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

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