Menu Close

decompose-inside-R-x-the-fraction-F-x-1-x-3-6-x-2-

Tinku Tara 0 Comments
Pin




Question Number 34614 by math khazana by abdo last updated on 09/May/18
decompose inside R(x) the fraction F(x)= (1/((x−3)^6 (x+2))) .
$${decompose}\:{inside}\:{R}\left({x}\right)\:{the}\:{fraction} \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{6}} \left({x}+\mathrm{2}\right)}\:. \\ $$
Commented by math khazana by abdo last updated on 09/May/18
changement x−3=t give F(x)=g(t)= (1/(t^6 (t+5))) let find D_5 (0) for f(t)= (1/(t+5)) ⇒f(t)=Σ_(k=0) ^5 ((f^((k)) (0))/(k!)) t^k +(t^6 /(6!))ξ(t) with ξ(t)_(t→0) →0 we have f^((k)) (t)= (((−1)^k k!)/((t+5)^(k+1) )) ⇒ f^((k)) (0) = (((−1)^k k!)/5^(k+1) ) ⇒f(t)= Σ_(k=0) ^5 (((−1)^k )/5^(k+1) ) t^k +(t^6 /(6!))ξ(t) ⇒g(t) = Σ_(k=0) ^5 (((−1)^k )/(5^(k+1) t^(6−k) )) +(1/(6!))ξ(t) = _(6−k =p) Σ_(p=1) ^6 (((−1)^(6−p) )/(5^(7−p) t^p )) +(1/(6!))ξ(t) from another side g(t) =Σ_(p=1) ^6 (λ_p /t^p ) +(a/(t+5)) ⇒ λ_p = (((−1)^(6−p) )/5^(7−p) ) a =lim_(t→−5) (t+5)g(t) = (1/((−5)^6 )) ⇒ g(t) = Σ_(p=1) ^6 (((−1)^(6−p) )/(5^(7−p) t^p )) + (1/(5^6 (t+5))) ⇒ F(x)= Σ_(p=1) ^6 (((−1)^(6−p) )/(5^(7−p) (x−3)^p )) + (1/(5^6 (x+2))) .
$${changement}\:{x}−\mathrm{3}={t}\:{give} \\ $$$${F}\left({x}\right)={g}\left({t}\right)=\:\:\frac{\mathrm{1}}{{t}^{\mathrm{6}} \left({t}+\mathrm{5}\right)}\:\:{let}\:{find}\:{D}_{\mathrm{5}} \left(\mathrm{0}\right)\:{for} \\ $$$${f}\left({t}\right)=\:\frac{\mathrm{1}}{{t}+\mathrm{5}}\:\Rightarrow{f}\left({t}\right)=\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:\:\frac{{f}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}\:{t}^{{k}} \:\:+\frac{{t}^{\mathrm{6}} }{\mathrm{6}!}\xi\left({t}\right)\:{with} \\ $$$$\xi\left({t}\right)_{{t}\rightarrow\mathrm{0}} \rightarrow\mathrm{0}\:\:\:{we}\:{have}\:{f}^{\left({k}\right)} \left({t}\right)=\:\frac{\left(−\mathrm{1}\right)^{{k}} \:{k}!}{\left({t}+\mathrm{5}\right)^{{k}+\mathrm{1}} }\:\Rightarrow \\ $$$${f}^{\left({k}\right)} \left(\mathrm{0}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{k}} \:{k}!}{\mathrm{5}^{{k}+\mathrm{1}} }\:\Rightarrow{f}\left({t}\right)=\:\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{5}^{{k}+\mathrm{1}} }\:{t}^{{k}} \:\:+\frac{{t}^{\mathrm{6}} }{\mathrm{6}!}\xi\left({t}\right) \\ $$$$\Rightarrow{g}\left({t}\right)\:=\:\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{5}^{{k}+\mathrm{1}} \:{t}^{\mathrm{6}−{k}} }\:\:+\frac{\mathrm{1}}{\mathrm{6}!}\xi\left({t}\right) \\ $$$$=\:_{\mathrm{6}−{k}\:={p}} \:\sum_{{p}=\mathrm{1}} ^{\mathrm{6}} \:\:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{6}−{p}} }{\mathrm{5}^{\mathrm{7}−{p}} \:{t}^{{p}} }\:\:+\frac{\mathrm{1}}{\mathrm{6}!}\xi\left({t}\right)\:\:{from}\:{another} \\ $$$${side}\:\:{g}\left({t}\right)\:=\sum_{{p}=\mathrm{1}} ^{\mathrm{6}} \:\:\frac{\lambda_{{p}} }{{t}^{{p}} }\:\:+\frac{{a}}{{t}+\mathrm{5}}\:\Rightarrow\:\lambda_{{p}} =\:\frac{\left(−\mathrm{1}\right)^{\mathrm{6}−{p}} }{\mathrm{5}^{\mathrm{7}−{p}} } \\ $$$${a}\:={lim}_{{t}\rightarrow−\mathrm{5}} \left({t}+\mathrm{5}\right){g}\left({t}\right)\:=\:\frac{\mathrm{1}}{\left(−\mathrm{5}\right)^{\mathrm{6}} }\:\Rightarrow \\ $$$${g}\left({t}\right)\:=\:\sum_{{p}=\mathrm{1}} ^{\mathrm{6}} \:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{6}−{p}} }{\mathrm{5}^{\mathrm{7}−{p}} \:{t}^{{p}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{6}} \left({t}+\mathrm{5}\right)}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\sum_{{p}=\mathrm{1}} ^{\mathrm{6}} \:\:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{6}−\boldsymbol{{p}}} }{\mathrm{5}^{\mathrm{7}−\boldsymbol{{p}}} \:\left(\boldsymbol{{x}}−\mathrm{3}\right)^{\boldsymbol{{p}}} }\:\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{6}} \left(\boldsymbol{{x}}+\mathrm{2}\right)}\:\:. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *