Science:Math Exam Resources/Courses/MATH257/December 2011/Question 03 (a)/Solution 1

We will start by computing the Fourier sine series. The Fourier sine series of $f(x)$ is given by

$Sf(x)=\sum _{n=0}^{\infty }b_{n}\sin({\frac {n\pi x}{L}})$
where
$b_{n}={\frac {2}{L}}\int _{0}^{L}f(x)\;\sin({\frac {n\pi x}{L}})\;dx\;\;$
for $n={\color {blue}1},2,3,\cdots$

Here, we have $f(x)=x$ and $L=1$. So

$b_{n}=2\int _{0}^{1}x\;\sin(n\pi x)\;dx.$

Integrate by parts using

$u=x\;\;\;\;\;\;dv=\sin(n\pi x)\;dx$

$du=dv\;\;\;\;v={\frac {-1}{n\pi }}\cos(n\pi x),$

we get that

${\begin{aligned}b_{n}&=2[{\frac {-x}{n\pi }}\cos(n\pi x)|_{0}^{1}+{\frac {1}{n\pi }}\int _{0}^{1}\cos(n\pi x)\;dx]\\&=2[{\color {blue}{\frac {-x}{n\pi }}\cos(n\pi x)|_{0}^{1}}+{\color {Brown}({\frac {1}{n\pi }})^{2}\sin(n\pi x)|_{0}^{1}}]\\&=2[{\color {blue}{\frac {-1}{n\pi }}\cos(n\pi )+0}+{\color {Brown}({\frac {1}{n\pi }})^{2}\sin(n\pi )-({\frac {1}{n\pi }})^{2}\sin(0)}].\end{aligned}}$

Since ${\color {Brown}\sin(n\pi )=0}$ and ${\color {Brown}\sin(0)=0}$, we get that

$b_{n}={\frac {-2}{n\pi }}\underbrace {\cos(n\pi )} _{=(-1)^{n}}={\frac {-2}{n\pi }}(-1)^{n}={\frac {2}{n\pi }}(-1)^{n+1}\;\;$

for $n=1,2,3,\cdots$
Therefore, the Fourier sine series of $f(x)$ is

$Sf(x)=\sum _{n=1}^{\infty }b_{n}\sin({\frac {n\pi x}{L}})=\sum _{n=1}^{\infty }{\frac {2}{n\pi }}(-1)^{n+1}\sin(n\pi x).$

Next, we will find the Fourier cosine series. The Fourier cosine series of $f(x)$ is given by

$Cf(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }a_{n}\cos({\frac {n\pi x}{L}})\;\;$

where
$a_{n}={\frac {2}{L}}\int _{0}^{L}f(x)\;\cos({\frac {n\pi x}{L}})\;dx\;\;$
for $n={\color {blue}0},1,2,\cdots$

With $f(x)=x$ and $L=1$,

$a_{n}=2\int _{0}^{1}x\;\cos(n\pi x)\;dx.$

When $n=0$ we have that $\cos(n\pi {x})$=1 and so we have two distinct cases for evaluating the integral; when $n>=1$ and when $n=0$. For the case where ${\color {red}n\geq 1}$, integrate by parts using

$u=x\;\;\;\;\;\;dv=\cos(n\pi x)\;dx$

$du=dv\;\;\;\;v={\frac {1}{n\pi }}\sin(n\pi x),$

we get that
${\begin{aligned}a_{n}&=2[{\frac {x}{n\pi }}\sin(n\pi x)|_{0}^{1}-{\frac {1}{n\pi }}\int _{0}^{1}\sin(n\pi x)\;dx]\\&=2[{\color {blue}{\frac {x}{n\pi }}\sin(n\pi x)|_{0}^{1}}+{\color {Brown}({\frac {1}{n\pi }})^{2}\cos(n\pi x)|_{0}^{1}}]\\&=2[{\color {blue}{\frac {x}{n\pi }}\sin(n\pi )+0}+{\color {Brown}({\frac {1}{n\pi }})^{2}\cos(n\pi )-({\frac {1}{n\pi }})^{2}\cos(0)}].\end{aligned}}$
Since ${\color {blue}\sin(n\pi )=0}$, we get that

$a_{n}=2[({\frac {1}{n\pi }})^{2}\underbrace {\cos(n\pi )} _{=(-1)^{n}}-({\frac {1}{n\pi }})^{2}\underbrace {\cos(0)} _{=1}]=2({\frac {1}{n\pi }})^{2}[(-1)^{n}-1]\;\;$

for $n={\color {red}1},2,3,\cdots$

Now for the case where ${\color {red}n=0}$, we get

$a_{0}=2\int _{0}^{1}x\;\cos(0)dx=2\int _{0}^{1}xdx=2({\frac {1}{2}}x^{2})|_{0}^{1}=1.$

Therefore, the Fourier cosine series of $f(x)$ is

$Cf(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }a_{n}\cos(n\pi x)={\frac {1}{2}}+\sum _{n=1}^{\infty }2({\frac {1}{n\pi }})^{2}[(-1)^{n}-1]\cos(n\pi x).$