Sine and Cosine Fourier series

**Klo** · June 26th 2010, 04:44 AM

Hi!
Im asked to write sine AND cosine series of $sin(x) ; x \in <0,\pi)$

*what does that mean? i thought the fourier series can exist for the sine part being 0 OR the cosine being 0 OR both sine and cosine being included in the formula.. how do i do then one SINE and another COSINE serie of the same function?

*Im confused with the condition: first of all it doesn't look like a periodic function to me, but im supposed to do a fourier transform on it...

if i imagine it's a periodic function(somehow) then:
*how does the function look like? is it the positive parts of sine one next to the other, thus with period = $\pi$ ... or is it the complete sine with a constant 0 put in between the positive parts thus the period would be $2\pi$ ?

*Also, how do i know what the function looks like on x < 0 ? (considering it is periodic, is the function even or odd or none of these?

Thanks!

**chisigma** · June 26th 2010, 05:11 AM

Probably is requested the Fourier expansion of the output of a 'full wave rectifier' ...

$f(x)= |\sin x|$ , $- \pi < x < \pi$ (1)

The function is 'even' so that there are only the terms in 'cosine' and is...

$\displaystyle f(x) = \frac{a_{0}}{2} + \sum_{n=1}^{\infty} a_{n}\ \cos nx$ (2)

... where...

$\displaystyle a_{0} = \frac{2}{\pi}\ \int_{0}^{\pi} \sin x \ dx = \frac{4}{\pi}$

$\displaystyle a_{n} = \frac{2}{\pi}\ \int_{0}^{\pi} \sin x \ \cos nx \ dx = \left\{\begin{array}{ll} - \frac{4}{\pi\ (n^{2}-1)} ,\,\, n \ even\\{}\\0 ,\,\, n \ odd \end{array}\right.$ (3)

Kind regards

$\chi$ $\sigma$

**Klo** · June 26th 2010, 05:26 AM

It definitely makes much more sense now... so i guess the question to find the sine fourier form would lead to $b_k = 0$ which would be the result - silly.
Thank you chisigma

**HallsofIvy** · June 26th 2010, 07:48 AM

No, the sine Fourier series for sin(x) would have all $b_k= 0$ except for k= 1: $b_1= 1$ because the sine Fourier sine series for sin(x) is precisely "sin(x)".

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