Partial fractions trick, repeated roots

Question

Do you know how can one extend this trick to find partial fractions coefficients when the roots of the denominator are repeated? From now, I'm just interested in the cases when the roots are algebraic.

Answer 1

I remember seeing a method for this in generatingfunctionology which I found interesting. You do the same idea of substituting convenient values of $x$, but when you have higher order terms, you can also take the derivative. For example, with $$\frac{x^2+1}{(x-1)^2(x+1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}$$ we want to solve for $A,B,C,D$ in the polynomial equation $$x^2+1=A(x-1)(x+1)^2+B(x+1)^2+C(x-1)^2(x+1)+D(x-1)^2$$ Plugging in $x=1$ and $x=-1$ gives the following equations respectively $$2=4B$$ $$2=4D$$ Now to solve for $A,C$ we can take the derivative and then plug in $x=1$ and $x=-1$. $$2x=2A(x-1)(x+1)+A(x+1)^2+2B(x+1)+2C(x-1)(x+1)+C(x-1)^2+2D(x-1)$$ We get the following equations $$2=4A+4B$$ $$-2=4C-4D$$ Solving gives $A,C=0$ and $B=D=\frac{1}{2}$. I guess this wasn't the most exciting example, but you can see that the method works and all the substitutions are still convenient ones that cancel a lot of terms. And maybe you will complain that when taking the derivative, you have to write out a lot more terms. But notice that with some foresight multiple terms will be $0$ in all cases so you can strategically skip writing those out.