Term-by-term differentiation of power series

An important but slightly technical result on power series states that if the power series f(z) = \sum_{n=0}^\infty a_n z^n has radius of convergence R then its derivative is \sum_{n=0}^\infty na_n z^{n-1}, where this series has radius of convergence at least R. An interesting post on Gowers’ Weblog gives a direct proof of this.

The main purpose of this post is to record a slightly different direct proof. As in Gowers’ proof we need to know that \sum_{n=0}^\infty na_n z^{n-1} has radius of convergence at least R. For completeness here is the standard proof: replace na_n z^{n-1} with a_nr^n (nz^{n-1}/r^n) where |z| < r < R and observe that \sum_{n=0}^\infty |a_nr^n| converges absolutely, while |nz^{n-1}/r^n| is bounded.

Now fix z such that |z| < R and choose s such that |z| + s < R. Since f(z) and \sum_{n=0}^\infty na_nz^{n-1} both converges absolutely and uniformly on the closed disc of radius |z| + s there exists N such that, whenever |h| < s, the errors in approximating

  1. \sum_{n=0}^\infty a_n (z+h)^n by \sum_{n=0}^N a_n (z+h)^n,
  2. \sum_{n=0}^\infty a_n z^n by \sum_{n=0}^N a_n z^n,
  3. \sum_{n=0}^\infty na_n z^{n-1} by \sum_{n=0}^N na_n z^{n-1}

are all < \epsilon. Since

\frac{1}{h} \sum_{n=0}^N (a_n(z+h)^n - a_nz^n - hna_n z^{n-1} ) = hg(h)

for some (complicated) polynomial g, the left-hand side is < \epsilon for all sufficiently small h. Hence

\bigl| \frac{1}{h} \sum_{n=0}^\infty (a_n(z+h)^n - a_nz^n - hna_n z^{n-1} )   \bigr| < 4\epsilon

for all sufficiently small h. The result follows.

I like this proof because all the heavy work is done by the absolute and uniform convergence of power series with their circle of convergence. Okay, this takes some work to prove, but the trick used above to show that \sum_{n=0}^\infty na_n z^{n-1} has radius of convergence at least R can be applied, or one can use the Weierstrass M-test.

There are several alternatives that are probably more conceptual, but depend on more technology. For example, there is a real variable proof which starts by showing that g(x) = \sum_{n=0}^\infty na_n x^{n-1} converges uniformly on a closed disc inside the circle of convergence, and then uses integration (of a uniform limit of continuous functions) to show that g has antiderivative f.

Here is an outline of a complex variable proof. Once all the ingredients are in place, it takes little work, but the taste is probably a bit artificial. Let r < R and let \overline{D}_r denote the closed unit disc of radius r in \mathbb{C}. Let f_M(z) = \sum_{n=1}^M a_n z^n. Then f_M(z) \rightarrow f(z) uniformly on \overline{D}_{r}. As a uniform limit of holomorphic functions, f is holomorphic (Morera's Theorem). By the derivative version of Cauchy's Integral Formula,

f'(z) = \frac{1}{2\pi \mathrm{i}} \oint_\gamma \frac{f(w)}{w^{2}} \mathrm{d} w

where \gamma is a circular contour about z contained in D_r. Since f_M(w)/w^2 converges uniformly to f(w)/w^2 on this contour as M \rightarrow \infty, we have

f'(z) = \lim_{M \rightarrow \infty} f'_M(z),

and the result follows.

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