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Question Number 33335 by prof Abdo imad last updated on 14/Apr/18

$$1)give?D_{n-1}\left(o\right)forf\left(x\right)=\frac{1}{x+2}$$$$2)drcomposeinsideR\left(x\right)thefraction$$$$F\left(x\right)=\frac{1}{x^n(x+2)}$$

Commented by prof Abdo imad last updated on 25/Apr/18

$$wehavef\left(x\right)=\sum _{k=0}^{n-1}\frac{f^{\left(k\right)}\left(0\right)}{k!}{x}^k+\frac{x^n}{n!}\xi \left(x\right)with$$$$\xi {\left(x\right)}_{x\to 0}\to 0butwehave{f}^{\left(k\right)}\left(x\right)=\frac{{(-1)}^{k}k!}{{(x+2)}^{k+1}}\Rightarrow$$$$f^{\left(k\right)}\left(0\right)=\frac{{(-1)}^{k}k!}{2^{k+1}}\Rightarrow$$$$f\left(x\right)=\sum _{k=0}^{n-1}\frac{{(-1)}^k}{2^{k+1}}{x}^k+\frac{x^n}{n!}\xi \left(x\right)$$$$2)F\left(x\right)=\sum _{k=1}^n\frac{{\lambda }_k}{x^k}+\frac{a}{x+2}$$$$a={lim}_{x\to -2}(x+2)F\left(x\right)=\frac{1}{{(-2)}^n}=\frac{{(-1)}^n}{2^n(x+2)}\Rightarrow$$$$F\left(x\right)=\sum _{k=1}^n\frac{{\lambda }_k}{x^k}+\frac{{(-1)}^n}{2^n(x+2)}=\sum _{k=1}^{n-1}\frac{{\lambda }_k}{x^k}+\frac{{\lambda }_n}{x^n}+\frac{{(-1)}^n}{2^n(x+2)}$$$${\lambda }_n={lim}_{x\to 0}{x}^{n}F\left(x\right)=\frac{1}{2}fromanotherside$$$$F\left(x\right)=\frac{1}{x^n}(\sum _{k=0}^{n-1}\frac{{(-1)}^k}{2^k}{x}^k+\frac{x^n}{n!}\xi \left(x\right))$$$$=\sum _{k=0}^{n-1}\frac{{(-1)}^k}{2^k{x}^{n-k}}+\frac{1}{n!}\xi \left(x\right)ch.ofindicen-k=p$$$$F\left(x\right)=\sum _{p=1}^n\frac{{(-1)}^{n-p}}{2^{n-p}{x}^p}+\frac{1}{n!}\xi \left(x\right)$$$$=\sum _{k=1}^n\frac{{(-1)}^{n-k}}{2^{n-k}{x}^k}+\frac{1}{n!}\xi \left(x\right)\Rightarrow {\lambda }_k=\frac{{(-1)}^{n-k}}{2^{n-k}}and$$$$F\left(x\right)=\sum _{k=1}^{n-1}\frac{{(-1)}^{n-k}}{2^{n-k}{x}^k}+\frac{1}{2x^n}+\frac{{(-1)}^n}{2^n(x+2)}.$$$$$$

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