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Math Help - Help with Maclaurin series!

  1. #1
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    Help with Maclaurin series!

    Find the Maclaurin series for [e^(x+1)^2 - 1]/(x+1)^2 EDIT: the lopsided parenthesis stands for ^( . I don't know why my latex code won't work.

    Here's how far I got...

    1 + (x+1)^2 + [(x+1)^4]/(2!) + [(x+1)^6]/(3!)...

    SUBTRACTED BY...

    1 - (x^2+2x) + (x^2+2x)^2 - (x^2+2x)^3...

    I'd like to know the summation notion (aka general term) of the series as well if that's possible.

    In case you were wondering, I split the expression into two parts with a common denominator and used the maclaurin series of e^x and 1/(1-x) as starting points.
    Last edited by Kaitosan; May 7th 2009 at 12:33 PM.
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  2. #2
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    Quote Originally Posted by Kaitosan View Post
    Find the Maclaurin series for [e^(x+1)^2 - 1]/(x+1)^2 EDIT: the lopsided parenthesis stands for ^( . I don't know why my latex code won't work.

    Here's how far I got...

    1 + (x+1)^2 + [(x+1)^4]/(2!) + [(x+1)^6]/(3!)...

    SUBTRACTED BY...

    1 - (x^2+2x) + (x^2+2x)^2 - (x^2+2x)^3...

    I'd like to know the summation notion (aka general term) of the series as well if that's possible.
    e^{(x+1)^2} - 1 = (x+1)^2 + \frac{(x+1)^4}{2!} + \frac{(x+1)^6}{3!} + ...

     <br />
\frac{e^{(x+1)^2} - 1}{(x+1)^2} = 1 + \frac{(x+1)^2}{2!} + \frac{(x+1)^4}{3!} + ... = \sum_{n=1}^{\infty} \frac{(x+1)^{2(n-1)}}{n!}<br />
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  3. #3
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    Thanks! Often I need someone to point me to the "obvious" lol.
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