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Math Help - Help with partial fractions?

  1. #1
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    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?
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  2. #2
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    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.
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     \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}
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    Quote Originally Posted by Ackbeet View Post
    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.
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  5. #5
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    Quote Originally Posted by TheCoffeeMachine View Post
    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.
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    That second post of yours helped me understand how to tackle something like this. Thank you!
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  7. #7
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    Quote Originally Posted by Lancet View Post
    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}
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    Quote Originally Posted by Prove It View Post

    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?
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  9. #9
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    The coefficients of the \displaystyle v term are equal, so \displaystyle A = -1, and the constant terms are equal, so \displaystyle A +B = 1.
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  10. #10
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    I get it, now. Thanks!
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