Question

The Fourier series expansion of the saw - toothed waveform $f\left(x\right)=x,intheintervel(-\pi ,\pi )$ of period $2\pi$ gives the series, $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+....=?$

• A. $\frac{\pi }{2}$
• B. $\frac{{\pi }^2}{4}$
• C. $\frac{{\pi }^2}{16}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D

$\frac{\pi }{4}$

$f\left(x\right)=x,(-\pi ,\pi )$
It is an odd function.
$a_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)dx=0$
$a_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\cos \text{nxdx}=0$
$b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\sin \text{nxdx}$
$=\frac{1}{\pi }{\int }_{-\pi }^{\pi }x\sin \text{nxdx}=\frac{2(-1{)}^{n+1}}{n}$
Hence Fourier series is,
$f\left(x\right)=\frac{a_0}{2}+\sum a_n\cos nx+\sum b_n\sin nx$
$f\left(x\right)={\sum }_{n=1}^{\infty }\frac{2(-1{)}^{n+1}}{n}\sin nx$
$x=2[1.\sin x-\frac{1}{2}\sin 2x+\frac{1}{3}\sin 3x-\frac{1}{4}\sin 4x+....]$
Put x $=\frac{\pi }{2}.$
$\frac{\pi }{2}=2[1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+....]$
$So,1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+....=\frac{\pi }{4}$