Find the power series expansion about the given point of the function:
z^3 + 6z^2 - 4z -3 about z0 = 1
Can someone show how to do this...
Thanks in advance
Is...
$\displaystyle f(z) = z^{3} + 6 z^{2} - 4 z - 3 = a_{3} (z-1)^{3} + a_{2} (z-1)^{2} + a_{1} (z-1) + a_{0}$ (1)
Because the coefficient of $\displaystyle z^{3}$ in (1) is $\displaystyle 1$ it will be $\displaystyle a_{3} = 1$ so that is...
$\displaystyle f(z) - (z-1)^{3} = 9 z^{2} - 7 z -2$ (2)
Because the coefficient of $\displaystyle z^{2}$ in (2) is $\displaystyle 9$ it will be $\displaystyle a_{2} = 9$ so that is...
$\displaystyle f(z) - (z-1)^{3} - 9 (z-1)^{2}= 11z -11 $ (3)
... so that is $\displaystyle a_{1} = 11$ and $\displaystyle a_{0}=0$. The answer to Your question is then...
$\displaystyle f(z) = z^{3} + 6 z^{2} - 4 z - 3 = (z-1)^{3} + 9 (z-1)^{2} + 11 (z-1) $ (4)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
or rewrite the polynomial
$\displaystyle z^3 + 6z^2 - 4z -3 = (z-1+1)^3 + 6(z-1+1)^2 - 4(z-1+1) - 3 $
$\displaystyle [(z-1)^3 + 3(z-1)^2 + 3(z-1) + 1 ] + 6[(z-1)^2 + 2(z-1) + 1] - 4[(z-1) + 1] -3 $
$\displaystyle (z-1)^3 + (z-1)^2 (3+6) + (z-1) (3 + 12 - 4 ) + 1 + 6 -4 -3 $
$\displaystyle = (z-1)^3 + 9(z-1)^2 + 11(z-1) $