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Question Number 40260 by maxmathsup by imad last updated on 17/Jul/18

$$letf\left(x\right)=sin\left(2x\right)$$$$1)find{f}^{\left(n\right)}\left(x\right)and{f}^{\left(n\right)}\left(0\right)$$$$2)developpfatintegrserie.$$

Commented by maxmathsup by imad last updated on 18/Jul/18

$$1)wehavef\left(x\right)=2sin\left(x\right)cos\left(x\right)\Rightarrow f^{\left(n\right)}\left(x\right)=2\sum _{k=0}^n{C}_n^k{\left(sinx\right)}^{\left(k\right)}{\left(cosx\right)}^{(n-k)}$$$$=2\sum _{k=0}^n{C}_n^{k}sin(x+\frac{k\pi }{2})cos(x+\frac{(n-k)\pi }{2})\Rightarrow$$$$f^{\left(n\right)}\left(0\right)=2\sum _{k=0}^n{C}_n^{k}sin\left(\frac{k\pi }{2}\right)cos\left(\frac{(n-k)\pi }{2}\right)$$$$=2\sum _{p=0}^{\left[\frac{n-1}{2}\right]}{C}_n^{2p+1}sin\left(\frac{(2p+1)\pi }{2}\right)cos\left(\frac{(n-2p-1)\pi }{2}\right)$$$$=2\sum _{p=0}^{\left[\frac{n-1}{2}\right]}{(-1)}^p{C}_n^{2p+1}cos(\frac{(n-1)\pi }{2}-p\pi )$$$$=2\sum _{p=0}^{\left[\frac{n-1}{2}\right]}{C}_n^{2p+1}cos\left(\frac{(n-1)\pi }{2}\right)$$$$2)f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{f^{\left(n\right)}\left(0\right)}{n!}{x}^n=\sum _{n=1}^{\mathrm{\infty }}\frac{f^{\left(n\right)}\left(0\right)}{n!}{x}^n$$$$f\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n!}cos\left(\frac{(n-1)\pi }{2}\right)\left(\sum _{p=0}^{\left[\frac{n-1}{2}\right]}{C}_n^{2p+1}\right)x^n$$$$$$$$$$

Commented by maxmathsup by imad last updated on 18/Jul/18

$$2)anothermethodwehavesint=Im\left(e^{it}\right)=Im\left(\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(it\right)}^n}{n!}\right)$$$$=Im(\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{t}^{2n}}{\left(2n\right)!}+i\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{t}^{2n+1}}{(2n+1)!})=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{t}^{2n+1}}{(2n+1)!}\Rightarrow$$$$sin\left(2x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{2}^{2n+1}{x}^{2n+1}}{(2n+1)!}andtheradiusofconvergenceisR=+\mathrm{\infty }$$$$$$

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