Fourier series of sinx

**tbyou87** · December 4th 2007, 04:26 PM

Find the full series of sinx on (-l,l) in its real and complex forms. Assume that l is not an integer multiple of $\pi$ . Hint: First find the series for $e^{ix}$ .

I don't see how finding the fourier series of $e^{ix}$ helps in finding the series for sinx.

THanks

**galactus** · December 4th 2007, 05:01 PM

The reason they said that is because:

$sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$

**tbyou87** · December 4th 2007, 06:14 PM

So I simplified the complex version for the coefficient of the fourier series to

$\frac{1}{-2i{\pi}n}(e^{-in\pi+il}-e^{in\pi-il})$

I'm not sure how to simplify it from here.

THanks

**TwistedOne151** · December 4th 2007, 06:36 PM

To simplify $\frac{1}{-2 \imath \pi n}(e^{-\imath n \pi+\imath l}-e^{\imath n\pi-\imath l})$ we can use the fact that $e^{\imath x} = \cos x + \imath\sin x$ to get that $\frac{e^{\imath x} - e^{-\imath x}}{2i} = sin x$

Thus $\frac{1}{-2 \imath \pi n} (e^{-\imath n \pi+\imath l}-e^{\imath n\pi- \imath l})=\frac{1}{2\imath \pi n}(e^{\imath (n\pi -l)}-e^{-\imath( n\pi -l)})$
$=\frac{1}{\pi n} \sin (n\pi -l)$

--Kevin C.

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