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Question Number 33843 by prof Abdo imad last updated on 25/Apr/18
decompose inside R(x) the fraction F(x)= (1/((x+3)^n (x+1))) with n integr .
$${decompose}\:{inside}\:{R}\left({x}\right)\:{the}\:{fraction} \\ $$$${F}\left({x}\right)=\:\:\frac{\mathrm{1}}{\left({x}+\mathrm{3}\right)^{{n}} \:\left({x}+\mathrm{1}\right)}\:{with}\:{n}\:{integr}\:. \\ $$
Commented by abdo imad last updated on 28/Apr/18
let use x+3=t ⇒F(x)=g(t)= (1/(t^n (t−2))) let find D_(n−1) (0)for g(t)= (1/(t−2)) we have g(t)=Σ_(k=0) ^(n−1) ((g^((k)) (0))/(k!))t^k +(t^n /(n!))ξ(t) with ξ(t)_(t→0) →0 but g^((k)) (t)= (((−1)^k k!)/((t−2)^(k+1) )) ⇒g^((k)) (0) =(((−1)^k k!)/((−1)^(k+1) 2^(k+1) )) =−((k!)/2^(k+1) ) ⇒g(t)=Σ_(k=0) ^(n−1) ( −(t^k /2^(k+1) ) ) +(t^n /(n!))ξ(t) g(t) = Σ_(k=1) ^(n−1) (λ_k /t^k ) +(λ_n /t^n ) + (a/(t−2)) a=lim_(t→2) (g−2)g(t) =(1/2^n ) λ_n =lim_(t→0) t^n g(t) =−(1/2) ⇒ g(t) =Σ_(k=1) ^n (λ_k /t^k ) −(1/(2t^n )) +(1/(2^n (t−2))) from another side g(t)=−(1/t^n )Σ_(k=0) ^(n−1) (t^k /2^(k+1) ) +(1/(n!)) ξ(t) =−Σ_(k=0) ^(n−1) (1/(t^(n−k) 2^(k+1) )) +(1/(n!))ξ(t) changement of indice n−k =p give g(t)=−Σ_(p=1) ^(n−1) (1/(t^p 2^(n−p +1) )) +(1/(n!)) ξ(t) =Σ_(k=1) ^(n−1) −(1/(2^(n−k +1) t^k )) +(1/(n!))ξ(t) ⇒λ_k =((−1)/2^(n−k+1) ) ⇒ g(t)=Σ_(k=1) ^(n−1) ((−1)/(2^(n−k+1) t^k )) −(1/(2t^n )) + (1/(2^n (t−2))) ⇒ F(x)=Σ_(k=1) ^(n−1) ((−1)/(2^(n−k+1) (x+3))) −(1/(2(x+3)^n )) +(1/(2^n (x+1))) .
$${let}\:{use}\:{x}+\mathrm{3}={t}\:\Rightarrow{F}\left({x}\right)={g}\left({t}\right)=\:\frac{\mathrm{1}}{{t}^{{n}} \left({t}−\mathrm{2}\right)}\:{let}\:{find}\:{D}_{{n}−\mathrm{1}} \left(\mathrm{0}\right){for} \\ $$$${g}\left({t}\right)=\:\frac{\mathrm{1}}{{t}−\mathrm{2}}\:{we}\:{have}\:{g}\left({t}\right)=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{{g}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}{t}^{{k}} \:+\frac{{t}^{{n}} }{{n}!}\xi\left({t}\right)\:{with} \\ $$$$\xi\left({t}\right)_{{t}\rightarrow\mathrm{0}} \rightarrow\mathrm{0}\:\:\:\:{but}\:\:{g}^{\left({k}\right)} \left({t}\right)=\:\frac{\left(−\mathrm{1}\right)^{{k}} {k}!}{\left({t}−\mathrm{2}\right)^{{k}+\mathrm{1}} }\:\Rightarrow{g}^{\left({k}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{k}} {k}!}{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \mathrm{2}^{{k}+\mathrm{1}} } \\ $$$$=−\frac{{k}!}{\mathrm{2}^{{k}+\mathrm{1}} }\:\Rightarrow{g}\left({t}\right)=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\:−\frac{{t}^{{k}} }{\mathrm{2}^{{k}+\mathrm{1}} }\:\right)\:+\frac{{t}^{{n}} }{{n}!}\xi\left({t}\right) \\ $$$${g}\left({t}\right)\:=\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\lambda_{{k}} }{{t}^{{k}} }\:\:+\frac{\lambda_{{n}} }{{t}^{{n}} }\:+\:\frac{{a}}{{t}−\mathrm{2}} \\ $$$${a}={lim}_{{t}\rightarrow\mathrm{2}} \left({g}−\mathrm{2}\right){g}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$$\lambda_{{n}} ={lim}_{{t}\rightarrow\mathrm{0}} \:{t}^{{n}} {g}\left({t}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{g}\left({t}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\lambda_{{k}} }{{t}^{{k}} }\:−\frac{\mathrm{1}}{\mathrm{2}{t}^{{n}} }\:+\frac{\mathrm{1}}{\mathrm{2}^{{n}} \left({t}−\mathrm{2}\right)} \\ $$$${from}\:{another}\:{side}\:{g}\left({t}\right)=−\frac{\mathrm{1}}{{t}^{{n}} }\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\frac{{t}^{{k}} }{\mathrm{2}^{{k}+\mathrm{1}} }\:+\frac{\mathrm{1}}{{n}!}\:\xi\left({t}\right) \\ $$$$=−\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{{t}^{{n}−{k}} \:\mathrm{2}^{{k}+\mathrm{1}} }\:+\frac{\mathrm{1}}{{n}!}\xi\left({t}\right)\:{changement}\:{of}\:{indice}\: \\ $$$${n}−{k}\:={p}\:{give}\:{g}\left({t}\right)=−\sum_{{p}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\:\frac{\mathrm{1}}{{t}^{{p}} \:\mathrm{2}^{{n}−{p}\:+\mathrm{1}} }\:+\frac{\mathrm{1}}{{n}!}\:\xi\left({t}\right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}^{{n}−{k}\:+\mathrm{1}} \:{t}^{{k}} }\:+\frac{\mathrm{1}}{{n}!}\xi\left({t}\right)\:\Rightarrow\lambda_{{k}} =\frac{−\mathrm{1}}{\mathrm{2}^{{n}−{k}+\mathrm{1}} }\:\Rightarrow \\ $$$${g}\left({t}\right)=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\:\frac{−\mathrm{1}}{\mathrm{2}^{{n}−{k}+\mathrm{1}} \:{t}^{{k}} }\:−\frac{\mathrm{1}}{\mathrm{2}{t}^{{n}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} \left({t}−\mathrm{2}\right)}\:\Rightarrow \\ $$$${F}\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{−\mathrm{1}}{\mathrm{2}^{{n}−{k}+\mathrm{1}} \left({x}+\mathrm{3}\right)}\:−\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{3}\right)^{{n}} }\:+\frac{\mathrm{1}}{\mathrm{2}^{{n}} \left({x}+\mathrm{1}\right)}\:. \\ $$$$ \\ $$
Commented by abdo imad last updated on 28/Apr/18
★F(x)= Σ_(k=1) ^(n−1) ((−1)/(2^(n−k+1) (x+3)^k )) −(1/(2(x+3)^n )) + (1/(2^n (x+1))) .★
$$\bigstar{F}\left({x}\right)=\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{−\mathrm{1}}{\mathrm{2}^{{n}−{k}+\mathrm{1}} \left({x}+\mathrm{3}\right)^{{k}} }\:−\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{3}\right)^{{n}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} \left({x}+\mathrm{1}\right)}\:.\bigstar \\ $$

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