# Partial fraction decomposition

1. Nov 5, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
Go here: http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1999s.pdf

You do not need to read the problem. In solution two for 1999 A3, I can do the partial fraction decomposition but I get something different from that... I am wondering whether that comes from the partial fraction decomposition or something else...

2. Relevant equations

3. The attempt at a solution

2. Nov 6, 2007

### HallsofIvy

Staff Emeritus
We don't need to read the problem and you won't tell us what you did- but you want someone to tell you if what you did was right? How, exactly?

3. Nov 6, 2007

### ehrenfest

It would take me like 6 lines of latex to show what I did, so I am asking whether it was the right choice of technique not whether I performed it right. I want to know how they arrived at the equality after "Note that".

4. Nov 6, 2007

### HallsofIvy

Staff Emeritus
1- 2x- x2= -(x2+ 2x+ 1-1)+1= 2- (x-1)2 so 1- 2x- x2= 0 has roots $x= 1\pm \sqrt{2}$ and so can 1-2x-x2 can be factored as $(x-1-\sqrt{2})(x-1+\sqrt{2})$. Using "partial fractions" then
$$\frac{1}{1-2x-x^2}= \frac{A}{x-1-\sqrt{2}}+ \frac{B}{x-1+\sqrt{2}}$$
gives A= $-1/2\sqrt{2}$ and B= $1/2\sqrt{2}$. The final form is got by dividing both numerator and denominator of the first fraction by $1+\sqrt{2}$ and the numerator and denominator of the second fraction by $1-\sqrt{2}$.

5. Nov 6, 2007

### ehrenfest

I tried that and then x is divided by what you want it to be multiplied by!

6. Nov 7, 2007

### ehrenfest

Can you explain how that gives the final form? I cannot figure out how to manipulate that any more.

EDIT: never mind I see

Last edited: Nov 7, 2007
7. Nov 7, 2007

### ehrenfest

You're factorization is wrong. The roots of it should be x = -1 +/- sqrt(2).