# Power Series

• May 10th 2010, 09:16 PM
jzellt
Power Series
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...

• May 10th 2010, 09:42 PM
chisigma
Is...

$f(z) = z^{3} + 6 z^{2} - 4 z - 3 = a_{3} (z-1)^{3} + a_{2} (z-1)^{2} + a_{1} (z-1) + a_{0}$ (1)

Because the coefficient of $z^{3}$ in (1) is $1$ it will be $a_{3} = 1$ so that is...

$f(z) - (z-1)^{3} = 9 z^{2} - 7 z -2$ (2)

Because the coefficient of $z^{2}$ in (2) is $9$ it will be $a_{2} = 9$ so that is...

$f(z) - (z-1)^{3} - 9 (z-1)^{2}= 11z -11$ (3)

... so that is $a_{1} = 11$ and $a_{0}=0$. The answer to Your question is then...

$f(z) = z^{3} + 6 z^{2} - 4 z - 3 = (z-1)^{3} + 9 (z-1)^{2} + 11 (z-1)$ (4)

Kind regards

$\chi$ $\sigma$
• May 10th 2010, 10:41 PM
simplependulum
or rewrite the polynomial

$z^3 + 6z^2 - 4z -3 = (z-1+1)^3 + 6(z-1+1)^2 - 4(z-1+1) - 3$

$[(z-1)^3 + 3(z-1)^2 + 3(z-1) + 1 ] + 6[(z-1)^2 + 2(z-1) + 1] - 4[(z-1) + 1] -3$

$(z-1)^3 + (z-1)^2 (3+6) + (z-1) (3 + 12 - 4 ) + 1 + 6 -4 -3$

$= (z-1)^3 + 9(z-1)^2 + 11(z-1)$