Expand \( f(x)=x \sin x, 0
Ask by Maxwell Newton. in India
Jan 08,2025
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Bonus Knowledge
To expand the function \( f(x) = x \sin x \) on the interval \( 0 < x < 2\pi \) as a Fourier series, we can use the standard form of a Fourier series: \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right) \] 1. **Determine the Fourier coefficients**: - The coefficient \( a_0 \) is given by: \[ a_0 = \frac{1}{\pi} \int_0^{2\pi} f(x) \, dx \] - The coefficients \( a_n \) and \( b_n \) are given by: \[ a_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos(nx) \, dx \] \[ b_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin(nx) \, dx \] 2. **Calculate \( a_0 \)**: \[ a_0 = \frac{1}{\pi} \int_0^{2\pi} x \sin x \, dx \] This integral can be calculated using integration by parts. 3. **Calculate \( a_n \)** and \( b_n \)**: Use integration by parts for both \( a_n \) and \( b_n \) to derive the coefficients. 4. **Combine the Fourier series**: After calculating \( a_0 \), \( a_n \), and \( b_n \), plug those values back into the Fourier series formula. In summary, expanding \( f(x) = x \sin x \) as a Fourier series involves computing its coefficients with integrals that represent the average value and the contributions from the sine and cosine terms over the specified interval. This provides a neat single series expression for the function over \( (0, 2\pi) \).