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I need to find and plot the fourier series of $\sin^{2}(x)$. I know that the Fourier Series for this function is clearly $\frac{1}{2} - \frac{1}{2} \cos(2x)$ which is the reduction formula for $\sin^2(x)$. but now how do i find the first, 5, 10 ... terms of the partial some and plot them?

Thanks in advance

Thomas Russell
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guthik
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    well well ... you must be very happy, because there are no such terms :) – S L Apr 25 '14 at 11:53
  • seriously?? so the first 10, 100, 500 ... terms of the partial sum are all zero O.O ? i mean. what would be the answer to that. i got pretty confused :D – guthik Apr 25 '14 at 12:08
  • may be i forgot to add something, the interval is [0, pi] – guthik Apr 25 '14 at 12:09
  • yep!! only two terms exist. $a_0$ and $a_2$ – S L Apr 25 '14 at 12:09
  • hold on ... wait!! – S L Apr 25 '14 at 12:11
  • something missing, right? – guthik Apr 25 '14 at 12:14
  • actually i don't get it. because the series is supoosed to converge more and more to the actual function, i think. so not having those terms... what exactly does it mean – guthik Apr 25 '14 at 16:18
  • on further consideration, i don't think $b_n$ = 0. clarify more pleaseee. – guthik Apr 25 '14 at 19:27
  • ... it converges, there is no sine term here, therefore $b_n = 0$ – S L Apr 26 '14 at 00:13
  • i did the integral, but i only got $b_n = 0 $ iff n = 2. otherwise i have terms present. precisely, these: $ 2cos(\pi n) \over {n^3 - 4n} $ for n != 2. on $(0, \pi)$. or is this wrong? – guthik Apr 26 '14 at 09:39
  • could you show your work?? I would like to comment on it. I think I should have been a little careful when you changed the interval :( – S L Apr 26 '14 at 16:01
  • yeah sure. why not? \begin{align} \int_{0}^{\pi}(sin(x))^2 sin(nx),dx, = 1\2\int_{0}^{\pi} (1-cos(2x))sin(nx),dx, \ \end{align} – guthik Apr 27 '14 at 11:01
  • then follows

    $1\2{ \int_{0}^{\pi}sin(nx),dx - $ $\int_{0}^{\pi}sin(nx)cos(2x),dx} $

     – guthik Apr 27 '14 at 11:10
  • After that, the result above is got. i don't see any zeros popping up, am afraid :(( . so, how can i plot these things. i really want to see the changes in graph due to the number of terms. can u help me with the overall thing? thanks :) – guthik Apr 27 '14 at 11:14
  • hmm ... I think you need, $$ \frac{2}{\pi} \int_0^\pi \sin^2(x) { \sin(2nx) , \mathrm{or}, \cos(2nx) } , dx$$ from the definition that the Fourier coefficients in the interval L. – S L Apr 27 '14 at 17:17
  • Check this out – S L Apr 27 '14 at 17:20
  • yeah, i left the $\frac{2}{\pi}$ out intentionally since it doesn't bring out the zero no matter what. why is it $sin(2nx)$ ? – guthik Apr 27 '14 at 18:24
  • No ... I also meant the inner $(2nx)$, that will give you, that fourier expansion has only 2 terms, $\frac 1 2 \cos(2nx)$ will be first term. – S L Apr 28 '14 at 05:24
  • now, i think am more confused :). so what is the final fourier expansion? – guthik Apr 28 '14 at 08:58
  • same lol :) ${{{{{}}}}}$ – S L Apr 28 '14 at 20:16
  • $1\over 2$ - $ 1\over 2 $ $cos(2x)$ right? – guthik Apr 29 '14 at 11:39
  • yep that's right ${{{}}}$ – S L Apr 29 '14 at 12:43
  • hey, i finally understand everything :)). thanks alot. it was me confusing myself here :D – guthik Apr 30 '14 at 13:00
  • you are welcome :) – S L Apr 30 '14 at 19:56
  • hey, can you put this as an answer so i can accept it? it was corredt after all and i just learnt how to approve correct answers :). – guthik May 25 '14 at 21:05
  • sure if you insist :D – S L May 26 '14 at 00:51

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The fourier expansion of $\sin^2(x)$ is $\frac 1 2 - \frac 1 2 \cos (2x)$ :D

S L
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