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Math Help - Finding a generating function

  1. #1
    Junior Member gusztav's Avatar
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    Finding a generating function

    Here's the problem:

    Let g(x) be a generating function for the sequence (a_n). Find a sequence (b_n) whose generating function is (1-x)g(x).

    ***

    We know that, given a sequence (a_n)_{n \in \mathbb{N}_0}, the generating function of that sequence is g(x)= \sum_{n=0}^{\infty} a_n x^n.
    Let's define h(x):=(1-x)g(x) and it is known that h(x) is a generating function for the sequence (b_n), i.e. h(x)=\sum_{n=0}^{\infty} b_n x^n=(1-x)g(x)=(1-x)\sum_{n=0}^{\infty} a_n x^n.
    Therefore we have the relationship \sum_{n=0}^{\infty} b_n x^n=(1-x)\sum_{n=0}^{\infty} a_n x^n and I'm a little lost finding what exactly is (b_n).


    Any help will be greatly appreciated!
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  2. #2
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    Quote Originally Posted by gusztav View Post
    Here's the problem:

    Let g(x) be a generating function for the sequence (a_n). Find a sequence (b_n) whose generating function is (1-x)g(x).

    ***

    We know that, given a sequence (a_n)_{n \in \mathbb{N}_0}, the generating function of that sequence is g(x)= \sum_{n=0}^{\infty} a_n x^n.
    Let's define h(x):=(1-x)g(x) and it is known that h(x) is a generating function for the sequence (b_n), i.e. h(x)=\sum_{n=0}^{\infty} b_n x^n=(1-x)g(x)=(1-x)\sum_{n=0}^{\infty} a_n x^n.
    Therefore we have the relationship \sum_{n=0}^{\infty} b_n x^n=(1-x)\sum_{n=0}^{\infty} a_n x^n and I'm a little lost finding what exactly is (b_n).


    Any help will be greatly appreciated!

    Observe \sum_{n=0}^{\infty} b_n x^n=(1-x)\sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^n - \sum_{n=0}^{\infty} a_n x^{n+1} = a_0 + \sum_{n=1}^{\infty} (a_n - a_{n-1}) x^{n}

    Thus \sum_{n=0}^{\infty} b_n x^n = a_0 + \sum_{n=1}^{\infty} (a_n - a_{n-1}) x^{n}
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  3. #3
    Moo
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    Hello,

    (1-x) \sum_{n=0}^\infty a_n x^n=\sum_{n=0}^\infty a_n x^n-\sum_{n=0}^\infty a_n x^{n+1}
    (by expanding)

    Change the indice of the second sum :

    =\sum_{n=0}^\infty a_n x^n-\sum_{n=1}^\infty a_{n-1} x^n

    Make the indices be the same :

    =a_0+\sum_{n=1}^\infty a_n x^n-\sum_{n=1}^\infty a_{n-1} x^n

    =a_0+\sum_{n=1}^\infty (a_n-a_{n-1}) x^n


    Hence b_0=a_0 and b_n=a_n-a_{n-1} for n \geq 1


    If you cannot visualize some steps, just write the first few terms of the sums




    Edit : too slow
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