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Math Help - Power series.

  1. #1
    Super Member Deadstar's Avatar
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    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...
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  2. #2
    MHF Contributor
    Jester's Avatar
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    Quote Originally Posted by Deadstar View Post
    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
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