The time is upon us. Finally, we may slot together our previous two proofs to achieve the generality we have sought since the beginning. I request only a little more of your patience.
By way of brief reminder, our purpose is to prove the existence of constant numerators in a partial fraction decomposition. Symbolically, this conjecture appears as the assertion that constant \(A\)s always exist, where \[ \frac{ \sum_{r=1}^m b_rx^{r-1}}{\prod_{r=1}^l(x+a_r)^{n_r}} \equiv \sum_{r=1}^l \sum_{p=1}^{n_r} \frac{A_{rp}}{(x+a_r)^p} \] \[ n,m,l \in \mathbb{N} \quad a,b \in \mathbb{R} \quad m = \sum_{r=1}^ln_r \textrm{.} \] We previously made use of the summation symbol (\(\sum\)) back in chapter III. However, there is still one as of yet unidentified figure: the \(\prod\). This is a capital pi, whose lowercase sibling \(\pi\) is no doubt a familiar friend. It denotes a product in much the same way as the summation symbol. Quite simply, you multiply the captive expression together for all integer values of the variable between and including the limits stated at the top and bottom of the symbol. This notation is very helpful as it saves us writing lots of dots... You may have noticed another unfamiliar expression, \(A_{rp}\). Here, the \(r\) and \(p\) are not digits but seperate subscripts. Imagine a grid of points, where a number is written on each point. Then, \((r,p)\) are the coordinates of a point, and \(A_{rp}\) is the number 'stored' on that point. Thus, each unique combination of \(r\) and \(p\) refers to a different variable.
Having recalled our overall aim, we now approach its proof. To begin, we use a slightly more limited statement, only considering a single expression in the numerator. Thus our aim is to prove constant \(A\)s always exist, where \[ \frac{x^{\sum m}}{\prod_{r=1}^l(x+a_r)^{n_r}}\equiv\sum_{r=1}^l\sum_{p=1}^{n_r}\frac{A_{rp}}{(x+a_r)^p} \] \[ \sum m \lt \sum n \] Notice that, for brevity, sometimes I have not included limits in sums. The limits in these cases are self-evident. For example, \(\sum m\) means the sum of all the different values denoted as subscripts of \(m\). With notation out of the way, I will now proceed to use induction a final time.
Of course, as with any proof by induction, we begin with our base case. We are inducing our statement by varying the number of distinct factors, consequently said base case deals with one distinct factor. Miraculously, we already proved this in chapter II! Therefore, we may simply state that there are constant \(A\)s where \[ \frac{x^{m_1}}{(x+a_1)^{n_1}}\equiv\sum_{r=1}^{n_1}\frac{A_{1r}}{(x+a_1)^r} \textrm{.} \] From here, we proceed to our inductive hypthosesis, allowing us the assumption that \(A\)s are constant where \[ \frac{x^{\sum_{r=1}^{k}{m_r}}} {\prod_{r=1}^k(x+a_r)^{n_r}}\equiv\sum_{r=1}^k\sum_{p=1}^{n_r}\frac{B_{rp}}{(x+a_r)^p} \] \[ \sum_{r=1}^{k}{m_r} \lt \sum_{r=1}^kn_r \textrm{.} \] Now, we will use our assumption, as well as previous proofs, to prove that the statement holds with \(k+1\) distinct factors.
To begin, we'll create an expression, and manipulate it to make use of our inductive assumption. \[ \frac{x^{\sum_{r=1}^{k+1}{m_r}}} {\prod_{r=1}^{k+1}(x+a_r)^{n_r}} \equiv \frac{x^{m_{k+1}}}{(x+a_{k+1})^{n_{k+1}}} \times \frac{x^{\sum_{r=1}^{k}{m_r}}}{\prod_{r=1}^k(x+a_r)^{n_r}} \] \[ \equiv\frac{x^{m_{k+1}}}{(x+a_{k+1})^{n_{k+1}}}\times\sum_{r=1}^k\sum_{p=1}^{n_r}\frac{B_{rp}}{(x+a_r)^p} \] Unfortunately proceeding from here is a little difficult. Despite appearances, the expression on left here is not identical to our base case. This is because, if our proof is to be valid for all numerator powers permitted by the original definition of partial fraction decomposition, we need to consider the possibility that \(m_{k+1} = n_{k+1}\).
To understand why this is necessary, we must consider adding inequalities, specifically those involving integers. Say we have four integers, \(a\), \(b\), \(c\) and \(d\), where \(a \lt b\) and \(c \lt d\). It follows that the greatest values of \(a\) and \(c\) are \(b-1\) and \(d-1\) respectively. Therefore, their sum's maximum should be \(b+d-2\). Let us now add the inequalities so that we have \(a+c \lt b+d\). If considered in isolation, this new inequality would suggest that \(a+c\) has a maximum of \(b+d-1\), which is one greater than our previously calculated maximum. Herein lies the problem: the inequality we want to 'follow' is the one of the sums. Therefore, we must consider the case where \(b=d\) in order to have covered all possibilities.
First, we will consider the case where \(m_{k+1} \lt n_{k+1}\). With this inequality in place, we may easily decompose the left hand fraction using our base case. \[ \frac{x^{m_{k+1}}}{(x+a_{k+1})^{n_{k+1}}}\times\sum_{r=1}^k\sum_{p=1}^{n_r}\frac{B_{rp}}{(x+a_r)^p} \] \[ \equiv\sum_{r=1}^{n_{k+1}}\frac{C_r}{(x+a_{k+1})^r}\times\sum_{r=1}^k\sum_{p=1}^{n_r}\frac{B_{rp}}{(x+a_r)^p} \] The results of expanding this product will only consist of terms with a constant numerator and two distinct factors in the denominator. The decomposition of this kind of expression was proved earlier in chapter III, so we can safely proceed to say that this case decomposes with constant numerators.
Moving on, we will now deal with our second case where \(m_{k+1}=n_{k+1}\). \[ \frac{x^{m_{k+1}}}{(x+a_{k+1})^{n_{k+1}}}\times\sum_{r=1}^k\sum_{p=1}^{n_r}\frac{B_{rp}}{(x+a_r)^p} \] \[ \equiv x\times\frac{x^{m_{k+1}-1}}{(x+a_{k+1})^{n_{k+1}}}\times\sum_{r=1}^k\sum_{p=1}^{n_r}\frac{B_{rp}}{(x+a_r)^p} \] The unexpanded fraction here is now compliant with our base case (the order of the numerator is lower than that of the denominator), so we can expand it as follows: \[ \equiv x\times\sum_{r=1}^{n_{k+1}}\frac{C_r}{(x+a_{k+1})^r}\times\sum_{r=1}^k\sum_{p=1}^{n_r}\frac{B_{rp}}{(x+a_r)^p} \] This is very similar to the situation faced in the first case, the only difference being that the expanded terms here will have numerators with a factor of \(x\), fortunately, the case of \(x\) in the numerator and two factors in the denominator was also proven in chapter III, so we can say that this, too, will decompose with constant numerators. This concludes our proof by induction for a single term in the numerator.
To make this such that we have a polynomial numerator,
we need only sum together many instances of this (with varying powers in the numerator), each multiplied by a constant.
This will obviously also have constant numerators, ending our proof.
QED.
\[
\square
\]
Throughout this proof, its been a fairly core principle that we've disregarded the values of constants, taking note only in certain cases, such as when they are zero. This begs the question: what happens when we don't ignore these numbers? What if we keep track of them as we build up the layers of generality? Will we get a general formula? The answer is yes, to an extent.
For example, we can find a formula for the decomposition of the kind of expression dealt with in chapter II using binomial expansion. \[ \frac{x^n}{(x+a)^m} \equiv \frac{([x+a]-a)^n}{(x+a)^m} \] \[ \equiv \sum_{r=0}^{n} \frac{{n\choose r} (-a)^{n-r}}{(x+a)^{m-r}} \] Continuing along a similar path, I managed to find, but not prove, a formula for an expression like the one covered in chapter III. \[ \frac {1} {(x+a)^n (x+b)^m} \equiv \sum_{r=1}^m \frac{ (-1)^{r+1} {{r+n-2} \choose {r-1}} A^{n+r-1}}{(x+b)^{m-r+1}} + (-1)^m \sum_{r=1}^n \frac{ {{r+m-2} \choose {m-1}} A^{r+m-1}}{ (x+a)^{n-r+1} } \] \[ \textrm{where} \quad A=\frac{1}{a-b} \] I believe it is likely possible to piece these together in order to derive a completely general formula but nested recursive summations are rather headache-inducing, so it will be a while before I return to this! One of the primary uses of finding such general formulas is that they may be used to easily compute values. In the next chapter, I will be looking into another approach to computing decompositions, including a program capable of decomposing fractions, hopefully nicely ending this series of posts on partial fractions!