answer to question25980 key of slution we develop the foction f(x) = sin(px) at fourier serie((f 2π periodic) f(x)= Σ_(n=1) ^(n=∝) a_n sin(nx) and a_n = 2/T ∫_([T]) sin(px)sin(nx)dx (T=2π) a_n =2π^(−1) ∫_0 ^π sin(px)sin(nx)dx−>a_n = (−1)^n sin(pπ).2n π^(−1) (n^2 − p^2 )^(−1) −−>sin(px)= 2 sin(pπ).π^(−1) Σ_(n=1) ^∝ n(−1)^(n−1) (n^2 −p^2 )^(−1) sin(nx) = 2π^(−1) sin(pπ)((1^2 −p^2 )^(−1) sin (x) −2(2^2 −p^2 )^(−1) sin(2x)+...)


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Question Number 26023 by abdo imad last updated on 17/Dec/17

$a n s w e r t o q u e s t i o n 25980 k e y o f s l u t i o n w e d e v e l o p t h e$ $f o c t i o n f (x) = s i n (p x) a t f o u r i e r s e r i e ((f 2 π p e r i o d i c)$ $f (x) = \sum_{n = 1}^{n =\propto} a_{n} s i n (n x) a n d a_{n} = 2 / T \int_{[T]} s i n (p x) s i n (n x) d x (T = 2 π)$ $a_{n} = 2 π^{- 1} \int_{0}^{π} s i n (p x) s i n (n x) d x - > a_{n} = {(- 1)}^{n} s i n (p π) . 2 n π^{- 1} {(n^{2} - p^{2})}^{- 1}$ $- - > s i n (p x) = 2 s i n (p π) . π^{- 1} \sum_{n = 1}^{\propto} n {(- 1)}^{n - 1} {(n^{2} - p^{2})}^{- 1} s i n (n x)$ $= 2 π^{- 1} s i n (p π) ({(1^{2} - p^{2})}^{- 1} s i n (x) - 2 {(2^{2} - p^{2})}^{- 1} s i n (2 x) + . . .)$
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