数学 - Python - 解析学 - 積分の計算 - 定積分の計算 - 三角関数(正弦と余弦)の積、加法定理、和、偶関数、奇関数

学習環境

解析入門(上) (松坂和夫数学入門シリーズ 4) (松坂和夫(著)、岩波書店)の第8章(積分の計算)、8.2(定積分の計算)、問題3の解答を求めてみる。

1. $\begin{array}{l} \int_{- π}^{π} \sin m x \sin n x dx \\ = \int_{- π}^{π} \frac{\cos (m - n) x - \cos (m + n) x}{2} d x \\ m = n \neq 0 \\ \int_{- π}^{π} \frac{1 - \cos (m + n) x}{2} dx \\ = \int_{0}^{π} (1 - \cos (m + n) x) dx \\ = {[x - \frac{\sin (m + n) x}{m + n}]}_{0}^{π} \\ = π \\ m \neq n \\ {[\frac{\sin (m - n) x}{m - n} - \frac{\sin (m + n) x}{m + n}]}_{0}^{π} \\ = 0 \end{array}$
2. $\begin{array}{l} \int_{- π}^{π} \sin m x \cos n x dx \\ = \int_{- π}^{π} \frac{\sin (m + n) x + \sin (m - n) x}{2} dx \\ = 0 \end{array}$
3. $\begin{array}{l} \int_{- π}^{π} \cos m x \cos n x dx \\ = \int_{- π}^{π} \frac{\cos (m + n) x + \cos (m - n) x}{2} dx \\ m = n \neq 0 \\ \int_{0}^{π} (\cos (m + n) x + 1) dx \\ = {[\frac{\sin (m + n) x}{m + n} + x]}_{0}^{π} \\ = π \\ m \neq n \\ 0 \\ m = n = 0 \\ \int_{- π}^{π} dx = 2 π \end{array}$

#!/usr/bin/env python3 from sympy import pprint, symbols, sin, cos, pi, Integral, plot print('3.') x = symbols('x') m, n = symbols('m, n', integer=True, nonnegative=True) fs = [sin(m * x) * sin(n * x), sin(m * x) * cos(n * x), cos(m * x) * cos(n * x)] for i, f in enumerate(fs, 1): print(f'({i})') I = Integral(f, (x, -pi, pi)) for o in [I, I.doit()]: pprint(o) print() p = plot(*[f.subs({m: m0, n: n0}) for f in fs for m0 in range(0, 2) for n0 in range(0, 2)], (x, -5, 5), ylim=(-5, 5), legend=True, show=False) colors = ['red', 'green', 'blue', 'brown', 'orange', 'purple', 'pink', 'gray', 'skyblue', 'yellow'] for o, color in zip(p, colors): o.line_color = color p.show() p.save('sample3.png')

% ./sample3.py 3. (1) π ⌠ ⎮ sin(m⋅x)⋅sin(n⋅x) dx ⌡ -π ⎧0 for (m = 0 ∧ n = 0) ∨ (m = 0 ∧ m = n ∧ n = 0) ∨ (m = 0 ∧ m = -n ∧ n = 0) ⎪ ⎪-π for (m = 0 ∧ m = -n) ∨ (m = -n ∧ m = n) ∨ (m = -n ∧ n = 0) ∨ (m = 0 ∧ m ⎨ ⎪π for (m = 0 ∧ m = n) ∨ (m = n ∧ n = ⎪ ⎩0 otherwise ∨ (m = 0 ∧ m = -n ∧ m = n ∧ n = 0) ∨ m = 0 ∨ n = 0 = -n ∧ m = n) ∨ (m = -n ∧ m = n ∧ n = 0) ∨ m = -n 0) ∨ m = n (2) π ⌠ ⎮ sin(m⋅x)⋅cos(n⋅x) dx ⌡ -π 0 (3) π ⌠ ⎮ cos(m⋅x)⋅cos(n⋅x) dx ⌡ -π ⎧2⋅π for (m = 0 ∧ n = 0) ∨ (m = 0 ∧ m = n ∧ n = ⎪ ⎨ π for (m = 0 ∧ m = -n) ∨ (m = 0 ∧ m = n) ∨ (m = -n ∧ m = n) ∨ (m = n ∧ n = ⎪ ⎩ 0 0) ∨ (m = 0 ∧ m = -n ∧ n = 0) ∨ (m = 0 ∧ m = -n ∧ m = n ∧ n = 0) 0) ∨ (m = -n ∧ n = 0) ∨ (m = 0 ∧ m = -n ∧ m = n) ∨ (m = -n ∧ m = n ∧ n = 0) ∨ otherwise m = -n ∨ m = n %

Kamimura's blog

2020年1月11日土曜日

数学 - Python - 解析学 - 積分の計算 - 定積分の計算 - 三角関数(正弦と余弦)の積、加法定理、和、偶関数、奇関数

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