For the function fx - -2- - - x - 02- 0 - x - -The value of an in the Fourier series expansin of fx is

A_n = 1π∫_ π^π f(x) cos nxdx = 1π [ ∫_ π^0 ( 2) cos nxdx + ∫_0^π (2) cos nx dx ] = 1π[( 2) (sin nxn )^0_ π + (2) (sin nxn )^π_0 ] = 0 ...

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Engineering Mathematics

Fourier Series 1 (Convergence Problems)

For the funct...

Question

For the function $f (x) = {\begin{matrix} - 2, & - π < x < 0 2, & 0 < x < π \end{matrix}$
The value of $a_{n}$ in the Fourier series expansin of f(x) is

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Solution

The correct option is C 0
$a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) cos n x d x$
$= \frac{1}{π} [\int_{- π}^{0} (- 2) cos n x d x + \int_{0}^{π} (2) cos n x d x]$
$= \frac{1}{π} [(- 2) {(\frac{sin n x}{n})}_{- π}^{0} + (2) {(\frac{sin n x}{n})}_{0}^{π}]$
= 0

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For the function f(x)={−2,−π<x<02,0<x<π The value of an in the Fourier series expansin of f(x) is

The correct option is C 0an=1π∫π−πf(x)cosnxdx =1π[∫0−π(−2)cosnxdx+∫π0(2)cosnxdx] =1π[(−2)(sinnxn)0−π+(2)(sinnxn)π0] = 0

For the function $f (x) = {\begin{matrix} - 2, & - π < x < 0 2, & 0 < x < π \end{matrix}$
The value of $a_{n}$ in the Fourier series expansin of f(x) is

The correct option is C 0
$a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) cos n x d x$
$= \frac{1}{π} [\int_{- π}^{0} (- 2) cos n x d x + \int_{0}^{π} (2) cos n x d x]$
$= \frac{1}{π} [(- 2) {(\frac{sin n x}{n})}_{- π}^{0} + (2) {(\frac{sin n x}{n})}_{0}^{π}]$
= 0