1. ## polynomial decomposition

I'm not able to show that this expression:

$H_1 = \frac{a^3(1+x^{-1})^3}{(1+a+(a-1)x^{-1})(1+a+a^2+(2a^2-2)x^{-1}+(1-a+a^2)x^{-2})}$

can be written

$H_2 = \frac{1}{2} \left( \frac{a-1+(1+a)x^{-1}}{1+a+(a-1)x^{-1}} + \frac{1-a+a^2+(2a^2-2)x^{-1}+(1+a+a^2)x^{-2}}{1+a+a^2+(2a^2-2)x^{-1}+(1-a+a^2)x^{-2}} \right)$

Partial fraction decomposition of $H_1$ gives

$H_3 = \frac{2(a^3x+a^3-ax+a)}{(a^2+a+1)(a^2x^2+2a^2x+a^2+ax^2-a+x^2-2x+1)} + \frac{a^3}{a^3+2a^2+2a+1} + \frac{2a}{(a+1)(ax+a+x-1)}$

I found $H_3$ by plugging $H_1$ into wolframalpha.

But I can't find a way to $H_2$ from $H_3$ either.

Any hints or things to try?

I have tried to write

$H_1 = \frac{A+Bx^{-1}}{1+a+(a-1)x^{-1}} + \frac{C+Dx^{-1}+Ex^{-2}}{1+a+a^2+(2a^2-2)x^{-1}+(1-a+a^2)x^{-2}}$

then multiplying through by the denominator gives

$a^3(1+x^{-1})^3 = (A+Bx^{-1})(1+a+a^2+(2a^2-2)x^{-1}+(1-a+a^2)x^{-2}) + (C+Dx^{-1}+Ex^{-2})(1+a+(a-1)x^{-1})$

and then evaluate at $x^{-1} = -\frac{1+a}{a-1}$ such that it reduces to

$a^3(1-\frac{1+a}{a-1})^3 = (A-B\frac{1+a}{a-1})(1+a+a^2-(2a^2-2)\frac{1+a}{a-1}+(1-a+a^2)(\frac{1+a}{a-1})^2)$

But that gives one equation with 2 unknowns and thus appear to be a dead end!

2. ## Re: polynomial decomposition

Hey niaren.

Are you allowed to go from partial fraction decomposition back to original decomposition or must you go from original to partial?

In a lot of proofs, people actually do the reverse if they can't do the original method.

3. ## Re: polynomial decomposition

Hi chiro, thanks for your reply. In going from $H_1$ to $H_2$ another guy told me it could be done by partial fraction decomposition. He said it was easy but tedious. He is quite experienced so I took his word for it. To start with I thought it sounded reasonable but now that I have look at it for some time I'm starting to doubt that going to $H_2$ is all that easy and moreover I doubt that going through partial fraction decomposition is maybe not more direct than some alternative method. So basically what I'm trying to find out is if there is some sequence of steps that can be applied starting from $H_1$ and have end result $H_2$.