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find the half range sine expansion Fourier series of the following function f(t)=t(π-t) 0≤t≤π find bn enter image description here

user82922
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    Please clarify the question, and show what you have tried so far. What do you mean by "half range sine of Fourier series"? – Mårten W Jul 01 '13 at 11:28
  • i mean find bn=2/p.integral(f(t).sin.n.pi.t/p) where the fourier series consists of sine and cosine range – user82922 Jul 01 '13 at 18:17

1 Answers1

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The FS for that range is

$$f(t) = \sum_{k=1}^{\infty} a_k \sin{k t}$$

where

$$a_k = \frac{\displaystyle \int_0^{\pi} dt\, f(t) \, \sin{k t}}{\displaystyle\int_0^{\pi} dt\, \sin^2{k t}} = \frac{2}{\pi} \int_0^{\pi} dt\, f(t) \, \sin{k t}$$

which, for $f(t) = t (\pi-t)$, is

$$\begin{align}a_k &= \frac{2}{\pi} \int_0^{\pi} dt\, t (\pi-t) \sin{k t}\\ &=2 \int_0^{\pi} dt\, t \sin{k t} - \frac{2}{\pi} \int_0^{\pi} dt\, t^2 \sin{k t} \\ &= 2 \pi \frac{(-1)^{k+1}}{k} + \frac{4 [1-(-1)^k)]}{\pi k^3} - 2 \pi \frac{(-1)^{k+1}}{k}\\ &= \frac{4 [1-(-1)^k)]}{\pi k^3}\end{align}$$

The integrals may be evaluated using integration by parts. Note that $a_k=0$ when $k$ is even. Therefore,

$$t (\pi-t) = \frac{8}{\pi} \sum_{k=0}^{\infty} \frac{\sin{(2 k+1)t}}{(2k+1)^3}$$

Ron Gordon
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