Answered: a. determine the Fourier coefficients.…

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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Related questions

Question

a. determine the Fourier coefficients.

b. Write the corresponding series.

c. Expand the Fourier series up to at least n = 5.

d. Write the convergence of the series

e. Use a graphing program to check that the Fourier series converges to the indicated function

g(t) = {-t+ 1, -n <t < 0
0 <t <n '
-t + 1, -n <t < 0
g(t + 2) = g(t)
%3D
l-t – 1,
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Expert Solution

Step 1

Given function is

$g (t) = \{\begin{array}{l} - t + 1, & - π < t < 0 \\ - t - 1, & 0 < t < π \end{array}, g (t + 2 π) = g (t)$

Solution of part a.:

Fourier coefficient of $g (t)$ are

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

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

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