Answer to Question #315437 in Complex Analysis for Roshan

Question #315437

obtain first three terms of taylors series for f(z) = sin z about 2=0


1
Expert's answer
2022-03-23T18:11:05-0400

f(0)=sin0=0f(0)=coszz=0=1f(0)=sinzz=0=0f(0)=coszz=0=1fv(0)=sinzz=0=0fv(0)=coszz=0=1First  3terms:f(z)=n=051n!f(n)(z)zn+o(z5)=z13!z3+15!z5+o(z5)==z16z3+1120z5+o(z5)f\left( 0 \right) =\sin 0=0\\f'\left( 0 \right) =\cos z|_{z=0}=1\\f''\left( 0 \right) =-\sin z|_{z=0}=0\\f'''\left( 0 \right) =-\cos z|_{z=0}=-1\\f'^v\left( 0 \right) =\sin z|_{z=0}=0\\f^v\left( 0 \right) =\cos z|_{z=0}=1\\First\,\,3terms:\\f\left( z \right) =\sum_{n=0}^5{\frac{1}{n!}f^{\left( n \right)}\left( z \right) z^n}+o\left( z^5 \right) =z-\frac{1}{3!}z^3+\frac{1}{5!}z^5+o\left( z^5 \right) =\\=z-\frac{1}{6}z^3+\frac{1}{120}z^5+o\left( z^5 \right)


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