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

find the taylor series of f(z)=sinz ,z=(π/4) in complex number

$${find}\:{the}\:{taylor}\:{series}\:{of}\:{f}\left({z}\right)={sinz}\:,{z}=\frac{\pi}{\mathrm{4}}\:{in}\:{complex}\:{number} \\ $$

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

f(z)=Σ_(n=0) ^∞  ((f^((n)) ((π/4)))/(n!))(z−(π/4))^n   we have f^((n)) (z)=sin(z+((nπ)/2)) ⇒  f^((n)) ((π/4))=sin((π/4)+((nπ)/2)) =sin((((2n+1)π)/4)) ⇒  sinz =Σ_(n=0) ^∞  (1/(n!))sin((((2n+1)π)/4))(z−(π/4))^n

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

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