MKU University B.Sc., Mathematics SMTJA51 Fourier Series and Laplace Transform Chapter 1

Table of Contents

Table of Contents
1 Fourier Series
 1.1 Periodic Functions
 1.2 Fourier Series - Full Range

Chapter 1
Fourier Series

In this chapter we discuss the basic concepts relating to Fourier series and obtain Fourier series development of several functions.

1.1 Periodic Functions

Definition 1.1.1. A function $f:R\to R$ is said to be periodic if there exists a positive number $\omega$ such that $f(x+\omega )=f\left(x\right)$ for all real numbers $x$ and $\omega$ is called a period of $f$. If a periodic function has a smallest positive period $\omega$, then $\omega$ is called the primitive period of $f.$

Example 1.1.2. The trigonometric functions $\sin <!--\; FUNCTION\; APPLICATION\; -->x$ and $\cos <!--\; FUNCTION\; APPLICATION\; -->x$ are periodic functions with primitive period $2\pi$ [since $\sin <!--\; FUNCTION\; APPLICATION\; -->(x+2\pi )=\sin <!--\; FUNCTION\; APPLICATION\; -->x$ and $\cos <!--\; FUNCTION\; APPLICATION\; -->(x+2\pi )=\cos <!--\; FUNCTION\; APPLICATION\; -->x]$.

Example 1.1.3. $\sin <!--\; FUNCTION\; APPLICATION\; -->2x$ and $\cos <!--\; FUNCTION\; APPLICATION\; -->2x$ are periodic functions with primitive period $\pi$ each.

Example 1.1.4. The constant function $f\left(x\right)=c$ is a periodic function. In fact every positive real number is a period of $f$ and hence this periodic function has no primitive period.

Example 1.1.5. Let $f:R\to R$ be a function defined by

$$f\left(x\right)=\{\begin{array}{cc}0,\hfill & \text{ if }x\text{ is rational }\hfill \\ 1,\hfill & \text{ if }x\text{ is irrational }\hfill \end{array}$$

Let $\omega$ be any rational number. If $x$ is rational then $x+\omega$ is also rational and if $x$ is irrational then $x+\omega$ is also irrational. Hence

$$\begin{array}{llll}\hfill f(x+\omega )& =\{\begin{array}{cc}0,\hfill & \text{ if }x\text{ is rational }\hfill \\ 1,\hfill & \text{ if }x\text{ is irrational }\hfill \end{array}& \hfill & \\ \hfill & =f\left(x\right)& \hfill & \end{array}$$

Hence every rational number is a period of $f$ and $f$ has no primitive period.

Remark 1.1.6. Let $f$ be a periodic function with period $\omega$. If the values of $f\left(x\right)$ we known in an interval of length $\omega$, then by periodicity $f\left(x\right)$ can be determined for all $x$. Hence the graph of a periodic function is obtained by periodic function of its graph in any interval of length $\omega$.

Example 1.1.7. The graph of the periodic function $f\left(x\right)=\sin <!--\; FUNCTION\; APPLICATION\; -->x$ and $f\left(x\right)=\cos <!--\; FUNCTION\; APPLICATION\; -->x$ is given below in

Figure 1.1: 

Example 1.1.8. Let $f$ be the periodic function defined by

$$f\left(x\right)=\{\begin{array}{cc}-1\hfill & \text{ if }-\pi \le x<0\hfill \\ 1\hfill & \text{ if }0\le x<\pi \hfill \end{array}\text{ and }f(x+2\pi )=f\left(x\right).$$

The graph of the periodic function $\sin <!--\; FUNCTION\; APPLICATION\; -->x$ is given below in Figure 1.2

Figure 1.2: 

Example 1.1.9. Let $f$ be a periodic function defined by $f\left(x\right)=\{\begin{array}{c}x\text{ if }-\pi <x<\pi \text{ and }\hfill \\ f(x+2\pi )=f\left(x\right)\hfill \end{array}$ (i.e) $f\left(x\right)=x$ if $-\pi <x<\pi$ and $f(x+2\pi )=f\left(x\right)$.

The graph of the periodic function sin $x$ is given below in Figure 1.3.

Figure 1.3: 

Solved Problems

Problem 1.1.10. Let $f$ and $g$ be periodic functions with period $\omega$ each and let $a$ and $b$ be real numbers. Prove that $\mathit{af}+\mathit{bg}$ is also a periodic function with period $\omega$.

Solution. Since $f$ and $g$ are periodic functions with period $\omega$ each we have for all $x,$

$$\begin{array}{lll}\hfill f(x+\omega )& =f\left(x\right)\text{ and }& \hfill \text{(1.1)}\\ \hfill g(x+\omega )& =g\left(x\right)& \hfill \text{(1.2)}\end{array}$$

Now,

$$\begin{array}{llll}\hfill (\mathit{af}+\mathit{bg})(x+\omega )& =\mathit{af}(x+\omega )+\mathit{bg}(x+\omega )& \hfill & \\ \hfill & =\mathit{af}\left(x\right)+\mathit{bg}\left(x\right)\text{ (by (}\text{1.1<!-- tex4ht:ref: p4eq1 -->}\text{) and (}\text{1.2<!-- tex4ht:ref: p4eq2 -->}\text{) }& \hfill & \\ \hfill & =(\mathit{af}+\mathit{bg})\left(x\right)& \hfill & \end{array}$$

Hence $\mathit{af}+\mathit{bg}$ is a periodic functions with period $\omega$.

Problem 1.1.11. If $\omega$ is a period of $f$ prove that $\mathit{n\omega }$ is also a period of $f$ where $n$ is any positive integer.

Solution. Let $n$ be any positive integer. Since $\omega$ is a period of $f$ we have

$$f\left(x\right)=f(x+\omega ).$$

Using this fact repeatedly we have

$$f\left(x\right)=f(x+\omega )=f(x+2\omega )=\dots <!--\; FUNCTION\; APPLICATION\; -->f(x+(n-1\left)\omega \right)=f(x+\mathit{n\omega })$$

It follows that $\mathit{n\omega }$ is a period of $f$.

Problem 1.1.12. Let $n$ be any positive integer. Prove that $\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$ is a periodic function with period $\frac{2\pi }{n}$.

Solution. Since $\sin <!--\; FUNCTION\; APPLICATION\; -->x$ is a periodic function with period $2\pi$. We have

$$\sin <!--\; FUNCTION\; APPLICATION\; -->(x+2\mathit{pi})=\sin <!--\; FUNCTION\; APPLICATION\; -->x\text{ for all }x.$$

Now, let $g\left(x\right)=\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$. Then

$$\begin{array}{llll}\hfill g\left(x+\frac{2\pi }{n}\right)& =\sin <!--\; FUNCTION\; APPLICATION\; -->\left[n\left(x+\frac{2\pi }{n}\right)\right]& \hfill & \\ \hfill & =\sin <!--\; FUNCTION\; APPLICATION\; -->(\mathit{nx}+2\pi )& \hfill & \\ \hfill & =\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}& \hfill & \\ \hfill & =\mathit{gx}& \hfill & \end{array}$$

Hence $\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$ is a periodic function with period $\frac{2\pi }{n}$.

Problem 1.1.13. Let $f\left(x\right)$ be a periodic function with period $\omega$. Prove that for any positive real number $a,f\left(\mathit{ax}\right)$ is a periodic function with period $\omega$.

Solution. Since $f\left(x\right)$ is a periodic function with period $\omega$, we have

$$f(x+\omega )=f\left(x\right)\forall <!--\; FUNCTION\; APPLICATION\; -->x.$$

Let $g\left(x\right)=f\left(\mathit{ax}\right)$. Now,

$$\begin{array}{llll}\hfill g\left(x+\frac{\omega }{a}\right)& =f\left[a\left(x+\frac{\omega }{a}\right)\right]& \hfill & \\ \hfill & =f(\mathit{ax}+\omega )=f\left(\mathit{ax}\right)& \hfill & \\ \hfill & =g\left(x\right)& \hfill & \end{array}$$

Hence $g\left(x\right)$ is a periodic function with period $\frac{\omega }{a}$.

1.2 Fourier Series - Full Range

In this section we are going to discuss the problem of representing various functions of period $2\pi$ (Full range) in terms of the simple functions namely constant function $c$ and some trigonometric functions $\sin <!--\; FUNCTION\; APPLICATION\; -->x,\cos <!--\; FUNCTION\; APPLICATION\; -->x,\sin <!--\; FUNCTION\; APPLICATION\; -->2x,\cos <!--\; FUNCTION\; APPLICATION\; -->2x,\dots <!--\; FUNCTION\; APPLICATION\; -->,\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$, $\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$ etc.

Definition 1.2.1 (Trigonometric Series). A series of the form

$$\begin{array}{llll}\hfill a_0& +a_1\cos <!--\; FUNCTION\; APPLICATION\; -->x+b_1\sin <!--\; FUNCTION\; APPLICATION\; -->x+a_2\cos <!--\; FUNCTION\; APPLICATION\; -->2x+b_2\sin <!--\; FUNCTION\; APPLICATION\; -->2x+\dots <!--\; FUNCTION\; APPLICATION\; -->\dots <!--\; FUNCTION\; APPLICATION\; -->+a_n\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}+\mathit{bn}\sin <!--\; FUNCTION\; APPLICATION\; -->x+\dots <!--\; FUNCTION\; APPLICATION\; -->& \hfill & \\ \hfill & =a_0+\underset{n=1}{\overset{\infty }{\sum ︁}}\left(a_n\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}+b_n\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}\right)& \hfill & \end{array}$$

where $a_n$ and $b_n$ are real constants is called a trigonometric series, $a_n$ and $b_n$ are called the coefficients of the series (Fourier coefficients).

Since each term of the trigonometric series is a function of period $2\pi$ it follows that if the series converges then the sum is also a function of period $2\pi$.

Definition 1.2.2 (Fourier series). Let $f\left(x\right)$ be a periodic function with period $2\pi$. Suppose $f\left(x\right)$ can be represented as a trigonometric series

$$f\left(x\right)=\frac{a_0}{2}+\underset{n=1}{\overset{\infty }{\sum ︁}}\left(a_n\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}+b_n\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}\right)$$

(1.3)

where,

$$\begin{array}{llllll}\hfill a_0& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\mathit{dx}& \hfill & & \hfill & \\ \hfill a_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & n=1,2,3,\dots <!--\; FUNCTION\; APPLICATION\; -->& \hfill & & \hfill & \\ \hfill b_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & n=1,2,3,\dots <!--\; FUNCTION\; APPLICATION\; -->& \hfill & & \hfill & \end{array}$$

Remark 1.2.3. The formulae for the coefficients $a_0,a_n,b_n$ given in the above definition are known as Euler’s Formulae.

Remark 1.2.4. If $f\left(x\right)$ is a periodic function with period $2\pi$ we can obtain the Fourier Series of $f\left(x\right)$ in any interval of length $2\pi$. If the interval is taken as $(c,c+2\pi$) then the Euler’s Formulae for Fourier coefficients are given by

$$\begin{array}{llllll}\hfill a_0& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_c^{c+2\pi }f\left(x\right)\mathit{dx}& \hfill & & \hfill & \\ \hfill a_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_c^{c+2\pi }f\left(x\right)\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & n=1,2,3,\dots <!--\; FUNCTION\; APPLICATION\; -->& \hfill & & \hfill & \\ \hfill b_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_c^{c+2\pi }f\left(x\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & n=1,2,3,\dots <!--\; FUNCTION\; APPLICATION\; -->& \hfill & & \hfill & \end{array}$$

Definition 1.2.5. A real function $f\left(x\right)$ is called an even function if $f(-x)=f\left(x\right)$ for all $x$.
The function $f\left(x\right)$ is called an odd function if $f(-x)=-f\left(x\right)$.

Example 1.2.6.

1. $\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$ is an even function.
2. $\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$ is an odd function.
3. $x^n$ is an odd function if $n$ is an odd integer and an even function if $n$ is an even integer.

Remark 1.2.7.

1. If $f\left(x\right)$ is an even function $\underset{-a}{\overset{a}{\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->}}f\left(x\right)\mathit{dx}=2\underset{0}{\overset{a}{\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->}}f\left(x\right)\mathit{dx}$.
2. If $f\left(x\right)$ is an even function $\underset{-a}{\overset{a}{\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->}}f\left(x\right)\mathit{dx}=0$.

Remark 1.2.8.

1. The product of two even functions is an even function.
2. The product of two odd functions is an even function.
3. The product of an even function and an odd function is an odd function.

Remark 1.2.9. If $f\left(x\right)$ is an odd function then $f\left(x\right)\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$ is also an odd function. Hence

$$a_0=\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\mathit{dx}=0\text{ and }{a}_n=\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-x}^{\pi }f\left(x\right)\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}=0.$$

Thus for an odd function the Fourier coefficients $a_0$ and $a_n$ are zero. Also

$$b_n=\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}=\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }f\left(x\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}$$

(Since $f\left(x\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$ is an even function.)

Remark 1.2.10. If $f\left(x\right)$ is an even function then $f\left(x\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$ is an odd function. Hence

$$b_n=\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}=0\forall <!--\; FUNCTION\; APPLICATION\; -->n.$$

Using the above remarks, working rules for calculating the Fourier coefficients of a periodic function with period $2\pi$ is given below.

Working Rules

Let $f\left(x\right)$ be a periodic function with period $2\pi$. Suppose the given interval is $(-\pi ,\pi )$.

1. Check whether $f\left(x\right)$ is an even function or an odd function.

2. 1. If $f\left(x\right)$ is an even function then $b_n=0$ for all $n$ and

      $$a_n=\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }f\left(x\right)\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}\text{ for all }n\ge 0$$

   2. If $f\left(x\right)$ is an odd function then $a_n=0$ for all $n\ge 0$ and

      $$b_n=\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }f\left(x\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}$$

3. If $f\left(x\right)$ is neither an even function nor an odd function in $(-\pi ,\pi )$ or if the given interval is not $(-\pi ,\pi )$, then calculate the Fourier coefficients by using Euler’s formulae.

The following results on integration will be useful in calculating the Fourier coefficients.

Result 1.2.11 (Bernoulli’s formula).

$$\begin{array}{llll}\hfill \int \; <!--\; nolimits\; -->\mathit{udv}& =\mathit{uv}-u^{\prime }{v}_1+u^{\mathit{\prime \prime }}{v}_2-u^{\mathit{\prime \prime }}{v}_3+\dots <!--\; FUNCTION\; APPLICATION\; -->\text{ where }& \hfill & \\ \hfill u^{\prime }& =\frac{\mathit{du}}{\mathit{dx}},u^{\prime }=\frac{d^{2}u}{d{x}^2},u^{\ast \prime }=\frac{d^{3}u}{d{x}^3}\text{ etc., and }& \hfill & \\ \hfill v_1& =\int \; <!--\; nolimits\; -->\mathit{vdx},v_2=\int \; <!--\; nolimits\; -->v_1\mathit{dx},v_3=\int \; <!--\; nolimits\; -->v_2\mathit{dx},\dots \text{ etc., }& \hfill & \end{array}$$

Result 1.2.12.

1. $\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->e^{\mathit{ax}}\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{bxdx}=\frac{e^{\mathit{ax}}}{a^2+b^2}[a\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{bx}-b\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{bx}]$
2. $\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->e^{\text{ar }}\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{bxdx}=\frac{e^{\mathit{ax}}}{a^2+b^2}[a\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{bx}+b\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{bx}]$.

Result 1.2.13. Now let $g\left(x\right)=\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$. Then

$$\begin{array}{llll}\hfill g\left(x+\frac{2\pi }{n}\right)& =\sin <!--\; FUNCTION\; APPLICATION\; -->\left[n\left(x+\frac{2\pi }{n}\right)\right]& \hfill & \\ \hfill & =\sin <!--\; FUNCTION\; APPLICATION\; -->(\mathit{nx}+2\pi )& \hfill & \\ \hfill & =\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}=g\left(x\right)& \hfill & \end{array}$$

Hence $g\left(x\right)=\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}$ is a periodic function with period $\frac{2\pi }{n}$ where $n$ is a positive integer.

Solved Problems

Problem 1.2.14. Determine the Fourier expansion of the function $f\left(x\right)=x$ where $-\pi \le x\le \pi$.

Solution. Given that $f\left(x\right)$ is a period function so I have to use the formula

$$f\left(x\right)=\frac{a_0}{2}+\underset{n=1}{\overset{\infty }{\sum ︁}}(a_n\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)+b_n\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right))$$

where,

$$\begin{array}{llll}\hfill a_0& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\mathit{dx}& \hfill & \\ \hfill a_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)\mathit{dx}& \hfill & \\ \hfill b_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)\mathit{dx}& \hfill & \end{array}$$

Obviously $f\left(x\right)=x$ is an odd function.
Hence $a_n=0$ for all $n\ge 0$. Now,

$$\begin{array}{llll}\hfill b_n& =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }x\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \end{array}$$

Taking $u=x$ and $\mathit{dv}=\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}$ and applying Bernoulli’s formula (Result 1.2.11) we get

$$\begin{array}{llll}\hfill b_n& =\frac{2}{\pi }{\left[\frac{-x\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n}+\frac{\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n^2}\right]}_0^{\pi }& \hfill & \\ \hfill & =-\frac{2}{\mathit{n\pi }}[\pi \cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{n\pi }]& \hfill & \\ \hfill & =\frac{-2{(-1)}^n}{n}& \hfill & \\ \hfill & =\frac{2{(-1)}^{n+1}}{n}& \hfill & \end{array}$$

Hence

$$\begin{array}{llll}\hfill f\left(x\right)=x& =\underset{n=1}{\overset{\infty }{\sum ︁}}{(-1)}^{n+1}\left(\frac{2}{n}\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}& \hfill & \\ \hfill \therefore x& =2\left[\frac{\sin <!--\; FUNCTION\; APPLICATION\; -->x}{1}-\frac{\sin <!--\; FUNCTION\; APPLICATION\; -->2x}{2}+\frac{\sin <!--\; FUNCTION\; APPLICATION\; -->3x}{3}+\dots <!--\; FUNCTION\; APPLICATION\; -->\right]& \hfill & \end{array}$$

Problem 1.2.15. Find the Fourier series for the function $f\left(x\right)=x^2$ where $-\pi \le x\le \pi$ and deduce that

1. $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->\dots <!--\; FUNCTION\; APPLICATION\; -->\dots <!--\; FUNCTION\; APPLICATION\; -->\dots <!--\; FUNCTION\; APPLICATION\; -->=\frac{{\pi }^2}{6}$
2. $\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\dots <!--\; FUNCTION\; APPLICATION\; -->\dots <!--\; FUNCTION\; APPLICATION\; -->\dots <!--\; FUNCTION\; APPLICATION\; -->\dots <!--\; FUNCTION\; APPLICATION\; -->.=\frac{{\pi }^2}{12}$
3. $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}-\dots <!--\; FUNCTION\; APPLICATION\; -->\dots <!--\; FUNCTION\; APPLICATION\; -->\dots <!--\; FUNCTION\; APPLICATION\; -->\dots <!--\; FUNCTION\; APPLICATION\; -->.=\frac{{\pi }^2}{8}$.

Solution. Let $f\left(x\right)=x^2$. We note that $f\left(x\right)$ is an even function. Hence,

$$\begin{array}{llll}\hfill a_0& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{x}f\left(x\right)\mathit{dx}=\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{x}^2\mathit{dx}& \hfill & \\ \hfill & =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }{x}^2\mathit{dx}(\because x^2\text{ is an even function})& \hfill & \\ \hfill & =\frac{2}{\pi }{\left[\frac{x^3}{3}\right]}_0^{\pi }=\frac{2{\pi }^2}{3}& \hfill & \end{array}$$

Now,

$$\begin{array}{llll}\hfill a_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{x}^2\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }{x}^2\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}\text{ (since }{x}^2\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}\text{ is an even function.)}& \hfill & \end{array}$$

Now by applying the Bernoulli’s formula

$$\int \; <!--\; nolimits\; -->\mathit{udv}=\mathit{uv}-u^{\prime }{v}_1+u^{\mathit{\prime \prime }}{v}_2-\dots$$

where $u=x^2$ and $\mathit{dv}=\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}$ so that $u^{\prime }=2x;u^{\mathit{\prime \prime }}=2;u^{◦}=0$.

$$\begin{array}{llll}\hfill a_n& =\frac{2}{\pi }{\left[\frac{2\pi \cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{n\pi }}{n^2}\right]}_0^{\pi }=\frac{4{(-1)}^n}{n^2}& \hfill & \\ \hfill b_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}=\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{x}^2\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & =0\text{ (since }{x}^2\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}\text{ is an odd function).}& \hfill & \end{array}$$

Hence

$$x^2=\frac{{\pi }^2}{3}+4\underset{n=1}{\overset{\infty }{\sum ︁}}\left(\frac{{(-1)}^n\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n^2}\right)$$

(1.4)

Deduction. 1 Put $x=\pi$ in (1.4) and we get

$$\begin{array}{llll}\hfill {\pi }^2& =\frac{{\pi }^2}{3}+4\underset{n=1}{\overset{\infty }{\sum ︁}}\left(\frac{{(-1)}^n\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{n\pi }}{n^2}\right)& \hfill & \\ \hfill {\pi }^2& =\frac{{\pi }^2}{3}+4\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->+\frac{1}{n^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->\right)& \hfill & \\ \hfill 4\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->+\frac{1}{n^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->\right)& ={\pi }^2-\frac{{\pi }^2}{3}=\frac{2{\pi }^2}{3}& \hfill & \\ \hfill \left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->+\frac{1}{3^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->\right)& =\frac{{\pi }^2}{6}& \hfill & \end{array}$$

Deduction. 2 Put $x=0$ in (1.4) and we get

$$\begin{array}{llll}\hfill \therefore 0& =\frac{{\pi }^2}{3}+4\left(-\frac{1}{1^2}+\frac{1}{2^2}-\frac{1}{3^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->\dots <!--\; FUNCTION\; APPLICATION\; -->\dots <!--\; FUNCTION\; APPLICATION\; -->+\frac{1}{n^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->\right)& \hfill & \\ \hfill \frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\dots <!--\; FUNCTION\; APPLICATION\; -->& =\frac{{\pi }^2}{12}& \hfill & \end{array}$$

Deduction. 3 Adding the results of Deduction.1 and Deduction.2 we get

$$\begin{array}{llll}\hfill 2\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->\right)& =\frac{{\pi }^2}{4}& \hfill & \\ \hfill \left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->\right)& =\frac{{\pi }^2}{8}& \hfill & \end{array}$$

Problem 1.2.16. Show that in the range $0$ to $2\pi$ the Fourier series expansion for $e^x$ is

$$\frac{e^{2\pi }-1}{\pi }\left[\frac{1}{2}+\underset{n=1}{\overset{\infty }{\sum ︁}}\left(\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n^2+1}\right)-\underset{n=1}{\overset{\infty }{\sum ︁}}\left(\frac{n\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n^2+1}\right)\right]$$

Solution. Let $f\left(x\right)=e^x$.

$$\begin{array}{llll}\hfill a_0& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_0^{2\pi }f\left(x\right)\mathit{dx}& \hfill & \\ \hfill & =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_0^{2\pi }{e}^x\mathit{dx}& \hfill & \\ \hfill & =\frac{1}{\pi }{\left[e^x\right]}_0^{2\pi }& \hfill & \\ \hfill & =\frac{e^{2\pi }-1}{\pi }& \hfill & \end{array}$$

Now taking,

$$\begin{array}{llll}\hfill I_n& ={\int \; <!--\; nolimits\; -->}_0^{2\pi }{e}^x\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & ={\left[e^x\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}\right]}_0^{2\pi }+n{\int \; <!--\; nolimits\; -->}_0^{2\pi }{e}^x\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & =\left(e^{2\pi }-1\right)+n\left[{\left\{e^x\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}\right\}}_0^{2\pi }-n{\int \; <!--\; nolimits\; -->}_0^{2\pi }{e}^x\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}\right]& \hfill & \\ \hfill \therefore I_n& =\left(e^{2\pi }-1\right)-n^2{I}_n& \hfill & \\ \hfill \therefore \left(n^2+1\right)I_n& =e^{2\pi }-1& \hfill & \\ \hfill \therefore I_n& =\left(\frac{e^{2\pi }-1}{n^2+1}\right)& \hfill & \\ \hfill a_n& =\frac{1}{\pi }\left(\frac{e^{2\pi }-1}{n^2+1}\right)& \hfill & \end{array}$$

Similarly we can prove that

$$\begin{array}{llll}\hfill b_n& =-\left(\frac{n\left(e^{2x}-1\right)}{\pi \left(n^2+1\right)}\right)& \hfill & \\ \hfill f\left(x\right)=e^x& =\frac{a_0}{2}+\underset{n=1}{\overset{\infty }{\sum ︁}}(a_n\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)+b_n\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right))& \hfill & \\ \hfill \therefore e^x& =\frac{e^{2\pi }-1}{\pi }\left[\frac{1}{2}+\underset{n=1}{\overset{\infty }{\sum ︁}}\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n^2+1}-\underset{n=1}{\overset{\infty }{\sum ︁}}\frac{n\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n^2+1}\right]& \hfill & \end{array}$$

Problem 1.2.17. If $f\left(x\right)=\{\begin{array}{cc}\hfill -x\hfill & \hfill \text{ if }-\pi <x<0\hfill \\ \hfill x\hfill & \hfill \text{ if }0<x<\pi \hfill \end{array}$ expand $f\left(x\right)$ as a Fourier series in the interval $(-\pi ,\pi )$.

Solution. Clearly $f(-x)=f\left(x\right)$ for all $x\in (-\pi ,\pi )$. Hence $f\left(x\right)$ is an even function in $(-\pi ,\pi )$. $\therefore b_n=0$. Hence the function can be expanded as a Fourier series of the form

$$\frac{a_0}{2}+\underset{n=1}{\overset{\infty }{\sum ︁}}{a}_n\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}\text{ where }{a}_0=\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\mathit{dx}.$$

Now,

$$\begin{array}{llll}\hfill a_0& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{x}f\left(x\right)\mathit{dx}=\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }f\left(x\right)\mathit{dx}\text{ (since }f\left(x\right)\text{ is an even function.)}& \hfill & \\ \hfill & =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^n\mathit{xdx}=\frac{2}{\pi }{\left[\frac{x^2}{2}\right]}_0^{\pi }=\frac{2}{\pi }\left[\frac{{\pi }^2}{2}\right]=\pi & \hfill & \\ \hfill \therefore a_n& =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }x\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & =\frac{2}{\pi }{\left[\frac{x\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n}\right]}_0^{\pi }-\frac{2}{\mathit{n\pi }}{\int \; <!--\; nolimits\; -->}_0^{\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & =\frac{2}{\pi n^2}{[\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}]}_0^{\pi }& \hfill & \\ \hfill & =\frac{2}{\pi n^2}\left[{(-1)}^n-1\right]& \hfill & \\ \hfill & =\{\begin{array}{cc}-\frac{4}{\pi n^2}\hfill & \text{ if }n\text{ is odd }\hfill \\ 0\hfill & \text{ if }n\text{ is even }\hfill \end{array}& \hfill & \end{array}$$

Hence,

$$\begin{array}{llll}\hfill f\left(x\right)& =\frac{\pi }{2}-\frac{4}{\pi }\sum ︁\left(\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n^2}\right)\text{ where }n\text{ is odd.}& \hfill & \\ \hfill \therefore f\left(x\right)& =\frac{\pi }{2}-\frac{4}{\pi }\left[\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->x}{1^2}+\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->3x}{3^2}+\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->5x}{5^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->\right]& \hfill & \end{array}$$

Note 1.2.18. This problem can also be re stated as $f\left(x\right)=\left|x\right|$ in the interval $-\pi <x<\pi .$

Problem 1.2.19. Find the Fourier series of the function

$$f\left(x\right)=\{\begin{array}{cc}\pi +2x\hfill & \text{ if }-\pi <x<0\hfill \\ \pi -2x\hfill & \text{ if }0\le x<\pi \hfill \end{array}$$

Hence deduce that

$$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->=\frac{{\pi }^2}{8}.$$

Solution. Here the given function is $f\left(x\right)=\pi -2\left|x\right|$ and hence it is an even function. Hence $b_n=0$ for all $n$. Now,

$$\begin{array}{llll}\hfill a_0& =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }(\pi -2x)\mathit{dx}& \hfill & \\ \hfill & =-\frac{1}{\pi }{\left[{\left(\pi -2x\right)}^2\right]}_0^{\pi }& \hfill & \\ \hfill & =-\frac{1}{\pi }\left[{(-\pi )}^2-{\pi }^2\right]=0& \hfill & \end{array}$$

Also

$$\begin{array}{llll}\hfill a_n& =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^x(\pi -2x)\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & =\frac{2}{\pi }{\left[(\pi -2x)\frac{\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n}\right]}_0^{\pi }+\frac{4}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & =0+\frac{4}{\pi n^2}{[-\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}]}_0^{\pi }& \hfill & \\ \hfill & =\frac{4}{\pi n^2}{[-\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}]}_0^{\pi }& \hfill & \\ \hfill & =\frac{4}{\pi n^2}[-{(-1)}^n+1]& \hfill & \\ \hfill & =\{\begin{array}{cc}0\hfill & \text{ if }n\text{ is odd }\hfill \\ \frac{8}{\pi n^2}\hfill & \text{ if }n\text{ is even }\hfill \end{array}& \hfill & \\ \hfill \therefore f\left(x\right)& =\frac{8}{\pi }\underset{n=1}{\overset{\infty }{\sum ︁}}\left(\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->(2n-1)\pi }{{(2n-1)}^2}\right)& \hfill & \end{array}$$

Putting $x=0$ in the above result we get

$$\begin{array}{llll}\hfill f\left(0\right)& =\frac{8}{\pi }\left[\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->\right]& \hfill & \\ \hfill \therefore \pi & =\frac{8}{\pi }\left[\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->\right]\left(f\right(0)=\pi ,\text{from given condition) }& \hfill & \\ \hfill \Rightarrow \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\dots <!--\; FUNCTION\; APPLICATION\; -->& =\frac{{\pi }^2}{8}.& \hfill & \end{array}$$

Problem 1.2.20. Find the Fourier series for $f\left(x\right)=|\sin <!--\; FUNCTION\; APPLICATION\; -->x|$ in $(-\pi ,\pi )$ of periodicity $2\pi$.

Solution. We note that $f\left(x\right)=|\sin <!--\; FUNCTION\; APPLICATION\; -->x|$ is an even function of $x$ though $\sin <!--\; FUNCTION\; APPLICATION\; -->x$ is an odd function. $\therefore b_n=0$. Hence $f\left(x\right)$ will contain only cosine terms in its Fourier series.

$$f\left(x\right)=\frac{a_0}{2}+\underset{n=1}{\overset{\infty }{\sum ︁}}{a}_n\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}.$$

Now,

$$\begin{array}{llll}\hfill a_0& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }|\sin <!--\; FUNCTION\; APPLICATION\; -->x|\mathit{dx}& \hfill & \\ \hfill & =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }|\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{xdx}\text{ (since }|\sin <!--\; FUNCTION\; APPLICATION\; -->x|\text{ is an even function.)}& \hfill & \\ \hfill & =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{xdx}& \hfill & \\ \hfill & =\frac{2}{\pi }{[-\cos <!--\; FUNCTION\; APPLICATION\; -->x]}_0^{\pi }& \hfill & \\ \hfill & =\frac{2}{\pi }[1+1]& \hfill & \\ \hfill a_0& =\frac{4}{\pi }& \hfill & \end{array}$$

Now,

$$\begin{array}{llll}\hfill a_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }|\sin <!--\; FUNCTION\; APPLICATION\; -->x|\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill a_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^0|\sin <!--\; FUNCTION\; APPLICATION\; -->x|\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}+\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_0^n|\sin <!--\; FUNCTION\; APPLICATION\; -->x|\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->x\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}(\because f(x)=|\sin <!--\; FUNCTION\; APPLICATION\; -->x|=\sin <!--\; FUNCTION\; APPLICATION\; -->x\in [-\pi ,\pi \left]\right)& \hfill & \\ \hfill & =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }[\sin <!--\; FUNCTION\; APPLICATION\; -->(n+1)x-\sin <!--\; FUNCTION\; APPLICATION\; -->(n-1\left)x\right]\mathit{dx}& \hfill & \\ \hfill & =\frac{1}{\pi }{\left[-\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->(n+1)x}{n+1}+\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->(n-1)x}{n-1}\right]}_0^{\pi }\text{ if }n\ne 1& \hfill & \\ \hfill & =\frac{1}{\pi }\left[-\frac{1}{n+1}\left|\left\{{(-1)}^{n+1}-1\right\}\right|+\frac{1}{n-1}\left\{{(-1)}^{n-1}-1\right\}\right]& \hfill & \\ \hfill & =\frac{1}{\pi }\left[\frac{1}{n+1}\left\{1+{(-1)}^n\right\}-\frac{1}{n-1}\left\{1+{(-1)}^n\right\}\right]& \hfill & \\ \hfill & =\frac{1}{\pi }\left(\frac{-2}{n^2-1}\right)\left\{1+{(-1)}^n\right\}& \hfill & \end{array}$$

For $n>1$,

$$a_n=\{\begin{array}{cc}0\hfill & \text{ if }n\text{ is odd }\hfill \\ -\frac{4}{\pi (n^2-1)}\hfill & \text{ if }n\text{ is even and }n\ne 1\hfill \end{array}$$

If $n=1$,

$$\begin{array}{llll}\hfill a_1& =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->x\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{xdx}& \hfill & \\ \hfill & =\frac{2}{\pi }{\int \; <!--\; nolimits\; -->}_0^{\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{xd}(\sin <!--\; FUNCTION\; APPLICATION\; -->x)& \hfill & \\ \hfill & =\frac{2}{\pi }{\left[\frac{{\sin <!--\; FUNCTION\; APPLICATION\; -->}^{2}x}{2}\right]}_0^{\pi }& \hfill & \\ \hfill a_1& =0& \hfill & \\ \hfill \therefore f\left(x\right)& =|\sin <!--\; FUNCTION\; APPLICATION\; -->x|=\frac{2}{\pi }-\frac{4}{\pi }\underset{n=1}{\overset{\infty }{\sum ︁}}\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->2\mathit{nx}}{\left(4n^2-1\right)}& \hfill & \\ \hfill & =\frac{2}{\pi }-\frac{4}{\pi }\underset{n=1}{\overset{\infty }{\sum ︁}}\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->2\mathit{nx}}{(2n-1)(2n+1)}& \hfill & \\ \hfill \therefore |\sin <!--\; FUNCTION\; APPLICATION\; -->x|& =\frac{2}{\pi }-\frac{4}{\pi }\underset{n=1}{\overset{\infty }{\sum ︁}}\left[\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->2\mathit{nx}}{(2n-1)(2n+1)}\right]& \hfill & \end{array}$$

Problem 1.2.21. If $f\left(x\right)=\{\begin{array}{cc}-\frac{\pi }{4}\hfill & \text{ if }-\mathit{pi}<x<0\hfill \\ \frac{\pi }{4}\hfill & \text{ if }0<x<\pi \hfill \end{array}$. Find the Fourier Series of $f\left(x\right)$ with period $2\pi$.

Solution. We note that $f\left(x\right)$ is a periodic function with period $2\pi$. Now,

$$\begin{array}{llll}\hfill a_0& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\mathit{dx}& \hfill & \\ \hfill & =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^0\left(-\frac{\pi }{4}\right)\mathit{dx}+\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_0^x\left(\frac{\pi }{4}\right)\mathit{dx}& \hfill & \\ \hfill & =\frac{1}{\pi }\left[\left(-\frac{\pi }{4}\right){\left[x\right]}_{-\pi }^0\right]+\frac{1}{\pi }\left[\left(\frac{\pi }{4}\right){\left[x\right]}_0^{\pi }\right]& \hfill & \\ \hfill & =-\frac{1}{\pi }\left(\frac{\pi }{4}\right)\left(\pi \right)+\frac{1}{\pi }\left(\frac{\pi }{4}\right)\pi & \hfill & \\ \hfill & =-\frac{\pi }{4}+\frac{\pi }{4}& \hfill & \\ \hfill a_0& =0& \hfill & \end{array}$$

Now,

$$\begin{array}{llll}\hfill a_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_{-\pi }^0\left(-\frac{\pi }{4}\right)\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}+{\int \; <!--\; nolimits\; -->}_0^{\pi }\left(\frac{\pi }{4}\right)\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & =-\frac{\pi }{4}{\left[\frac{\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n}\right]}_{-\pi }^0+\frac{\pi }{4}{\left[\frac{\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n}\right]}_0^{\pi }& \hfill & \\ \hfill & =-\frac{\pi }{4}[0-0]+\frac{\pi }{4}[0-0]& \hfill & \\ \hfill a_n& =0& \hfill & \end{array}$$

Now, for $n>1$

$$\begin{array}{llll}\hfill b_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}& \hfill & \\ \hfill & =\frac{1}{\pi }\left[{\int \; <!--\; nolimits\; -->}_{-\pi }^0\left(-\frac{\pi }{4}\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}+{\int \; <!--\; nolimits\; -->}_0^{\pi }\left(\frac{\pi }{4}\right)\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}\right]& \hfill & \\ \hfill & =\frac{1}{\pi }\frac{\pi }{4}\left[{\int \; <!--\; nolimits\; -->}_{-\pi }^0-\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}+{\int \; <!--\; nolimits\; -->}_0^{\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}\right]& \hfill & \\ \hfill & =\frac{1}{4}\left[{\int \; <!--\; nolimits\; -->}_0^{-\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}+{\int \; <!--\; nolimits\; -->}_0^{\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}\right]& \hfill & \\ \hfill & =\frac{1}{4}\left({\left[-\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n}\right]}_0^{-\pi }+{\left[\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}}{n}\right]}_0^{\pi }\right)& \hfill & \\ \hfill & =\frac{1}{4n}\left[{\int \; <!--\; nolimits\; -->}_0^{-\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}+{\int \; <!--\; nolimits\; -->}_0^{\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->\mathit{nxdx}\right]& \hfill & \\ \hfill & =\frac{1}{4n}\left({\left[-\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}\right]}_0^{-\pi }+{\left[\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{nx}\right]}_0^{\pi }\right)& \hfill & \\ \hfill & =\frac{1}{4n}(-\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{n\pi }+1-\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{n\pi }+1)& \hfill & \\ \hfill & =\frac{1}{4n}(2-2\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{n\pi })& \hfill & \\ \hfill & =\frac{1-\cos <!--\; FUNCTION\; APPLICATION\; -->\mathit{n\pi }}{2n}& \hfill & \\ \hfill b_n& =\frac{1-{(-1)}^n}{2n}& \hfill & \\ \hfill b_n& =\{\begin{array}{cc}\frac{1}{n}\hfill & \text{ if }n\text{ is odd }\hfill \\ 0\hfill & \text{ if }n\text{ is even }\hfill \\ \hfill \end{array}& \hfill & \end{array}$$

Hence,

$$f\left(x\right)=\sin <!--\; FUNCTION\; APPLICATION\; -->x+\frac{1}{3}\sin <!--\; FUNCTION\; APPLICATION\; -->3x+\frac{1}{5}\sin <!--\; FUNCTION\; APPLICATION\; -->5x+\cdots$$

Problem 1.2.22. Find the Fourier series for defined in $f\left(x\right)=e^x$ defined in $[-\pi ,\pi ]$.

Solution. The Fourier series is given as

$$f\left(x\right)=\frac{a_0}{2}+\underset{n=1}{\overset{\infty }{\sum ︁}}\left(a_n\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)+b_n\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)\right)$$

$$\begin{array}{llll}\hfill a_0& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{e}^x\mathrm{ d}x& \hfill & \\ \hfill & =\frac{1}{\pi }\left({e^x|}_{-\pi }^{\pi }\right)& \hfill & \\ \hfill & =\frac{1}{\pi }\left(e^{\pi }-e^{-\pi }\right)& \hfill & \\ \hfill a_0& =\frac{2\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right)}{\pi }& \hfill & \\ \hfill a_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{e}^x\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)dx& \hfill & \\ \hfill & =\frac{1}{\mathit{n\pi }}{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{e}^x\mathrm{ d}(\sin <!--\; FUNCTION\; APPLICATION\; -->(\mathit{nx}\left)\right)& \hfill & \\ \hfill & =\frac{1}{\mathit{n\pi }}\left({e^x\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)|}_{-\pi }^{\pi }-{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{e}^x\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)dx\right)& \hfill & \\ \hfill a_n& =\frac{1}{\mathit{n\pi }}\left(e^{\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{n\pi }\right)-e^{\pi }\sin <!--\; FUNCTION\; APPLICATION\; -->(-\mathit{n\pi })+\frac{1}{n}{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{e}^x\mathrm{ d}(\cos <!--\; FUNCTION\; APPLICATION\; -->(\mathit{nx}\left)\right)\right)& \hfill & \\ \hfill \sin <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{n\pi }\right)=0\forall <!--\; FUNCTION\; APPLICATION\; -->n\in Z& & \hfill \\ \hfill \therefore a_n& =\frac{1}{\mathit{n\pi }}\left(\frac{1}{n}{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{e}^x\mathrm{ d}(\cos <!--\; FUNCTION\; APPLICATION\; -->(\mathit{nx}\left)\right)\right)& \hfill & \\ \hfill & =\frac{1}{n^2\pi }\left({e^x\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)|}_{-\pi }^{\pi }-{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{e}^x\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)dx\right)& \hfill & \\ \hfill & =\frac{1}{n^2\pi }\left(e^{\pi }\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{n\pi }\right)-e^{\pi }\cos <!--\; FUNCTION\; APPLICATION\; -->(-\mathit{n\pi })-\pi a_n\right)& \hfill & \\ \hfill \cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{n\pi }\right)={(-1)}^n\forall <!--\; FUNCTION\; APPLICATION\; -->n\in Z& & \hfill \\ \hfill \therefore a_n& =\frac{1}{n^2\pi }\left(\left(e^{\pi }-e^{\pi }\right){(-1)}^n-\pi a_n\right)& \hfill & \\ \hfill & =\frac{1}{n^2\pi }\left(2\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right){(-1)}^n-\pi a_n\right)& \hfill & \\ \hfill a_n+a_n\frac{1}{n^2}& =\frac{2\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right){(-1)}^n}{n^2\pi }& \hfill & \\ \hfill a_n\left(\frac{1+n^2}{n^2}\right)& =\frac{2\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right){(-1)}^n}{n^2\pi }& \hfill & \\ \hfill \Rightarrow a_n& =\frac{2\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right){(-1)}^n}{\left(n^2+1\right)\pi }& \hfill & \\ \hfill b_n& =\frac{1}{\pi }{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{e}^x\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)dx& \hfill & \\ \hfill & =-\frac{1}{\mathit{n\pi }}{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{e}^x\mathrm{ d}(\cos <!--\; FUNCTION\; APPLICATION\; -->(\mathit{nx}\left)\right)& \hfill & \\ \hfill & =-\frac{1}{\mathit{n\pi }}\left({e^x\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)|}_{-\pi }^{\pi }-{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{e}^x\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)dx\right)& \hfill & \\ \hfill & =-\frac{1}{\mathit{n\pi }}\left(e^{\pi }\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{n\pi }\right)-e^{\pi }\cos <!--\; FUNCTION\; APPLICATION\; -->(-\mathit{n\pi })-{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{e}^x\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)dx\right)& \hfill & \\ \hfill & =-\frac{1}{\mathit{n\pi }}\left(2\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right){(-1)}^n-{\int \; <!--\; nolimits\; -->}_{-\pi }^{\pi }{e}^x\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)dx\right)& \hfill & \\ \hfill & =-\frac{1}{\mathit{n\pi }}\left(2\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right){(-1)}^n-\pi a_n\right)& \hfill & \\ \hfill & =-\frac{2\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right){(-1)}^n}{\mathit{n\pi }}+\frac{a_n}{n}& \hfill & \\ \hfill & =-\frac{2\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right){(-1)}^n}{\mathit{n\pi }}+\frac{2\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right){(-1)}^n}{n\left(n^2+1\right)\pi }& \hfill & \\ \hfill & =\frac{2\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right){(-1)}^n}{\mathit{n\pi }}\left(-1+\frac{1}{n^2+1}\right)& \hfill & \\ \hfill \Rightarrow b_n& =\frac{2n{(-1)}^{n+1}\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right)}{\left(n^2+1\right)\pi }& \hfill & \end{array}$$

$\therefore$ The Fourier series for the function is

$$f\left(x\right)=\frac{\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right)}{\pi }+2\underset{n=1}{\overset{\infty }{\sum ︁}}\frac{{(-1)}^n\sinh <!--\; FUNCTION\; APPLICATION\; -->\left(\pi \right)}{\pi }\left(\frac{\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)}{n^2+1}-\frac{n\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\mathit{nx}\right)}{n^2+1}\right)$$