Mathematically Correct; Pedagogically Questionable

One important aspect of the ongoing struggle over K-12 math education is over how much rigor is desirable. I’m among those who believe the typical U.S. high school goes way overboard in emphasizing rigor for the vast majority of students. The (college) sophomore calculus textbook I used as a student long ago has something to say about this that I think is well worth bearing in mind. This is from the Preface for the Teacher to A. W. Goodman’s 1969 Analytic Geometry and the Calculus, 2nd ed.  (I corrected a typo in Theorem R: the righthand side was printed as “f (u0) u’(x0)”.)

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…The student who is well prepared and who is interested in pure mathematics for its own sake may be able to understand and appreciate a rigorous course in the calculus. But the majority of students are still a little insecure in their algebra and trigonometry, and are far more interested in learning what the calculus can do and where it is going than in following a purely logical argument… I contend that one should not try to state all the hypotheses in a theorem, because the statement can become so long as to be incomprehensible to the average student… As an illustration, consider the following:

Theorem S.  If  y = f(u)  and  u = g(x) , then the derivative of the composite function y = f (g(x))  is given by

dy/dx = dy/du • du/dx

Here is a statement that is brief and simple, and the average student has a reasonable chance of understanding it. Now let is look at the same theorem when stated in a rigorous fashion.

Theorem R. Let f and g be two real-valued functions of a real variable and suppose that the range of g is a subset of the domain of f. Let  h = fg  be the composite function defined over the domain of g by setting  h(x) = f(g(x)) for each x in the domain of  g. If x0 is an interior point of the domain of g, and g is a differentiable function at x0, and if f is a differentiable function at u0 = g(x0), where u0 is an interior point of the domain of f, then h is a differentiable function at x0, and further the derivative is given by the formula

h’(x0) = f(u0) g’(x0)

There is no doubt that R is the correct statement and S is full of gaps. However, the average student can learn and use S, but when R is presented he will either fall asleep or totally ignore it. It is just too complicated for him to master at this stage of his mathematics study. The presentation of R rather than S does real harm because it serves to repel many students who are originally attracted to mathematics and who might turn out to be capable technicians or teachers (perhaps even creative mathematicians) if they are given a reasonable chance to develop.

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It’s interesting to compare the above with what appears in the 4th edition of Applied Calculus by Hughes-Hallett et al, a popular textbook these days for “brief surveys” of calculus for non-science students. Applied Calculus simply states the above theorem as the “Chain Rule”: it does not call it a theorem, nor does it provide any more than a vague intuitive argument for it. In fact, I don’t think it includes anything beyond handwaving arguments for anything. In other words, Applied Calculus is even less rigorous than Goodman’s Theorem S approach. Yet, based on my own experience, it’s all most college business majors (for example) can handle—and yes, an intuition for what calculus can do really is useful to them! The other extreme, as represented by Theorem R, is undoubtedly perfect for some students, but I’m sure Goodman is correct: it won’t do anything for “the average student” except convince them calculus isn’t worth the trouble.

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3 thoughts on “Mathematically Correct; Pedagogically Questionable

  1. Most college students serious about math would have a hard time with version R. I know I would fall asleep or look for another book to explain it.
    It’s possible to derive version S without too much difficulty, it seems to me, using the limit definition of a derivative.
    It also seems to me that using something like Geometer’s sketchpad we can graphically show what the derivative is of f(g(x)) by constructing a moving tangent to the curve, graphing the value of the slope, then calculating the values of dy/du and du/dx and graphing their product. (I know it sounds hard but it’s not) At least this will imply that the answer is not completely full of crap.

  2. Good point about Geometer’s Sketchpad. (I suppose Geogebra could do the same thing?) Of course it wouldn’t add rigor, but it’d be a great way to make the “chain rule” more intuitive! Would you care to implement it? I’d love to see that.

  3. The chain rule is important as the foundation of many other demonstrations and proofs. As there are many students who will not accept what is not demonstrated as true, a demonstration of the general case should be given. It is sometimes hard to wrap one’s mind around the chain rule. How the demonstration is presented is important, so it doesn’t appear as a self-evident manipulation of differential fractions. All the formal jargon associated with a formal proof can be dropped, I agree. But don’t dump the proof.

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