Question

Find the value of\( \frac{1}{1^4} +\frac{1}{3^4} +\frac{1}{5^4} +\frac{1}{7^4} \) +….by finding the half range Fourier cosine series of the function f(x) = x in the interval 0<x<l.

(a) \(\frac{\pi^4}{12} \)

(b) \(\frac{\pi^4}{48} \)

(c) \(\frac{\pi^4}{24} \)

(d) \(\frac{\pi^4}{96} \)

I have been asked this question during an online exam.

My enquiry is from Fourier Half Range Series in section Fourier Series of Engineering Mathematics

Answer 1

Right choice is (d) \(\frac{\pi^4}{96} \)

Easiest explanation: Using Parseval’s relation,

\(\int_{-l}^l(f(x))^2 dx=l[\frac{a_0^2}{2}+\sum_{n=1}^{∞}(a_n^2+b_n^2 ) ] \)

L.H.S. = \(\int_0^l x^2 dx= \frac{l^3}{3} \)

a0 = \(\frac{2}{l} \int_0^l xdx= l \)

an = \(\frac{2}{l} \int_0^lxcos(\frac{nπx}{l})dx \)

= \(\frac{-4l}{\pi^2}  (\frac{1}{1^2} +\frac{1}{3^2} +\frac{1}{5^2} +…) \)

R.H.S. = \(\frac{l}{2} (\frac{l^2}{2}+\left(-4\frac{l}{\pi}\right)^2 (\frac{1}{1^4} +\frac{1}{3^4} +\frac{1}{5^4} +…..)) \)

Therefore,

\(\frac{1}{1^4} +\frac{1}{3^4} +\frac{1}{5^4} +\frac{1}{7^4} +….= (\frac{l^3}{3}-\frac{l^3}{4})\left(2\frac{\pi^4}{(16l^3 )}\right) \)

= \(\frac{\pi^4}{96}.\)