91Ó°ÊÓ

The given function is $f\left(x\right)={\left(x-1\right)}^2$ and the sum is $\underset{}{\overset{}{∑}}1/n^4$

A Fourier series is an infinite sum of sines and cosines expansion of a periodic function. The orthogonality relationships of the sine and cosine functions are used in the Fourier Series.

Parseval's theorem states that a signal's energy can be represented as the average energy of its frequency components.

The function is even about x=1 , so only cosine terms will be present.

The coefficients are then:

$$\begin{gathered} a_0=\frac{2}{T}{∫}_0^T{\left(x-1\right)}^{2}d\left(x-1\right) \\ =\frac{1}{3}{\overline{){\left(x-1\right)}^3}}_0^2 \\ =\frac{2}{3} \end{gathered}$$

Find the value of $a_n$

$$\begin{gathered} a_n=\frac{2}{T}{∫}_0^T{\left(x-1\right)}^2\cos \frac{2n\mathrm{Ï€³\ae }}{T}dx \\ ={∫}_0^2{\left(x-1\right)}^2\cos \left(\mathrm{Ï€²Ô³\ae }\right)dx \\ ={∫}_0^2\left(x^2-2x+1\right)\cos \left(\mathrm{Ï€²Ô³\ae }\right)dx \\ =I_1+I_2+I_3 \end{gathered}$$

Find the value of $I_1$

$$\begin{gathered} I_1={∫}_0^2{x}^2\cos \left(\mathrm{Ï€²Ô²Ô³\ae }\right)dx \\ ={\left[\frac{x^2}{n\mathrm{Ï€}}\sin \left(\mathrm{Ï€²Ô³\ae }\right)+\frac{2x}{n^2{\mathrm{Ï€}}^2}\cos \left(n\mathrm{Ï€³\ae }\right)-\frac{2}{n^3{\mathrm{Ï€}}^3}\sin \left(n\mathrm{Ï€³\ae }\right)\right]}_0^2 \\ =\frac{4}{n^2{\mathrm{Ï€}}^2} \end{gathered}$$

Find the value of $I_2$

$$\begin{gathered} I_2=-2{∫}_0^{2}x\cos \left(\mathrm{Ï€²Ô}x\right)dx \\ =-2{\overline{)\left[\frac{x}{n\mathrm{Ï€}}\sin \left(n\mathrm{Ï€³\ae }\right)+\frac{1}{n^2{\mathrm{Ï€}}^2}\cos \left(n\mathrm{Ï€}x\right)\right]}}_0^2 \\ =0 \end{gathered}$$

$$\begin{gathered} I_3={∫}_0^2\cos \left(n\mathrm{π}x\right)dx \\ ={\overline{)\frac{\sin \cos \left(n\mathrm{π}x\right)}{n\mathrm{π}}}}_0^2 \\ =0 \end{gathered}$$

Thus, the function is equal to:

$f\left(x\right)=\frac{1}{3}+\frac{4}{{\mathrm{π}}^2}\underset{n=1}{\overset{∞}{∑}}\frac{\cos \left(n\mathrm{π}x\right)}{n^2}$

The average of the square of the function over the interval is equal to:

$$\begin{gathered} \frac{1}{2}{∫}_0^2{\left(x-1\right)}^{4}dx=\frac{1}{2}{∫}_0^2{\left(x-1\right)}^{4}d\left(x-1\right) \\ =\frac{1}{10}{\left(x-1\right)}^5{∫}_0^{2}dx \\ =\frac{1}{5} \end{gathered}$$

Thus, by the Parseval theorem:

$$\begin{gathered} \frac{1}{5}=\frac{1}{9}+\frac{1}{2}\underset{n=1}{\overset{∞}{∑}}\frac{16}{{\mathrm{π}}^2}\frac{1}{n^4} \\ \underset{n=1}{\overset{∞}{∑}}\frac{1}{n^4}=\frac{\mathrm{π}}{90} \end{gathered}$$

Part (a) the Fourier series is $f\left(x\right)=\frac{1}{3}+\frac{4}{{\mathrm{Ï€}}^4}\underset{n=1}{\overset{∞}{∑}}\frac{\cos \left(n\mathrm{Ï€³\ae }\right)}{n^2}$

Part (b) the sum is $\underset{n=1}{\overset{∞}{∑}}\frac{1}{n^4}=\frac{{\mathrm{π}}^4}{90}$