Question

Not yet answered Marked out of 200 P Hog question Periodic function f(0) = 02 in the range of give by the following figure: <O< with the period = 27 is f(0) 4 f(0) = 02 72 -27 - 10 I 25 Fourier coefficients for the above is obtained as: An = cos(nn) (Note that for odd and even n, cos(nn) has different values bn=0 When Fourier series is obtained with above coefficients and when 0 = 7, what is value of ?

Answer 1

Solution:

Given $2\pi-$ periodic function is: $f(\theta) =\theta^2,~-\pi\leq \theta\leq\pi$ ​​​​​​.

Since the function
f(0) = f(-o)
is an $2\pi-$ periodic even function, hence it's Fourier series is a Fourier cosine series. Therefore,

$a_0=\frac{2} {\pi} \int_0^\pi f( \theta) d\theta=\frac{2} {\pi} \int_0^\pi \theta^2 d\theta=\frac {2\pi^2}{3}, \\ ~~~a_n=\frac{2} {\pi} \int_0^\pi f( \theta) \cos(n\theta) d\theta=\frac{2} {\pi} \int_0^\pi \theta^2\cos(n\theta) d\theta= \frac {4}{n^2}\cos(n\pi),~n=1,2,3,....$

But given that

$a_0=\frac {\pi^2}{3}.$

Hence the Fourier cosine series of the function must have the form:

$f(\theta) =\frac{2a_0}{2}+\sum_{n=1}^\infty a_n\cos(n\theta)~(\because 2 a_0=\frac {2\pi^2}{3}=2\int_0^\pi f( \theta) d\theta) \\ ~~~~~~~~~~~=\frac{\pi^2}{3}+\sum_{n=1}^\infty \frac{4\cos(n\pi)} {n^2} \cos(n\theta)\\ \therefore f(\theta) =\frac{\pi^2}{3}+4\sum_{n=1}^\infty \frac{(-1)^n} {n^2} \cos(n\theta) ~~(\because\cos(n\pi) =(-1)^n)$

Now, since the function is continuous on $[-\pi, \pi],$ therefore we have

$f(\pi) =\frac{\pi^2}{3}+4\sum_{n=1}^\infty \frac{(-1)^n} {n^2} \cos(n\pi)\\ ~~~~~\Rightarrow \pi^2-\frac{\pi^2}{3}=4\sum_{n=1}^\infty \frac{(-1)^n} {n^2}(-1)^n\\ ~~~~~\Rightarrow \frac{2\pi^2}{3}=4\sum_{n=1}^\infty \frac{1} {n^2}\\ \therefore {\color{Red} \frac{\pi^2}{6}=\sum_{n=1}^\infty \frac{1} {n^2}}.$

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