Simplify the function

\[f(x) = \dfrac{(x-1)-(x-2)}{(x-1)(x-2)} + \dfrac{(x-2)-(x-3)}{(x-2)(x-3)} + \dfrac{(x-3)-(x-4)}{(x-3)(x-4)} + \dfrac{(x-4)-(x-5)}{(x-4)(x-5)}.\]

We will explore the possible starting points given in the suggestion:

### Common denominator

When we add fractions we need a common denominator. In this case it would be \((x-1)(x-2)(x-3)(x-4)(x-5).\) The first term would become \[\dfrac{\big((x-1)-(x-2)\big)(x-3)(x-4)(x-5)}{(x-1)(x-2)(x-3)(x-4)(x-5)}.\]

Writing out the whole function would be very long, so we will introduce notation that will help us write \(f(x)\) in a shorter way, and should make it easier to spot any structure or symmetry that occurs. \[(x - a_1)(x-a_2)...(x-a_n) = [a_1a_2...a_n]\] So \((x-1)(x-2)(x-3)\) would be written \([123]\).

\[f(x) = \dfrac{([1]-[2])[345] + ([2]-[3])[145] + ([3]-[4])[125] + ([4]-[5])[123]}{[12345]}\]

### Simplifying the numerator

It turns out that the numerator for each fraction simplifies to \(1\). This seems helpful but then we are left with \[f(x) = \dfrac{1}{(x-1)(x-2)} + \dfrac{1}{(x-2)(x-3)} + \dfrac{1}{(x-3)(x-4)} + \dfrac{1}{(x-4)(x-5)}.\] Where might we go from here?

### Simplifying terms

Each term can be split into two fractions, e.g. \[\dfrac{(x-1) - (x-2)}{(x-1)(x-2)} = \dfrac{(x-1)}{(x-1)(x-2)} - \dfrac{(x-2)}{(x-1)(x-2)},\] which gives us fractions with common factors that can be simplified.

\[\begin{align*} f(x) &= \dfrac{(x-1)}{(x-1)(x-2)} - \dfrac{(x-2)}{(x-1)(x-2)} + \cdots + \dfrac{(x-4)}{(x-4)(x-5)}- \dfrac{(x-5)}{(x-4)(x-5)} \\ &= \dfrac{1}{(x-2)} - \dfrac{1}{(x-1)} + \dfrac{1}{(x-3)} - \dfrac{1}{(x-2)}+ \dfrac{1}{(x-4)} - \dfrac{1}{(x-3)}+ \dfrac{1}{(x-5)}- \dfrac{1}{(x-4)} \end{align*}\]We can see that many of the terms appear twice, in fact most terms have a pair with which they will cancel. We call this a telescoping sum. This leaves us with

\[f(x) = \dfrac{1}{x-5} - \dfrac{1}{x-1}.\]

This method is based around the idea of splitting a fraction up into the sum of simpler fractions which is called partial fraction decomposition.

We ended up with \(f(x)\) written in two different forms, let’s call them \(f_1\) and \(f_2\): \[f_1(x) = \dfrac{1}{x-5} - \dfrac{1}{x-1} \text{ and } f_2(x) = \dfrac{4}{(x-1)(x-5)}.\]

- How would we show that \(f_1\) can be rewritten as \(f_2\)?
- Starting with \(f_2\), how would we show it can be rewritten as \(f_1\)?

Is it possible to split each of the following fractions up into the sum of two fractions?

- \(\dfrac{1}{(x-2)(x-3)}\)

Depending on the approach taken to simplify \(f(x)\), we may have already noticed that \(\dfrac{1}{(x-2)(x-3)} = \dfrac{(x-2)-(x-3)}{(x-2)(x-3)}.\)

Therefore we can use the simplifying terms approach from above and split this into two fractions: \[\begin{align*} \dfrac{1}{(x-2)(x-3)} &= \dfrac{(x-2)-(x-3)}{(x-2)(x-3)} \\ &= \dfrac{(x-2)}{(x-2)(x-3)} - \dfrac{(x-3)}{(x-2)(x-3)} \\ &= \dfrac{1}{(x-3)} - \dfrac{1}{(x-2)} \end{align*}\]How are the examples below different from this? How are they the same?

- \(\dfrac{7}{(x-1)(x-8)}\)

- \(\dfrac{1}{(x-2)(x-8)}\)

This is different in that we can’t just rewrite the numerator in terms of the brackets that appear in the denominator. Or can we? What if we have multiples of each bracket? \[\dfrac{1}{(x-2)(x-8)} = \dfrac{A(x-2) - B(x-8)}{(x-2)(x-8)}\]

This would give us \(Ax - 2A - Bx + 8B = 1\) and so we can write down two equations, \(A-B = 0\) and \(-2A + 8B = 1\), which solve to give \(A = \frac{1}{6}\) and \(B = \frac{1}{6}\).

\[\dfrac{1}{(x-2)(x-8)}= \dfrac{1/6}{(x-8)} - \dfrac{1/6}{(x-2)}\]

Could we have worked this out in a different way if we thought about what \(\dfrac{6}{(x-2)(x-8)}\) could be rewritten as?

Is it possible to write all fractions in the form \(\dfrac{p}{(x-q)(x-r)}\) as \(\dfrac{A}{(x-q)} + \dfrac{B}{(x-r)},\) where \(p, q, r, A\) and \(B\) are constants? What happens if \(q = r\)?