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Math Help - Series expansion LHS = RHS

  1. #1
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    Series expansion LHS = RHS

    Hello everyone im new to this site so hope I am posting in the right place etc. Im struggling with the following question, I just really dont know how to get started, any help would be appreciated!

    By considering appropriate series expansions, prove that:

    ex * e(x^2)/2 * e(x^3)/3 * .... = 1 + x + x2 + .... when |x| < 1.

    By expanding each individual exponential term on the left-hand side and multiplying out, show that the coefficient of x19 has the form:

    1/19! + 1/19 + r/s, where 19 does not divide s.

    Thanks

    LexiBelle
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  2. #2
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    Re: Series expansion LHS = RHS

    Quote Originally Posted by LexiBelle View Post
    Hello everyone im new to this site so hope I am posting in the right place etc. Im struggling with the following question, I just really dont know how to get started, any help would be appreciated!

    By considering appropriate series expansions, prove that:

    ex * e(x^2)/2 * e(x^3)/3 * .... = 1 + x + x2 + .... when |x| < 1.
    The right hand side is a geometric series that sums to 1/(1- x).
    The exponetials on the left can be combined to give e^{x+ x^2/2+ x^3/3+ \cdot\cdot\cdot} and you should recogize that exponent as the Taylor's series for ln(x) about x= 1.

    By expanding each individual exponential term on the left-hand side and multiplying out, show that the coefficient of x19 has the form:

    1/19! + 1/19 + r/s, where 19 does not divide s.

    Thanks

    LexiBelle
    Thanks from LexiBelle
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  3. #3
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    Re: Series expansion LHS = RHS

    Ok im having a couple of problems here. Isnt this slightly different to the Taylor series for ln(x) about x = 1? As that one has negative signs in and this one is all positive? I still cant seem to see where to go from here ?

    Thanks

    LexiBelle
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