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Partial fraction decomposition

  1. Mar 21, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the partial fraction decomposition of ##\displaystyle \frac{1}{x^4 + 2x^2 \cosh (2 \alpha) + 1}##

    2. Relevant equations


    3. The attempt at a solution
    Using the identity ##\displaystyle \cosh (2 \alpha) = \frac{e^{2 \alpha} + e^{- 2\alpha}}{2}##, we can get the fraction to the form ##\displaystyle \frac{1}{(x^2 + e^{2 \alpha})(x^2 + e^{- 2 \alpha})}##. Since we have two irreducible quadratic factors, it would seem that we would now try to find A, B C, and D such that ##\displaystyle \frac{1}{(x^2 + e^{2 \alpha})(x^2 + e^{- 2 \alpha})} = \frac{Ax + B}{x^2 + e^{2 \alpha}} + \frac{Cx + D}{x^2 + e^{-2 \alpha}} ##. But in the solutions to my book, it just directly says to find A and B such that ##\displaystyle \frac{1}{(x^2 + e^{2 \alpha})(x^2 + e^{- 2 \alpha})} = \frac{A}{x^2 + e^{2 \alpha}} + \frac{B}{x^2 + e^{-2 \alpha}} ##. Why am I wrong in assuming that the numerators must be linear factors?
     
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  3. Mar 21, 2017 #2

    ehild

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    It is not wrong but not necessary to assume linear terms. You can take y=x2 the variable, then you have the fraction as ##\frac{1}{(y + e^{2 \alpha})(y + e^{- 2 \alpha})}##
     
  4. Mar 21, 2017 #3
    Why isn't it necessary to assume linear terms? For example, if we were to do the PFD of ##\displaystyle \frac{1}{x(x^2 + 1)}##, we have to assume that ##x^2 + 1## has a linear term in the numerator. Why would these two cases be different?
     
  5. Mar 21, 2017 #4

    ehild

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    The fraction has only powers of x2.But try to do as you started, having linear and constant terms in the numerator and see what you get.
     
  6. Mar 21, 2017 #5
    I get the same answer as I would if I had just constants. In other words, the coefficients of x in each numerator are just 0... What's the general theory behind this? I always thought that quadratic terms had to have linear factors in the numerator, at least that's what it says in articles on partial fraction decomposition. Are they wrong?
     
    Last edited: Mar 21, 2017
  7. Mar 22, 2017 #6

    ehild

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    No, they have to have linear factors, but the coefficients can be zero.
    As I wrote in Post#2, the fraction does not have linear terms in x. You can take y=x2 the variable, then you have the fraction as ##\frac{1}{(y+e^{2α})(y+e^{-2α})} ##, and it has only constant term in the numerator of the partial fractions.
     
  8. Mar 23, 2017 #7
    So are you saying that since the denominator just has factors with squares of x, it's a special case in which the linear terms in the numerator will have coefficients of 0 for x?
     
  9. Mar 23, 2017 #8

    ehild

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    Yes.
     
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