# need a help to simplify

• Sep 11th 2008, 11:46 PM
dimuk
need a help to simplify
$\displaystyle a_n(G)=\frac{t_n(G)}{(n-1)!}$ and $\displaystyle h_n(G)=\sum _{k=1}^n {n-1 \choose k-1} t_k(G)h_{n-k}(G).$ Then need to remove $\displaystyle t_k(G)$ and obtain $\displaystyle a_n(G)=\frac{1}{(n-1)!} h_n(G)-\sum _{k=1}^n \frac{1}{(n-k)!}h_{n-k}(G)a_k(G).$
• Sep 12th 2008, 01:05 AM
bkarpuz
Quote:

Originally Posted by dimuk
$\displaystyle a_n(G)=\frac{t_n(G)}{(n-1)!}$ and $\displaystyle h_n(G)=\sum _{k=1}^n {n-1 \choose k-1} t_k(G)h_{n-k}(G).$ Then need to remove $\displaystyle t_k(G)$ and obtain $\displaystyle a_n(G)=\frac{1}{(n-1)!} h_n(G)-\sum _{k=1}^n \frac{1}{(n-k)!}h_{n-k}(G)a_k(G).$

Under the condition $\displaystyle h_{0}(G)=1$, I got
$\displaystyle a_{n}(G)=\frac{1}{(n-1)!}h_{n}(G)-\sum\limits_{k=1}^{{\color{magenta}n-1}}\frac{1}{(n-k)!}h_{n-k}(G)a_{k}(G).$

Are there any assumptions we should know?
• Sep 12th 2008, 01:30 AM
dimuk
need a help to simplify
I think I made a mistake. Your answer is correct. Let me know how did u get it.

Thanks.
• Sep 12th 2008, 01:44 AM
bkarpuz
Quote:

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
$\displaystyle {n\choose k}=\frac{n!}{k!(n-k)!},$
then isolate $\displaystyle t_{n}$ and get
$\displaystyle t_{n}=(n-1)! a_{n}.$
Substitute this into the other equation, and get
$\displaystyle h_{n}=\sum\limits_{k=1}^{n}\frac{(n-1)!}{(k-1)!(n-k)!}(k-1)!a_{k}h_{n-k}$
.....$\displaystyle =(n-1)!\sum\limits_{k=1}^{n}\frac{1}{(n-k)!}a_{k}h_{n-k}$
.....$\displaystyle =(n-1)!\Bigg[\bigg(\sum\limits_{k=1}^{n-1}\frac{1}{(n-k)!}a_{k}h_{n-k}\bigg)+a_{n}h_{0}\Bigg].$
By arranging the last expression, we easily get the desired result under the assumption
$\displaystyle h_{0}=1$. (Wink)