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 $[0,1]$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

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 $a_{n}={\frac {2}{L}}\int _{0}^{L}f(x)\cos \left({\frac {n\pi {x}}{L}}\right).$

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

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

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, MER Tag Fourier series, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag