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Question Number 31546 by prof Abdo imad last updated on 10/Mar/18

$$letconsiderthenumrticalfunction$$$$f\left(x\right)=\frac{1}{x^2+x+1}calculate{f}^{\left(n\right)}\left(x\right)thengive$$$$f^{\left(n\right)}\left(0\right).$$

Answered by prof Abdo imad last updated on 10/Apr/18

$$letdecomposef\left(x\right)insideC\left(x\right)$$$$rootsof{x}^2+x+1=0$$$$\mathrm{\Delta }=1-4=-3={\left(i\sqrt{3}\right)}^2\Rightarrow x_1=\frac{-1+i\sqrt{3}}{2}=j$$$$x_2=\frac{-1-i\sqrt{3}}{2}=\stackrel{-}{j}so$$$$f\left(x\right)=\frac{1}{(x-j)(x-\stackrel{-}{j})}=\frac{a}{x-j}+\frac{b}{x-\stackrel{-}{j}}$$$$a=\frac{1}{j-\stackrel{-}{j}}=\frac{1}{i\sqrt{3}},b=\frac{1}{\stackrel{-}{j}-j}=\frac{1}{-i\sqrt{3}}\Rightarrow$$$$f\left(x\right)=\frac{1}{i\sqrt{3}}(\frac{1}{x-j}-\frac{1}{x-\stackrel{-}{j}})\Rightarrow$$$$f^{\left(n\right)}\left(x\right)=\frac{1}{i\sqrt{3}}({\left(\frac{1}{x-j}\right)}^n-{\left(\frac{1}{x-\stackrel{-}{j}}\right)}^n)$$$$=\frac{1}{i\sqrt{3}}(\frac{{(-1)}^{n}n!}{{(x-j)}^{n+1}}-\frac{{(-1)}^{n}n!}{{(x-\stackrel{-}{j})}^{n+1}})$$$$=\frac{{(-1)}^{n}n!}{i\sqrt{3}}(\frac{1}{{(x-j)}^{n+1}}-\frac{1}{{(x-\stackrel{-}{j})}^{n+1}})\Rightarrow$$$$f^{\left(n\right)}\left(0\right)=\frac{{(-1)}^{n}n!}{i\sqrt{3}}(\frac{{(-1)}^{n+1}}{j^{n+1}}-\frac{{(-1)}^{n+1}}{\stackrel{{-}^{n+1}}{j}})$$$$=\frac{-n!}{i\sqrt{3}}\left(\frac{-j^{n+1}+\stackrel{-n+1}{j}}{1}\right)=\frac{n!}{i\sqrt{3}}(j^{n+1}-\stackrel{{-}^{n+1}}{j})$$$$=\frac{n!}{i\sqrt{3}}(2iIm\left(j^{n+1}\right)=\frac{2n!}{\sqrt{3}}Im\left(e^{2i(n+1)\frac{\pi }{3}}\right)\Rightarrow$$$$f^{\left(n\right)}\left(0\right)=\frac{2(n!)}{\sqrt{3}}sin\left(\frac{2(n+1)\pi }{3}\right).$$$$$$

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