# Help with partial fractions?

• June 11th 2011, 04:34 PM
Lancet
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?
• June 11th 2011, 04:41 PM
Ackbeet
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.
• June 11th 2011, 04:52 PM
TheCoffeeMachine
\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}
• June 11th 2011, 05:14 PM
TheCoffeeMachine
Quote:

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. (Thinking)
• June 11th 2011, 05:25 PM
Ackbeet
Quote:

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. (Thinking)

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.
• June 11th 2011, 05:33 PM
Lancet
That second post of yours helped me understand how to tackle something like this. Thank you!
• June 11th 2011, 08:05 PM
Prove It
Quote:

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

$\displaystyle \frac{1 - v}{(v + 1)^2} = -\frac{1}{v + 1} + \frac{2}{(v + 1)^2}$
• June 11th 2011, 08:23 PM
Lancet
Quote:

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?
• June 11th 2011, 08:27 PM
Prove It
The coefficients of the $\displaystyle v$ term are equal, so $\displaystyle A = -1$, and the constant terms are equal, so $\displaystyle A +B = 1$.
• June 11th 2011, 08:34 PM
Lancet
I get it, now. Thanks!