I have been reading the most awful book about mathematics, one which assumes that the reader knows the same vast amount as the author, a man who explains nothing (because you already have that knowledge, assumed). So what is the point, if all he si doing is making comment on the common knowledge  Yet, the point of maths is what he purports to be on about. And I think he fails, chapter by chapter.

However, I did glean a small pointer that could be offered as an extension here.

The expansion (a+b)² = a²+2ab+b² was discussed at some length in Year 8 Numeracy. However, if b is small in comparison to a, perhaps ∂a,  we have

(a+∂a)² = a²+ 2a∂a+ (∂a)².        If ∂a really is small compared to a then its square is trivially small and we can see that a small change in a, to a+∂a produces a change in  a² closely related to 2a, or, in algebra,  (a+∂a)²  - a² = 2a∂a+ (∂a)²  then see that     (∂a)²  is going to vanish and so as ∂a shrinks, so the change in the square is more obviously twice the value at the time. One helpful way to write this, is

(a+∂a)²  - a²     =    2a∂a+ (∂a)²     =    2a+∂a        so that the change in the function

∂a                              ∂a                              is 2a as ∂a becomes irrelevant or ignorable

Suppose we're looking at numbers a little bigger than one. We could use any value but 1 will do very nicely. Thenthen 1.1² = 1.21,   1.01² = 1.0201, 1.001² = 1.002001 and the change between 1²=1 and the 1.000x² result is .000x doubled. Newton saw this — yes, way back around the time of the Black Death and the Fire of London;  so did many much earlier mathematicians, looking at the effects of trying to square numbers. So if you consider the change rather than the square, the movement rather than the static situation, you can see how the square changes.

Now, in the spirit of A-level maths and the heady days of coursework, extend this idea.

If we have a cubic, (a+∂a)³ = a³+ 3a²∂a+ 3a(∂a)² + ∂a³

I think it is quite clear that the equivalent to 2a for the quadratic is going to be 3a² for the cubic.

I think we can jump to seeing that the equivalent for index n will be       n aⁿ⁻¹.

This was Newton's early grasp of differentiation (and yes, he began from looking at why squaring a number looked as if there were some clear shortcuts).

The A-level course, determined to include calculus in its content, defines a function f(x) and its differential f'(x) and explores how we can use this process. Among the many ideas this spawns there is the occasion where one is seeking a root of a function (so that function is continuous and you can differentiate it). Suppose you think a root is near x₀; then for some nearby point x₁,

f'(x₀) = f(x₀) / (x₁-x₀) is an approximation. If you prefer, estimate the tangent to the curve y = f(x) at x₀ and you will soon show this is y = f'(x₀) (x₁-x₀) +f(x₀). Where y=0 (at the root) you will have a simpler form that leads to   x₁-x₀  = - f(x₀) / f'(x₀) and from there x₁ =  x₀ - f(x₀) / f'(x₀).

And many of you will be well beyond the 'So what?' point. Yet if you start at a chosen x₀ you can generate a value for x₁ which will be nearer your root. And if you can do that, you can do it again. And again. It is a surprise how well this works and how soon your calculator can do no better.

Example: the hunt for root 2, √2. Let f(x) = x²-2, why not start with x₀=1,5. Obviously, from what we wrote earlier, f'(x) = 2x (even if we know no differentiation), so  x₁ =  x₀ -  (x₀²-2)/2x₀

After just 3 iterations we have 14 d.p. and no further correction available. Starting with a decent stab at the root rarely takes a 4th line.

I show the Excel formulae for the square root and the nth root, a fairly simple extension.For example, the 20th root of 125, x₀=2 had zero correction at 14 lines, x₄ = 1.27305011554642

Can we extend this idea further still? Provided the functions are continuous and differentiable, it is quite amazing how far this can be taken.

Example; if you have a number of linear equations represented by a matrix we have a revised formula roughly like this, where x is now a vector:      x₁ =  x₀ - M⁻¹ f(x₀).

Serious FM students will see uses for matrices and routes to avoiding finding matrix inverses explicitly.

Example: suppose x is complex, then we probably switch to a different letter set and write

z ↦ z - p(z) / p'(z) .   You might enjoy reading about Julia sets and nova fractals. It would be particularly good to be able to produce these oneself, a sort of extended A-level coursework.

Of course, at some point you wonder "Qui Carat?", which was where I reached in the awful book I described at the top of this page. A valid question to ask at school, equivalent to "Why are we doing this?" or "What is mathematics for?". An existential question for some, but at a practical level I say that we need some ways of finding answers and the precision offered by mathematics gives us solutions. We then take these overly precise results back to the relatively imprecise and more exacting practical world and do our best to make these answers fit. The problem is, fundamentally, that our models are always too simplistic and that our results, such as physical material constructions, have been modelled as one-dimensional where in practice this is far from the truth. So, to an extent, we must either make our modelling far more sophisticated or accept that, having a simplified solution, we understand some of the model results, enough to avoid some of the potential errors.The construction and refinement of models as illustrated in sixth form applied is relevant. I could wish this was introduced at school early on a course such as CDT; if this was then also shown in a maths lesson, we might open some eyes to the benefits of modelling in both the practical and theoretical senses.

DJS 20200824

Qui carat = Who cares