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Question Number 50551 by pooja24 last updated on 17/Dec/18

$$findtherealrootof$$$$x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}+\frac{x^9}{216}-\frac{x^{11}}{1320}+\dots \dots =.443$$$$pleaseifanysimpleexpansionofit$$$$thenplztellme$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

$$f\left(x\right)=x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}+\frac{x^9}{216}-\frac{x^{11}}{1320}+...$$$$\frac{df\left(x\right)}{dx}=1-x^2+\frac{x^4}{2}-\frac{x^6}{6}+\frac{x^8}{24}-\frac{x^{10}}{120}+...$$$$let=k=x^2$$$$Righthandside$$$$1-k+\frac{k^2}{2!}-\frac{k^3}{3!}+\frac{k^4}{4!}-\frac{k^5}{5!}+...$$$$=e^{-k}$$$$=e^{-x^2}$$$$so\frac{df\left(x\right)}{dx}=e^{-x^2}$$$$f\left(x\right)=\int e^{-x^2}dx$$$$$$$$[fromtable{e}^{-0.812}\approx 0.443$$$$$$$$$$$$$$

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