Answer to Question #201556 in Math for Asad Ali

Question

Expand Fourier series f(x)=cos x limit -pi/2 to pi/2

Answer 1

$$f(x)=\dfrac{a_0}{2}+\displaystyle\sum_{n=1}^{\infin}\bigg(a_n\cos (\dfrac{n\pi x}{L})+b_n\sin (\dfrac{n\pi x}{L})\bigg)$$

$$a_0=\dfrac{1}{L}\displaystyle\int_{-L}^{L}f(x)dx$$

$$a_n=\dfrac{1}{L}\displaystyle\int_{-L}^{L}f(x)\cos (\dfrac{n\pi x}{L})dx$$

$$b_n=\dfrac{1}{L}\displaystyle\int_{-L}^{L}f(x)\sin(\dfrac{n\pi x}{L})dx$$

$$a_0=\dfrac{2}{\pi}\displaystyle\int_{-\pi/2}^{\pi/2}\cos x dx=\dfrac{2}{\pi}\big[\sin x\big]\begin{matrix} \pi/2 \\ - \pi/2 \end{matrix}=\dfrac{4}{\pi}$$

$$a_n=\dfrac{2}{\pi}\displaystyle\int_{-\pi/2}^{\pi/2}\cos x\cos (2nx)dx$$

$$=\dfrac{2}{\pi}\big[-\dfrac{\cos((2n+1)x)}{2(2n+1)}-\dfrac{\cos((2n-1)x)}{2(2n-1)}\big]\begin{matrix} \pi/2 \\ - \pi/2 \end{matrix}=0$$

$$=\dfrac{2}{\pi}\big[\dfrac{(-1)^n}{2n+1}-\dfrac{(-1)^n}{2n-1}\big]=\dfrac{4(-1)^{n+1}}{\pi(4n^2-1)}$$

$$b_n=\dfrac{2}{\pi}\displaystyle\int_{-\pi/2}^{\pi/2}\sin x\cos (2nx)dx=0$$

$$f(x)=\dfrac{2}{\pi}+\displaystyle\sum_{n=1}^{\infin}\big(\dfrac{4(-1)^{n+1}}{\pi(4n^2-1)}\cos (2nx)\big)$$