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I would like to ask some help on complex fourier series. For this fourier series, its quite hard for me because it look confusing when solving it. This is my question

1) Find the complex form of the Fourier Series expansion of

$$f(x)=\cos ax, \quad -\pi< x <\pi. $$

Someone please show me some work done and steps to solve it so that I can understand better. I'm a beginner in maths.

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    Is that meant to be $\cos (ax)$?2012-09-17
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    yes sir. Could you please help me sir2012-09-17

3 Answers 3

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Expand $\cos (ax) = \frac{1}{2} (e^{iax}+e^{-iax})$.

Then $\hat{f_n} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \cos (ax) e^{-inx} dx = \frac{1}{2 \pi} \frac{1}{2}\int_{-\pi}^{\pi} (e^{iax}+e^{-iax}) e^{-inx} dx = \frac{1}{4 \pi} \int_{-\pi}^{\pi} (e^{i(a-n)x}+e^{-i(a+n)x}) dx$.

This is a straightforward integration. Take care when $a$ is an integer (in which case the Fourier series is trivially obtained from the above expansion).

If $a \in \mathbb{Z}$, then the formula for $\cos$ gives $f_n = \frac{1}{2} \delta_{|n||a|}$ (ie, $f_n = \frac{1}{2}$ iff $n = \pm a$).

If $a \notin \mathbb{Z}$, then $\hat{f_n} = \frac{1}{4 \pi} \int_{-\pi}^{\pi} (e^{i(a-n)x}+e^{-i(a+n)x}) dx = \frac{1}{4 \pi}( \left.\frac{e^{i(a-n) x}}{i (a-n)}\right|_{-\pi}^{\pi} + \left.\frac{e^{-i(a+n) x}}{-i (a+n)}\right|_{-\pi}^{\pi})$, noting that $e^{i n \pi} = (-1)^n$, we have

$$ \hat{f_n} = \frac{1}{4 \pi i} (-1)^n( \frac{e^{i a\pi} - e^{-i a\pi}}{a-n} - \frac{e^{-i a\pi} - e^{i a\pi}}{a+n}) \\ = \frac{1}{4 \pi i} (-1)^n (e^{i a\pi} - e^{-i a\pi})(\frac{1}{a-n} + \frac{1}{a+n})\\ = \frac{1}{2 \pi}(-1)^n \sin ( a \pi) \frac{2 a}{a^2-n^2} \\ = \frac{1}{ \pi}(-1)^{n+1} \sin ( a \pi) \frac{a}{n^2-a^2} $$

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    yep i got to this step and then i lost because i dont know the intergaration. Could you please show me the intergration?2012-09-17
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    $\int_a^b e^{c t} dt = \frac{1}{c}(e^{c b}-e^{c a})$.2012-09-17
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    I still unable to get it copper hat. I making again the equation bigger. Could you show me how you done?? Please copper.hat2012-09-17
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    the exponential killing me and the constant a2012-09-17
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    Try harder. Its just a matter of substituting values and simplifying.2012-09-17
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    yep i got it almost correctly but not same as my first approach method. but however i will not try on this method because the exponential confused me2012-09-18
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    @David: I had the wrong integration limits (originally $[0,2 \pi)$) in the integrals above which must have thrown you off. I am sorry about that.2012-09-18
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    Ah its ok my friend. BTW thanks for reminding me copper.hat. I might try now and see whether i got it right or not2012-09-18
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$$c_n=\frac{1}{2\pi}\int_{-\pi}^\pi\cos ax\,\,e^{-inx}dx$$

We make integration by parts twice to

$$I:=\int_{-\pi}^\pi\cos ax\,\,e^{-inx}dx=\left.\frac{i}{n}\cos ax\,\,e^{-inx}\right|_{-\pi}^\pi+\frac{ai}{n}\int_{-\pi}^\pi\sin ax\,\,e^{-inx}dx=$$

$$=\left.-\frac{a}{n^2}\sin ax\,\,e^{-inx}\right|_{-\pi}^\pi+\frac{a^2}{n^2}I$$

I think you can handle it from here.

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    Thanks Don. I already got my answer perfectly on this method approach. Now i trying on 2nd method which mentioned by copper.hat but i not able to get it bcoz the exponential intergration2012-09-17
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Complex form of Fourier series is given by $$f(x) \sim \sum\limits_{n=-\infty}^{\infty}c_n(f)e^{inx}.$$ You need to find Fourier coefficients $$c_n(f)=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}f(x)e^{-inx}\;dx\,.$$ Thus, simply integrate by parts (twice)$$c_n(f)=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}f(x)e^{-inx}\;dx=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}\mathrm{cos}(ax)e^{-inx}\;dx.$$

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    I already got my answer perfectly using this approach. Its only i figuring out how to do it on the way that copper.hat mention it2012-09-17