017
4.794822074091237
%