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Question Number 39038 by maxmathsup by imad last updated on 01/Jul/18
let f(z) = (z/(z^2 −z+2)) developp f at integr serie.
$${let}\:{f}\left({z}\right)\:=\:\frac{{z}}{{z}^{\mathrm{2}} \:−{z}+\mathrm{2}} \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$
Commented by prof Abdo imad last updated on 02/Jul/18
let decompose f(x) ondide C(x) Δ=1−4(2)=−7=(i(√7))^2 ⇒ z_1 =((1+i(√7))/2) and z_2 =((1−i(√7))/2) ⇒ f(z) = (z/((z−z_1 )(z−z_2 ))) = (a/(z−z_1 )) +(b/(z−z_2 )) a= (z_1 /(z_1 −z_2 )) = (z_1 /(i(√7))) =((−iz_1 )/( (√7))) b = (z_2 /(z_2 −z_1 )) = (z_2 /(−i(√7))) = ((iz_2 )/( (√7))) ⇒ f(x)= ((−iz_1 )/( (√7)(z−z_1 ))) +((iz_2 )/( (√7)(z−z_2 ))) ⇒ f^((n)) (x) =((−iz_1 )/( (√7))){ (((−1)^n n!)/((z−z_1 )^(n+1) ))} +((iz_2 )/( (√7))){ (((−1)^n n!)/((z−z_2 )^(n+1) ))} = (((−1)^n n!)/( (√7))){ (z_2 /((z−z_2 )^(n+1) )) −(z_1 /((z−z_1 )^(n+1) ))} ⇒ f^((n)) (0)=(((−1)^n n!)/( (√7))){ (z_2 /((−1)^(n+1) z_2 ^(n+1) ))−(z_1 /((−1)^(n+1) z_1 ^(n+1) ))} = ((n!)/( (√7))){ ((−z_2 )/z_2 ^(n+1) ) +(z_1 /z_1 ^(n+1) )} finally f(z) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) z^n =Σ_(n=0) ^∞ (1/( (√7)))( (1/z_1 ^n ) −(1/z_2 ^n ))z^n
$${let}\:{decompose}\:{f}\left({x}\right)\:{ondide}\:{C}\left({x}\right) \\ $$$$\Delta=\mathrm{1}−\mathrm{4}\left(\mathrm{2}\right)=−\mathrm{7}=\left({i}\sqrt{\mathrm{7}}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${z}_{\mathrm{1}} =\frac{\mathrm{1}+{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:{and}\:{z}_{\mathrm{2}} =\frac{\mathrm{1}−{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:\Rightarrow \\ $$$${f}\left({z}\right)\:=\:\frac{{z}}{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:=\:\frac{{a}}{{z}−{z}_{\mathrm{1}} }\:+\frac{{b}}{{z}−{z}_{\mathrm{2}} } \\ $$$${a}=\:\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }\:=\:\frac{{z}_{\mathrm{1}} }{{i}\sqrt{\mathrm{7}}}\:=\frac{−{iz}_{\mathrm{1}} }{\:\sqrt{\mathrm{7}}} \\ $$$${b}\:=\:\frac{{z}_{\mathrm{2}} }{{z}_{\mathrm{2}} −{z}_{\mathrm{1}} }\:=\:\frac{{z}_{\mathrm{2}} }{−{i}\sqrt{\mathrm{7}}}\:=\:\frac{{iz}_{\mathrm{2}} }{\:\sqrt{\mathrm{7}}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\:\frac{−{iz}_{\mathrm{1}} }{\:\sqrt{\mathrm{7}}\left({z}−{z}_{\mathrm{1}} \right)}\:+\frac{{iz}_{\mathrm{2}} }{\:\sqrt{\mathrm{7}}\left({z}−{z}_{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\frac{−{iz}_{\mathrm{1}} }{\:\sqrt{\mathrm{7}}}\left\{\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({z}−{z}_{\mathrm{1}} \right)^{{n}+\mathrm{1}} }\right\}\:+\frac{{iz}_{\mathrm{2}} }{\:\sqrt{\mathrm{7}}}\left\{\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({z}−{z}_{\mathrm{2}} \right)^{{n}+\mathrm{1}} }\right\} \\ $$$$=\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\:\sqrt{\mathrm{7}}}\left\{\:\:\frac{{z}_{\mathrm{2}} }{\left({z}−{z}_{\mathrm{2}} \right)^{{n}+\mathrm{1}} }\:−\frac{{z}_{\mathrm{1}} }{\left({z}−{z}_{\mathrm{1}} \right)^{{n}+\mathrm{1}} }\right\}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\:\sqrt{\mathrm{7}}}\left\{\:\frac{{z}_{\mathrm{2}} }{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{z}_{\mathrm{2}} ^{{n}+\mathrm{1}} }−\frac{{z}_{\mathrm{1}} }{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{z}_{\mathrm{1}} ^{{n}+\mathrm{1}} }\right\} \\ $$$$=\:\frac{{n}!}{\:\sqrt{\mathrm{7}}}\left\{\:\frac{−{z}_{\mathrm{2}} }{{z}_{\mathrm{2}} ^{{n}+\mathrm{1}} }\:+\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{1}} ^{{n}+\mathrm{1}} }\right\}\:\:{finally}\: \\ $$$${f}\left({z}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{z}^{{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\left(\:\frac{\mathrm{1}}{{z}_{\mathrm{1}} ^{{n}} }\:−\frac{\mathrm{1}}{{z}_{\mathrm{2}} ^{{n}} }\right){z}^{{n}} \\ $$
Commented by prof Abdo imad last updated on 02/Jul/18
let simplify (1/z_1 ^n ) −(1/z_2 ^n ) = ((z_2 ^n −z_1 ^n )/((z_1 z_2 )^n )) = (((((1−i(√7))/2))^n −( ((1+i(√7))/2))^n )/2) = −(1/2^(n+1) ) { (1+i(√7))^n −(1−i(√7))^n } =−(1/2^(n+1) ) { Σ_(k=0) ^n C_n ^k (i(√7))^k −Σ_(k=0) ^n C_n ^k (−i(√7))^k } =((−1)/2^(n+1) ) Σ_(k=0) ^n C_n ^k { (i(√7))^k −(−i(√7))^k } =((−1)/2^(n+1) ) Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) { 2i(−1)^p 7^p (√7) } =((−i)/2^n ) (√7)Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) (−7)^p ⇒ f(z)=Σ_(n=0) ^∞ ((−i)/2^n ){ Σ_(p=0) ^([((n−1)/2)]) (−7)^(p ) C_n ^(2p+1) }z^n
$${let}\:{simplify}\:\:\frac{\mathrm{1}}{{z}_{\mathrm{1}} ^{{n}} }\:−\frac{\mathrm{1}}{{z}_{\mathrm{2}} ^{{n}} } \\ $$$$=\:\frac{{z}_{\mathrm{2}} ^{{n}} \:−{z}_{\mathrm{1}} ^{{n}} }{\left({z}_{\mathrm{1}} {z}_{\mathrm{2}} \right)^{{n}} }\:\:=\:\frac{\left(\frac{\mathrm{1}−{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\right)^{{n}} \:−\left(\:\frac{\mathrm{1}+{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\right)^{{n}} }{\mathrm{2}} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\:\left\{\:\left(\mathrm{1}+{i}\sqrt{\mathrm{7}}\right)^{{n}} \:−\left(\mathrm{1}−{i}\sqrt{\mathrm{7}}\right)^{{n}} \right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\:\left\{\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({i}\sqrt{\mathrm{7}}\right)^{{k}} \:−\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(−{i}\sqrt{\mathrm{7}}\right)^{{k}} \right\} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\left\{\:\left({i}\sqrt{\mathrm{7}}\right)^{{k}} \:−\left(−{i}\sqrt{\mathrm{7}}\right)^{{k}} \right\} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:\left\{\:\mathrm{2}{i}\left(−\mathrm{1}\right)^{{p}} \:\mathrm{7}^{{p}} \sqrt{\mathrm{7}}\:\:\right\} \\ $$$$=\frac{−{i}}{\mathrm{2}^{{n}} }\:\sqrt{\mathrm{7}}\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:\left(−\mathrm{7}\right)^{{p}} \:\Rightarrow \\ $$$${f}\left({z}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{−{i}}{\mathrm{2}^{{n}} }\left\{\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\left(−\mathrm{7}\right)^{\boldsymbol{{p}}\:} \:\boldsymbol{{C}}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \right\}{z}^{{n}} \\ $$

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