# sequence

• Sep 21st 2011, 01:02 PM
surjective
sequence
Hello,

I have the following:

$\displaystyle A(x)=(x+1) + \frac{1}{2x^{2}-3x+1}$

How do I find a sequence $\displaystyle a_{0},a_{1},a_{2},...$ such that $\displaystyle A(x)=\sum_{i=0}^{\infty} a_{i}x^{i}$

Thanks.
• Sep 21st 2011, 01:09 PM
alexmahone
Re: sequence
Quote:

Originally Posted by surjective
Hello,

I have the following:

$\displaystyle A(x)=(x+1) + \frac{1}{2x^{2}-3x+1}$

How do I find a sequence $\displaystyle a_{0},a_{1},a_{2},...$ such that $\displaystyle A(x)=\sum_{i=0}^{\infty} a_{i}x^{i}$

Thanks.

$\displaystyle \frac{1}{2x^{2}-3x+1}=\frac{1}{(1-x)(1-2x)}=(1-x)^{-1}(1-2x)^{-1}$

You could now expand using the binomial theorem.
• Sep 21st 2011, 01:31 PM
surjective
Re: sequence
Could you elaborate a bit. It's not quite clear. I mean, when expanded how do I include x+1 ?? Assistance would be appreciated!!!
• Sep 21st 2011, 01:49 PM
alexmahone
Re: sequence
Quote:

Originally Posted by surjective
Could you elaborate a bit. It's not quite clear. I mean, when expanded how do I include x+1 ?? Assistance would be appreciated!!!

$\displaystyle (1-x)^{-1}(1-2x)^{-1}=(1+x+x^2+...)(1+2x+4x^2+...)$

After simplifying the above expression, simply add $\displaystyle x+1$ to obtain the required power series.