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Math Help - need a help to simplify

  1. #1
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    need a help to simplify

    a_n(G)=\frac{t_n(G)}{(n-1)!} and h_n(G)=\sum _{k=1}^n {n-1 \choose k-1} t_k(G)h_{n-k}(G). Then need to remove t_k(G) and obtain a_n(G)=\frac{1}{(n-1)!} h_n(G)-\sum _{k=1}^n \frac{1}{(n-k)!}h_{n-k}(G)a_k(G).
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  2. #2
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by dimuk View Post
    a_n(G)=\frac{t_n(G)}{(n-1)!} and h_n(G)=\sum _{k=1}^n {n-1 \choose k-1} t_k(G)h_{n-k}(G). Then need to remove t_k(G) and obtain a_n(G)=\frac{1}{(n-1)!} h_n(G)-\sum _{k=1}^n \frac{1}{(n-k)!}h_{n-k}(G)a_k(G).
    Under the condition h_{0}(G)=1, I got
    <br />
a_{n}(G)=\frac{1}{(n-1)!}h_{n}(G)-\sum\limits_{k=1}^{{\color{magenta}n-1}}\frac{1}{(n-k)!}h_{n-k}(G)a_{k}(G).<br />

    Are there any assumptions we should know?
    Last edited by bkarpuz; September 12th 2008 at 02:15 AM. Reason: Coloured the upper bound of the sum.
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  3. #3
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    need a help to simplify

    I think I made a mistake. Your answer is correct. Let me know how did u get it.

    Thanks.
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  4. #4
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by dimuk View Post
    I think I made a mistake. Your answer is correct. Let me know how did u get it.

    Thanks.
    It is so simple, first note that
    <br />
{n\choose k}=\frac{n!}{k!(n-k)!},<br />
    then isolate t_{n} and get
    <br />
t_{n}=(n-1)! a_{n}.<br />
    Substitute this into the other equation, and get
    h_{n}=\sum\limits_{k=1}^{n}\frac{(n-1)!}{(k-1)!(n-k)!}(k-1)!a_{k}h_{n-k}
    ..... =(n-1)!\sum\limits_{k=1}^{n}\frac{1}{(n-k)!}a_{k}h_{n-k}
    ..... =(n-1)!\Bigg[\bigg(\sum\limits_{k=1}^{n-1}\frac{1}{(n-k)!}a_{k}h_{n-k}\bigg)+a_{n}h_{0}\Bigg].
    By arranging the last expression, we easily get the desired result under the assumption
    h_{0}=1.


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