# Thread: Power series.

1. ## Power series.

Find the power-series expansions about 0 of...

$(1-2z)^{\frac{1}{2}}$.

K. So the way i did this was to use the binomial theorem but it came out with the wrong answer. Could someone tell me where about im making the mistake...

Here's it set out so as you can see my though process... sort of.

$(1-2z)^{\frac{1}{2}} = 1 + (\frac{1}{2})(-2z) - (\frac{1}{2})^2(-2z)^2(\frac{1}{2!})$ $+ (\frac{1}{2})^3(-2z)^3(\frac{1}{3!}) -(\frac{1}{2})^4(-2z)^4(\frac{1}{4!})$ $= 1 - z - \frac{z^2}{2} - \frac{z^3}{3!} - \frac{z^4}{4!}$...

But the answer has a 15 multiplying the z^4 term so somethings not right...

Any help? Pretty sure i've just gotten the binomial theorem in a complete mess...

2. Originally Posted by Deadstar
Find the power-series expansions about 0 of...

$(1-2z)^{\frac{1}{2}}$.

K. So the way i did this was to use the binomial theorem but it came out with the wrong answer. Could someone tell me where about im making the mistake...

Here's it set out so as you can see my though process... sort of.

$(1-2z)^{\frac{1}{2}} = 1 + (\frac{1}{2})(-2z) - (\frac{1}{2})^2(-2z)^2(\frac{1}{2!})$ $+ (\frac{1}{2})^3(-2z)^3(\frac{1}{3!}) -(\frac{1}{2})^4(-2z)^4(\frac{1}{4!})$ $= 1 - z - \frac{z^2}{2} - \frac{z^3}{3!} - \frac{z^4}{4!}$...

But the answer has a 15 multiplying the z^4 term so somethings not right...

Any help? Pretty sure i've just gotten the binomial theorem in a complete mess...
Binomial expansion

$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 \ cdots$

Not sure why you're squaring, cubing etc the 1/2