The Fundamental Theorem of Calculus (all together now)

So we’ve seen two sides of the FToC: the first part, which says that given a continuous function $f:\left[a,b\right]\rightarrow\mathbb{R}$ we can integrate and differentiate to get our function back:

$\displaystyle\frac{d}{dx}\int\limits_a^xf(t)dt=f(x)$

and the second part, which says that given a differentiable function $F:\left[a,b\right]\rightarrow\mathbb{R}$ whose derivative is the continuous function $f$ , we can integrate to get (part of) our function back again:

$\displaystyle F(b)-F(a)=\int\limits_a^bf(x)dx$

Now, we proved these two sides in very different ways, but it turns out that we can get from one to the other.

Let’s assume the first part holds. Then we take the function $F$ and define $f(x)=F'(x)$ as its derivative. The first part of the theorem tells us that we know a function whose derivative is $f$ : the function defined by $G(x)=\int_a^xf(t)dt$ . And we know that any two functions with the same derivative must differ by a constant! That is, there is some real number $C$ with $F(x)=G(x)+C$ . Using this to evaluate $F(b)-F(a)$ we find:

$\displaystyle F(b)-F(a)=(G(b)+C)-(G(a)+C)=G(b)-G(a)=$
$\displaystyle\int\limits_a^bf(x)dx-\int\limits_a^af(x)dx=\int\limits_a^bf(x)dx$

Which gives us the second part of the theorem.

On the other hand, what if we assume the second part of the theorem holds? Then we start with a continuous function $f:\left[a,b\right]\rightarrow\mathbb{R}$ . Given $x\in\left[a,b\right]$ , the function is continuous on the subinterval $\left[a,x\right]$ , and so the second part of the FToC says that $F(x)-F(a)=\int_a^xf(t)dt$ . That is, the integral in the first part of the FToC differs by a constant ( $F(a)$ ) from the function $F$ we assumed to be an antiderivative of $f$ . Thus it must itself be an antiderivative of $f$ .

So each half of the Fundamental Theorem implies the other, and we can prove either one first before immediately deriving the other.

February 15, 2008 - Posted by John Armstrong | Analysis, Calculus | | 123 Comments

123 Comments »

Disclaimer #1: FTC is independent from the mean value theorem for continuous functions. It can be derived from the positivity of the Riemann integral, i.e. the fact that the integral of a positive function is positive.
Disclaimer #2: The increasing function theorem (that says that any function on an interval with positive derivative is increasing) is independent from the mean vamue theorem (also known as the Lagrange theorem), it can be proven directly by using the existence of the least upper bound.
Everybody who is serious about learning calculus and elementary analysis should know this.

Comment by Michael Livshits | February 15, 2008 | Reply
You’re already on the record that everyone who doesn’t use exactly your approach is either not serious or an idiot. Can’t we just take that as read by now?

Comment by John Armstrong | February 15, 2008 | Reply
PLease don’t words into my mouth, John. I haven’t called anybody an idiot yet and I never said that you have to follow my approach exactly to earn my approval. I just wanted to stress a couple of points that are important for clear understanding of the subject. What’s wrong with that? I hope you have read the article by Mark Bridger in AMM that discusses the issue at some lenght. I recommend it to anybody interested in the subject.

Comment by Michael Livshits | February 15, 2008 | Reply
Forgot “yes but” again, sorry.

Comment by Michael Livshits | February 15, 2008 | Reply
I haven’t tried to follow all your mathematical points in detail, Michael, because your general tone has been so shrill and inflammatory and rude, but can you tell me without resorting to insult: how do you prove that a continuous function is Riemann-integrable?

In case it isn’t already clear: I know how to prove this. I’m wondering how you would prove this.

I would again encourage you to get all this out of your system by opening your own blog, and showing the world how you think it should be done. Are you up to it?

Comment by Todd Trimble | February 15, 2008 | Reply
To avoid a silly quibble, I’m asking how you prove that if f is defined and continuous on an interval [a, b], then the Riemann integral $\int_{a}^{b} f(x) d x$ exists.

Comment by Todd Trimble | February 15, 2008 | Reply
First of all, $f$ is continuous on [a,b], therefore it’s uniformly continuous there. Then the Riemann sums for any two fine enough, i.e., each interval shorter than d, partitions the corresponding Riemann sums will differ by less than e(b-a). So we have bunch of consistent approximations, as accurate as we want, and you can say that there will be “an accumulation point” that is the “exact value” of the integral.
Is it good enough?

Comment by Michael Livshits | February 15, 2008 | Reply
Yes, but this pushes the question back to showing that a continuous function on [a, b] is uniformly continuous. Proof?

Comment by Todd Trimble | February 15, 2008 | Reply
I don’t use pointwise notions, I use uniform estimates, so I don’t need this theorem. But if you really want it, here is one. Assume that $f$ is not uniformly continuous, then there will be some positive e and 2 sequences of points $x_n$ and $y_n$ in our segment, such that $|x_n -y_n| \rightarrow 0$ , but $|f(x_n)-f(y_n)| \geq e$ .
Now, we can assume that the sequences converge to c (otherwise we can take a convergent subsequence), then both $f(x_n)$ and $f(y_n)$ will converge to $f(c)$ , and we get a contradiction.

Comment by Michael Livshits | February 16, 2008 | Reply
There are some other proofs, relying on existence of finite subcovers of open covers (compactness) and their Lebesque numbers, for example, but all these proofs use compactness in one form or another, and there is no getting away from it.
So you also think that I’m just a crank, Todd?

Comment by Michael Livshits | February 16, 2008 | Reply
Actually, if you follow the references, you will find that elementary calculus can be greatly simplified and made more accessible to the students. In fact all the theory can be reduced to bunch of not very hard problem sets.

Comment by Michael Livshits | February 16, 2008 | Reply
And I never said you were “just” a crank. You’re cranky, sure, but your approach isn’t wrong.

You see, that’s the biggest difference between you and me (besides the fact that I’m willing to take the leap and undertake a project like this one, while you can only snipe from the sidelines): I consistently point out that we have alternative approaches, each with its own strengths and weaknesses, while you stridently insist that my approach is flat-out wrong.

The worst I’ve ever said about your mathematics is that I find the “derivative is a quotient” position to be an oversimplification at best, which is harmful in the long run. Even so, in the context of single-variable calculus teaching it has its own strengths.

What I find so objectionable about you is the whiny way in which you consistently return to insist that your way is right and my way is wrong. You have the pretentious tone of the eternal critic: he who fails to create of himself, and must subsist on tearing down the work of others. Bonus points for that classic hit, “you’ve been corrupted by the System”.

You see, when it comes down to it, your little quibbles don’t matter in the end. All along I grant you that you have a valid point, but I choose a different path. Anything beyond that is just you throwing a temper tantrum in thread after thread after thread.

Comment by John Armstrong | February 16, 2008 | Reply
I don’t use pointwise notions, I use uniform estimates, so I don’t need this theorem.

Well, John was proving a theorem [FTC] that holds for continuous functions, not just (a priori) uniformly continuous functions. So until you prove the result presently in question, you are proving a weaker theorem. That’s may be your choice of course, but let’s at least put our cards on the table, so that readers are aware of some subtle but important differences in choice of hypotheses.

On the other hand, you at least agreed on the statement of FTC, and that the hypotheses (which could be weakened further, of course) involve continuous functions. In your own approach, you have presupposed existence of the Riemann integral. Hence it seems fair to press you for details on that. Let’s continue then.

Now, we can assume that the sequences converge to c (otherwise we can take a convergent subsequence)

Proof?

Comment by Todd Trimble | February 16, 2008 | Reply
So you also think that I’m just a crank, Todd?

Oh no. That would be far too harsh a condemnation. You strike me as having some competence.

I am pressing you not to give you a hard time, but to make a serious point.

There are some other proofs, relying on existence of finite subcovers of open covers (compactness) and their Lebesque numbers, for example, but all these proofs use compactness in one form or another, and there is no getting away from it.

And now we’re in the neighborhood of the point I’m trying to make. So you agree that there is no getting around issues of compactness of [a, b].

Thus, the underlying mathematical infrastructure, revolving around the topology of the real line and its attendant consequences, including MVT, is something there’s no getting away from.

What I’m trying to get you to acknowledge is that compactness of closed bounded intervals, the mean value theorem, existence of the Riemann integral, etc., are really all part of the same fundamental circle of ideas residing at the base of the calculus. And this circle of ideas is what John has been pointing to.

Comment by Todd Trimble | February 16, 2008 | Reply
…In your own approach, you have presupposed existence of the Riemann integral.
No, I haven’t, I constructed the definite by approximationa-la Darboux. The classical proof goes by bisection. Let’s divide our segment in 2 equal parts. Then at least one contains an infinite number of elements of our sunsequence. Pick this part and and pick $x_{n_1}$ from this part iterate the process, on each step picking $n_k$ greater than $n_{k-1}$ . The sequence $x_{n_k}$ will converge to thepoint of intersection of all our subinterval. The “existence” of such a point can be taken as an axiom for the reals or proven if we take other axioms to start with.

Comment by Michael Livshits | February 16, 2008 | Reply
Actually, I’d like to explore this a bit. I’m assuming that Michael accepts my proof of the Heine-Cantor Theorem since he didn’t throw a fit when I made that post.

So can’t we just take as read that continuous functions on closed intervals are uniformly continuous? Or are we assuming that Michael has a problem with invoking “connected compacta are closed intervals”?

Ultimately, though, Michael’s proofs are all invoking a whole lot of limits, given how offensive he seemed to find the notion of limits a day or so ago.

Comment by John Armstrong | February 16, 2008 | Reply
No, I haven’t, I constructed the definite by approximationa-la Darboux.

You constructed what where?

It’s cool if you want to use the Darboux integral. According to the Wikipedia article, this requires that you consider absolute maxima and minima of functions on subintervals (specified by paritions). We come back then to the Extreme Value Theorem for continuous functions on closed bounded intervals — back to the consequences of compactness.

Once again: Heine-Borel, Extreme Value Theorem, Existence of Darboux Intergal, Mean Value Theorem: they are all facets of a beautiful circle of ideas, and it is this big picture that should really be in focus throughout this discussion.

The “existence” of such a point can be taken as an axiom for the reals or proven if we take other axioms to start with.

Yes, (modulo some complaints one could make about presentation, but I won’t cavil here and now) you just ran through the idea behind a nice classical proof of sequential compactness of [a, b]. (For those into logic, I’ll point out Michael’s use of Dependent Choice.] But this last part again raises some interesting issues of just how you want to set up the theory (e.g., how one defines the real numbers). One of myriad foundational decisions one must face up to [and that John has].

But in light of the fact that both of you, when it gets down to brass tacks, need to appeal to the same fundamental circle of ideas, I don’t really understand why you take John so hard to task. Don’t you agree that Mean Value Theorem and Extreme Value Theorem and Existence of Darboux Integral are all closely related, when you get to the bottom of things?

Comment by Todd Trimble | February 16, 2008 | Reply
Ultimately, though, Michael’s proofs are all invoking a whole lot of limits, given how offensive he seemed to find the notion of limits a day or so ago.

Well, any competent writer can avoid the word “limit”, but the basic underlying ideas involved are pretty much the same in the end. So I wonder about the vehemence of the discussion. It’s strange to me.

Comment by Todd Trimble | February 16, 2008 | Reply
…You constructed what where?
I constructed the definite integral by approximation a-la Darboux. In my notes and my talk slides, the ones that Todd didn’t care to read.
Now about compactness and uniform continuity. Uniform continuity continuity can be usually checked directly. Nobody forces us to to start with pointwise notions and use compactness to prove uniform continuity. In fact the proof of the existence of a convergent subsequence that I gave is non-constructive, on each step of the process we have to examine an infinite number of elements of our sequence in order to decide whic half of the segment to pick. You can encode some unsolved number theory problem into this decision perocess and then you will not be able to decide before you solve it. The convergent subsequence will be only in your dreams. Likewise, MVT is not constructive either, you have to throw in some additional assumptions in order to find where this mean value hit by your function. On the other hand, positivity of integral is a computational fact, as the FTC is, to use non-constructive theorem in proving a constructive one is a bad idea, it distorts the logical structure of the theory and it distorts the readers’ understanding of whet is needed and what is not needed for the theorem to be true.

Comment by Michael Livshits | February 16, 2008 | Reply
I constructed the definite integral by approximation a-la Darboux. In my notes and my talk slides, the ones that Todd didn’t care to read.

Careful. In fact, I did read them, some weeks ago. (What I didn’t read in detail are all the numerous complaints and “bitching and moaning” comments [your words] made in comments to John these past few weeks — they are unpleasant to read. This is why you should really get your own blog, Michael. Leave John in peace.)

And, I don’t mind telling you that I found quite a bit of your writings interesting, and I did pick on the fact that philosophically you are coming at this from a constructive point of view (à la Bishop and his modern-day disciples). And that’s perfectly fine. Not a thing wrong with that. In fact, such an approach is commendable for many reasons, some of which you are touching upon now.

But what is unfair, in my view, is that you are attacking John’s proof of a statement that he explicitly laid out, when in fact the proof demands a certain amount of machinery which in one sense or another is unavoidable, as you admitted. If you want to say instead, if we strengthen the hypotheses, we get easier proofs, fine — say that. John would not disagree, I’m pretty sure. But, in these past few posts, you didn’t at all attack the actual statements, just the proofs. And that’s why I put all those questions to you above, to get you to acknowledge that.

And I hope you can acknowledge that you’ve been pretty nasty about it, too.

Why don’t you get your own blog, I keep asking you?

Comment by Todd Trimble | February 16, 2008 | Reply
Again, I’m not totally against, the general notion of compactness, MVT, Heine-Borel and such. But we have to be cognizant that the existence in these theorems is only the existence in princile. These theorems may be good for mathematical theorizing, but they are not practical. Likewise, the pointwise notions are not practical, we can’t have an initite, let alone a continuum of different rules fo find d such that |x-a|<e implies f(x)-f(a)|<e, it’s a purely mathematical fantasy, and we better be earnest about it, at least with ourselves. Pushing these notions upon students who are mostly interested in practical applications and don’t even know how to use the elementary calculus is ridiculous.

Comment by Michael Livshits | February 16, 2008 | Reply
In the previous post it should be: … existence in principle,
…such that |x-a|<d implies |f(x)-f(a)|<e sorry for the typos.

Comment by Michael Livshits | February 16, 2008 | Reply
Dear Michael, without getting into thorny ontological issues of mathematical existence, I do take your point, as I think I made clear in comment 20. But now that the underlying philosophical differences between you and John are beginning to be laid bare, and now that you are raising the specter of “practicality”, let me gently remind you that John’s blog is not primarily addressed to students of engineering or of other practical applications, but to those who are interested in some modern-day mathematics and its inner structure.

Be cognizant, then, that the differences between you and John are fundamentally philosophical. But then, it is highly improper that you should be calling into question his competence or insight, as you have been recently. He has simply taken a different path than what you would take, at a fundamental philosophical level, as he has been telling you ad nauseam.

Get a blog, Michael!

Comment by Todd Trimble | February 16, 2008 | Reply
Pushing these notions upon students who are mostly interested in practical applications

And here’s where I pounce and say the same thing again:

I’M NOT WRITING LECTURE NOTES HERE. THIS IS NOT HOW I TEACH CALCULUS.

I never claimed that this was even remotely what I would say in a calculus class. Still, you think you’ve got me skewered, either because you can’t imagine a purpose other than teaching a calculus class, or because you can’t imagine that I’d be able to present the same ideas in two different ways depending on context.

Comment by John Armstrong | February 16, 2008 | Reply
I attacked the proofs because they misrepresent the mathematics invoved. They use a difficult theorems to prove a simple theorem. They give the people an impression that MVT is needed to prove FTC, and it is not true. The fact that you need compactness to prove uniform continuity of a poinwise continuous function doesn’t change the matter. My criticism is still valid. By the way, I tried to post the links to my constructions of the definite integral and the posts keep disappearing, somebody must be deleting them.

Comment by Michael Livshits | February 16, 2008 | Reply
That’d likely be Akismet. It can be amazingly astute.

Comment by John Armstrong | February 16, 2008 | Reply
O.K., I will not bother you any more, still usinf MVT to prove FTC is a prime example of trashy mathematics, calculus or not calculus, it demonstrates a definite lack of good taste.

Comment by Michael Livshits | February 16, 2008 | Reply
Look, the Mean Value Theorem is not a difficult theorem; it (or Rolle’s theorem) is a straightforward consequence of the Extreme Value Theorem, which is straightforward from considerations of compactness. Existence of Darboux integral, which you use in your approach, is also a straightforward consequence evolving from considerations of compactness. It’s a circle of ideas, Michael. Your brain is big enough to encompass that circle.

Try to keep the bigger picture in focus.

As far as impressions that people get, hard to say. I bet that a lot of people reading this blog can wrap their heads around the idea that there are lots of possible approaches. This is really, or should be, more about the big picture.

Get a blog!

the posts keep disappearing, somebody must be deleting them

Now, now.

Comment by Todd Trimble | February 16, 2008 | Reply
still usinf MVT to prove FTC is a prime example of trashy mathematics, calculus or not calculus, it demonstrates a definite lack of good taste

There you go again. Resorting to insult and inflammatory comments. You are not doing yourself any favors here.

Best that you get your own blog.

Comment by Todd Trimble | February 16, 2008 | Reply
Thanks for telling me about that Akismet, John. Sorry, I wnated to answer a Todd’s comment. He says: “But what is unfair, in my view, is that you are attacking John’s proof of a statement that he explicitly laid out, when in fact the proof demands a certain amount of machinery which in one sense or another is unavoidable, as you admitted.” Now, I admitted that compactness is unavoidable to derive uniform continuity from pointwise continuity. That has no relevance to the MVT. And what is wrong about having an opinion about merits and demerits of this or that proof? Isn’t it what the blog is about, having a discussion or an argument? Who is whining now? I never said that the proof was not correct, I just said that it obscures the true meaning of the theorem by dragging in some irrelevant information. What’s wrong with that?

Comment by Michael Livshits | February 16, 2008 | Reply
What’s wrong with that?

See: shrill
See: inflammatory
See: rude

It’s all in the tone, Michael. You don’t say, “you could also look at it this way”. You say that my approach is wrong. You say it’s misleading because it leads the reader somewhere other than where you want to go. You don’t understand that there are not only different proofs of the same theorem, but that there are different motivations, and different intentions than your own.

Comment by John Armstrong | February 16, 2008 | Reply
Sorry, I saw this proof so many times in so many books, and every time I see it my stomach turns. John, I didn’t mean to insult you, I meant to insult your proof, which is not really yours, so you don’t have to take it personally. For what I know, you are probably one of the nicest people, working for that trashy and corrupt establishment called calculus. Evevrybody complains about it, everybody admits that it’s the worst part of the undergraduate mathematics, but somehow very little is done go fix it, and it makes me sad, especially when you give me a cold shoulder when I try to share some constructive (no pun intended) ideas on how to improve the situation.

Comment by Michael Livshits | February 16, 2008 | Reply
You don’t present them constructively. And it’s not just that proof. Ever since I started in on calculus, you’ve been on almost every post, decrying how horrible my coverage is. Really, I just want you to shut up and take your crusade somewhere else. You’re making no converts here.

Comment by John Armstrong | February 16, 2008 | Reply
…I meant “very little is done to fix it…” Well, I think that it is the wrong approach to this theorem, and I think that the standard approach to calculus is wrong too. Sorry for the shrill and rudeness and inflamatory remarks, but why are you so touchy? Maybe it’s because there is some truth in what I say?

Comment by Michael Livshits | February 16, 2008 | Reply
Actually, Todd found my ideas interesting, it’s a bit short of becoming a convert, but it’s a start.

Comment by Michael Livshits | February 16, 2008 | Reply
I think they’re valid ideas, and they constitute a worthwhile alternative with its own strengths and weaknesses. I keep saying that over and over and over again, and yet you ignore it. You love to feel as if I’m oppressing you (crank sign) along with the rest of The Establishment (crank sign), when really I don’t much care one way or the other.

Set up your own weblog to expound your own ideas, and stop being a parasite on mine.

Comment by John Armstrong | February 16, 2008 | Reply
I never said that the proof was not correct, I just said that it obscures the true meaning of the theorem

In fact, you said

usinf MVT to prove FTC is a prime example of trashy mathematics, calculus or not calculus, it demonstrates a definite lack of good taste

See the difference? You also said

O.K., I will not bother you any more

and yet, here we are again.

Anyway, let’s talk about the mathematics a little. First off, compactness is totally relevant to the Mean Value Theorem. Don’t you know how to prove the MVT from the Extreme Value Theorem? It’s not hard.

The existence of the Darboux integral for continuous functions also follows from considerations of compactness and the Extreme Value Theorem (one needs the boundedness of continuous functions on [a, b] in order to get upper and lower sums, and that’s where EVT comes in).

So the point (which I’ve made several times) is that these various results, each being just a few short steps away from compactness, therefore cannot be all that far apart! But so far, all I’ve pressed you on is about existence of the integral. You still haven’t proved the Fundamental Theorem of Calculus here in comments; all you’ve said here, over and over in various ways, is that John’s proof “sucks”. So maybe that’s the next step. Let’s bring it out in the open: show us your proof. Then maybe we can discover together whether the path you take is so very divergent from John’s. I think it will be part of the same basic circle of ideas.

Actually, Todd found my ideas interesting, it’s a bit short of becoming a convert

Well, I’m a generous and broadminded guy. But I’m not likely to get religion here.

Comment by Todd Trimble | February 16, 2008 | Reply
Well you and Todd are probably right that I should get my own blog, this way I can just put links on your blog instead of being shrill, so people could see both approaches and decide for themselves. I must admit that I’m a bit disappointed that you offered me less feedback, mathematically speaking, than I hoped for, that you bacame so defensive so soon. Oh, well, I guess I should work on my manners.

Comment by Michael Livshits | February 16, 2008 | Reply
Well you and Todd are probably right that I should get my own blog, this way I can just put links on your blog instead of being shrill, so people could see both approaches and decide for themselves.

Whoo-hoo! Finally…

“Oh, well, I guess I should work on my manners.” Uh, yeah. Do unto others, and all that. Let that be your byword as you construct your blog, and best wishes.

Comment by Todd Trimble | February 16, 2008 | Reply
The proof of FTC that I suggested in in comments 25-28 for FTCI, I got a bit bogged down with typesetting there, but the idea is thath the absolute value of the $\int_{x_0}^x f(u)-f(x_0)du$ can be estimated by $e|x-x_0|$ where $e \rightarrow 0$ when $x \rightarrow x_0$ when $f$ is continuous at $x_0$ .
Actually, I never saw a derivation of MVT from EVT, I’ll think about it, but they look like rather different animals to me, EVT is about compactness and MVT is about connectedness, MVT is also true for open intervals, where EVT fails.
As for Darboux integral, EVT is not needed for it, all you need is uniform continuity. You don’t really care where thses minima and maxima actually attained on the subintervals, and that’s the hard part of EVT.

Comment by Michael Livshits | February 16, 2008 | Reply
… I meant “whether these minima and maxima are actually attained…”

Comment by Michael Livshits | February 16, 2008 | Reply
Ha! The proof doen’t have to be wrong to be trashy or in bad taste.

Comment by Michael Livshits | February 16, 2008 | Reply
By MVT I meant the one for continuous function, that it hits the zero if it changes sign. Is it the one you were talking about, or you were talking abouth the Lagrange theorem (= MVT for the derivative)? I’m a bit confused. Well, either way it’s not too important.

Comment by Michael Livshits | February 16, 2008 | Reply
Ha yourself. You said, “I just said that it obscures the true meaning of the theorem”. Oh yes, that’s all you said.

Not. You’ve said much, much more, that is shrill, inflammatory, and rude. E.g., the bit about “trashy” and “in bad taste”.

Anyway, John proved MVT from EVT under his post “Differential Mean Value Theorem”. How do you prove MVT?

Comment by Todd Trimble | February 16, 2008 | Reply
Wait.. This whole time you’ve been ranting against my use of the mean value theorem, and you don’t even know which theorem I mean? And now you say that the difference isn’t important?!

You haven’t read my proofs. You haven’t read anything! You’re just spouting off your little rants completely independent of context. Seriously, just go away before you embarrass yourself further.

Comment by John Armstrong | February 16, 2008 | Reply
(I meant to bold-face just the part “I just said”.)

I think you’re talking about the Intermediate Value Theorem. Look again carefully at John’s post on the Differential Mean Value Theorem. Big difference. I’m not quite sure what you mean “MVT for the derivative”.

Comment by Todd Trimble | February 16, 2008 | Reply
Nah, they are both rather difficult and totally unnecessary for FTC, my proof is simpler, more and more natural. The differential MVF (Lagrange) theorem or rather its precursor Rolle’s theorem is indeed usually derived from EVT. Anyway, I’m tired of these semi-vacuous arguments already, so you won.
Conngratulations!

Comment by Michael Livshits | February 16, 2008 | Reply
I meant .. more direct, more natural amd more general….

Comment by Michael Livshits | February 16, 2008 | Reply
Well, Michael, after all your fulminations against the mean value theorem, and in light of the fact that after all this time you completely misunderstood which theorem we’re talking about (and then lamely say, “it doesn’t matter”!!), it’s time to be blunt with you. You’ve made a fool of yourself. A complete and utter disgrace.

Since your horrendous misunderstanding has come to light, you can hardly expect anyone to take seriously your pronouncement that it (the MVT) is “rather difficult”. I’ll say again, for anyone else reading and trying to follow the discussion, that it’s actually easy. The crucial point was covered in John’s post The Fermat Theorem, the proof of which is very intuitive and simple once one draws a picture (and a picture would have been nice there).

Let me ameliorate the harsh judgment slightly by saying to you, Michael: I do think you’re an intelligent person, and therefore I do hold out some hope that you will see one day that the mean value theorem (whatever it is!!) is really no big deal in terms of difficulty. It just seems crazy to me to be getting so emotional about it!

You should go on teaching calculus the way you see best (not here! get your own blog), but Michael, for your own mental health: please, calm the f**k down. Take a deep breath. Listen a little before rushing to judgment. Relax…

And please, get your own blog. Leave John in peace.

Comment by Todd Trimble | February 16, 2008 | Reply
You had a flame war about the Fundamental Theorem of Calculus, and I missed it? That’s the perils of using an RSS feed.

Comment by Walt | February 16, 2008 | Reply
Well the comments are all still here, Walt. Enjoy

Comment by John Armstrong | February 16, 2008 | Reply
I am enjoying every bit of it!

All this is precious! The flame war should be preserved till eternity.

Comment by Vishal | February 17, 2008 | Reply
… I meant .. it was only a part of it…, sorry for the typos and sloppy editing.

Comment by Michael Livshits | February 17, 2008 | Reply
Damn that Askimet (if that’s what it is), again it removed my comment, in which I tried to answer a character assassination that John posted (see the next posting by him). By the way, he blocked the comment box, preventing me from answering his ridiculous accusations.

Here is one: “Now it’s plain as day that Michael is a crank, pushing his pet theories while remaining so embittered to the “system” that “indoctrinates” students against him.”

The origin of it a comment (see comments 21 and 22) that I made in jest for The Fundamental Theorem of Calculus I).

He also writes: “This whole time he hasn’t actually been reading a single thing about my proof, and evidently he hasn’t read the proofs in the calculus textbooks he so despises. He doesn’t even know which theorem I’m invoking! And it is important, because the different theorems say vastly different things.”

Only the last sentence deserves any comment. If you take a look at the proof that John gives for the intagral mean value theorem, you can see that it uses the intrmediate value theorem for continuous functions. So I was only formally wrong, since one of he theorems is an easy corollary of the other. Now, the positivity of integral is a much more elementary fact which follows from the positivity of the Riemann (or Darboux or whatever) sums, while any of these mean- or intermediate value theorems require rather subtle tools, like completeness and compactness.

So, John’s proof (of FTC) uses complicated things to explain simple things, and it’s bad mathematical taste. It’s even worse to put such a proof in a calculus book, since the reader may think that MVT is needed to prove the FTC, and it’s clearly not the case. It’s this indiscriminate and unnecessary appeal to the advamced machinery to explain simple things that makes most of the calculus books so trashy.

Finally, John, thanks for letting me use your blog as a soap-box, and happy blogging. By the way, I have never questioned your charcter (although I have my doubts about it now), I just questioned your mathematics, and it looks like you can’t handle it. For everybody else who enjoyed the FTC flame war, you can look at The Chain Rule for the opening shots.

Comment by Michael Livshits | February 17, 2008 | Reply
I get busy and let posts build up in my reader for a little while, and this is what I come back to? Dammit, it seems that I need to pay more attention to calculus…

Comment by Charles | February 17, 2008 | Reply
Now, the positivity of integral is a much more elementary fact which follows from the positivity of the Riemann (or Darboux or whatever) sums, while any of these mean- or intermediate value theorems require rather subtle tools, like completeness and compactness.

We’ve already covered this point. You’ve already admitted (at comment 10) that that proving the very existence of the Riemann or Darboux integral for continuous functions also requires arguments stemming from compactness. Now, no doubt you will say that you’re only interested in uniformly continuous functions, but this is a point that has also been covered (see comment 13): insofar as John is discussing theorems whose statements you didn’t dispute, the proofs require considerations of compactness. I repeat: you’ve already admitted that. No use trying to wriggle out of it now.

It’s this indiscriminate and unnecessary appeal to the advamced machinery to explain simple things that makes most of the calculus books so trashy.

It’s not clear which calculus books you’re talking about. But the point is moot if your beef is with calculus books addressed to college freshman, as opposed to what we’re discussing here. Let’s stay on topic, shall we?

Since you’re still here, there is just one more point which I’d like you to discuss. I agree that your sketched proof of the FToCI did not invoke the MVT (integral version), and that’s fine. It’s a different presentation which some people may appreciate, although I still think you’re making much too big a deal about the differences. But anyway, let’s now turn to FToCII, which in one version says that if F’ = f, where f is continuous, then $\int_{a}^{b} f(x) dx = F(b) - F(a)$ . In your comment 7 to the post on FToCII, you mentioned that the difference between any two primitives is a constant. John mentioned this as one of the consequences of the MVT. How do you prove this?

Comment by Todd Trimble | February 17, 2008 | Reply
To Charles: Good idea, Charley, you may see something here that will give your brain a healthy workout

Now to you Todd, and I appreciate that you are still willing to continue a civilized discussion instead of throwing epithets at me. Let me start with answering your question first. As, I said before, I only use the uniform estimates to simplify the subject, and I give a couple of proofs of the montonicity theorem, that is also called an increasing function theorem. The theorem says that the functions with nonnegative derivatives are non-decreasing. It follows that the functions with non-positive derivatives are non-incresing, so the functions withh zero deritaives must be constant. To see the easiest of the proofs, you can click on my name here to get to my web page, then click on “My Calculus Project” then click on “Slides” at the bottom of the page and leaf to pages 11 and 12. To see the other proof, you click on “Calculus writeup” on the top of “My Calculus Project” page and leaf to page 30 for a more leusurely discussion, the other proof appears on page 33.
This second proof is due to Hermann Karcher and appears on page 5 of his essay that I mentioned in my comment #40 at the beginning of the flame war on John’s page on the chain rule. The issue is discussed at length in the article “Uniform Calculus and the Law of Bounded Change” by Mark Bridger, it’s available from his home page at NEU, click on “Mark Bridger” on “My Calculus Project” or on the link that I included into my comments #32 and 33 on John’s “The Chain Rule” page to get directly to it. See “The Law of Bounded Change” on page 4 with a proof on page 5, it is a generalization of the Karcher’s proof to uniformly differentiable functions.

Now, if the derivative of a function (on a closed interval) is continuous, then the function will be uniformly differentiable on this interval, i.e., for evey e there is d such that when |x-y| is less than d |g(x)-g(y))-(x-y)g’(y)| is less than e|x-y|. This serves the FTCII. Compactness only enters in passing from pointwise differentiability to uniform differentiability.

If you don’t like this proof, here is another one, that does not use continuity of the derivative or compactness, it only uses the exstence of the lowest upper bound, i.e., completeness. Assume $g$ to be defined on [A,B] and $g'>0$ . We want to prove that $g(A) \leq g(B)$ .
Let $c <B> 0$ , (from the usual definition of the derivative) there will be some $x > c$ such that $g(c) < g(x)$ , so $c$ can not be the LUB unless $c=B$ . Now we can weaken the assumption of this theorem to @latex g’ \geq 0$ by applying this result to the function $g(x)+ex$ for some positive $e$ . We see that $g(B)-g(A) \geq e(A-B)$ , and since $e$ is arbitrary, we must have $g(A) \leq g(B)$ .
So we proved the monotonicity theorem, and therefore the fact that the functions with zero derivatives are constants.

Comment by Michael Livshits | February 18, 2008 | Reply
Now, Todd, I want to answer the other points you make. The fact that comactess is used to pass from pointwise to uniform continuity doesn’t justify the use of the integral mean value theorem to prove FTCI at all. And unfortunately I saw it in almost every book on calculus or introductory analysis. My guess is that this proof goes back at least to Lagrange, and people kept copying it ever since without giving it a second thought. Now, there are some computational problems with the intermediate value theorem for continuous functions (with is used in the proof of IMVT) that makes the existence of that intermediate valus rather shadowy. You can look at pages 317-320 of Foundations of Mathematical Analysis by J. K. Truss, OUP 1997 (they are availabe on google) for a nice example.

But, to put it bluntly, IMVT has nothing to do with FTC. Let me explain again. What is the Riemann integral? It is the limit of the Riemann sums over the net of the tugged partitions of our interval, patrially ordered by relation “one partition is a refinement of the other.” So, if the function is ono-negative, the integral is non-negative, that’s all there is to it. It is a computational fact, the average of a bunch of positive munbers is positive, damn it, what does it have to do with IMVT, IVT, DMVT or any of these?! The answer is: “Nothing.” FTCI follows DIRECTLY from positivity of the Riemann integral and the definition of continuity, as I explained already and John clarified in his posting “FToC Flame War Wrap-Up.” I rest my case, Your Honor.

So, in my opinion, I haven’t made a fool of myself at all, maybe I have exposed a lot of yes-men from the Church of Limitology as fools, but it was not my primary goal.

Comment by Michael Livshits | February 18, 2008 | Reply
My guess is that this proof goes back at least to Lagrange, and people kept copying it ever since without giving it a second thought.

You guess because it never occurs to you that someone might have considered the points you raise and decided to go a different way.

And again, as I’ve said over and over, I never said that your proof was wrong. I never said that the Mean Value Theorem was essential. This conspiracy is entirely in your little paranoid head. I have seen your proof before and decided to go another way. I even assessed your proof in parallel with mine in a post.

I didn’t decide which proof to present because I haven’t seen any other way. I simply find my approach to be more elegant. And as this is my own exposition, it’s my choice to make. I’ve never attacked your mathematical taste or ability, though you consistently attack mine. Face it: you just can’t stand the idea that anyone disagrees with you, even if they steadfastly refuse to contradict you.

Comment by John Armstrong | February 18, 2008 | Reply
“comactess”=compactness, “with is used”=which is used, “munbers”=numbers, sorry for the typos. Damn, I wish this site had a preview option for comments.

Comment by Michael Livshits | February 18, 2008 | Reply
What’s so elegant in going from Bronx to Downtown Manhattan through Boston?

Comment by Michael Livshits | February 18, 2008 | Reply
Then either send me cash to host this thing and hire a tech to maintain it, or start your own weblog with all the bells and whistles. The comment script is part and parcel of the free WordPress package, and not something I chose. Seriously, stop whining about it because there’s nothing I can do to change it without spending money I don’t have.

Comment by John Armstrong | February 18, 2008 | Reply
Again, you resort to personal attacks instead of arguing for the merits of your proof, except that “it’s your choice” and “you think it’s elegant.” And I never accused you of not seing it any other way, get over it.

Comment by Michael Livshits | February 18, 2008 | Reply
You seem to think that there is a unique, objective standard of elegance. There isn’t. I disagree with you, and that’s all there is to it. You’re not wrong, but neither am I. It’s an aesthetic choice.

To you, my approach seems like a detour, analogous to the side-trip to Boston you mention. To me, that’s not an analogous situation at all.

The proof you gave establishes an equivalent limit, yes, but it obscures (at least to me) exactly what connection the limit has to a derivative. At the end your limit can be unpacked and shown to be equivalent, but my proof keeps the notion of a rate of change in the forefront from beginning to end. Yes, I sacrifice some generality in my proof, but my point is not to achieve full generality. I already gave this argument for the merits, and for how it ties in with the rest of my presentation. You’ve ignored it because it doesn’t support your position.

And you’ve analogized me to the textbook writers who so offend you so many times that an attack on them is effectively an attack on me. It’s disingenuous to pretend otherwise.

Comment by John Armstrong | February 18, 2008 | Reply
The main reason I’m adding these comments is your lowly attempt to tar-and-feather me. The other is that some readers of your blog find the topic interesting.

Comment by Michael Livshits | February 18, 2008 | Reply
That’s crap and you know it. You’ve been trying to tar-and-feather my arguments from the get-go. I’ve been trying to make the point that the difference is largely aesthetic, but you’ve been offensive and rude the whole way through.

Comment by John Armstrong | February 18, 2008 | Reply
I disagree, it’s more of a mathematical rather than the aesthetic choice. And how does my prrof obscures the connection? Just draw a picture of the narrow vertical strip between $x$ and $x_0$ , bounded on top by $y-f(x)$ . The upper edge of the strip is below $y=f(x_0)+e$ and above $y=f(x_0)-e$ , $e$ as small as we want for the strip narrow enough, isn’t it obvious?

Comment by Michael Livshits | February 18, 2008 | Reply
So what does that strip have to do with the rate of change of area? Yes, you can draw the connection, just like you can unpack the limit. I just don’t find it to be as clear.

Comment by John Armstrong | February 18, 2008 | Reply
$y-f(x)$ is $y=f(x)$ in the previous post, I hope it’s clear.

Comment by Michael Livshits | February 18, 2008 | Reply
Oh, (the area of the strip)/(its width)=the average rate of change of the area.

Comment by Michael Livshits | February 18, 2008 | Reply
You see now, my proof uses the definitions only + positivity of the Riemann integral.

Comment by Michael Livshits | February 18, 2008 | Reply
Yes, it is. And that’s exactly what I meant by drawing the connection or unpacking the limit to show the equivalence. But that’s an extra conceptual step, and one that I think detracts from the proof.

I tend to prefer higher-level, more conceptual proofs, while you prefer lower-level proofs that use as few tools as possible. It’s an aesthetic choice.

Comment by John Armstrong | February 18, 2008 | Reply
Cmon, John, brighten-up, I don’t attack you, I want to save your soul

Comment by Michael Livshits | February 18, 2008 | Reply
You write at the beginning of your blog: “The main-line exposition should be accessible to the “Generally Interested Lay Audience.” It’s this audience that I worry about, people may end up thinking that MVT is necessary, and it’s a fallacy that is very common.

Comment by Michael Livshits | February 18, 2008 | Reply
And I’ve said from the beginning that you’re right: it doesn’t require the MVT. I never claimed it did, anyway. But I prefer the road that goes through the MVT. When you present the proofs, whether to a class or to on your own weblog (which I still see no evidence of), you’re welcome to use whatever approach you want.

Comment by John Armstrong | February 18, 2008 | Reply
Unfortunately “higher-level, more conceptual proofs” all too often obscure the true meaning of theorems and the logical structure of the theory. I understand that you like them, as a child that likes his new toys (no insult intended), but I hope you grow out of it. Calculus and analysis, from the practical point of view, are sleek ways to derive approximate answers from approximate data. In fact, all these differential euations are usually gross approximations of what they model, as you undoubtedly know. So, why don’t we chill out?

Comment by Michael Livshits | February 18, 2008 | Reply
You call me “childish”, and you wonder why I think you’re being rude and inflammatory? “No insult intended” really doesn’t cut it, when the statement is inherently insulting.

I’m not an analyst, I’m an algebraist and a topologist. What you see as the true meaning, I see as niggling details.

Comment by John Armstrong | February 18, 2008 | Reply
Yes, you said it to ME, but you haven’t said it to the readers of you blog explicitly. They will have to read the comments to get to that, and maybe you should give me some credit for making it clear.

Comment by Michael Livshits | February 18, 2008 | Reply
I didn’t call you childish, cmon, I meant it in a positive way, I think it’s indearing to love your new toys at any age.
As for being an algebraist and a topologist I understand that, but when you teach analysis, you better do it as an analyst, not as someone who looks down on it.

Comment by Michael Livshits | February 18, 2008 | Reply
I don’t care how you say you meant it (I don’t even quite believe your excuse), but it’s still patronizing and insulting on its face. And it’s the same sort of tone you’ve had since you started commenting here.

And I never said I look down on analysis. I just choose to emphasize different aspects of the subject when I cover it. When you cover it, you get to emphasize the parts that you prefer to.

Again: quit complaining about how I do things and get your own space. You’ll have to in.. about three hours.

Comment by John Armstrong | February 18, 2008 | Reply
And being an algebraist, why didn’t you like the idea of differentiation as division in a ring of functions? Beats me.

Comment by Michael Livshits | February 18, 2008 | Reply
I already explained that back on that post. You’re extremely sloppy about the “ring”, mostly about exactly which functions are in the ring. To what extent the quotient concept can be shored up, it’s extremely misleading when we move to functions of more than one variable.

Comment by John Armstrong | February 18, 2008 | Reply
Yeah, it’s time to say goodbye. As Newton sadd, tact is the ability to make a point without making an enemy, Ho-hum…

Comment by Michael Livshits | February 18, 2008 | Reply
Then I suppose you’ve proven yourself as tactless as I am.

Comment by John Armstrong | February 18, 2008 | Reply
Not quite so if you see division as factoring, and it was your idea to drag division as it is into many variables. You probably have heard of the Malgrange preparation theorem. It’s a primary example of the insight you can get from the algebraic point of view.

Comment by Michael Livshits | February 18, 2008 | Reply
That’s not division, that’s factoring. You’re not multiplying by an inverse. I don’t know anyone who thinks the two are the same thing at all.

I’m not going to get into the viewpoint I prefer here and now. I’ll get to that in due time. And, luckily, two and a half hours from now you’ll be able to read on in blessed silence.

Comment by John Armstrong | February 18, 2008 | Reply
Probably more so, but it was you who blew up first with “what do you want?!” and “I’ve seen all this so many times” disparaging remarks. I know it’s difficult to separate the message from the messenger, but it’s a valuable skill. As Confucius said, “a real gentleman can learn even from a fool. Come to think about it, crazy people often have rather interesting ideas.

Comment by Michael Livshits | February 18, 2008 | Reply
Thanks, but I’ll work on that skill once I finish up the job (and then tenure) hunt, my long-range research program, my expository work here, and the many superficially nonmathematical side projects I have going on.

Comment by John Armstrong | February 18, 2008 | Reply
#86 sounds really strange. Doesn’t a/b=c and a=bc are the same, by and large?

Comment by Michael Livshits | February 18, 2008 | Reply
Most definitely not! Even if the ring is commutative (which it has to be to make fractional notation well-defined), there’s no reason to assume that any element has an inverse, or that factoring behaves well. In rings of functions, zero divisors abound, and that’s only the start of the difficulties.

Comment by John Armstrong | February 18, 2008 | Reply
Good luck with becoming a stallwart of the evil establishment

Comment by Michael Livshits | February 18, 2008 | Reply
But differentiability in one variable is divisibility of f(x)-f(a) by x-a. I never said you can always divide or ther were no zero dividers, you are just putting words in my mouth again. And by the way, you say very often “you think this, you think that…” This is very rude, I never said such a thing to you. I just said “your proof sucks,” it’s much milder offence.

Comment by Michael Livshits | February 18, 2008 | Reply
But “divisibility” isn’t the same thing as “dividing”. To even write $\frac{a}{b}$ when the ring isn’t commutative or when $b$ has no inverse is nonsensical.

Comment by John Armstrong | February 18, 2008 | Reply
When I get my own space, will you visit and leave some nasty remarks? Please do. Bye.

Comment by Michael Livshits | February 18, 2008 | Reply
But 0/0 is nonsensical, and making sense of it is differentiation. By and large, a lot of interesting mathematics comes from trying to make sense out of nonsensical things, like delta-function, for example.

Comment by Michael Livshits | February 18, 2008 | Reply
Making sense of it involves a limit. Making sense of it involves all that messy 19th-century rigorization you hate so much.

Comment by John Armstrong | February 18, 2008 | Reply
x-a is not invertible in the ring of polynomials, but p(x)-p(a) can be divided by x-a, and that’s differentiation.

Comment by Michael Livshits | February 18, 2008 | Reply
It’s divisible, but the result isn’t a “quotient” in the algebraic sense.

Comment by John Armstrong | February 18, 2008 | Reply
You talk like choir boy from the Church of Limitology. Newton and Euler did calculus too.

Comment by Michael Livshits | February 18, 2008 | Reply
Man, you are too hung-up on the terminology. What is the result then, when we divide evenly one polynomial by the other?

Comment by Michael Livshits | February 18, 2008 | Reply
Yes, and they left gaps in their logic. I’m not trying to convince you that theoretical rigor is something to be valued, because I know that’s not going to happen. But you’re not going to convince me that it’s worth throwing away, either. And you’re especially not going to do it by painting me as a religious zealot. See: shrill. See: rude. See: inflammatory.

Comment by John Armstrong | February 18, 2008 | Reply
It’s not division. You’re not dividing in an algebraist’s sense.

Look, you reopened this by saying that I should, as an algebraist, like your approach to differentiation. You don’t get to pick and choose what I like about algebra. Again: I’ll get to the more expansive view on derivatives in due time.

Comment by John Armstrong | February 18, 2008 | Reply
And speaking about limits, space-time breaks down for small distances, what does is do to your limits?
Derivatives are just neat asymptotic expressions for the finite differences, no more than that.

Comment by Michael Livshits | February 18, 2008 | Reply
Why don’t you answer the question? How do call p(x)/q(x) when q divides p evenly?

Comment by Michael Livshits | February 18, 2008 | Reply
How do you call p/q? What is a quotient one polynomial by the other that divited is evenly “in the algebraic sense” as you put it? Can you give me a reference?

Comment by Michael Livshits | February 18, 2008 | Reply
Whoever brought physics into this? Who ever said anything about space-time? Keep looking back into the archives: I’ve built the theory of real numbers all the way up from the Peano axioms for the natural numbers! I’m not nearly talking about real-world applications.

Now, I know from your comments that you enjoy the real-world applications and problem-solving aspects of mathematics, but that’s just not what I’m here for. Again: my house, my motivations.

As for what you’re calling a quotient, it all depends on the ring. In this context, I’d first restrict to a domain where q(x) is nonzero (I know you hate pointwise arguments, but it’s my house), and then I’d divide now that I’m in a ring where that operation makes sense. Then I can take a limit (again: you hate. again: my house) to find the unique continuous extension to a missing point in the domain, if it exists.

Comment by John Armstrong | February 18, 2008 | Reply
Rigor is not a thing in itself, it’s a method of reasoning. If it doesn’t work, it’s time to reconsider your assumptions. And yes, your attitude towards limits is the one of a religious fanatic, I hate to say.

Comment by Michael Livshits | February 18, 2008 | Reply
Ha-ha! But you are dividing the VALUES of the polynomials, not the polynomials as expressions, it’s not algebraic, you think like an analyst! As for continuum, please put Hermann Weyl’s “The Continuum” on your reading list.

Comment by Michael Livshits | February 18, 2008 | Reply
I don’t see how it’s not working. And I don’t see how I’m a fanatic here. How many times have I said that your approach is also correct, but that I choose a different one?

The only point where I hold firm to limits is that I’m not willing to be sloppy and say that “dividing by $x-a$ ” is perfectly well-defined. It can be made sensible in certain situations, but the distinction is analytic, not algebraic. In the ring of polynomials, $x-a$ has no inverse, and you cannot divide by it.

Comment by John Armstrong | February 18, 2008 | Reply
But you are dividing the VALUES of the polynomials, not the polynomials as expressions, it’s not algebraic, you think like an analyst!

See, here’s where you think you’re so clever to use a moving target. You’ve asked me to talk about this process of division, which I see as inherently analytic, and then you find fault with my talking in terms of analysis.

The algebraic view on derivatives that I’m thinking of actually goes far deeper than these mere rings. But again, it’s not time for that yet.

Comment by John Armstrong | February 18, 2008 | Reply
I think physics can teach you a lesson or two about mathematics, besides being a major motivation and a good source of nice problems. As V.I. Arnold said, mathematics is a part of physics where experiments are cheap. There is a nice book by eldovich and Yaglom called “High Meth for Beginners,” with a preface by S.P. Novikov. You can download it as a djvu file P2P, and I got it on my veb page. Want a link?

Comment by Michael Livshits | February 18, 2008 | Reply
And again: context and scope escape you. I do know plenty of physics. I’m just not talking about physics here yet. And it’s extremely disingenuous to use such ill-defined bleeding-edge physics as quantization of space-time itself as arguments for how to discuss a field of mathematics that, though originally inspired by physics, can be defined purely on its own terms.

Comment by John Armstrong | February 18, 2008 | Reply
But we get the same result, whether it’s your method or mine, even in the ring of continouous functions! So why are you so adamant about limits? It’s just a fashoinable formalism, and nothing else.

Comment by Michael Livshits | February 18, 2008 | Reply
Because when you leave the realm of single-valued functions of a single variable we don’t get the same result. My viewpoint generalizes, while yours has to be retooled to get more generality.

Comment by John Armstrong | February 18, 2008 | Reply
“And again: context and scope escape you” see “rude,” see “touchy,” see “offensive.”

Comment by Michael Livshits | February 18, 2008 | Reply
Yes, I am being rude. I’m sick of you, and I have been for a long time. But I gave a cutoff of midnight, so I’llhonor that and keep playing this game until you hit the killfile in 40 minutes.

Comment by John Armstrong | February 18, 2008 | Reply
Is it a good way to start with the general notion? Why not start with the distributions then? How much do I have to retool? All the rules of differentiation hold, all the formulas hold, the only thing that changes is mathematical mantras, from explicit devision to uniform estimates. And when you calculate your limits you also have to prove some inequalities. Aand what’s wrong with retooling? Different tools work best in different situations. Lipschitz and Holder estimates abound in analysis, it gives flexibility. Classical definitions break down too, and you have to retool when you get to serious mathematics or serious applications.

Comment by Michael Livshits | February 18, 2008 | Reply
There’s a balance to be struck. On the one hand, an intuition must be built up before moving to the most general case. On the other hand, too much focus on the special case leads to artifacts of the specialization that must be unlearned later.

I’ve made my choice to strike this balance at a point before rigor breaks down, and before the special case looks significantly different than the general.

Comment by John Armstrong | February 18, 2008 | Reply
I’d just like to point out from Michael’s comment #83 attempting to use Newton to discuss tact and avoiding making enemies…do you know ANYTHING about Newton? The man made enemies as easily as breathing, and was well known for failing to have tact regularly. He’s one of the worst possible “experts” to cite when discussing etiquette…

Comment by Charles | February 18, 2008 | Reply
The problem with the differentiation as dividing is that it only works in one variable. In two variables, x/y really can’t be given a definition.

Comment by Walt | February 18, 2008 | Reply
very, very funny!! I think next time we will read “er… MVT means Music Vision Tele”

Comment by JuanPablo | March 4, 2008 | Reply
The mean value theorem IS the fundamental theorem of calculus. See my website for explanation of this. Nothing cranky about this. The only cranks are the ignorant mathematics professors and teachers who do not understand what they are talking about.

Comment by John Gabriel | March 29, 2009 | Reply
John, don’t encourage him any further.

Comment by John Armstrong | March 29, 2009 | Reply

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Mathematics for the interested outsider