Science:Math Exam Resources/Courses/MATH307/December 2012/Question 04 (b)

MATH307 December 2012

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Question 04 (b)
Consider the Fourier series $t^{2}-t=\sum _{n=-\infty }^{\infty }c_{n}e^{2\pi int}$ for $0\leq t\leq 1$. (b) The coefficient cn can be written as an inner product $c_{n}=\langle f,g\rangle$. What are ƒ and g?

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Hint

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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. To compute the coefficient, start with $f(t)=\sum _{n=-\infty }^{\infty }c_{n}e_{n}(t)$ where $\displaystyle f(t)=t^{2}-t$ and $\displaystyle e_{n}(t)=e^{2\pi int}$. Also, we have L=1-0=1. Take the inner product of ƒ(t) with $e_{m}$. The only term in the infinite sum that remains is the one with $n=m$, and in this case $\langle e_{n},e_{n}\rangle =1$. Thus $\left\langle f,e_{m}\right\rangle =\sum _{n=-\infty }^{\infty }c_{n}\left\langle e_{n},e_{m}\right\rangle =c_{m}$ and we get the formula $c_{m}=\langle f,e_{m}\rangle =\int _{0}^{1}f(t)e^{-2\pi imt}\,dt=\int _{0}^{1}f(t){\overline {g(t)}}\,dt$ Since $\displaystyle {\overline {g(t)}}=e^{-2\pi imt}$ we have $\displaystyle g(t)=e^{2\pi imt}$, and so we overall find that $\displaystyle c_{n}=\langle t^{2}-t,e^{2\pi int}\rangle$

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