Finding a generating function

**gusztav** · December 19th 2008, 12:47 AM

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!

**Isomorphism** · December 19th 2008, 01:10 AM

Originally Posted by gusztav

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

**Moo** · December 19th 2008, 01:12 AM

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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