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Differential Coefficient

  1. Aug 22, 2007 #1
    1. The problem statement, all variables and given/known data
    Im supposed to find the nth differential coeff. of [tex]\frac{x^2}{(x-a)(x-b)}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    Using partial fractions, I simplified it to:


    To simplify matters, I assumed [tex]t=\frac{1}{x-a}-\frac{1}{x-b}[/tex].

    This expression now becomes [tex]\frac{1}{b-a}x^2t[/tex].

    Differentiating wrt x , [tex]\frac{1}{b-a}(2xt+x^2\frac{dt}{dx})[/tex]

    I did this two more times, and it seems to be something like this:


    Now, since [tex]t=\frac{1}{x-a}-\frac{1}{x-b}[/tex], similarly,


    Substituting this in the above equation, I get another expression which is quite messed up. So far, is what I've done right? And is there another simpler way to do this? It seems to me that I've missed something somewhere which would make this problem a lot simpler, but I just cant find it. I dont think brute force is the only way to do this, and a push in the right direction would really be very welcome.
  2. jcsd
  3. Aug 22, 2007 #2


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    The partial fraction stuff isn't going well. First divide x^2 by the denominator. That will give you 1+(Ax+B)/((x-a)*(x-b)). Find A and B. Now with a little work you can find constants C and D so that this becomes 1+C/(x-a)+D/(x-b). Now THAT'S partial fractions.
  4. Aug 23, 2007 #3
    I dont get it. What did I do wrong? What I did pans out. I was just hoping there was a more elegant solution to the problem.
  5. Aug 23, 2007 #4
    One thing I was thinking of was taking x to be a complex number of the form [tex]x=cos\theta + isin\theta[/tex], and then assuming [tex]cos\theta -a = rcos\phi[/tex] and [tex]sin\theta =rsin\phi[/tex] and going on from there if the current situation is an incomplete solution. Any insights would be highly appreciated.
  6. Aug 23, 2007 #5


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    You didn't do anything wrong - you just didn't 'partial fraction' as much as you could. You still have an expression that gets overly complicated as you take more and more derivatives. As I said, you could reduce the expression to something of the form A+B/(x-a)+C/(x-b) which is manageable.
  7. Aug 24, 2007 #6
    Ok. Ill try that.
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