# need a help to simplify

• September 11th 2008, 11:46 PM
dimuk
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 $a_n(G)=\frac{1}{(n-1)!} h_n(G)-\sum _{k=1}^n \frac{1}{(n-k)!}h_{n-k}(G)a_k(G).$
• September 12th 2008, 01:05 AM
bkarpuz
Quote:

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?
• September 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.
• September 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
$
{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
$h_{0}=1$. (Wink)