Question #75766, Math / Differential Equations
Obtain the Fourier series expansion for the following periodic function which has a period of π \pi π f ( x ) = { ( 4 / π ) for 0 < x < π / 2 } f(x) = \{(4 / \pi) \text{ for } 0 < x < \pi / 2\} f ( x ) = {( 4/ π ) for 0 < x < π /2 }
Answer.
f ( x ) = { 4 π , 0 ≤ x < π / 2 0 , − π 2 ≤ x < 0 f(x) = \begin{cases} \dfrac{4}{\pi}, & 0 \leq x < \pi/2 \\ 0, & -\frac{\pi}{2} \leq x < 0 \end{cases} f ( x ) = ⎩ ⎨ ⎧ π 4 , 0 , 0 ≤ x < π /2 − 2 π ≤ x < 0 f ( x ) = a 0 2 + ∑ n = 1 ∞ ( a n cos n x + b n sin n x ) f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos nx + b_n \sin nx \right) f ( x ) = 2 a 0 + n = 1 ∑ ∞ ( a n cos n x + b n sin n x ) a n = 1 π ∫ − π 2 π 2 f ( x ) d x = 1 π ∫ 0 π 2 4 π d x = 4 π 2 x ∣ x = 0 x = π 2 = 2 π . a_n = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \, dx = \left. \frac{1}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{4}{\pi} \, dx = \frac{4}{\pi^2} x \right|_{x=0}^{x=\frac{\pi}{2}} = \frac{2}{\pi}. a n = π 1 ∫ − 2 π 2 π f ( x ) d x = π 1 ∫ 0 2 π π 4 d x = π 2 4 x ∣ ∣ x = 0 x = 2 π = π 2 . a n = 1 π ∫ − π 2 π 2 f ( x ) cos n x d x = 1 π ∫ 0 π 2 4 π cos n x d x = 4 π 2 1 n sin n x ∣ x = 0 x = π 2 = 4 π 2 1 n sin π n 2 a_n = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \cos nx \, dx = \left. \frac{1}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{4}{\pi} \cos nx \, dx = \frac{4}{\pi^2} \frac{1}{n} \sin nx \right|_{x=0}^{x=\frac{\pi}{2}} = \frac{4}{\pi^2} \frac{1}{n} \sin \frac{\pi n}{2} a n = π 1 ∫ − 2 π 2 π f ( x ) cos n x d x = π 1 ∫ 0 2 π π 4 cos n x d x = π 2 4 n 1 sin n x ∣ ∣ x = 0 x = 2 π = π 2 4 n 1 sin 2 πn b n = 1 π ∫ − π 2 π 2 f ( x ) sin n x d x = 1 π ∫ 0 π 2 4 π sin n x d x = − 4 π 2 1 n cos n x ∣ x = 0 x = π 2 = b_n = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \sin nx \, dx = \left. \frac{1}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{4}{\pi} \sin nx \, dx = -\frac{4}{\pi^2} \frac{1}{n} \cos nx \right|_{x=0}^{x=\frac{\pi}{2}} = b n = π 1 ∫ − 2 π 2 π f ( x ) sin n x d x = π 1 ∫ 0 2 π π 4 sin n x d x = − π 2 4 n 1 cos n x ∣ ∣ x = 0 x = 2 π = = − 4 π 2 1 n cos π n 2 + 4 π 2 n = -\frac{4}{\pi^2} \frac{1}{n} \cos \frac{\pi n}{2} + \frac{4}{\pi^2 n} = − π 2 4 n 1 cos 2 πn + π 2 n 4 f ( x ) = 2 π + 4 π 2 ∑ n = 1 ∞ 1 n [ sin π n 2 cos n x − ( cos π n 2 − 1 ) sin n x ] = f(x) = \frac{2}{\pi} + \frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n} \left[ \sin \frac{\pi n}{2} \cos nx - \left( \cos \frac{\pi n}{2} - 1 \right) \sin nx \right] = f ( x ) = π 2 + π 2 4 n = 1 ∑ ∞ n 1 [ sin 2 πn cos n x − ( cos 2 πn − 1 ) sin n x ] = = 2 π + 4 π 2 ∑ n = 1 ∞ 1 n [ sin ( π n 2 − n x ) + sin n x ] = \frac{2}{\pi} + \frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n} \left[ \sin \left( \frac{\pi n}{2} - nx \right) + \sin nx \right] = π 2 + π 2 4 n = 1 ∑ ∞ n 1 [ sin ( 2 πn − n x ) + sin n x ]
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#339914 on Dec 2023