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Question Number 37354 by math khazana by abdo last updated on 12/Jun/18
let f(z) = ((z^2 +1)/((z^2 −1)(z^2 −4))) developp f at integr serie .
$${let}\:{f}\left({z}\right)\:=\:\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left({z}^{\mathrm{2}} \:−\mathrm{1}\right)\left({z}^{\mathrm{2}} \:−\mathrm{4}\right)} \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$
Commented by math khazana by abdo last updated on 12/Jun/18
we have f(z)= ((z^2 −1 +2)/((z^2 −1)(z^2 −4))) = (1/(z^2 −4)) +(2/((z^2 −1)(z^2 −4))) =(1/(z^2 −4)) −(2/3){ (1/(z^2 −1)) −(1/(z^2 −4))} = (1/(z^2 −4)) +(2/(3(z^2 −4))) +(2/3) (1/(1−z^2 )) =(5/3) (1/(z^2 −4)) +(2/3) (1/(1−z^2 )) = ((−5)/(12)) (1/(1−((z/2))^2 )) +(2/3) (1/(1−z^2 )) so if ∣z∣<1 we get f(z)= −(5/(12)) Σ_(n=0) ^∞ ((z/2))^(2n) +(2/3)Σ_(n=0) ^∞ z^(2n) =−(5/(12)) Σ_(n=0) ^∞ (z^(2n) /4^n ) +(2/3)Σ_(n=0) ^∞ z^(2n) .
$${we}\:{have}\:{f}\left({z}\right)=\:\:\frac{{z}^{\mathrm{2}} \:−\mathrm{1}\:+\mathrm{2}}{\left({z}^{\mathrm{2}} −\mathrm{1}\right)\left({z}^{\mathrm{2}} −\mathrm{4}\right)} \\ $$$$=\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{4}}\:\:+\frac{\mathrm{2}}{\left({z}^{\mathrm{2}} −\mathrm{1}\right)\left({z}^{\mathrm{2}} −\mathrm{4}\right)} \\ $$$$=\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{4}}\:−\frac{\mathrm{2}}{\mathrm{3}}\left\{\:\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{1}}\:−\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{4}}\right\} \\ $$$$=\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{4}}\:\:+\frac{\mathrm{2}}{\mathrm{3}\left({z}^{\mathrm{2}} −\mathrm{4}\right)}\:\:+\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\mathrm{1}}{\mathrm{1}−{z}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{5}}{\mathrm{3}}\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{4}}\:\:+\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\mathrm{1}}{\mathrm{1}−{z}^{\mathrm{2}} } \\ $$$$=\:\frac{−\mathrm{5}}{\mathrm{12}}\:\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}} }\:+\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\mathrm{1}}{\mathrm{1}−{z}^{\mathrm{2}} }\:{so}\:{if}\:\mid{z}\mid<\mathrm{1}\:{we}\:{get} \\ $$$${f}\left({z}\right)=\:−\frac{\mathrm{5}}{\mathrm{12}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{n}} \:\:+\frac{\mathrm{2}}{\mathrm{3}}\sum_{{n}=\mathrm{0}} ^{\infty} \:{z}^{\mathrm{2}{n}} \\ $$$$=−\frac{\mathrm{5}}{\mathrm{12}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{z}^{\mathrm{2}{n}} }{\mathrm{4}^{{n}} }\:+\frac{\mathrm{2}}{\mathrm{3}}\sum_{{n}=\mathrm{0}} ^{\infty} \:{z}^{\mathrm{2}{n}} \:\:. \\ $$

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