Question about Fourier Series

**Paymemoney** · March 4th 2011, 03:50 PM

Hi

Can someone explain these two cases

i have a final Fourier series that is:
$f(t) \sim \frac{\pi}{2} + \frac{4}{\pi} \sum_{i=1}^\infty \frac{cos(2n+1)t}{(2n+1)^2}$ which was simplified from $\frac{1}{\pi}[\frac{2}{n^2} - \frac{2(-1)^{n}}{n^2}]$ .

when n -> odd $\frac{4}{\pi n^2}$
when n -> even 0

and other one $f(t) \sim \frac{8}{\pi} \sum_{i=1}^\infty \frac{sin(2n-1)t}{(2n-1)}$ which was simplified from $\frac{2}{\pi n}(2-2(-1)^{n})$

when n -> even 0
when n -> odd $\frac{8}{\pi n}$

when do i use (2n+1) to replace n? and when do i use (2n-1) to replace n?

P.S

**Opalg** · March 5th 2011, 11:34 AM

For a start, if you are summing a function of n from 1 to infinity then the notation should say $\sum_{n=1}^\infty\,,$ not $\sum_{i=1}^\infty$ .

If you want to sum over all the odd numbers, then you can either call them 2n–1 and start at n=1, or you can use 2n+1 but then you must start at n=0.

So in those Fourier series, the cosine sum should have the summation $\sum_{n=0}^\infty\,,$ and the sine sum should have the summation $\sum_{n=1}^\infty$ .

**Paymemoney** · March 5th 2011, 07:45 PM

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**Paymemoney** · March 5th 2011, 08:00 PM

Originally Posted by Opalg

So in those Fourier series, the cosine sum should have the summation $\sum_{n=0}^\infty\,,$ and the sine sum should have the summation $\sum_{n=1}^\infty$ .

Is this always the case or are there times were this does not apply?

Ok, theres another question i have done and i don't understand why the answer i got for $b_n$ when n is odd is
$\frac{-1}{\pi} \sum_{n=1}^\infty \frac{1}{2n} sin(2n\pi t)$

whereas my answer is:
$\frac{-1}{\pi} \sum_{n=1}^\infty \frac{1}{n} sin(n\pi t)$

The question is

$y = \left\{{-1 ---> {\text -1 \leq t < 0}\atop<br /> {t -------->{\text-1 \leq t < 0}\atop<br /> {f(t+2) ------> {\text {for all t}}}\right$

P.S

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