Question Number 140141 by mnjuly1970 last updated on 04/May/21
$$ \\ $$$$\:\:\:\:\:\:\:\:\:……{advanced}\:\:{calculus}…… \\ $$$$\:\:\:\:{when}\:\:\:\mid{z}\mid<\mathrm{1}\:{and}:: \\ $$$$\:\Omega:=\frac{{sin}\left({x}\right)}{{z}^{\mathrm{2}} +\mathrm{2}{z}\:{cos}\left({x}\right)+\mathrm{1}}\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} {z}^{{n}} \\ $$$$\:{are}\:{satisfied}\:,\:{then}\:{solve}\:,\:\:{a}_{{n}} \:… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:……………… \\ $$$$ \\ $$
Answered by qaz last updated on 04/May/21
![Ω=((e^(ix) −e^(−ix) )/(2i[z^2 +z(e^(ix) +e^(−ix) )+1])) =((e^(ix) −e^(−ix) )/(2i(ze^(ix) +1)(ze^(−ix) +1))) =(1/(2iz))[(1/(ze^(−ix) +1))−(1/(ze^(ix) +1))] =(1/(2iz)){Σ_(n=0) ^∞ (−1)^n z^n (e^(−inx) −e^(inx) )} =Σ_(n=0) ^∞ (−1)^(n+1) z^(n−1) sin (nx) ⇒a_n =(((−1)^(n+1) )/z)sin (nx)](https://www.tinkutara.com/question/Q140158.png)
$$\Omega=\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}\left[{z}^{\mathrm{2}} +{z}\left({e}^{{ix}} +{e}^{−{ix}} \right)+\mathrm{1}\right]} \\ $$$$=\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}\left({ze}^{{ix}} +\mathrm{1}\right)\left({ze}^{−{ix}} +\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{iz}}\left[\frac{\mathrm{1}}{{ze}^{−{ix}} +\mathrm{1}}−\frac{\mathrm{1}}{{ze}^{{ix}} +\mathrm{1}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{iz}}\left\{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {z}^{{n}} \left({e}^{−{inx}} −{e}^{{inx}} \right)\right\} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {z}^{{n}−\mathrm{1}} \mathrm{sin}\:\left({nx}\right) \\ $$$$\Rightarrow{a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{z}}\mathrm{sin}\:\left({nx}\right) \\ $$
Commented by mnjuly1970 last updated on 04/May/21
$${nice}\:{very}\:{nice}… \\ $$