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Math Help - Maclaurin Series

  1. #1
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    Maclaurin Series

    I have to find the Maclaurin series of this:
    f(x)=x^2/(sqrt(2+x))
    I'm not really sure where to start. Any help would be appreciated.
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  2. #2
    o_O
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    f(x) = x^2 \left(2+x\right)^{-\frac{1}{2}} = 2^{-\frac{1}{2}}x^2 {\color{blue}\left(1 + \frac{x}{2}\right)^{-\frac{1}{2}}}

    Here, we can use binomial series:
    {\color{blue} \left(1+\frac{x}{2}\right)^{-\frac{1}{2}}} = 1 \ + \ \left(-\frac{1}{2}\right)\frac{x}{2} \ + \ \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!} \left(\frac{x}{2}\right)^2 \ + \ \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!} \left(\frac{x}{2}\right)^3  \ + \ \cdots {\color{white}.}  + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots \left(-\frac{1}{2} - n + 1\right)}{n!} \left(\frac{x}{2}\right)^n + \cdots ..... \text{if } \left| \frac{x}{2} \right| < 1

    Now simplify ...

    You could've went down the route by taking derivatives. You would still get the same coefficients (since they are unique).
    Last edited by o_O; December 4th 2008 at 10:51 PM.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by o_O View Post
    f(x) = x^2 \left(2+x\right)^{-\frac{1}{2}} = 2x^2 {\color{blue}\left(1 + \frac{x}{2}\right)^{-\frac{1}{2}}}

    Here, we can use binomial series:
    {\color{blue} \left(1+\frac{x}{2}\right)^{-\frac{1}{2}}} = 1 \ + \ \left(-\frac{1}{2}\right)\frac{x}{2} \ + \ \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!} \left(\frac{x}{2}\right)^2 \ + \ \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!} \left(\frac{x}{2}\right)^3 \ + \ \cdots {\color{white}.} + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots \left(-\frac{1}{2} - n + 1\right)}{n!} \left(\frac{x}{2}\right)^n + \cdots ..... \text{if } \left| \frac{x}{2} \right| < 1

    Now simplify ...

    You could've went down the route by taking derivatives. You would still get the same coefficients (since they are unique).
    This is the preferred method, but when o_O suggested if there was another method I was intrigued and I found one. Lets give it a go...it is completely impractical but interesting

    \int\frac{dx}{\sqrt{1+x^2}}=\text{arcsinh}(x)

    And it can easily be shown that \begin{aligned}\text{arcsinh}(x)&=\ln\left(x+\sqrt  {1+x^2}\right)\\<br />
&=\frac{1}{2}\ln\left(\left(x+\sqrt{1+x^2}\right)^  2\right)\\<br />
&=\frac{1}{2}\ln\left(1+2x^2+2x\sqrt{1+x^2}\right)  \end{aligned}

    So we know that

    \forall{x}\backepsilon|x|<1~~\ln(1+x)=-\sum_{n=0}^{\infty}\frac{(-1)^nx^{n}}{n}

    So

    \forall{x}\backepsilon\left|2x^2+2x\sqrt{1+x^2}\ri  ght|<1~~\frac{1}{2} \ln\left(1+2x^2+2x\sqrt{1+x^2}\right)=\frac{-1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n\left(2x^2+2x\sqrt{1+x^2}\right)^{n}}{n}

    So

    \forall{x}\backepsilon\left|2x^2+2x\sqrt{1+x^2}\ri  ght|<1 \frac{1}{\sqrt{1+x^2}}=\frac{-1}{2}\sum_{n=1}^{\infty}(-1)^n\left(2x^2+2x\sqrt{1+x^2}\right)^{n-1}\left(2x^2+2x\sqrt{1+x^2}\right)'

    And finally

    \forall{x}\backepsilon\left|2x+2\sqrt{x+x^2}\right  |<1~ \frac{1}{\sqrt{1+x}}=\frac{-1}{2}\sum_{n=1}^{\infty}(-1)^n\left(2x+2\sqrt{x+x^2}\right)^{n-1}\left(2x^2+2x\sqrt{1+x^2}\right)'\bigg|_{x=\sqrt  {x}}

    I might have made a computational error..its late

    THis was jsut for fun since your question was already answered
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  4. #4
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    Thank you both so much. Wow Mathstud that looks complex.
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