4
$\begingroup$

Daniel Velleman has written Calculus: A Rigorous First Course.

It seems to me that it's questionable whether such a course should be used for all but a small number of students, not just because most will never appreciate logical rigor, but because there's so much stuff they may appreciate that you won't be telling them about because you'll be explaining logical rigor instead. However, skip this paragraph and resume reading below.

Velleman introduces an unconventional notation: $f(x)\to 6$ as $x\to2^\ne.$ The superscript $\text{“}\ne\text{''}$ serves as a reminder that although $f(x)$ is not forbidden to be equal to $6$ as it is approaching $6,$ nonetheless $x$ is forbidden to be equal to $2$ as it is approaching $2.$

Besides understanding concepts in a logically rigorous way (as in $\varepsilon$-$\delta$ arguments and some other occasions), it seems to me the principal place where this is useful is in understanding why a proof of the chain rule is problematic in a way in which proofs of other differentiation rules are not. \begin{align} \frac{dy}{dx} = {} & \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \cdots\overbrace{\cdots\cdots\cdots\cdots\cdots}^{\large\text{What goes here?}}\cdots = \frac{dy}{du} \cdot \frac{du}{dx}. \\[12pt] & \lim_{\Delta x\to0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x\to0} \frac{\Delta y}{\Delta u}\cdot\frac{\Delta u}{\Delta x} = \text{?} \end{align} We have $\Delta x\to0^\ne.$ And so $\Delta u\to0.$ (And if you want to be logically rigorous, here we have relied on a theorem that says differentiable functions are continuous.)

But we have only $\Delta u\to0,$ not $\Delta u\to0^\ne,$ and that is the difficulty to be overcome.

So my question is whether, besides the proof of the chain rule, there is any other place in the conventional less-than-rigorous calculus course where Velleman's novel notation is so useful for efficiently stating the point? (Addendum in response to a posted answer: I'd have thought the following was obvious, but I guess it wasn't: I was not proposing that this helps in actually proving the chain rule.)

$\endgroup$
  • 1
    $\begingroup$ My opinion as a student is that one fundamental aspect of mathematics is that we have a universally agreed upon system to work in, at least up to some level. In theory, everyone agrees what it means when we say $\lim_{x\to a} f(x) = \alpha = f(a)$. We agree upon such a system, and we draw logical conclusions from that system, in a way that is not open to debate - to parallel physics, we may say that let's agree that the universe is an Einstein or Newton universe, what conclusions may then follow. We then compare our conclusion to our "observations", in case of mathematics, "intuition". $\endgroup$ – Project Book Oct 22 '17 at 3:17
  • 1
    $\begingroup$ @ProjectBook : "Intuition" seems to be a catch-all term, used when you don't know what to call it instead of "intuition". Anyway, your comment seems somewhat tangential. $\endgroup$ – Michael Hardy Oct 22 '17 at 3:26
  • $\begingroup$ Based on the facts you mention in question, I think that the notation does highlight that limit as $x\to a$ is something very different from what happens at $x=a$. I think this will help students who are so tempted to plug the value of $x$ (or the differentiate and plug devotees of L'Hospital's Rule). It takes ages to understand that plugging does not work everywhere but it does work in most places as most common functions have a special feature called continuity. See my answer math.stackexchange.com/a/1822706/72031 $\endgroup$ – Paramanand Singh Oct 22 '17 at 3:35
  • 3
    $\begingroup$ I don't see any harm in a new notation if it is explained properly (and compared with other prevalent notations). Math is not about symbols and notations, but rather about the ideas which are described by these symbols. Unfortunately most modern textbooks have replaced rigor in mathematics with excessive use of symbols and jargons (formalism). One must always prefer substance (rigor) over form (notation). $\endgroup$ – Paramanand Singh Oct 22 '17 at 3:46
  • 1
    $\begingroup$ You can say that. Let's say it is the substance of a logical deduction. $\endgroup$ – Paramanand Singh Oct 22 '17 at 6:44
4
$\begingroup$

Here is another place where the notation is important: If $\lim_{x \to a} f(x) = b$ and $\lim_{u \to b} g(u) = c$, does it follow that $\lim_{x \to a} g(f(x)) = c$? Of course, the answer is no, but why? Here's an answer using the new notation: Let $u = f(x)$ and $y = g(u) = g(f(x))$. Then we have: as $x \to a^\ne$, $u \to b$, and as $u \to b^\ne$, $y \to c$. But the mismatch between $u \to b$ and $u \to b^\ne$ means that these cannot be combined to conclude that as $x \to a^\ne$, $y \to c$. If $g$ is continuous at $b$, then we can say that as $u \to b$ (no superscript), $y \to c$, and then we can conclude that as $x \to a^\ne$, $y \to c$.

Most calculus books have a theorem about limits of sums, differences, products, and quotients, but no systematic way to work out limits of compositions. The reason, in my view, is that they don't have the notation necessary for talking about limits of compositions. This new notation fills this gap. The new notation is used in a number of places throughout the book. (Similar notation even comes up in the proof of the fundamental theorem of calculus.)

The use of a superscript is consistent with other standard notations that use a superscript to indicate how a variable approaches a limit, such as $x \to a^+$.

In response to the second paragraph of the original post: I think the comments in this paragraph apply to a book like Spivak, but not my book. Spivak covers a lot of topics that belong in an analysis course, and as a result there are lots of standard calculus topics that he doesn't cover. But that's not true of my book, which is really a calculus book and not an analysis book. The arrow notation I have introduced is no more complicated or difficult to use than standard arrow notation, but it makes informal (non $\epsilon$-$\delta$) reasoning about limits more reliable. Isn't that what good calculus notation should do?

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Counterexamples to the proposition about compositions would appear to be $(1)$ cases in which $g(b)$ exists and differs from $c$ and $(2)$ cases in which $f(x)$ has values outside the domain of $g.$ Maybe case $(2)$ is relevant outside of a Spivak-like context. On the whole I like the new notation, except that $(1)$ I would rather write $x \mathrel{\downarrow} a$ than $x \to a^+$ and similarly $x \mathrel{\underset\ne\to} a. \qquad$ $\endgroup$ – Michael Hardy Oct 25 '17 at 19:03
  • $\begingroup$ What experience do you have teaching in typical flagship state universities? $\endgroup$ – Michael Hardy Oct 25 '17 at 19:06
  • 1
    $\begingroup$ I'd be happy with the notation $x \underset{\ne}{\to} a$. But $x \to a^+$ is pretty standard in calculus books, and I was trying to stay close to standard notation. That's why I used $x \to a^{\ne}$. $\endgroup$ – Dan Velleman Oct 25 '17 at 20:13
  • 1
    $\begingroup$ Experience at state universities: not much, although I did teach for 3 years at Univ. of Texas in the 1980s. So I don't know how my approach will go over with typical state university students. But I think people sometimes underestimate the value of rigor in introductory courses. In some ways, rigor makes things easier for students (by making it clear what the rules are), and it is handwaving that makes math challenging for them. $\endgroup$ – Dan Velleman Oct 25 '17 at 20:17
  • $\begingroup$ That probably depends on what kind of students you've got. The way I learned math, starting in 8th grade there was emphasis on how to know independently of authorities that things you were learning are true, and before 8th grade I was doing that to some extent although it wasn't explicitly expected. But students who show up in calculus, or who have passed calculus, often have never suspect that that is a thing. And some of them pride themselves on working hard and following instructions and always getting an "A+" by doing that, and they don't suspect that they$\,\ldots\qquad$ $\endgroup$ – Michael Hardy Oct 25 '17 at 23:17
1
$\begingroup$

Note that in the French tradition limits of functions are not defined the same way as in America: If the function $f$ happens to be defined at the point $a$ (in most cases it is not) then according to the French the $\lim_{x\to a} f(x)$ only exists, if $f$ is continuous at $a$, whereas according to Baby Rudin the value $f(a)$ is not taken into consideration for the $\lim_{x\to a} f(x)$. The following Wikipedia entries show this difference:

https://en.wikipedia.org/wiki/Limit_of_a_function

https://fr.wikipedia.org/wiki/Limite_(mathématiques)

Now Vellemann wants to make this distinction visible in his limit notation. For my part I'm less than happy with his proposal, because the $\ne$ command does not affect (or operate on) the constant $2$ in any way. Instead the command concerns the freedom the point $x$ has when approximating $2$. Therefore I'd write $x\to_{\ne}2$ (or similar) instead.

As an aside: I don't think that the special notation introduced here is helping the "Calculus 101" proof of the chain rule. The correct procedure is to avoid denominators in this proof altogether. See my answer to this question.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ When one writes $\displaystyle f'(x) = \lim_{h\to0} \frac{f(x+h) - f(x)} h,$ the function of $h$ whose limit is taken is always undefined at $h=0. \qquad$ $\endgroup$ – Michael Hardy Oct 22 '17 at 18:30
  • 1
    $\begingroup$ You miss the point about the chain rule. I never said it helps with the proof; I said it helps the reader understand why a proof of the chain rule is problematic in a way in which proofs of other differentiation rules are not. It is a simple way to explain what's different about that one. $\endgroup$ – Michael Hardy Oct 22 '17 at 18:37
  • 1
    $\begingroup$ How much experience do you have teaching first-year calculus? What kinds of students did you have? (Many who have taught calculus for decades don't know what kinds of students they have had and think they do know, so maybe that last question doesn't amount to much.) $\endgroup$ – Michael Hardy Oct 22 '17 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.