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Thread: Partial Fractions

  1. #1

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    Partial Fractions

    Hello good VB people,

    Now I know that you can do things like

    5/(x+4)(x-3) = A/(x+4) + B/(x-3)

    or

    3/(x²+2)(x+4) = (Ax + B)/(x²+2) + C(x+4)

    but can you apply the same idea to sin(ax)/sqrt[(x-k)(x+k)] ??

    For example

    sin(ax)/sqrt[(x-k)(x+k)] = Asin(ax)/sqrt(x-k) + Bsin(ax)/sqrt(x+k)

    Even if you can apply the same theory, I wonder what form the RHS should take. Perhaps

    sin(ax)/sqrt[(x-k)(x+k)] = (Ax + B)sin(ax)/sqrt(x-k) + (Cx + D)sin(ax)/sqrt(x+k) ???




    Any imput would be appreciated, as I may be about to make a major breakthrough with the integral

    INT[ sin(ax)/sqrt{(x-c)² + b} ] dx

    which has thus far eluded my friend and I. It looks like it should be easy, but not even Mathematica or Maple produce an answer in terms of kown functions.


    Ok cheers, Mattywoo

  2. #2
    Frenzied Member zaza's Avatar
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    Re: Partial Fractions

    Can't you do a substitution on that for x-c and then use a trig function to sort out the square root and solve the whole thing by parts?
    I use VB 6, VB.Net 2003 and Office 2010



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  3. #3

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    Re: Partial Fractions

    Tried it. It doesn't work.

    The new integral that appears on the RHS when you do it by parts includes arsinh(x) or arctan(x) and so it beaks down.

    If you simply make the trig substitution the new integral that you get looks like sin(sinh(x)) or sin(tan(x)) if you rationalise the denominator first.

    It's definitely not as easy as that. But thanks for the imput.

    Can anyone answer my query on partial fractions?

    All the best, Mattywoo

  4. #4
    Addicted Member Glaysher's Avatar
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    Re: Partial Fractions

    Tried it but didn't get anything useful

    Only reduced the problem to 1/sqrt[(x-k)(x+k)] by bringing the sin(ax) to the front of the fraction.

  5. #5

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    Re: Partial Fractions

    Sure, but can we legally do partial fractions on it?

    If so I may be able to solve the integral.

  6. #6
    Addicted Member Glaysher's Avatar
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    Re: Partial Fractions

    I tried squaring it. Doing partial fractions on what I got and then square rooting but that led to something that was hard to integrate. Other way leads to:

    1/sqrt[(x-k)(x+k)] = A/sqrt[(x-k)] + B/sqrt[(x+k)]

    1 = Asqrt(x+k) + Bsqrt(x-k)

    x=-k

    1 = Bsqrt(-2k)

    x=k

    1 = Asqrt(2k)

    Unless k = 0 this means one of the square roots will be of a negative number which would bring in complex numbers and a whole heap of problems

    So unless you can think of different numerators for A and B that will eliminate this problem I can't see how this would work

  7. #7

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    Re: Partial Fractions

    The last time I did par fracs for "something over sqrt something" the numerator looked like Ax + B, not just A.

    But thanks for your help all the same

  8. #8
    Addicted Member Glaysher's Avatar
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    Re: Partial Fractions

    Standard ones are:

    Linear factors

    eg 1/[(x + 1)(x + 2)] = A/(x + 1) + B/(x + 2)

    Repeated factors

    eg 1/[(x + 1)^2(x + 2)] = A/(x + 1) + B/(x + 1)^2 + C/(x + 2)

    Quadratic factors

    eg 1/[(x^2 + 1)(x + 2)] = (Ax + B)/(x^2 + 1) +c/(x + 2)

    I would say your problem is most like repeated factors but can't see how it would work in the same way. My reasoning was that in all cases the possibility that only constants remain on top of the fractions is there so it was worth working through to see if it worked which it didn't.

    eg Repeated factors would be assuming A = 0
    eg Quadratic factors would be assuming A = 0

    Long shot but might have worked

    Problem is that there are infinitely many ways of building the fractions. There is always another way. The above forms are useful because they always give an answer. I don't know of any more that always do.

  9. #9

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    Re: Partial Fractions

    Okay cheers. I just wondered if it could work, but I was clutching at staws admittedly!

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