Expand \( f(x)=x \sin x, 0

Ask by Maxwell Newton. in India

Jan 08,2025

To expand the function \( f(x) = x \sin x \) for \( 0 < x < 2\pi \) into its Fourier series, we'll follow the standard procedure for finding the Fourier series on the interval \( [0, 2\pi] \). ### Fourier Series Representation The general Fourier series for a function \( f(x) \) defined on \( [0, 2\pi] \) is given by: \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right) \] where the coefficients are calculated as: \[ a_0 = \frac{1}{\pi} \int_0^{2\pi} f(x) \, dx \] \[ a_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos(nx) \, dx \quad \text{for } n \geq 1 \] \[ b_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin(nx) \, dx \quad \text{for } n \geq 1 \] ### Calculating the Coefficients 1. **Calculating \( a_0 \):** \[ a_0 = \frac{1}{\pi} \int_0^{2\pi} x \sin x \, dx \] Using integration by parts: \[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x \] Evaluating from \( 0 \) to \( 2\pi \): \[ a_0 = \frac{1}{\pi} \left[ -2\pi \cos(2\pi) + \sin(2\pi) - (0 + \sin 0) \right] = \frac{1}{\pi} (-2\pi) = -2 \] 2. **Calculating \( a_n \):** \[ a_n = \frac{1}{\pi} \int_0^{2\pi} x \sin x \cos(nx) \, dx \] Using the product-to-sum identity: \[ \sin x \cos(nx) = \frac{\sin((n+1)x) + \sin((1-n)x)}{2} \] Thus, \[ a_n = \frac{1}{2\pi} \left[ \int_0^{2\pi} x \sin((n+1)x) \, dx + \int_0^{2\pi} x \sin((1-n)x) \, dx \right] \] Evaluating these integrals (noting that for integer \( n \), the integrals simplify due to periodicity): \[ a_n = \frac{2}{n^2 - 1} \quad \text{for } n \geq 2 \] *Note:* For \( n = 1 \), the coefficient \( a_1 \) would be undefined (division by zero), which indicates that the cosine term for \( n = 1 \) is not present in the Fourier series. 3. **Calculating \( b_n \):** \[ b_n = \frac{1}{\pi} \int_0^{2\pi} x \sin x \sin(nx) \, dx \] Again, using the product-to-sum identity: \[ \sin x \sin(nx) = \frac{\cos((n-1)x) - \cos((n+1)x)}{2} \] Thus, \[ b_n = \frac{1}{2\pi} \left[ \int_0^{2\pi} x \cos((n-1)x) \, dx - \int_0^{2\pi} x \cos((n+1)x) \, dx \right] \] Evaluating these integrals: - For \( n = 1 \), one of the integrals becomes \( \int_0^{2\pi} x \, dx = 2\pi^2 \), and the other term vanishes due to periodicity. \[ b_1 = \frac{1}{2\pi} \cdot 2\pi^2 = \pi \] - For \( n \geq 2 \), both integrals vanish due to the periodicity of the cosine function. \[ b_n = 0 \quad \text{for } n \geq 2 \] ### Final Fourier Series Combining the coefficients, the Fourier series expansion for \( f(x) = x \sin x \) on \( 0 < x < 2\pi \) is: \[ f(x) = -1 + \pi \sin x + \sum_{n=2}^{\infty} \frac{2}{n^2 - 1} \cos(nx) \] ### Summary \[ \boxed{\,f(x) = -1 + \pi\,\sin x\; +\; \sum_{n=2}^{\,\infty}\frac{2}{\,n^{2} - 1\,}\cos(nx)\,} \]