Ask Uncle Colin: About Partial Fractions
Dear Uncle Colin,
When I express $\frac{x^2}{(x-1)(x-4)^2}$ in partial fractions, why do I need to use three separate fractions? I accept that that’s how it works, I just want to know why!
Pupils Absolutely Raging: “Two Is Adequate, Like!”
Hi, PARTIAL, and thanks for your message!
I’ve got a handwavy explanation and one that goes into a bit more depth.
Wave your hands in the air like you just don’t care
As a general principle, you need the top of a partial fraction to be one degree lower than the bottom. The $(x-4)^2$ part is quadratic, so it needs to have a linear expression – something like $ax+b$ – on the top.
However, you can rewrite this as $a(x-4) + (b+4a)$ or $a(x-4) + c$, which gives you the three fractions you were expecting.
Take your hands out of the air and seek a small black square
The top of any (proper) fraction with $(x-1)(x-4)^2$ as a denominator could be any quadratic expression.
An arbitrary quadratic expression has three coefficients, so we will need to solve for three constants.
More specifically, we’re trying to change the basis in our quadratic. At the moment, it’s written as $x^2$, but we’d ideally like to write it in a way that’s more convenient for cancelling down fractions.
The space of quadratic expressions is three-dimensional, so we need three linearly independent expressions, picked as conveniently as possible – by which I mean, so that they will cancel with things on the bottom to leave relatively nice fractions.
Two obvious candidates are $(x-1)$ and $(x-4)^2$, which leave $(x-4)^2$ and $(x-1)$ respectively – but we need a third to span the space.
I stress that we can pick whatever we like here, so long as it’s not a linear combination of the other two, but the most convenient choice is $(x-1)(x-4)$, which just leaves an $x-4$ on the bottom.
So, we can rewrite any quadratic expression as $a(x-1) + b(x-4)^2 + c(x-4)(x-1)$, and anything in that form, when divided through by the original bottom, is relatively simple to integrate, differentiate or apply the binomial expansion to.
Hope that helps!
- Uncle Colin