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

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)+...)

$${answer}\:{to}\:{question}\mathrm{25980}\:{key}\:{of}\:{slution}\:{we}\:{develop}\:\:{the} \\ $$$${foction}\:{f}\left({x}\right)\:=\:{sin}\left({px}\right)\:{at}\:{fourier}\:{serie}\left(\left({f}\:\mathrm{2}\pi\:{periodic}\right)\right. \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \:{a}_{{n}} {sin}\left({nx}\right)\:{and}\:\:{a}_{{n}} =\:\mathrm{2}/{T}\:\int_{\left[{T}\right]} {sin}\left({px}\right){sin}\left({nx}\right){dx}\:\:\:\left({T}=\mathrm{2}\pi\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${a}_{{n}} =\mathrm{2}\pi^{−\mathrm{1}} \:\int_{\mathrm{0}} ^{\pi} {sin}\left({px}\right){sin}\left({nx}\right){dx}−>{a}_{{n}} =\:\left(−\mathrm{1}\right)^{{n}} \:{sin}\left({p}\pi\right).\mathrm{2}{n}\:\pi^{−\mathrm{1}} \left({n}^{\mathrm{2}} \:−\:{p}^{\mathrm{2}} \right)^{−\mathrm{1}} \\ $$$$−−>{sin}\left({px}\right)=\:\mathrm{2}\:{sin}\left({p}\pi\right).\pi^{−\mathrm{1}} \:\sum_{{n}=\mathrm{1}} ^{\propto} \:{n}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{−\mathrm{1}} \:\:{sin}\left({nx}\right) \\ $$$$=\:\mathrm{2}\pi^{−\mathrm{1}} \:{sin}\left({p}\pi\right)\left(\left(\mathrm{1}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{−\mathrm{1}} \:{sin}\:\left({x}\right)\:−\mathrm{2}\left(\mathrm{2}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{−\mathrm{1}} {sin}\left(\mathrm{2}{x}\right)+...\right) \\ $$

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