# Integration and continuum

1. Aug 10, 2014

### bobie

We know that in order to be integrated a function must be continuous.
Does this imply that space and time must be a continuum?
If they were considered discrete, say at the level of Planck's unit, would this affect the integrability of functions?
It it would not, would it affect the precision of the result?

2. Aug 10, 2014

### pwsnafu

False. Continuity is sufficient for integrability, but not necessary.

No. We model space-time with a continuum. We don't know if the real world is a continuum.

No in the slightest. Integrability of functions does not rest on anything about the real world.

I don't know what you are asking. Result of what?

3. Aug 10, 2014

### bobie

Can we derive and integrate a function that is not continuous?
If you integrate a physical function on time and time cannot go to infinitesimal bu stops a Planck's unit Tp, do you get an absolutely identical result?

4. Aug 10, 2014

### haruspex

Integrate yes, differentiate no (except perhaps piecewise). Conditions for integrability depend on the theoretical basis you care to choose, e.g. Riemann-Stieltjes or Lebesgue. None require continuity.
No, but telling the difference could be beyond any feasible accuracy.

5. Aug 10, 2014

### bobie

Can you confirm that also for differentiation the fact that we consider time and space discrete is irrelevant?
The concept of continuousness is not at all related by discreteness?
Can you suggest some links to expand on this?
Thanks.

Last edited: Aug 10, 2014
6. Aug 10, 2014

### pwsnafu

It is extraordinarily unlikely that we will ever create an experiment that can distinguish between a continuous model that uses standard derivatives, and a discrete model using discrete calculus and Plank's distance. Plank's distance is that small.

Note that even if a function is not differentiable in the standard sense we can find weak derivatives or distributional derivatives of any continuous function we want. So we can still use calculus anyway.

7. Aug 10, 2014

### haruspex

Yes and no. In strict mathematical terms differentiation becomes inappropriate for step functions. But let's take a real physical example, a field generated by a potential gradient. If the potential is really a step function then the field at a point becomes meaningless. But you never can measure a field at a point - you can only measure its average value over a short distance. So long as the measuring distance is much larger than the underlying granularity no problem arises.

8. Aug 10, 2014

### bobie

Make me understand this:
let's take y= x2

if we consider the world of x discrete at the level not of Planck, but of integers, we get the same results with the only difference that the curve is not smooth but grows in little blocks.

(I read that when calculus was discovered there was a heatd discussion (as usual) between Newton and Leibniz an that the latter supported the view of calculus by blocks, then the other view prevailed).

Coming back to my example, nothing would change, we still have a derivative slighly different, perhaps (instead od 2x , 2x+1) and the results would be nearly the same-

Of course at Plancks level the difference is negligible, near to 0

So when do serious problems really arise?

Last edited: Aug 10, 2014
9. Aug 10, 2014

### HallsofIvy

Staff Emeritus
What people have been trying to tell you is that "mathematics" and "physics" are NOT the same. Mathematics has NO necessary relation to the "real world". We can use mathematics by creating MODELS that represent the "real world" but are not the real world itself- and there are always idealizations or inaccuracies in those models. The fact that we can or cannot do something in mathematics does not depend on what we can or cannot do in the "real world".

(I notice that you also posted a question on the "precision of integrals" where you seemed to think that integration is necessarily only an "approximation". You were told there that this is incorrect- integrals of (integrable) functions are exact. Perhaps you were reading about numerical approximations to integrals. The area of a circle of radius 1 is exactly $\pi$. Numerically, that is approximately 3.141592.)