Question Number 26023 by abdo imad last updated on 17/Dec/17 | ||
$${answer}\:{to}\:{question}\mathrm{25980}\:{key}\:{of}\:{slution}\:{we}\:{develop}\:\:{the} \\ $$$${foction}\:{f}\left({x}\right)\:=\:{sin}\left({px}\right)\:{at}\:{fourier}\:{serie}\left(\left({f}\:\mathrm{2}\pi\:{periodic}\right)\right. \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \:{a}_{{n}} {sin}\left({nx}\right)\:{and}\:\:{a}_{{n}} =\:\mathrm{2}/{T}\:\int_{\left[{T}\right]} {sin}\left({px}\right){sin}\left({nx}\right){dx}\:\:\:\left({T}=\mathrm{2}\pi\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${a}_{{n}} =\mathrm{2}\pi^{−\mathrm{1}} \:\int_{\mathrm{0}} ^{\pi} {sin}\left({px}\right){sin}\left({nx}\right){dx}−>{a}_{{n}} =\:\left(−\mathrm{1}\right)^{{n}} \:{sin}\left({p}\pi\right).\mathrm{2}{n}\:\pi^{−\mathrm{1}} \left({n}^{\mathrm{2}} \:−\:{p}^{\mathrm{2}} \right)^{−\mathrm{1}} \\ $$$$−−>{sin}\left({px}\right)=\:\mathrm{2}\:{sin}\left({p}\pi\right).\pi^{−\mathrm{1}} \:\sum_{{n}=\mathrm{1}} ^{\propto} \:{n}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{−\mathrm{1}} \:\:{sin}\left({nx}\right) \\ $$$$=\:\mathrm{2}\pi^{−\mathrm{1}} \:{sin}\left({p}\pi\right)\left(\left(\mathrm{1}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{−\mathrm{1}} \:{sin}\:\left({x}\right)\:−\mathrm{2}\left(\mathrm{2}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{−\mathrm{1}} {sin}\left(\mathrm{2}{x}\right)+...\right) \\ $$ | ||