Question #223832

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

Expert's answer

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)}\\\\\nf'=\\frac{d}{dz}\\left(\\frac{4z^2+3z}{\\left(z-1\\right)^2\\left(z+4\\right)\\left(z-3\\right)}\\right)\\\\\n=\\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}\\\\\n=-\\frac{8z^4+13z^3+7z^2-132z-36}{\\left(z-1\\right)^3\\left(z+4\\right)^2\\left(z-3\\right)^2}\\\\\n\\begin{bmatrix}z=-1\\end{bmatrix}\\\\\nf=\\frac{4\\left(-1\\right)^2+3\\left(-1\\right)}{\\left(-1-1\\right)^2\\left(-1+4\\right)\\left(-1-3\\right)}\\\\\nf=-\\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}\\\\\nf''= \\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}\\\\\nf''=-\\frac{25}{3456}"

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

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