# Math Help - Help with partial fractions?

1. ## Help with partial fractions?

I'm familiar with the cover up method of dealing with partial fractions. That being said, this is what I'm looking at:

$\frac{1 - v}{(v+1)^2}$

Obviously, the cover up method won't work here, as we would end up with a zero in the denominator.

So, how is something like this handled? Looking elsewhere, I see some very drawn out methods using matrices - is there not a simple way to address a problem like this?

2. Just take a look at the wiki article - it's quite good. You can use a matrix method to solve the system of equations you get. Or you can use the residue method. I wouldn't say there was a "simple" way to do this one - all the methods will be a decent amount of work.

3. \begin{aligned}\displaystyle \frac{1-v}{(v+1)^2} & = \frac{1}{(v+1)^2}-\frac{v}{(v+1)^2} \\& = \frac{1}{(v+1)^2}-\frac{(v+1)-1}{(v+1)^2} \\& = \frac{1}{(v+1)^2}-\frac{1}{v+1}+\frac{1}{(v+1)^2} \\& = \frac{2}{(v+1)^2}-\frac{1}{v+1}.\end{aligned}

4. Originally Posted by Ackbeet
You can use a matrix method to solve the system of equations you get. Or you can use the residue method. I wouldn't say there was a "simple" way to do this one - all the methods will be a decent amount of work.
I could be wrong, but I think the equations that arise are far too simple to require the residue/matrix method at all.

5. Originally Posted by TheCoffeeMachine
I could be wrong, but I think the equations that arise are far too simple to require the residue/matrix method at all.
Perhaps. I only meant that the Heaviside cover-up method was not applicable here - undoubtedly the easiest and fastest way to do partial fractions when they're all simple poles.

6. That second post of yours helped me understand how to tackle something like this. Thank you!

7. Originally Posted by Lancet
I'm familiar with the cover up method of dealing with partial fractions. That being said, this is what I'm looking at:

$\frac{1 - v}{(v+1)^2}$

Obviously, the cover up method won't work here, as we would end up with a zero in the denominator.

So, how is something like this handled? Looking elsewhere, I see some very drawn out methods using matrices - is there not a simple way to address a problem like this?
Since none of the posts have mentioned the standard method for partial fractions with a repeated factor in the denominator...

\displaystyle \begin{align*} \frac{A}{v + 1} + \frac{B}{(v + 1)^2} &= \frac{1 - v}{(v + 1)^2} \\ \frac{A(v + 1) + B}{(v + 1)^2} &= \frac{1 - v}{(v + 1)^2} \\ A(v + 1) + B &= 1 - v \\ Av + A + B &= -v + 1 \end{align*}

So $\displaystyle A = -1$ and $\displaystyle A+B = 1 \implies -1 + B = 1 \implies B = 2$

So the partial fraction decomposition is

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

8. Originally Posted by Prove It

$Av + A + B = -v + 1$

$\displaystyle A = -1$ and $\displaystyle A+B = 1$

Can you clarify how you get from that first line to the second?

9. The coefficients of the $\displaystyle v$ term are equal, so $\displaystyle A = -1$, and the constant terms are equal, so $\displaystyle A +B = 1$.

10. I get it, now. Thanks!