# Partial fraction decomposition

1. Mar 21, 2017

### Mr Davis 97

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?

2. Mar 21, 2017

### ehild

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. Mar 21, 2017

### Mr Davis 97

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. Mar 21, 2017

### ehild

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. Mar 21, 2017

### Mr Davis 97

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
6. Mar 22, 2017

### ehild

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. Mar 23, 2017

### Mr Davis 97

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. Mar 23, 2017

Yes.