Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 33843 by prof Abdo imad last updated on 25/Apr/18

$$decomposeinsideR\left(x\right)thefraction$$$$F\left(x\right)=\frac{1}{{(x+3)}^n(x+1)}withnintegr.$$

Commented by abdo imad last updated on 28/Apr/18

$$letusex+3=t\Rightarrow F\left(x\right)=g\left(t\right)=\frac{1}{t^n(t-2)}letfind{D}_{n-1}\left(0\right)for$$$$g\left(t\right)=\frac{1}{t-2}wehaveg\left(t\right)=\sum _{k=0}^{n-1}\frac{g^{\left(k\right)}\left(0\right)}{k!}{t}^k+\frac{t^n}{n!}\xi \left(t\right)with$$$$\xi {\left(t\right)}_{t\to 0}\to 0but{g}^{\left(k\right)}\left(t\right)=\frac{{(-1)}^{k}k!}{{(t-2)}^{k+1}}\Rightarrow g^{\left(k\right)}\left(0\right)=\frac{{(-1)}^{k}k!}{{(-1)}^{k+1}{2}^{k+1}}$$$$=-\frac{k!}{2^{k+1}}\Rightarrow g\left(t\right)=\sum _{k=0}^{n-1}(-\frac{t^k}{2^{k+1}})+\frac{t^n}{n!}\xi \left(t\right)$$$$g\left(t\right)=\sum _{k=1}^{n-1}\frac{{\lambda }_k}{t^k}+\frac{{\lambda }_n}{t^n}+\frac{a}{t-2}$$$$a={lim}_{t\to 2}(g-2)g\left(t\right)=\frac{1}{2^n}$$$${\lambda }_n={lim}_{t\to 0}{t}^{n}g\left(t\right)=-\frac{1}{2}\Rightarrow g\left(t\right)=\sum _{k=1}^n\frac{{\lambda }_k}{t^k}-\frac{1}{2t^n}+\frac{1}{2^n(t-2)}$$$$fromanothersideg\left(t\right)=-\frac{1}{t^n}\sum _{k=0}^{n-1}\frac{t^k}{2^{k+1}}+\frac{1}{n!}\xi \left(t\right)$$$$=-\sum _{k=0}^{n-1}\frac{1}{t^{n-k}{2}^{k+1}}+\frac{1}{n!}\xi \left(t\right)changementofindice$$$$n-k=pgiveg\left(t\right)=-\sum _{p=1}^{n-1}\frac{1}{t^p{2}^{n-p+1}}+\frac{1}{n!}\xi \left(t\right)$$$$=\sum _{k=1}^{n-1}-\frac{1}{2^{n-k+1}{t}^k}+\frac{1}{n!}\xi \left(t\right)\Rightarrow {\lambda }_k=\frac{-1}{2^{n-k+1}}\Rightarrow$$$$g\left(t\right)=\sum _{k=1}^{n-1}\frac{-1}{2^{n-k+1}{t}^k}-\frac{1}{2t^n}+\frac{1}{2^n(t-2)}\Rightarrow$$$$F\left(x\right)=\sum _{k=1}^{n-1}\frac{-1}{2^{n-k+1}(x+3)}-\frac{1}{2{(x+3)}^n}+\frac{1}{2^n(x+1)}.$$$$$$

Commented by abdo imad last updated on 28/Apr/18

$$★F\left(x\right)=\sum _{k=1}^{n-1}\frac{-1}{2^{n-k+1}{(x+3)}^k}-\frac{1}{2{(x+3)}^n}+\frac{1}{2^n(x+1)}.★$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com