need a help to simplify

**dimuk** · September 11th 2008, 11:46 PM

$a_n(G)=\frac{t_n(G)}{(n-1)!}$ 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 $a_n(G)=\frac{1}{(n-1)!} h_n(G)-\sum _{k=1}^n \frac{1}{(n-k)!}h_{n-k}(G)a_k(G).$

**bkarpuz** · September 12th 2008, 01:05 AM

Originally Posted by dimuk

$a_n(G)=\frac{t_n(G)}{(n-1)!}$ 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
$ 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). $

Are there any assumptions we should know?

**dimuk** · September 12th 2008, 01:30 AM

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

Thanks.

**bkarpuz** · September 12th 2008, 01:44 AM

Originally Posted by dimuk

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
$ {n\choose k}=\frac{n!}{k!(n-k)!}, $
then isolate $t_{n}$ and get
$ t_{n}=(n-1)! a_{n}. $
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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