Answered: Obtain the fourier series expansion of…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Obtain the fourier series expansion of the function of period 2π defined by

f ( t ) =

-π -π<t<0

t 0<t<π

Expert Solution

Step 1

Given: $f (x) = \{\begin{cases} - π - π < t < 0 \\ t 0 < t < π \end{cases}$

we know the Fourier series expansion is

$f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos n t + \sum_{n = 1}^{\infty} b_{n} \sin n t . . . . (i)$

where

$a_{0} = \frac{1}{π} \int_{- π}^{π} f (t) d t a_{0} = \frac{1}{π} [\int_{- π}^{0} - π d t + \int_{0}^{π} t d t] a_{0} = \frac{1}{π} [- π {[t]}_{- π}^{0} + {[\frac{t^{2}}{2}]}_{0}^{π}] a_{0} = \frac{1}{π} [- π^{2} + \frac{π^{2}}{2}] a_{0} = - \frac{1}{π} [\frac{π^{2}}{2}] = - \frac{π}{2}$

Step 2

$a_{n} = \frac{1}{π} \int_{- π}^{π} f (t) \cos n t d t a_{n} = \frac{1}{π} [- π \int_{- π}^{0} \cos n t d t + \int_{0}^{π} t \cos n t d t] a_{n} = \frac{1}{π} [- π \times 0 + \frac{1}{n^{2}} [{(- 1)}^{n} - 1]] a_{n} = \frac{1}{π} (\frac{1}{n^{2}} [{(- 1)}^{n} - 1])$

Now,

$b_{n} = \frac{1}{π} \int_{- π}^{π} f (t) \sin n t d t b_{n} = \frac{1}{π} [- π \int_{- π}^{0} \sin n t d t + \int_{0}^{π} t \sin n t d t] b_{n} = \frac{1}{π} [- π \frac{1}{n} (- 1 + {(- 1)}^{n}) + (- {(- 1)}^{n} \frac{π}{n})] b_{n} = \frac{1}{π} [\frac{π - 2 π {(- 1)}^{n}}{n}] b_{n} = \frac{1 - 2 {(- 1)}^{n}}{n}$

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