Partial fraction decomposition

  • #1
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Homework Statement


Find the partial fraction decomposition of ##\displaystyle \frac{1}{x^4 + 2x^2 \cosh (2 \alpha) + 1}##

Homework Equations




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?
 

Answers and Replies

  • #2
ehild
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Homework Statement


Find the partial fraction decomposition of ##\displaystyle \frac{1}{x^4 + 2x^2 \cosh (2 \alpha) + 1}##

Homework Equations




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?
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})}##
 
  • #3
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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})}##
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?
 
  • #4
ehild
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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?
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.
 
  • #5
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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.
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:
  • #6
ehild
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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?
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.
 
  • #7
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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.
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?
 
  • #8
ehild
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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?
Yes.
 

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