Question

The radius of convergence of the Taylor series expansion of the function f(z)=4z^{2}+3z/(z-1)^{2}(z+4)(z-3) abou z=-1 is

Answer 1

We know that Taylor series expansion of function at Z_o is

$f(z)= f(z_0)+(z-z_0)f'(z_0)+\frac{(z-z_0)^2}{2!}f''(z_0)+\frac{(z-z_0)^3}{3!}f'''(z_0)+...$

$f(z)=\frac{4z^2+3z}{\left(z-1\right)^2\left(z+4\right)\left(z-3\right)}\\ f'=\frac{d}{dz}\left(\frac{4z^2+3z}{\left(z-1\right)^2\left(z+4\right)\left(z-3\right)}\right)\\ =\frac{\frac{d}{dz}\left(4z^2+3z\right)\left(z-1\right)^2\left(z+4\right)\left(z-3\right)-\frac{d}{dz}\left(\left(z-1\right)^2\left(z+4\right)\left(z-3\right)\right)\left(4z^2+3z\right)}{\left(\left(z-1\right)^2\left(z+4\right)\left(z-3\right)\right)^2}\\ =-\frac{8z^4+13z^3+7z^2-132z-36}{\left(z-1\right)^3\left(z+4\right)^2\left(z-3\right)^2}\\ \begin{bmatrix}z=-1\end{bmatrix}\\ f=\frac{4\left(-1\right)^2+3\left(-1\right)}{\left(-1-1\right)^2\left(-1+4\right)\left(-1-3\right)}\\ f=-\frac{1}{48}\\$

$f''=\begin{bmatrix}f=-\frac{8z^4+13z^3+7z^2-132z-36}{\left(z-1\right)^3\left(z+4\right)^2\left(z-3\right)^2}\\ z=-1\end{bmatrix}\\ f''= \frac{\left(32z^3+39z^2+14z-132\right)\left(z-1\right)^3\left(z+4\right)^2\left(z-3\right)^2-\left(3\left(z-1\right)^2\left(z+4\right)^2\left(z-3\right)^2+\left(4z^3+6z^2-46z-24\right)\left(z-1\right)^3\right)\left(8z^4+13z^3+7z^2-132z-36\right)}{\left(z-1\right)^6\left(z+4\right)^4\left(z-3\right)^4}\\ f''=-\frac{25}{3456}$

$f(z)= -\frac{1}{\sqrt{48}}+(z+1)(-\frac{1}{{48}})+(z+1)^2* \frac{1}{2!}*\frac{4*25}{3456}+...$