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Question Number 34849 by math khazana by abdo last updated on 11/May/18

$$letf\left(x\right)=\frac{e^{-x}}{2+x}$$$$developpfatintegrserie.$$

Commented by math khazana by abdo last updated on 13/May/18

$$wehavef\left(x\right)=\frac{e^{-x}}{2(1+\frac{x}{2})}\Rightarrow for\mid x\mid <2$$$$2f\left(x\right)=\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\left(\frac{x}{2}\right)}^n\right).\left(\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^n}{n!}\right)$$$$=\left(\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2^n}{x}^n\right).\left(\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}{x}^n\right)$$$$=\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^{n}with{c}_n=\sum _{i+j=n}{a}_i{b}_j$$$$=\sum _{i+j=n}\frac{{(-1)}^i}{2^i}\frac{{(-1)}^j}{j!}=\sum _{i=0}^n{a}_i{b}_{n-i}$$$$=\sum _{i=0}^n\frac{{(-1)}^i}{2^i}\frac{{(-1)}^{n-i}}{(n-i)!}=\sum _{i=0}^n\frac{{(-1)}^n}{(n-i)!2^i}\Rightarrow$$$$f\left(x\right)=\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{i=0}^n\frac{{(-1)}^n}{2^i(n-i)!}\right)x^n.$$$$$$

Commented by math khazana by abdo last updated on 13/May/18

$$anothermethod$$$$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{\left(n\right)}\left(0\right)}{n!}{x}^{n}letcalculate{f}^{\left(n\right)}\left(0\right)$$$$leibnitzformulagive{f}^{\left(n\right)}\left(x\right)=\sum _{k=0}^n{C}_n^k{\left(\frac{1}{2+x}\right)}^{\left(k\right)}{\left(e^{-x}\right)}^{(n-k)}$$$$=\sum _{k=0}^n{C}_n^k\frac{{(-1)}^{k}k!}{{(x+2)}^{k+1}}{(-1)}^{n-k}{e}^{-x}\Rightarrow$$$$f^{\left(n\right)}\left(0\right)=\sum _{k=0}^n{C}_n^k\frac{k!{(-1)}^n}{2^{k+1}}\Rightarrow$$$$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{k=0}^n{C}_n^k\frac{k!}{n!}\frac{{(-1)}^n}{2^{k+1}}\right)x^{n}but$$$$C_n^k.\frac{k!}{n!}=\frac{n!}{k!(n-k)!}\frac{k!}{n!}=\frac{1}{(n-k)!}\Rightarrow$$$$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{k=0}^n\frac{{(-1)}^n}{(n-k)!2^{k+1}}\right)x^n.$$

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