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Math Help - Binomial Series to find a Maclaurin Series

  1. #1
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    Binomial Series to find a Maclaurin Series

    I'm having a tough time finding the Maclaurin series (we have to use the binomial series) for
    <br />
f(x)=\frac{1}{\sqrt [3]{1+x^2}}<br />
    Am I supposed to try to make it fit the form  (1+x)^k ? If so, I'm confused about how to do so. Any help walking me through this step by step would be greatly appreciated. Thanks!
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  2. #2
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    Quote Originally Posted by psu1024 View Post
    I'm having a tough time finding the Maclaurin series (we have to use the binomial series) for
    <br />
f(x)=\frac{1}{\sqrt [3]{1+x^2}}<br />
    Am I supposed to try to make it fit the form  (1+x)^k ? If so, I'm confused about how to do so. Any help walking me through this step by step would be greatly appreciated. Thanks!
    Use the formula for (1+y)^{-1/3} and then substitute y=x^2.
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    So does the equation become this:
    <br />
(1+y)^{-1/3} = 1-\frac{y}{3}+\frac {2y^2}{3^22!}-\frac {28y^3}{3^33!}+...<br />
    How do I substitute the  x^2 back into the equation though?
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  4. #4
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    Quote Originally Posted by psu1024 View Post
    So does the equation become this:
    <br />
(1+y)^{-1/3} = 1-\frac{y}{3}+\frac {2y^2}{3^22!}-\frac {28y^3}{3^33!}+...<br />
    How do I substitute the  x^2 back into the equation though?
    That's correct. Now replace y by x^2, and you get \frac1{\sqrt[3]{1+x^2}} = 1-\frac{x^2}{3}+\frac {2x^4}{3^22!}-\frac {28x^6}{3^33!}+\ldots.
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    Thanks! That's how I ended up doing the problem. I just wasn't sure if there was some weird substitution technique or something. Thanks for your help.
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