Previous to this, we made the first of the two puzzle pieces that we will soon slot together to form our proof. Prior to unification, however, we must create the second. Once again, we are dealing with two variables, however, rather than being the powers of the numerator and denominator, this time, they are both in the denominator, for we are going after the proof of our statement where there are two distinct factors beneath the fraction. Expressed in symbols, our aim is to prove that constant \(A,B\)s exist where \[ \frac{1}{(x+a)^n(x+b)^m}\equiv\sum_{r=1}^n\frac{A_r}{(x+a)^r}+\sum_{r=1}^m\frac{B_r}{(x+b)^r} \] \[ \textrm{where} \quad n,m\in\mathbb{N} \quad \textrm{and} \quad a\neq b \textrm{.} \] The \(\sum\) symbol means to sum together the expression inside for all values between those given on the top and bottom (inclusive). Note also that as \(n\) and \(m\) are natural numbers, they are nonzero, so both factors are to a power of at least one. Additionally, since \(a\neq b\) we know the factors are definitely different. This means we don't have to worry about edge cases where the form looks different to what is above.
As you may have guessed, we are once again using induction here. Though, this time we are forced to use it separately for each variable. No quick rearranging to save the day. Beyond that misfortune though, we are lucky, as the application of induction here is far more straightforward, with no sideproofs peppering the process.
For the first half of this proof, we will take care of \(n\). Thus our goal is \[ \frac{1}{(x+a)^n(x+b)}\equiv\frac{D}{x+b}+\frac{C_1}{x+a}+\frac{C_2}{(x+a)^2}+\dots+\frac{C_n}{(x+a)^n} \] where all all \(C\)s, as well as \(D\) are constant. Given that we are using induction, we must first prove our base case. In this case, we want prove that constant \(A,B\) always exist, where: \[ \frac{1}{(x+a)(x+b)} \equiv \frac{A}{x+a} + \frac{B}{x+b} \] \[ \textrm{and} \quad a \neq b \textrm{.} \] Since we only want to prove that they exist, we need only provide one example of a pair \(A,B\) that satisfy the identity. To obtain this 'example', we simply use the method shown in chapter 1. Having used it, I assert that \[ A=-\frac{1}{a-b} \quad \textrm{and} \quad B=\frac{1}{a-b} \textrm{.} \] To prove that these values are indeed correct, we need only substitute them into one side of our identity, and produce the other side. Note that I am using the multiplied out form of said identity. \[ A(x+b)+B(x+a) \] \[ \equiv-\frac{1}{a-b}(x+b)+\frac{1}{a-b}(x+a) \] \[ \equiv\frac{(x+a)-(x+b)}{a-b} \] \[ \equiv\frac{a-b}{a-b} \] \[ \equiv1 \] This proves our base case. Though, the keen eyed might notice that our values of \(A\) and \(B\) might not be valid for all \(a,b\), as they would be undefined where \(a-b=0\). However, \(a-b=0\) is the same as \(a=b\), which is beyond the scope of this proof, as we stated that \(a \neq b\) in our original proposition. So, no contradiction is present.
Moving on to the inductive step, we now vary the power of the first denominator factor, making our inductive hypothesis the existence of constant \(E,F\)s where \[ \frac{1}{(x+a)^k(x+b)}\equiv\frac{F}{x+b}+\frac{E_1}{x+a}+\frac{E_2}{(x+a)^2}+\dots+\frac{E_n}{(x+a)^k} \textrm{.} \] Our goal is to prove RHS numerator constancy for a power of \((k+1)\), so we'll start by expressing that in terms of a power of \(k\), enabling us to substitute in our inductive hypothesis: \[ \frac{1}{(x+a)^{k+1}(x+b)}\equiv\frac{1}{x+a}\times\frac{1}{(x+a)^k(x+b)} \] \[ \equiv\frac{1}{x+a}\times\left(\frac{F}{x+b}+\frac{E_1}{x+a}+\frac{E_2}{(x+a)^2}+\dots+\frac{E_n}{(x+a)^k}\right) \] \[ \equiv\frac{F}{(x+a)(x+b)}+\frac{E_1}{(x+a)^2}+\frac{E_2}{(x+a)^3}+\dots+\frac{E_n}{(x+a)^{k+1}} \textrm{.} \] Notice that the first term here only has denominator factors with powers of one, and so is identical to the result proved in our base case; we will substitute accordingly, leaving \[ \equiv\frac{F_1}{(x+a)}+\frac{F_2}{(x+b)}+\frac{E_1}{(x+a)^2}+\frac{E_2}{(x+a)^3}+\dots+\frac{E_n}{(x+a)^{k+1}} \textrm{.} \] All terms remaining are inkeeping with our conjecture, so this concludes the first half of this section's proof.
Before proceeding to \(m\), let's consider some additional utility that partial fraction decomposition displays when we're dealing with two denominator factors. It gives us the ability to take an expression that is graphically quite opaque, an algebraic division, and peel back the curtain by expressing it as the sum of reciprocals, which are relatively simple functions.
Even without decomposition, we are able to locate the asymptotes of constituent reciprocals using the denominator's factors (equate them to \(0\)). Upon separating the fraction out however, we also able to obtain the coefficient that each reciprocal is multiplied by. Considering two factors to the power of one in the denominator, our fraction is the sum of just two reciprocals. Those two graphs can be drawn out, and the original expression's graph eyeballed as their sum. I recommend following along from here on desmos.
Where we keep the distance between the asymptotes constant (denominators stay the same) but vary the coefficients, we notice that one factor will assert dominance over the other. The dominant factor will bring the sum graph closer to itself than the other factor. Now varying the distance between asymptotes as well, we see that dominance is a function of both this difference, and the scale by which one coefficient is larger. Without partial fraction decomposition, we would have no idea about this battle being waged beneath the surface!
Returning to our proof, and with \(n\) out of the way, we turn our attention to \(m\). Thus, our aim is now the overall aim of this section, constant \(A,B\) where \[ \frac{1}{(x+a)^n(x+b)^m}\equiv\sum_{r=1}^n\frac{A_r}{(x+a)^r}+\sum_{r=1}^m\frac{B_r}{(x+b)^r} \textrm{.} \] We are again going to use induction, and we see that we have already proved our base case in the preceding half of this section, \[ \frac{1}{(x+a)^n(x+b)}\equiv\frac{D}{x+b}+\frac{C_1}{x+a}+\frac{C_2}{(x+a)^2}+\dots+\frac{C_n}{(x+a)^n} \textrm{.} \] Moving on to the inductive step, our assumption here is constant \(E,F\)s where \[ \frac{1}{(x+a)^n(x+b)^k}\equiv\sum_{r=1}^n\frac{E_r}{(x+a)^r}+\sum_{r=1}^k\frac{F_r}{(x+b)^r} \textrm{.} \] We will now show that when the power is \((k+1)\), the numerators are still constant, via substituting known expressions in, as we did in the first half: \[ \frac{1}{(x+a)^n(x+b)^{k+1}}\equiv\frac{1}{x+b}\times\frac{1}{(x+a)^n(x+b)^k} \] \[ \equiv \frac{1}{x+b} \times \left( \sum_{r=1}^n\frac{E_r}{(x+a)^r}+\sum_{r=1}^k\frac{F_r}{(x+b)^{r}} \right) \] \[ \equiv\sum_{r=1}^n\frac{E_r}{(x+a)^r(x+b)}+\sum_{r=1}^k\frac{F_r}{(x+b)^{r+1}} \textrm{.} \] The contents of the first sum is identical to our base case, so we will substitute that in: \[ \equiv\sum_{r=1}^nE_r\left(\frac{D}{x+b}+\frac{C_1}{x+a}+\frac{C_2}{(x+a)^2}+\dots+\frac{C_r}{(x+a)^r}\right)+\sum_{r=1}^k\frac{F_r}{(x+b)^{r+1}} \] At this point it is clear the only terms left remaining after expansion will be those compliant with our proposition, so we may be lazy and just proceed straight to: \[ \frac{1}{(x+a)^n(x+b)^m}\equiv\sum_{r=1}^n\frac{A_r}{(x+a)^r}+\sum_{r=1}^m\frac{B_r}{(x+b)^r} \] As promised, this was a very straightforward, textbook use of induction: we just substituted our inductive hypotheses and base cases in and we were there! We have proved our aim, but before concluding this chapter, there is one thing left to do.
We also need to prove an additional but very similar proposition, \[ \frac{x}{(x+a)^n(x+b)^m}\equiv\sum_{r=1}^n\frac{A_r}{(x+a)^r}+\sum_{r=1}^m\frac{B_r}{(x+b)^r} \] \[ \textrm{where} \quad n,m\in\mathbb{N} \quad \textrm{and} \quad a\neq b \textrm{.} \] This is almost identical to the just proved case. However, there is now an \(x\) in the numerator. The proof is essentially the same. We take the base case as before to be each factor to the first power, \[ \frac{x}{(x+a)(x+b)}\equiv\frac{A}{x+a}+\frac{B}{x+b} \textrm{.} \] Then, we take values for \(A\) and \(B\), \[ A=\frac{a}{a-b} \textrm{,} \quad -\frac{b}{a-b} \textrm{.} \] We will now, as before substitute these in for a proof by example: \[ A(x+b)+B(x+a) \] \[ \equiv\frac{a}{a-b}(x+b)-\frac{b}{a-b}(x+a) \] \[ \equiv\frac{a(x+b)-b(x+a)}{a-b} \] \[ \equiv\frac{x(a-b)}{a-b} \] \[ \equiv x \] This proves the base case.
The rest of the proof is identical. In fact, you could go to the top of this page, putting that one \(x\) in the numerator where necessary, and follow the proof through. However, for those who want to see this copied out with the \(x\), it's available here. For the rest, though, that is this chapter completed. Unification awaits...