Question Number 5525 by sanusihammed last updated on 17/May/16
$${Expand}\:\:{f}\left({z}\right)\:=\:{cos}\left({z}\right) \\ $$$${about}\:{point}\:\:{z}\:=\:\Pi/\mathrm{3} \\ $$$$ \\ $$$$ \\ $$
Commented by Yozzii last updated on 18/May/16
$${Taylor}\:{expansion}\:{of}\:{f}\left({z}\right)\:{near}\:{z}=\pi/\mathrm{3}: \\ $$$${f}\left({z}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\frac{{f}^{\left({i}\right)} \left(\pi/\mathrm{3}\right)\left({x}−\frac{\pi}{\mathrm{3}}\right)^{{i}} }{{i}!}\right\} \\ $$$$ \\ $$$${f}^{\left(\mathrm{0}\right)} \left({z}\right)={cosz} \\ $$$${f}^{\mathrm{1}} \left({z}\right)=−{sinz} \\ $$$${f}^{\left(\mathrm{2}\right)} \left({z}\right)=−{cosz} \\ $$$${f}^{\left(\mathrm{3}\right)} \left({z}\right)={sinz} \\ $$$${f}^{\left(\mathrm{4}\right)} \left({z}\right)={cosz} \\ $$$${f}^{\left(\mathrm{5}\right)} \left({z}\right)=−{sinz} \\ $$$${f}^{\left(\mathrm{2}{n}−\mathrm{1}\right)} \left({z}\right)=\left(−\mathrm{1}\right)^{{n}} {sinz} \\ $$$${f}^{\left(\mathrm{2}{n}\right)} \left({z}\right)=\left(−\mathrm{1}\right)^{{n}} {cosz} \\ $$