Question

A function with a period $2\pi$ is shown below. The Fourier series fo the function is given by

• A. $f\left(x\right)=\frac{1}{2}+{\sum }_{n=1}^{\infty }\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)\cos nx$
• B. $f\left(x\right)={\sum }_{n=1}^{\infty }\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)\cos nx$
• C. $f\left(x\right)=\frac{1}{2}+{\sum }_{n=1}^{\infty }\frac{2}{n\pi }\sin \frac{n\pi }{2}\sin nx$
• D. $f\left(x\right)={\sum }_{n=1}^{\infty }\frac{2}{n\pi }\sin \frac{n\pi }{2}\sin nx$

Answer 1

The correct option is A $f\left(x\right)=\frac{1}{2}+{\sum }_{n=1}^{\infty }\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)\cos nx$

$f\left(x\right)=\{\begin{array}{c}1,-\frac{\pi }{2}\le \times \le \frac{\pi }{2}\\ 0,otherwise\end{array}$
It is given that T = $2\pi$
so, $a_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)dx$
$=\frac{1}{\pi }{\int }_{-\pi /2}^{\pi /2}\left(1\right)dx=1$
$a_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\cos nxdx$
$=\frac{2}{x\pi }\sin \frac{n\pi }{2}$
$b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\sin nxdx=0$
$(\therefore$ integrand is an odd function)
So, Fourier series representation is
$f\left(x\right)=\frac{a_0}{2}+{\sum }_{n=1}^{\infty }{a}_n\cos nx+{\sum }_{n=1}^{\infty }{b}_n\sin nx$
$=\frac{1}{2}+{\sum }_{n=1}^{\infty }\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)\cos nx$