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Question Number 147569 by Sozan last updated on 21/Jul/21

$$findthetaylorseriesoff\left(z\right)=sinz,z=\frac{\pi }{4}incomplexnumber$$

Answered by mathmax by abdo last updated on 22/Jul/21

$$\mathrm{f}\left(\mathrm{z}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\frac{\pi }{4}\right)}{\mathrm{n}!}{(\mathrm{z}-\frac{\pi }{4})}^{\mathrm{n}}\mathrm{we}\mathrm{have}{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{z}\right)=\sin (\mathrm{z}+\frac{\mathrm{n}\pi }{2})\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\frac{\pi }{4}\right)=\sin (\frac{\pi }{4}+\frac{\mathrm{n}\pi }{2})=\sin \left(\frac{(2\mathrm{n}+1)\pi }{4}\right)\Rightarrow$$$$\mathrm{sinz}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}!}\sin \left(\frac{(2\mathrm{n}+1)\pi }{4}\right){(\mathrm{z}-\frac{\pi }{4})}^{\mathrm{n}}$$

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