3 Comments

1. Jun15

   Dr. Zen Harper

   Hi Kevin,

   I too thought about and taught this myself a few years ago, and the easiest proof I came up with used the Mean Value Theorem (but only in the special case of polynomials, which presumably can be simplified further in this special case, but I haven’t thought about that).

   I directly considered

   $| (f(z+h)-f(z))/h – g(z) |$,

   for z fixed and |h| sufficiently small [but nonzero of course!], with g(z) being the FORMAL power series for f'(z); as you say, the tricky part is to show that f'(z) really does exist and does equal g(z), but I’m assuming we already know that g(z) does converge.

   Take modulus signs inside the sum, using the triangle inequality, and you end up having to estimate

   $|p_n(z,h)| = | ((z+h)^n – z^n)/h – nz^{n-1} |$,

   for each positive integer n.

   Now, trick (1): by the Binomial Theorem, this is a polynomial in z and h with POSITIVE coefficients, so it is less than the same polynomial but with |z| and |h| instead of z and h, i.e. $|p_n(z,h)| \leq p_n(|z|, |h|)$.

   Now trick (2): you can hit it with the Mean Value Theorem (twice!); or if you prefer with Taylor’s Theorem with remainder; or if you fancy your algebraic skills, come up with a direct calculus-free estimate [which must be possible, but I haven’t thought about it].

   The rest of the details are easy [and fun?!] to fill in, showing that this does indeed tend to zero as |h| tends to zero [which actually follows from the same term-by-term differentiation result applied to g(z): the formal power series for g'(z) does also converge absolutely, and you take out a factor of |h|. This is trick (3). But, we really do need absolute convergence here.]

   I’d be surprised and also interested if this could be simplified much further; the Mean Value Theorem must be “algebraically removable” in the special case of polynomials (an exercise for the reader…). Of course we don’t even need the full Mean Value Theorem, an inequality up to a constant would do here if it is easier.

   Reply
2. Jun18

   Dr. Zen Harper

   …hi again, sorry the TeX didn’t seem to work in my last comment. The following also came to mind, although it seems decidedly less “cool” than the other proof to me; but the method [instead of differentiating term-by-term, try integrating the formal derivative term-by-term!] is useful in many other contexts so is maybe worth mentioning.

   If you don’t like the Mean Value Theorem, another way is to show that f(z+h)-f(z) is a line integral of g, for z fixed and |h| sufficiently small as before, by integrating g term-by-term [from z to z+h] and use the nice convergence properties of the power series (uniformly absolute convergence). This works very well if you are only doing real variables. For complex variables, it is a bit more tricky, but I think it still works without needing any general Cauchy’s Theorem for analytic functions; you only need to evaluate the line integral of polynomials, which can even be directly proved just with real variables, real/imaginary parts, and basic calculus, although it would be messy!

   Of course this is a cheat; to be rigorous, you need to know (1) the Riemann integral for continuous functions, and (2) the easy form of the Fundamental Theorem of Calculus, i.e. differentiating the integral gives the original function, if the original function is continuous [because we end up with f being the integral of g plus a constant, and g is known to be continuous.]

   Strangely enough, the Mean Value Theorem appears to have been completely bypassed by this approach [proving that the Riemann integral exists for uniformly continuous functions needs almost no analysis, just completeness of R and little else. Evaluating the integral of a polynomial by the harder Fundamental Theorem of Calculus, that the integral of the derivative is the function, implicitly uses the Mean Value Theorem, but this can be done directly for polynomials. Come to think of it, this ends up almost being the same as my above comment about proving the Mean Value Theorem directly for polynomials!]

   But I think the level of “coolness” for this proof has dropped below a certain minimum level by now, so I’ll stop!

   Reply
   • Jun19

     Kevin Houston

     Hi Zen,
     Thanks for this and the other comment on cool proofs. I had considered both approaches, MVT and integrating the formal derivative. No method I’ve looked at seems to fulfill my criteria that it needs to be obvious why we are taking that approach unless it is a really cool trick.
     My final solution will be more like the MVT version you gave as it is easier to follow. The trouble with all these proofs is that students don’t see the necessity of having to prove the result and the integrating the formal derivative approach suffers from the fact that it appears to them that you are using a circular argument (although they wouldn’t put it that way). In the end it comes down to do I do something a little bit harder to stretch them by making them work or do I do something which I know they will understand more. (The latter is not always the best option, sometimes it is best to make them work of course!)

     By the way I think the Latex should come out correctly if you do $latex [some latex here and finish with a dollar sign].

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