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

Ask by Harmon Lowe. in India

Jan 08,2025

To expand \( f(x) = x \sin x \) as a Fourier series on the interval \( 0 < x < 2\pi \), we first recognize that this function is periodic and can be expressed as a Fourier series of the form: \[ f(x) = a_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)) \] where the coefficients \( a_0 \), \( a_n \), and \( b_n \) are calculated as follows: - The constant term \( a_0 \) is given by: \[ a_0 = \frac{1}{2\pi} \int_{0}^{2\pi} f(x) \, dx = \frac{1}{2\pi} \int_{0}^{2\pi} x \sin x \, dx \] - The cosine coefficients \( a_n \) are given by: \[ a_n = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \cos(nx) \, dx = \frac{1}{\pi} \int_{0}^{2\pi} x \sin x \cos(nx) \, dx \] - The sine coefficients \( b_n \) are given by: \[ b_n = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \sin(nx) \, dx = \frac{1}{\pi} \int_{0}^{2\pi} x \sin^2(x) \, dx \] Calculating these integrals can be quite a task and often involves integration by parts and the use of trigonometric identities. For \( b_n \), using \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \) can greatly simplify the calculation. Ultimately, you would arrive at specific values for \( a_0 \), \( a_n \), and \( b_n \), enabling you to write the Fourier series for \( f(x) = x \sin x \) over the specified interval. However, without direct computation, you would have an expression involving those coefficients. The Fourier series would then be: \[ f(x) = \text{(computed coefficient terms)} \] It’s important to remember that the Fourier series will converge to \( f(x) \) almost everywhere in the interval, except at points of discontinuity (if any occur). Feel free to perform the integrals for exact coefficients or let me know if you'd like help with specific calculations!