Math Problem Statement
Find the Fourier sine series on [0, π] of the function f(x) = {0, 0 ≤ x < π/2; π - x, π/2 ≤ x ≤ π}
Solution
To find the Fourier sine series for the given piecewise function on the interval , we need to follow these steps:
1. Fourier Sine Series General Formula
For a function defined on , the Fourier sine series is given by:
where
In this case, the function is defined as:
\begin{cases} 0 & 0 \leq x < \frac{\pi}{2} \\ \pi - x & \frac{\pi}{2} \leq x \leq \pi \end{cases}$$ ### 2. Calculate $$b_n$$ We will break the integral into two parts based on the piecewise function. $$b_n = \frac{2}{\pi} \left( \int_0^{\frac{\pi}{2}} 0 \cdot \sin\left(\frac{n\pi x}{\pi}\right) dx + \int_{\frac{\pi}{2}}^{\pi} (\pi - x) \sin\left(\frac{n\pi x}{\pi}\right) dx \right)$$ The first integral evaluates to zero because the function is zero over the interval $$[0, \frac{\pi}{2}]$$. Thus, we are left with: $$b_n = \frac{2}{\pi} \int_{\frac{\pi}{2}}^{\pi} (\pi - x) \sin\left(n x\right) dx$$ ### 3. Solve the Integral We now need to compute this integral. Let me calculate this step for you. The integral evaluates to the following result for $$b_n$$: $$b_n = \frac{2}{\pi} \left( \frac{\pi \cos\left(\frac{n\pi}{2}\right)}{2n} + \frac{\sin\left(\frac{n\pi}{2}\right)}{n^2} - \frac{\sin(n\pi)}{n^2} \right)$$ Since $$\sin(n\pi) = 0$$ for all integers $$n$$, this term vanishes. The final expression for $$b_n$$ simplifies to: $$b_n = \frac{2}{\pi} \left( \frac{\pi \cos\left(\frac{n\pi}{2}\right)}{2n} + \frac{\sin\left(\frac{n\pi}{2}\right)}{n^2} \right)$$ This gives the coefficients for the Fourier sine series of the given piecewise function. Would you like further details on this solution or clarification on any steps? Here are some related questions: 1. How do Fourier series coefficients change for different boundary conditions? 2. Can we apply this method to other piecewise functions? 3. What is the physical interpretation of the Fourier sine series? 4. How does the series converge to the original function? 5. What are the orthogonality conditions for sine functions in Fourier series? **Tip**: When solving Fourier series problems, symmetry in the function can often simplify calculations, so always check for even or odd symmetries first!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Fourier Series
Piecewise Functions
Trigonometry
Formulas
Fourier sine series: f(x) ~ Σ(b_n sin(nπx/π))
Fourier coefficient: b_n = (2/π) ∫[0,π] f(x) sin(nπx/π) dx
Theorems
Fourier series expansion for periodic functions
Suitable Grade Level
University (Calculus/Advanced Mathematics)