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Math Help - Confused on Exponential Generating function

  1. #1
    Member oldguynewstudent's Avatar
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    Confused on Exponential Generating function

    Given the following notation: \left[\left[f(x)\right]\right]_{x^{k}} = the coefficient of x^{k} in f(x).

    Also for an exponential generating function  \left[\left[e^{x}\right]\right]_{\frac{x^{k}}{k!}}= \left[\left[\sum_{k\geq0}\frac{x^{k}}{k!}\right]\right]_{\frac{x^{k}}{k!}}=1.

    We have an example in Combinatorics A Guided Tour by David R. Mazur p. 129-130 as follows:

    Find a formula for the n-th term of the sequence \{d_{n}\}_{n\geq0} defined by d_{0}=1\,\, d_{n}=nd_{n-1}+1\,\,\, n\geq1.

    After a short derivation we have d_{n}=\left[\left[e^{x}*\frac{1}{1-x}\right]\right]_{\frac{x^{n}}{n!}}= \sum_{j=0}^{n}\left({n\atop j}\right)1*(n-j)!= \sum_{j=0}^{n}\left({n\atop j}\right)(n-j)!

    I am familiar with (1+x)^{n}=\sum_{k\geq0}\left({n\atop k}\right)x^{k}. Also I know \frac{1}{1-x}=\sum_{k\geq0}k!\frac{x^{k}}{k!}. But I just do not see how the binomial fits into the above summation.

    I have been working almost nonstop on this course and applied statistics this summer. Any help with how the binomial gets into the above summation would be greatly appreciated. By the way, this is the most fascinating course I have ever taken even though I seem to be struggling just a little.
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  2. #2
    Junior Member Renji Rodrigo's Avatar
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    In this case you can use the cauchy product of the series


    Given two series   \sum\limits^{\infty}_{k=0}a_{k}

    and  \sum\limits^{\infty}_{k=0}b_{k} , the Cauchy product is

     \sum\limits^{\infty}_{k=0}c_{k}

    where

     c_{n}= \sum\limits^{n}_{k=0}a_{k}b_{n-k}\;\;n \in N.
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    Hi oldguynewstudent,

    I think the theorem you need is that if
    f(x) = \sum_{n=0}^{\infty} \frac{a_n}{n!} x^n
    and
    g(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n
    then the coefficient of \frac{x^n}{n!} in f(x) \cdot g(x)
    is
    \sum_r \binom{n}{r} a_r b_{n-r}
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  4. #4
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by awkward View Post
    Hi oldguynewstudent,

    I think the theorem you need is that if
    f(x) = \sum_{n=0}^{\infty} \frac{a_n}{n!} x^n
    and
    g(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n
    then the coefficient of \frac{x^n}{n!} in f(x) \cdot g(x)
    is
    \sum_r \binom{n}{r} a_r b_{n-r}
    Yes, that is what I was looking for!
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