Question

(a) Find the Fourier series of period 2$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\left(x-1\right)}^{\mathbf{2}}$on (0,2)$$" width="9" height="19" role="math">$$" width="9" height="19" role="math">" width="9" height="19" role="math">

(b) Use your result in (a) to evaluate$\mathbf{\sum }_{}^{}\mathbf{1}\mathbf{/}{\mathit{n}}^{\mathbf{4}}$.

Short answer

Part (a) the Fourier series is $f\left(x\right)=\frac{1}{3}+\frac{4}{{\mathrm{\pi }}^2}\sum _{n=1}^{\infty }\frac{\cos \left(n\mathrm{\pi x}\right)}{n^2}$

Part (b) the sum is $\sum _{n=1}^{\infty }\frac{1}{n^4}=\frac{{\mathrm{\pi }}^4}{90}$

Step-by-step solution

Given information

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

Meaning of the Fourier Series and Definition of Parseval theorem.

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.

Part (a)Find the Fourier series

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

The coefficients are then:

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

Find the value of $a_n$

$$\begin{gathered} a_n=\frac{2}{T}{\int }_0^T{\left(x-1\right)}^2\cos \frac{2n\mathrm{\pi x}}{T}dx \\ ={\int }_0^2{\left(x-1\right)}^2\cos \left(\mathrm{\pi nx}\right)dx \\ ={\int }_0^2\left(x^2-2x+1\right)\cos \left(\mathrm{\pi nx}\right)dx \\ =I_1+I_2+I_3 \end{gathered}$$

Find the value of $I_1$

$$\begin{gathered} I_1={\int }_0^2{x}^2\cos \left(\mathrm{\pi nnx}\right)dx \\ ={\left[\frac{x^2}{n\mathrm{\pi }}\sin \left(\mathrm{\pi nx}\right)+\frac{2x}{n^2{\mathrm{\pi }}^2}\cos \left(n\mathrm{\pi x}\right)-\frac{2}{n^3{\mathrm{\pi }}^3}\sin \left(n\mathrm{\pi x}\right)\right]}_0^2 \\ =\frac{4}{n^2{\mathrm{\pi }}^2} \end{gathered}$$

Find the value of $I_2$

$$\begin{gathered} I_2=-2{\int }_0^{2}x\cos \left(\mathrm{\pi n}x\right)dx \\ =-2{\overline{)\left[\frac{x}{n\mathrm{\pi }}\sin \left(n\mathrm{\pi x}\right)+\frac{1}{n^2{\mathrm{\pi }}^2}\cos \left(n\mathrm{\pi }x\right)\right]}}_0^2 \\ =0 \end{gathered}$$

Find the value of $I_3$

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

Thus, the function is equal to:

$f\left(x\right)=\frac{1}{3}+\frac{4}{{\mathrm{\pi }}^2}\sum _{n=1}^{\infty }\frac{\cos \left(n\mathrm{\pi }x\right)}{n^2}$

Part (b) Evaluate the sum ∑1/n4

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

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

Thus, by the Parseval theorem:

$$\begin{gathered} \frac{1}{5}=\frac{1}{9}+\frac{1}{2}\sum _{n=1}^{\infty }\frac{16}{{\mathrm{\pi }}^2}\frac{1}{n^4} \\ \sum _{n=1}^{\infty }\frac{1}{n^4}=\frac{\mathrm{\pi }}{90} \end{gathered}$$

Part (a) the Fourier series is $f\left(x\right)=\frac{1}{3}+\frac{4}{{\mathrm{\pi }}^4}\sum _{n=1}^{\infty }\frac{\cos \left(n\mathrm{\pi x}\right)}{n^2}$

Part (b) the sum is $\sum _{n=1}^{\infty }\frac{1}{n^4}=\frac{{\mathrm{\pi }}^4}{90}$