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

MATH257 December 2011

• Q1 • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) •

Other MATH257 Exams

• December 2011 •

Question 03 (a)
(a) Find both the Fourier sine series and the Fourier cosine series of the function $f(x)=x$ on the interval $[0,1]$ .

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Hint
Recall that the Fourier coefficients of a sine series of a function $f(x)$ are given by, $b_{n}={\frac {2}{L}}\int _{0}^{L}f(x)\sin \left({\frac {n\pi {x}}{L}}\right)$ and the coefficients of a cosine series in the form ${\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }{}a_{n}\cos \left({\frac {n\pi {x}}{L}}\right)$ are given by $a_{n}={\frac {2}{L}}\int _{0}^{L}f(x)\cos \left({\frac {n\pi {x}}{L}}\right).$

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $Cf(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }a_{n}\cos(n\pi x)={\frac {1}{2}}+\sum _{n=1}^{\infty }2({\frac {1}{n\pi }})^{2}[(-1)^{n}-1]\cos(n\pi x).$