For a convergent series the sum is defined as the limit. There is no residual “infinitely small difference” between the sum and the limit — Banno

Ok.

For a convergent series the sum is defined as the limit. There is no residual “infinitely small difference” between the sum and the limit. The sum is the limit. Partial sums are less than the limit, but their difference goes to zero in the standard real number system. — Banno

This is all from proofs by Cauchy that I don't understand. Do you understand it?

well I haven’t had an exam on it in 50 years…

Not just Cauchy.


Tell me where I’m wrong if you can.

Tell me where I’m wrong if you can. — Banno

According to Zvi Rosen, the sum and the limit are not equal (according to Cauchy). They're just as close as we "want" them to be. This is Cauchy's definition of a limit:

When the values successively attributed to the same variable approach indefinitely a fixed value, eventually differing from it by as little as one could wish, that fixed value is called the limit of all the others. — Cauchy

So it's true that the idea of an infinitesimal was removed, but the idea of infinitely small remained, and we added the idea of "as small as we want."

We'll see what the other's said.

Well, you can play with all that if you like - some of what you say here looks muddled. The salient bit today is that a limit is not a rounding off.

To which we might add, as a corollary, The limit is not “almost” the value.

The limit is not “almost” the value. — Banno

That's incorrect.

For (2) to be possible, I must be offering you the actual value. — Srap Tasmaner

Sorry Srap, I can't see how you make this conclusion. 'Within a specified tolerance' does not indicate "the actual value" has been given. It just indicates that the value is within a specified tolerance. In neither case is the value which is being rounded off, actually specified. If "the actual value" was specified the procedure would be unnecessary. So I don't see any significant difference between the two, just two different forms of rounding off.

But an electron is conceived as a point. — frank

I don't think so, electrons are conceived as a cloud of probability, with a variable density.

Isn't that the same as the idea of an infinitesimal in math? — frank

An electron could not be infinitely small, because this would reduce the probability of them having any location to practically zero. And that is contrary to what is observed and verified by the cloud of probability conception.

The issue is actually quite complex, because "point particle" is really just a conception of convenience. It's not meant to actually indicate the physical properties of the supposed particle. Rather it's a convenient way to conceptualize interactions. Compare this to the concept of "centre of gravity" for example. This is meant to represent a point which indicates where a body's weight or mass is centred around. But it's just a conception of convenience which helps to model interactions, it doesn't indicate a real point that the body is centring itself around. Nor does the "point particle" concept indicate a real point where an electron is located. They are both conceptions of convenience, intended to facilitate the representation of interactions.

According to Zvi Rosen, the sum and the limit are not equal (according to Cauchy). They're just as close as we "want" them to be. — frank

It's just a matter of definition. Notice what you say, that they are as close as we want them to be. Banno wants them to be equal, and so he defines them that way. But in the context of this discussion such a stipulation is really meaningless.

The salient bit today is that a limit is not a rounding off. — Banno

Then why did you say to@jgill, "a more intricate form of 'rounding off'"? That really looks like "rounding off" to me. The point being, that applying a limit to that which is limitless (infinite), is nothing other than a form of rounding off. it's really no different from saying that pi is 3.14, or that it is 3.14159, or however you want to round it off. You are apply a limit to what is limitless, and that is a form of rounding off.

Ok. Details?

Simple example of a limit with an exact value
Consider the sequence
$a_n = \frac{1}{n}$

Claim
$\lim_{n \to \infty} \frac{1}{n} = 0$

Proof (ε–N)
Let $\varepsilon \gt 0$ be arbitrary.

Choose $N \gt \frac{1}{\varepsilon}$.

Then for all $n \ge N$,
$\left| \frac{1}{n} - 0 \right| = \frac{1}{n} \lt \varepsilon$

Since this holds for every positive $\varepsilon$, the difference between $\frac{1}{n}$ and $0$ can be made smaller than any positive real number.

Therefore,
$\lim_{n \to \infty} \frac{1}{n} = 0$

Conclusion
There is no “infinitely small but non-zero” remainder. In ℝ, being smaller than every positive real number forces equality with zero.

With the help of ChatGPT. Let me know if it's wrong. Looks OK to me.

Then why did you say to@jgill, "a more intricate form of 'rounding off'"? — Metaphysician Undercover

Because he was looking at Numerical Analysis not Real Analysis.

Sorry Srap, I can't see how you make this conclusion. — Metaphysician Undercover

It was a short post, making a single point, which answers exactly this question.

That's incorrect. — frank

It's also an answer to this, I think.

Ok. Details? — Banno

Yea, you're wrong.

It's also an answer to this, I think. — Srap Tasmaner

How so?

But an electron is conceived as a point.
— frank

I don't think so — Metaphysician Undercover

It is. An electron is a point particle.

, I just gave a proof involving a sequence that gives the exact value of the limit: zero. $\lim_{n \to \infty} \frac{1}{n} = 0$

This is a counter instance to your insistence. The "=" is not an approximation.

So if you would keep your credibility, show your working.

Damn keyboard keeps sticking.

But the point I made is that "point particle" is a conception of convenience, designed for the purpose of representing interactions. It does not represent how the electron is actually conceived as existing. The electron is modeled as a "point particle", but it does not exist that way, the probability cloud is a better representation (though still very inadequate) of how electrons exist.

You're right that per Cauchy, the sum of the series is the limit. However, the devil is in the details. The sum and the limit are never equal. see here.

I'm not concerned about credibility or showing that I'm working. :grin:

But the point I made is that "point particle" is a conception of convenience, designed for the purpose of representing interactions. It does not represent how the electron is actually conceived as existing. The electron is modeled as a "point particle", but it does not exist that way, the probability cloud is a better representation (though still very inadequate) of how electrons exist. — Metaphysician Undercover

I actually took a deep dive on this at one point. The electron is, in fact, conceived by scientists as a point. It's startling, but true.

I'm not concerned about credibility or showing that I'm working. — frank

Banno's proofs continue to be a matter of begging the question. Stipulate that the limit is the value, then use that as a premise in proving an instance of this.

The electron is, in fact, conceived by scientists as a point. It's startling, but true. — frank

That's half true, because the electron is also conceived as a probability cloud. Hence the wave/particle duality.

The sum and the limit are never equal. see here. — frank

Were it says

If, for increasing values of n, the sum Sn approaches a certain limit S, the series will be called convergent and the limit in question will be called the sum of the series.

The limit will be called the sum of the series.

:meh:

Stipulate that the limit is the value, then use that as a premise in proving an instance of this. — Metaphysician Undercover

You misread.
What is stipulated is what is meant by a limit:

Definition (limit of a sequence)
$L$ is the limit of the sequence $(a_n)$ iff
for every $\varepsilon \gt 0$
there exists $N$ such that for all $n \ge N$,
$\lvert a_n - L \rvert \lt \varepsilon$

If ∣x∣<ε for every ε>0, then x=0 is not a stipulation about limits; it is a theorem about the real numbers, derived from the order structure of ℝ.

The structure of the argument is:

• Define the limit (ε–N).
• Assume: the ε–N condition holds for some L.
• Introduce the independent fact about ℝ: no nonzero real can be smaller than every positive real.
• Conclude: therefore the difference is zero → exact equality

We are nto doing numerical analysis.

The limit will be called the sum of the series. — Banno

Yes. I don't think either of us is interested in arguing about what that actually means. Let's leave it.

The limit will be called the sum of the series. — Banno

The meaning of of this was just given.
$L$ is the limit of the sequence $(a_n)$ iff
for every $\varepsilon \gt 0$
there exists $N$ such that for all $n \ge N$,
$\lvert a_n - L \rvert \lt \varepsilon$

it says: The terms of the sequence can be made as close as you like to L by going far enough out in the sequence.

Importantly, there is no little bit left over because in the real numbers there is no positive number smaller than every positive number. So if the difference between $a_n$ and L can be made smaller than any positive number you choose, the difference must actually be zero.

But yes, I am getting a bit sick of working on the tags... especially since folk seem to ignore them.

If ∣x∣<ε for every ε>0, then x=0 is not a stipulation about limits; it is a theorem about the real numbers, derived from the order structure of ℝ. — Banno

The conclusion "x=0" is not valid without a further stipulation that there can be nothing between the least ε and zero. But we know there is no least ε and there will always be another lesser ε . Therefore x has no place in that number system, and is wrongly inserted as a category mistake. What is x? And how is it allowed to fit into the number line in this way, when it is not itself a number?.

Since what constitutes "the real numbers" is a matter of stipulation, you are wrong to say it's not a stipulation. You have inserted, through a category mistake, something called x which is not a number, but somehow you claim that it is equal to a number, zero in this case. That is a stipulation.

The conclusion "x=0" is not valid without a further stipulation that there can be nothing between the least ε and zero. — Metaphysician Undercover

That stipulation is what ℝ is. It is not an extra, and it does not make the argument that there is a limit circular.

It is not a stipulation about limits.

The intuition goes: Given that there are real numbers, and given that our sequence can get as close as we like to some number, let's call that number the limit of the sequence.

Added: the pedagogic problem - it's not a mathematical problem - is how to dissipate the notion that the limit is "a little bit more" than the sequence? Notice that the limit is set out in terms of the sequence - the limit is provided by the sequence alone! so the limit results form the sequence. But it need not be one of the elements of the sequence. It's not something the sequence reaches toward — it is a property of the sequence itself. The limit isn't something the sequence is trying to get to; it's a concise description of how the sequence behaves. The sequence doesn't "know about" or "aim for" its limit - the limit is simply our label for a pattern in the sequence's terms.

look at what "as close as we like to some number" means. It means there is no limit to how close we can get to that number. That is how you define "limit" a specified number for which there is no limit to how close we can get to it.

Added: the pedagogic problem - it's not a mathematical problem - is how to dissipate the notion that the limit is "a little bit more" than the sequence? — Banno

Obviously, there is always "a little but more" in terms of how close we can get to the limit. that is implied by your definition of "limit". If we'd like to get closer to the limit, than any previously proposed closeness, we can do that, and get closer to that limit. This is what your definition indicates. Therefore, to "dissipate the notion" that there is always more, would be a big mistake, contrary to the definition. Why would you aspire to do this?.

Notice that the limit is set out in terms of the sequence - the limit is provided by the sequence alone! so the limit results form the sequence. But it need not be one of the elements of the sequence. It's not something the sequence reaches toward — it is a property of the sequence itself. — Banno

So here is where your mistake lies. The limit is the condition for the sequence, the sequence is derived from it, as a formula, a repetition of "half the value between this point and the specified limit". Therefore your mistake is in saying "the limit results form the sequence". The limit is necessarily specified prior to producing the sequence. Then, the sequence is produced from the way that "limit" was defined. We can always get closer to the limit, if that is what we want to do.

Because the limit is prior to the sequence it is not "a property of the sequence itself". The limit preexists the sequence as a necessary condition for it. So if one is to be said to be a property of the other, the sequence is a property of the limit. By this mistake, what you say which follows, is all wrong.

The limit isn't something the sequence is trying to get to; it's a concise description of how the sequence behaves. The sequence doesn't "know about" or "aim for" its limit - the limit is simply our label for a pattern in the sequence's terms. — Banno

The sequence is designed, and produced from the limit. Therefore knowing the limit, and aiming for it, in this way of getting ever closer to it, is an essential aspect of the sequence. the sequence is derived from the nature of "the limit". And this is clear from the way you define "limit". We know "some number", and we also know that there is no limit to how close we can get (we can get as close as we like) to it. The sequence is derived from the specified number.

This is all from proofs by Cauchy that I don't understand. Do you understand it? — frank

You are making this sound more esoteric than it is. This is freshman calculus that has been studied by generations upon generations of students. In our class you were expected to understand the theorems and their proofs and to be able to sketch some of them from memory. (Of course, most of this has been thoroughly forgotten decades ago...)

Mathematics as an academic discipline is more like science and less like philosophy, in that the focus is not so much on authors and their original texts. The formulations and proofs that you find in modern textbooks are often not the same as those given by their original discoverers, even when they bear the names of Cauchy, d'Alembert, Weierstrass, etc., because clearer, simpler, more robust or more general versions have been developed. The history of mathematics is a worthy subject in itself, but that is not the topic here.

Obviously, there is always "a little but more" in terms of how close we can get to the limit. that is implied by your definition of "limit". — Metaphysician Undercover

Being obvious to Meta is not a proof.

Always keep in mind that Meta argues that there are no numbers between 1 and 3.

The sequence is designed, and produced from the limit. — Metaphysician Undercover

This is exactly arse about. The limit is a result of the sequence. Those who care to look can see exactly that in the proofs offered earlier.

The history of mathematics is a worthy subject in itself, but that is not the topic here. — SophistiCat

Cauchy, Weierstrass, and Riemann saved calculus from mounting criticism that it doesn't make sense. So, that did become the topic. We've finished talking about it now, though. :grin:

This is exactly arse about. The limit is a result of the sequence. Those who care to look can see exactly that in the proofs offered earlier. — Banno

Take a look at the quoted sequence:

he key is that an infinite sequence may have a finite sum: ½ + ¼ + ⅛ ... = 1 — Banno

Quite clearly, the limit must be assumed prior to taking half of it.

We set out the sequence $a_n = (1/2)^n$, or the sum $\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^n$, then find that the limit is 1.

One might set the limit to one then look for a sequence, but of course there is more than one such sequence... quite a few more.

Your posts make less and less sense as we proceed.