Question

The Fourier series expansion of a symmetric and even function, f(x) where
$f\left(x\right)=\{\begin{array}{c}1+2x/\pi ,-\pi \le \times \le 0\\ 1-2x/\pi ,0\le \times \le \pi \end{array}$

• A. ${\sum }_{n=1}^{\infty }\frac{4}{{\pi }^2{n}^2}(1+\cos n\pi )$
• B. ${\sum }_{n=1}^{\infty }\frac{4}{{\pi }^2{n}^2}(1-\cos n\pi )$
• C. ${\sum }_{n=1}^{\infty }\frac{4}{{\pi }^2{n}^2}(1-\sin n\pi$)
• D. ${\sum }_{n=1}^{\infty }\frac{4}{{\pi }^2{n}^2}(1+\sin n\pi$)

Answer 1

The correct option is B ${\sum }_{n=1}^{\infty }\frac{4}{{\pi }^2{n}^2}(1-\cos n\pi )$

Given $f\left(x\right)=\{\begin{array}{c}1+\left(\frac{2x}{\pi }\right);-\pi \le \times \le 0\\ 1-\left(\frac{2x}{\pi }\right);0\le \times \le \pi \end{array}$
Fourier series for an even function is an
$=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\cos nxdx,$
$=\frac{1}{\pi }\left\{2{\int }_0^{\pi }(1-\frac{2x}{\pi })\cos nxdx\right\}$
$=\frac{2}{\pi }[{\left\{(1-\frac{2x}{\pi })\left(\frac{1}{n}\sin nx\right)\right\}}_0^{\pi }-{\int }_0^{\pi }(-\frac{2}{\pi })\left(\frac{1}{n}\sin nx\right)dx]$
$=\frac{2}{\pi }\left[\right(1-2\left)\right(0)-(1-0\left)\right(0)+\frac{2}{\pi n}{(-\frac{1}{n}\cos nx)}_0^{\pi }$
$=\frac{2}{\pi }[0+\frac{2}{\pi n^2}(1-\cos n\pi \left)\right]$
$a_n=\frac{4}{{\pi }^2{n}^2}(-1\cos n\pi )$
$a_0=0$
Fourier expansion of $f\left(x\right)={\sum }_{n=1}^{\infty }\frac{4}{{\pi }^2{n}^2}(1-\cos n\pi )$