Question Number 130399 by gowsalya last updated on 25/Jan/21
Answered by Dwaipayan Shikari last updated on 25/Jan/21
$${f}\left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} {x}^{{n}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +{a}_{\mathrm{3}} {x}^{\mathrm{3}} +… \\ $$$${f}\left(\mathrm{0}\right)={a}_{\mathrm{0}} \\ $$$${f}'\left(\mathrm{0}\right)={a}_{\mathrm{1}} \\ $$$${f}''\left(\mathrm{0}\right)=\mathrm{2}{a}_{\mathrm{2}} \\ $$$${f}'''\left(\mathrm{0}\right)=\mathrm{6}{a}_{\mathrm{3}} \\ $$$${f}^{{iv}} \left(\mathrm{0}\right)=\mathrm{24}{a}_{\mathrm{4}} \\ $$$$..{f}^{{n}} \left(\mathrm{0}\right)={n}!{a}_{{n}} \:\Rightarrow\frac{{f}^{{n}} \left(\mathrm{0}\right)}{{n}!}={a}_{{n}} \Rightarrow\frac{{f}^{{n}} \left(\mathrm{0}\right)}{{n}!}{x}^{{n}} ={a}_{{n}} {x}^{{n}} \\ $$$$\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{f}^{{n}} \left(\mathrm{0}\right)}{{n}!}{x}^{{n}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} {x}^{{n}} ={f}\left({x}\right) \\ $$