“He who sees things in their growth and first origins will obtain the clearest view of them” (Aristotle)

Our most basic ‘mathematical’ impressions are neither numerical nor geometrical. We situate an object (desired or feared) relative to certain familiar objects whose position we know or think we know. We position the new object by means of words such as ‘near’, ‘under’, ‘on top of’, ‘alongside’ and so forth. Babies and toddlers are not alone in working this way : even adults (including mathematicians and scientists) fall back on this method when confronted by a real life situation as opposed to a situation in a laboratory). If I ask where the nearest church or Post Office is, no one turns on their mobile phone complete with SATNAP (?) and tells me the latitude and longitude or even  the exact distance and orientation with respect to the present location. He or she would say, “Go right, then left, it’s the other side of the Church”, where the church is an easily recognizable landmark. If I ask where my pen or pencil is, I won’t be given the co-ordinates relative to the bottom left hand corner of the room; I will be told “It’s under the table”.
Now these relations are neither numerical or geometrical, even less algebraic : they are, as Bohm pertinently pointed out in an interview ‘topological’. Topology is that branch of geometry that deals with proximity and connectivity to the exclusion of metrical concepts. Perhaps we ought to make topology the first ‘science’ learned at school, though it probably does not need to be taught at this stage. Regrettably, topology is an exceedingly abstruse (and largely useless) branch of mathematics that only specialists study.
After ‘topological’ concepts, we have numerical ones, ‘How many of this?’ ‘How many of that?’  So-called primitive tribes get on perfectly well in their environment without numbers and arithmetic operations but the latter become necessary when we have a commercial and bureaucratic society such as developed in the Middle East in Assyria, Babylon and so on. Geometry was a somewhat later development, becoming necessary only when it was vital to assess accurately the areas of plots of land (for taxation purposes and also when State buildings became so large and complex that the coup d’oeil and rule of thumb of the craftsman/builder was no longer adequate. You can raise a pretty good straight wall without any calculation or mathematics but if you want to build a pyramid orientated in a particular way with regard to the stars, you require both a relatively advanced arithmetic and geometry.
After geometry came algebra (unknown to the Greeks) and algebra has so swamped mathematics that even Euclidian geometry, itself thoroughly idealized and remote from the natural world, seems homely and ‘concrete’ today since you can at least draw lines and, if you wish, even make spheres out of Blu-tack and play around with them.
A problem that was posed in the mathematical magazine M500 prompted some reflections on the gulf between geometry and Calculus and the even greater gulf between both of these and physical reality. The problem was about proving the basic limit     lim   (sin θ)/θ  = 1
θ → 0 . As a mathematical fundamentalist, I view sines and cosines as essentially ratios between line segments rather than infinite series.

lim sin θ)/θ  = 1  as  θ → 0 If  angle PÔD = θ  in radians, PE, the arc subtended by the angle θ, is and  > PD = r sin θ. Also, QE = r tan θ > rθ> PD

So   r sin θ  <  rθ  < r tan θ

This inequality holds for any circle with r > 0 and all angles θ for which sin θ, cos θ and tan θ are defined. We take θ as positive (clockwise from the x axis) and, since we are only concerned with small angles,  0 < θ < π/2.

Dividing by r sin θ which is positive and non-zero we have

1  <  θ/(sin θ)   <  1/cos θ
1 has limit 1 since it is never anything else.
If we can show that the limit of 1/cos θ is 1 as θ → 0  the expression θ/(sin θ)   will be squeezed between two limits.
What can at once be deduced from the diagram is :

1.  r cos θ must be smaller than the radius and so, for unit radius,  0 < cos θ < 1

2. As θ decreases, cos θ increases, or “If f < θ, cos f > cos θ

1/cos θ is thus monotonic decreasing and has a lower limit of 1 which is sufficient to establish convergence. If one wants to apply the canonical test, we have to find a δ  such that, for any ε  > 0  whenever 0 < θ <  δ

1– 1/cos θ)  < e
With δ < cos–1 ((1/(1+ ε)) we should be home and, applying the ‘sandwich principle’ for limits, we have

lim        θ/(sin θ)  = 1
θ → 0

Turning this on its head, we finally obtain
lim   (sin θ)/θ  = 1
θ → 0

Note, however, that θ is the independent variable — sin θ depends on θ and not the reverse.

From here we can find the  derivative of sin θ  in a straightforward manner by using nothing more than the definition of the derivative and the ‘Double Angle’  formula sin (A + B) = sin A cos B + cos A sin B which can be easily proved geometrically for all angles A, B  where
0 < A < π/2 and 0 < B <π/2  (Note 1).

However, what’s all this got to do with the well-known power series?

sin θ = θ – θ3/3! + θ5/5!  – θ7/7! + ……

Define a convergent power series f(x) with the convenient property that d2 f(x)/dx2 = – f(x)

Setting A0 = 0, A1 = 1 and equating coefficients we eventually end up with

f(x) = x –x3/3! +  x5/5! – x7/7!  ……+ (–1)n x2n+1 /(2n–1)!…

But I’m none too happy about identifying the above series with sin x (and its derivative with cos x). For, if sin x and cos x are  geometric relations between line segments, when there is no triangle, there can be no sine or cosine. For me, geometric sin x is undefined at x = 0 (and likewise at x = π/2 &c.) although the limit of sin x as x → 0 is certainly 0. (It is distressing how often it seems necessary to point out, even to mathematicians, that the existence of a limit does not in any way guarantee that this limit is actually attained.)
All in all, I would feel a lot easier if the ‘sin x power series’ were derived (or defined) recursively term by term along with a demonstration that the difference between f(x) and sin x is always decreasing as we add more terms with limit zero. For sin x is, in  my eyes, itself the limit of a power series as n increases without bound, i.e. If  0 < x < π/2

lim  f(x) = x –x3/3! +  x5/5! …+ (–1)n x2n+1/(2n–1)!   =  sin x
n →

Realirty is Discrete

A more general point needs to be made. Practically all proofs in analysis and Calculus depend on the assumption that the independent variable (in this case the angle θ) can be made arbitrarily small. This is quite legitimate if we restrict ourselves to pure mathematics. But Calculus was invented by Newton and Leibnitz to elucidate problems in physics. Translated into physical terms, the basic assumption of Calculus, is equivalent to the presumption that space and time are ‘infinitely divisible’. But I do not believe they are for both logical and observational reasons. There is a growing (but still minority) view amongst theoretical physicists that Space/Time is ‘grainy’, i.e. that there are minimal distances and minimal intervals of time just as there are minimal transfers of energy (quanta). If this proves to be the case, Calculus and a lot else besides constitutes a very misleading model of a reality that is essentially discrete. The great majority of differential equations are, in any case, unsolvable analytically and increasingly the trend is to slog things out iteratively with high-speed computers taking things to the level of precision required by the conditions of the problem and then stopping. Dreadful to say so, but it looks like Calculus’s reign, like that of the dinosaurs, is drawing to a close and that the future will go to algorithmic methods, genetic or otherwise.

Sebastian Hayes

Note 1   Theorem  d(sin x)   =  cos x
dx

Proof     By the definition of the derivative we have

d(sin x)   =  lim       sin (x + h) – sin x
dx          h → 0                    h

=  lim       (sin x) (cos h) + (cos x)(sin h) – sin x
h → 0                               h

=  lim    (sin x) (cos h – 1)   +     lim   (cos x)(sin h)
h → 0                  h                   h → 0       h

It is not obvious that (cos h – 1)/h  has a limit but (after consulting a Calculus textbook) I substitute
– 2 sin2 (h/2)  for  (cos h – 1)  and, employing the limit we have already found ((sin θ)/θ = 1) we have

lim    (sin x) (cos h – 1)   =  sin x  lim  {– (sin h/2)(sin h/2)}
h → 0                 h                             h → 0                            h/2

=   sin x  lim  (– (sin h/2)) (1)    =  sin x × 0 = 0
h → 0

So all we have left is       lim   (cos x)(sin h)
h → 0                 h

Employing once again the ubiquitous lim sin θ/θ = 1 as θ → 1   I end up with the desired
d(sin x)   =  cos x  lim   (sin h)/h    =   cos x
dx                    h → 0