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Math Help - McLaurin series

  1. #1
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    McLaurin series

    Ok, so I've got the following problem:

    Find an expression for the following McLaurin-series:

    Σ (((n^2) - n +1) / n!) * x^n



    (from n = 0 to infinity).



    I decided to start with the known McLaurin series for e^x:


    e^x = Σ (x^n) / n!


    I then differentiate and get:


    e^x = Σ (n*(x^(n-1))) / n!


    I differentiate one more time and get:


    e^x = Σ (((n^2) - n) / n!) * x^(n-2)


    I can't get any further than this though. So what I need now is to get the (+1) in the numerator of the n-term, and x^(n-2) has to be reversed back to x^n. Can I multiply e^x with x^2 to get the x taken care of? And how do I get the (+1) term?


    Any tips would be greatly appreciated!
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  2. #2
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    Quote Originally Posted by krje1980 View Post
    Ok, so I've got the following problem:

    Find an expression for the following McLaurin-series:

    Σ (((n^2) - n +1) / n!) * x^n



    (from n = 0 to infinity).



    I decided to start with the known McLaurin series for e^x:


    e^x = Σ (x^n) / n!


    I then differentiate and get:


    e^x = Σ (n*(x^(n-1))) / n!


    I differentiate one more time and get:


    e^x = Σ (((n^2) - n) / n!) * x^(n-2)


    I can't get any further than this though. So what I need now is to get the (+1) in the numerator of the n-term, and x^(n-2) has to be reversed back to x^n. Can I multiply e^x with x^2 to get the x taken care of? And how do I get the (+1) term?


    Any tips would be greatly appreciated!

    So after differentiating twice you got e^x=\sum\frac{n^2-n}{n!}x^{n-2} = \sum\frac{x^{n-2}}{(n-2)!} .

    But now pay attention that \sum\frac{n^2-n+1}{n!}x^n=\sum\frac{n^2-n}{n!}x^n+\sum\frac{1}{n!}x^n , and you can split the series this way because everything converges absolutely here for all x.

    Tonio
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  3. #3
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    Thanks a lot . So the correct answer will then be:

    (x^2)(e^x) + (e^x)

    (or (e^x)((x^2) + 1))

    ?
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