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Question Number 94334 by mathmax by abdo last updated on 18/May/20

$$letf\left(x\right)=\frac{sinx}{x}ifx\ne 0andf\left(0\right)=1$$$$1){findf}^{\left(n\right)}\left(x\right)and{f}^{\left(n\right)}\left(0\right)$$$$2)developpfatintegrseriest{x}_0=0and{x}_0=\frac{\pi }{2}$$

Answered by abdomathmax last updated on 18/May/20

$$1)\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{sinx}}{\mathrm{x}}\Rightarrow {\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\left(\frac{1}{\mathrm{x}}\right)}^{\left(\mathrm{k}\right)}{\left(\mathrm{sinx}\right)}^{(\mathrm{n}-\mathrm{k})}$$$$=\frac{1}{\mathrm{x}}\sin (\mathrm{x}+\frac{\mathrm{n}\pi }{2})+\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\frac{{(-1)}^{\mathrm{k}}\mathrm{k}!}{{\mathrm{x}}^{\mathrm{k}+1}}\sin (\mathrm{x}+\frac{(\mathrm{n}-\mathrm{k})\pi }{2})$$$$2)\mathrm{we}\mathrm{have}\mathrm{sinx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1)!}{\mathrm{x}}^{2\mathrm{n}+1}\Rightarrow$$$$\frac{\mathrm{sinx}}{\mathrm{x}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1)!}{\mathrm{x}}^{2\mathrm{n}}$$

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