Chapter 11: Problem 26
(a) Suppose \(f(-L)=f(L), f^{\prime}(-L)=f^{\prime}(L), f^{\prime}\) is continuous, and \(f^{\prime \prime}\) is piecewise continuous on \([-L, L]\). Use Theorem 11.2 .4 and integration by parts to show that $$ f(x)=a_{0}+\sum_{n=1}^{\infty}\left(a_{n} \cos \frac{n \pi x}{L}+b_{n} \sin \frac{n \pi x}{L}\right), \quad-L \leq x \leq L $$ with $$ \begin{array}{c} a_{0}=\frac{1}{2 L} \int_{-L}^{L} f(x) d x \\ a_{n}=-\frac{L}{n^{2} \pi^{2}} \int_{-L}^{L} f^{\prime \prime}(x) \cos \frac{n \pi x}{L} d x, \quad \text { and } \quad b_{n}=-\frac{L}{n^{2} \pi^{2}} \int_{-L}^{L} f^{\prime \prime}(x) \sin \frac{n \pi x}{L} d x, n \geq 1 \end{array} $$ (b) Show that if, in addition to the assumptions in (a), \(f^{\prime \prime}\) is continuous and \(f^{\prime \prime \prime}\) is piecewise continuous on \([-L, L],\) then $$ a_{n}=\frac{L^{2}}{n^{3} \pi^{3}} \int_{-L}^{L} f^{\prime \prime \prime}(x) \sin \frac{n \pi x}{L} d x . $$
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.