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# Cool proofs again

In between marking exams, marking essays, external examining, getting my two PhD students to submit their theses, and, well you get the idea, I’ve been trying to come up with a cool proof of a theorem in complex analysis:

Let be a power series with radius of convergence . Then, .

I.e., the derivative of is what you expect it to be, the termwise differentiation of the series.

I’ve not been able to find a good proof of this. My plan of attack has been to show first via the root test that the formal derivative has radius of convergence . The main sticking point is then to show that this formal derivative really is equal to . The clearest proof I found was in James Pierpoint‘s book *Functions of a Complex Variable* which you can find online here. Like many old books (this is from 1914) I’ve found that, if you allow for slightly odd notation and nomenclature, it contains excellent material that is currently passed over. (Does anyone still teach Raabe’s test?)

Anyhow, the proof is on pages 172-173 and relies on pages 84-85. For me the proof is very clear because of equation 3 on page 84: the expansion of via rows can be turned into a power series in by summing the columns. My problem is that when I try to turn everything into a rigorous proof this becomes obscure. Also, we have to prove that *all* the formal derivatives of have radius of convergence . This in itself is not difficult but it means we have to set up the notion of all formal derivatives rather than just the first one. For me this may be too much extraneous clutter for students.

The hunt continues…

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.

…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!

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].