Science:Math Exam Resources/Courses/MATH307/December 2010/Question 06 (f)/Solution 1

If we extend the domain from $[0,1]$ to $[0,2]$, we can write the new Fourier coefficients $d_{n}$ using the techniques from part (a) as

$d_{n}={\frac {1}{2}}\int _{0}^{2}xe^{-2\pi {}inx/2}{\textrm {d}}x.$

We could go and directly compute these integrals but we want to relate them to the coefficients $c_{n}$ from part (a) computed on a domain $[0,1]$. To do this, we can use a substitution and let $u={\frac {x}{2}}$. Then $\displaystyle {}dx=2du$. With this transformation we have $u(0)=0$ and $u(2)=1$ and so putting everything together,

$d_{n}=2\int _{0}^{1}ue^{-2\pi {}inu}{\textrm {d}}u=2c_{n}$

where we have recognized that the new integral is precisely what we computed to get the coefficients on a period $[0,1]$. Therefore on $[0,2]$ the Fourier coefficients double and

$d_{n}={\begin{cases}2{\frac {1}{2}}=1,\qquad {}n=0\\2{\frac {i}{2\pi {}n}}={\frac {i}{\pi {}n}},\qquad {}n\neq 0\end{cases}}.$

Therefore we have that $\displaystyle {}|d_{n}|=2|c_{n}|$ and the points to plot for the frequency amplitude are $\displaystyle {}(n,2|c_{n}|)$.