# Determine the fourier cosine intergral representation of F(x) Given F(x) = 10< x< 1, -11< x< 2,...

## Question:

Determine the fourier cosine intergral representation of F(x)

Given{eq}F(x) {/eq} = {eq}10< x< 1\\-11< x< 2\\0> x> 2 {/eq}

## Fourier Cosine Series:

The Fourier cosine series of a function {eq}f(x) {/eq} defined on the interval {eq}[0, \ l] {/eq} is given as:

{eq}\displaystyle f(x)=a_0 + \sum_{n=1}^{\infty} a_n \cos(\frac{n \pi x}{l}),\\ \displaystyle a_0=\frac 1 l \int_0^l \ f(x) \ dx\\ \displaystyle a_n= \frac 2 l \int_0^l f(x) \cos(\frac{n \pi x}{l}) \ dx . {/eq}

## Answer and Explanation:

The Fourier cosine series is given as

{eq}\displaystyle f(x)=a_0 + \sum_{n=1}^{\infty} a_n \cos(\frac{n \pi x}{l}),\\ \displaystyle a_0=\frac 1 l \int_0^l \ f(x) \ dx\\ \displaystyle a_n= \frac 2 l \int_0^l f(x) \cos(\frac{n \pi x}{l}) \ dx . {/eq}

Here,

{eq}\begin{align} F(x) = & 1, \ 0< x< 1\\ & -1, \ 1< x< 2\\ & 0, \ x> 2 \end{align} {/eq}

{eq}\begin{align} \displaystyle a_0 &=\frac 1 2 \int_0^2 \ F(x) \ dx\\ \displaystyle &=\frac 1 2 \left [ \int_0^1 \ 1 \ dx+\int_1^2 \ (-1) \ dx \right ] \\ \displaystyle &=\frac 1 2 \left [ x \right ]_0^1-\frac 1 2 \left [ x \right ]_1^2 \\ \displaystyle &=\frac 1 2 \left [ 1-0 \right ]-\frac 1 2 \left [2-1 \right ] \ \ \ \ \ \ \ \ \ \text{ (Using Fundamental Theorem of Calculus ) }\\ \displaystyle &=0 \end{align} {/eq}

{eq}\begin{align} \displaystyle a_n &= \frac 2 2 \int_0^2 F(x) \cos(\frac{n \pi x}{2}) \ dx \\ \displaystyle &= \int_0^1 \cos(\frac{n \pi x}{2}) \ dx - \int_1^2 \cos(\frac{n \pi x}{2}) \ dx \\ \displaystyle &= \left [ \frac{2}{n \pi} \sin(\frac{n \pi x}{2}) \right ]_0^1 - \left [ \frac{2}{n \pi} \sin(\frac{n \pi x}{2}) \right ]_1^2 \\ \displaystyle &= \frac{2}{n \pi} \left [ \sin(\frac{n \pi}{2})- \sin 0 \right ] - \frac{2}{n \pi} \left [ \sin(\frac{n \pi 2}{2})-\sin(\frac{n \pi}{2}) \right ] \ \ \ \ \ \ \ \ \ \text{ (Using Fundamental Theorem of Calculus ) }\\ \displaystyle &= \frac{2}{n \pi} \left [ \sin(\frac{n \pi}{2}) \right ] - \frac{2}{n \pi} \left [ \sin(n \pi )-\sin(\frac{n \pi}{2}) \right ] \\ \displaystyle &= \frac{2}{n \pi} \left [ \sin(\frac{n \pi}{2}) \right ] + \frac{2}{n \pi} \left [ \sin(\frac{n \pi}{2}) \right ] \\ \displaystyle &= \frac{4}{n \pi} \left ( \sin(\frac{n \pi}{2}) \right ) \end{align} {/eq}

Hence, the resultant series is

{eq}\displaystyle f(x)=0 + \sum_{n=1}^{\infty} \frac{4}{n \pi} \left ( \sin(\frac{n \pi}{2}) \right ) \cos \left (\frac{n \pi x}{2} \right )\\ \displaystyle f(x)=\color{blue}{ \frac{4}{ \pi} \sum_{n=1}^{\infty} \frac{1}{n} \sin \left ( \frac{n \pi}{2} \right ) \cos \left (\frac{n \pi x}{2} \right ) }. {/eq}