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Calculate the Sine Fourier series expansion for $\displaystyle f(t) = t^2 $ in $\displaystyle 0 < t < 2.$

I know I need to use $\displaystyle ∑ B_n \sin\left(\frac{nπt}{2}\right)$

I know the answer for $\displaystyle B_n$ is $\displaystyle -\frac{8n}{nπ}$ for even $\displaystyle n$ and $\displaystyle \frac{8}{nπ}-\frac{32}{n^3π^3}$ for odd $\displaystyle n$, but I have no clue how to get there. Thanks for the help.

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  • $\begingroup$ My apologies for the messy formatting. I am working on improving it right now - I'm new to this coding, sorry! $\endgroup$
    – Outlaw94
    Apr 24, 2013 at 3:13
  • $\begingroup$ have you studied fourier series before? also, are you familiar with the concept of orthogonal functions? $\endgroup$
    – DanZimm
    Apr 24, 2013 at 3:15
  • $\begingroup$ I've just started to learn fourier series, and orthogonal functions doesn't ring a bell - have I missed something? I know that I need to do something different for a period of 2L, and I know I need to use an equation I'm going to edit into the problem right now. $\endgroup$
    – Outlaw94
    Apr 24, 2013 at 3:22
  • $\begingroup$ @Outlaw94: I edited your post. Make sure that's what you mean to ask. $\endgroup$ Apr 24, 2013 at 3:52
  • $\begingroup$ For odd $n$, it should be $ \frac{8}{nπ}-\frac{32}{n^3π^3} $. $\endgroup$ Apr 24, 2013 at 4:01

1 Answer 1

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Here is a general term for $B_n$

$$ B_n = 8\,{\frac { \left( -1 \right) ^{n+1}{n}^{2}{\pi }^{2}+2\, \left( -1 \right) ^{n}-2}{{n}^{3}{\pi }^{3}}}.$$

Now, the Fourier series is given by

$$ t^2 = 8\sum_{n=1}^{\infty}\,{\frac { \left( -1 \right) ^{n+1}{n}^{2}{\pi }^{2}+2\, \left( -1 \right) ^{n}-2}{{n}^{3}{\pi }^{3}}} \sin\left( \frac{n\pi t}{2}\right). $$

You can split the above series for even and odd $n$ if you want.

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  • $\begingroup$ Ah, I see where I made my mistake when I attempted the problem on my own. Thank you for your assistance! $\endgroup$
    – Outlaw94
    Apr 24, 2013 at 7:33
  • $\begingroup$ @Outlaw94: You are welcome. $\endgroup$ Apr 24, 2013 at 7:35
  • $\begingroup$ Thanks Dr. for that remarking point on my aswer.+1 $\endgroup$
    – Mikasa
    Apr 24, 2013 at 8:02
  • $\begingroup$ @BabakS.: You are very welcome. $\endgroup$ Apr 24, 2013 at 10:44

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