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October 21st, 2018, 08:16 PM   #21
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I'm trying to understand your reasoning man, I really am. I just don't seem to get it.

Quote:
 Originally Posted by v8archie Yes, but that's rather different in most contexts - we don't generally discard the imaginary part of complex numbers in producing our solutions, so consistency matters.
What does "discarding imaginary parts" have to do with consistency, or with consistency mattering?

And how do you know that the complex numbers are consistent? How would you show it?

October 21st, 2018, 09:12 PM   #22
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Quote:
 Originally Posted by Micrm@ss What does "discarding imaginary parts" have to do with consistency, or with consistency mattering?
Contrast with the hyperreals. There we literally discard any non-real part of the answer to get the answer in the reals. It doesn't matter what the infinitesimal part of the answer is, or whether a different method produces a different infinitesimal term in the answer, because we throw it away.

We don't do that with the complex numbers. Where we require a real number answer, we demand that the imaginary part of the solution be zero - we don't just discard whatever it happens to be.

Quote:
 Originally Posted by Micrm@ss And how do you know that the complex numbers are consistent? How would you show it?
I don't know that they are. You asked if I believe them to be. Nobody has yet discovered an inconsistency that I've heard of. If I understand Gödel correctly, proving it to be consistent may very well be impossible. Perhaps a contingent proof (they are consistent if the reals are) would be possible.

 October 21st, 2018, 09:49 PM #23 Senior Member   Joined: Oct 2009 Posts: 905 Thanks: 351 My apologies for being unclear. Whenever I say consistent, I always mean relative consistent. Meaning: consistent if ZF is consistent. That's the best we can do for many systems. So with this clarification in mind, what do you think about the consistency of the complex numbers and the hyperreals? Are they consistent? I also don't really see what this has to do with "discarding infinitesimal parts". I'll think about it some more to see if I can make sense of it.
 October 21st, 2018, 10:13 PM #24 Senior Member   Joined: Sep 2016 From: USA Posts: 680 Thanks: 454 Math Focus: Dynamical systems, analytic function theory, numerics As someone who only has passing familiarity with hyperreals, I can say I've never been motivated to bother any serious study since they don't appear to equip you with any tools which aren't already straight-forward applications of limits. Can you (anyone) give an example of some application, technique, or concept which is only possible or significantly easier/more powerful/etc when using infinitesimals?
October 21st, 2018, 10:34 PM   #25
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 Originally Posted by Micrm@ss I also don't really see what this has to do with "discarding infinitesimal parts". I'll think about it some more to see if I can make sense of it.
Derivative of $f(x)=x^2$ with infinitesimals ($\epsilon$ is infinitesimal):
\begin{align} f'(x) &= \frac{f(x+\epsilon) - f(x)}{\epsilon} \\ &= \frac{(x^2 + 2\epsilon x + \epsilon^2) - x^2}{\epsilon} \\ &= \frac{2\epsilon x + \epsilon^2}{\epsilon} \\ &= 2x + \epsilon \end{align}
Now we discard the infinitesimal part of the answer to get the derivative in the reals. Thus $$f'(x)=2x$$

I've talked about consistency - I don't think I can make myself more clear.

October 21st, 2018, 10:50 PM   #26
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 Originally Posted by v8archie Derivative of $f(x)=x^2$ with infinitesimals ($\epsilon$ is infinitesimal): \begin{align} f'(x) &= \frac{f(x+\epsilon) - f(x)}{\epsilon} \\ &= \frac{(x^2 + 2\epsilon x + \epsilon^2) - x^2}{\epsilon} \\ &= \frac{2\epsilon x + \epsilon^2}{\epsilon} \\ &= 2x + \epsilon \end{align} Now we discard the infinitesimal part of the answer to get the derivative in the reals. Thus $$f'(x)=2x$$ I've talked about consistency - I don't think I can make myself more clear.
As far as I know (though I am far from an expert), the result here isn't obtained by "discarding" the infinitesimal part. Instead, it is obtained from the result that $\epsilon^2 = 0$ for any infinitesimal. Note that in the reals this would imply that $\epsilon = 0$ but this is not the case here.

From this, your equation actually simplifies from applying this identify and simply cancelling as
$\frac{2 \epsilon x + \epsilon^2}{\epsilon} = \frac{2 \epsilon x}{\epsilon} = 2x$

October 21st, 2018, 11:07 PM   #27
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Quote:
 Originally Posted by SDK As someone who only has passing familiarity with hyperreals, I can say I've never been motivated to bother any serious study since they don't appear to equip you with any tools which aren't already straight-forward applications of limits. Can you (anyone) give an example of some application, technique, or concept which is only possible or significantly easier/more powerful/etc when using infinitesimals?
No, there's never been such an application.

BUT in math you never know what obscure curiosity will turn out to be important. If tomorrow morning professor so-and-so in Helskinki proves P = NP using nonstandard analysis, overnight everyone will care about it. Till that happens, it's just a technical curiousity, but a very interesting one.

Also Professor Tao is using it to solve notational problems in higher analysis. Tao's one of the smartest mathematicians in the world and has won the Fields medal. Infinitesimals may come back into style.

That's a philosophical point too. It's true that once a theorem is proved, it's proved forever. But whether people care about that theorem is a matter of trend, fashion, and politics. Ideas go in and out of style. Infinitesimals being a great example.

October 21st, 2018, 11:15 PM   #28
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 Originally Posted by SDK As far as I know (though I am far from an expert), the result here isn't obtained by "discarding" the infinitesimal part. Instead, it is obtained from the result that $\epsilon^2 = 0$ for any infinitesimal.
I don't think that's right. But even if it is, consider the derivative of $\sqrt{x}$:
\begin{align} f'(x) &= \frac{f(x+\epsilon) - f(x)}{\epsilon} \\ &= \frac{\sqrt{x + \epsilon} - \sqrt{x}}{\epsilon} \\ &= \frac{\big(\sqrt{x+\epsilon}-\sqrt{x}\big)\big(\sqrt{x+\epsilon}+\sqrt{x}\big)} {\epsilon\big(\sqrt{x+\epsilon} + \sqrt{x}\big)} \\ &= \frac{(x+\epsilon)-x}{\epsilon\big(\sqrt{x+\epsilon} + \sqrt{x}\big)} \\ &= \frac{\epsilon}{\epsilon\big(\sqrt{x+\epsilon} + \sqrt{x}\big)} \\ &= \frac{1}{\sqrt{x+\epsilon}+\sqrt{x}} \end{align}
You might say that we should use a Taylor series for $\sqrt{x+\epsilon} = \sqrt{x}\sqrt{1+\frac \epsilon x}$, but whether you do or not, we are discarding terms in $\epsilon$ or $\sqrt{\epsilon}$, so it's not only larger powers of $\epsilon$ that disappear when we simplify to get $$f'(x) = \frac{1}{2\sqrt x}$$

Edit: I suppose that if you use the Taylor series before cancelling, you might be able to argue that only terms of $\epsilon^2$ and greater are disappearing. But I'm not happy with the idea of using Taylor series to determine derivatives anyway - it seems rather circular.

Last edited by v8archie; October 21st, 2018 at 11:53 PM.

October 21st, 2018, 11:16 PM   #29
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Quote:
 Originally Posted by v8archie I've talked about consistency - I don't think I can make myself more clear.
Archie I think I understand your point of view a little better. If I was overly critical earlier (ok, no if about it) I apologize.

* First, there are two different levels or approaches to the hyperreals. One is via calculus, for example Keisler's famous book. In that approach, the technical construction of the hyperreals is not even discussed or is put off to the end of the course.

It's perfectly analogous to the way basic calculus books assume the properties of the real numbers and never mention Dedekind cuts. Calculus students don't need to know how specialists construct the reals within set theory. Likewise students of nonstandard calculus don't need to know the technical construction, which is way beyond the level of calculus.

So it's unfair of me to hit you over the head with ultrafilters. In fact I haven't spent much time studying nonstandard calculus, so you probably understand that part of it better than I do.

I think you must have read a nonstandard calculus book so we're all kind of talking apples and oranges. You don't need to know anything about the construction to work with the hyperreals.

* The second point is that when you said we assume the hyperreals and I said no, we construct it, there is a kernel of truth in what you say. To construct the hyperreals we need a set-theoretic gadget called a nonprincipal ultrafilter on the natural numbers. And you can't prove that such a thing exists in ZF. You need a weak form of the axiom of choice. So yes, there is a bit of pulling a rabbit out of a hat if you regard AC with suspicion. But in modern math AC is assumed. There is no "controversy." It gives better theorems and cleans up infinite cardinalities so we use it.

Hope some of this made sense.

Last edited by Maschke; October 21st, 2018 at 11:21 PM.

October 21st, 2018, 11:26 PM   #30
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Quote:
 Originally Posted by Maschke I think you must have read a nonstandard calculus book so we're all kind of talking apples and oranges.
Correct. I've heard a couple of times about ultrafilters, but never been motivated to find out what they might be.

Are you saying that the Hyperreals weren't constructed purely to formalise Newton/Leibnitz early attempts at calculus? Or just that the construction can be made more rigorously that just assuming the existence of infinitesimals? If the first is true, was there any particular reason for doing it?

Also, does your use of that particular quote at the top of your post mean that you can shed some light on the consistency question?

Last edited by v8archie; October 21st, 2018 at 11:30 PM.

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