Section 4

Identically Vanishing Power Series

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Problem 3.248

The sequence a0,a1,a2,…,an,…a_{0},a_{1},a_{2},\ldots ,a_{n},\ldots is assumed to satisfy the condition

lim sup⁡n→∞log⁡∣an∣n=−h,h>0. \limsup _{n\to \infty } \frac{\log \left|a_{n}\right|}{\sqrt{n}} = -h, \quad h>0.

Then the series

Φ(s)=2a0+a1(es+e−s)+a2(e2s+e−2s)+⋯+an(ens+e−ns)+⋯ \Phi (s) = 2a_{0} + a_{1}(e^{s}+e^{-s}) + a_{2}\left(e^{\sqrt{2}s}+e^{-\sqrt{2}s}\right) + \cdots + a_{n}\left(e^{\sqrt{n}s}+e^{-\sqrt{n}s}\right) + \cdots

converges in the infinite strip

−h<Re(s)<h -h < \mathrm{Re}\left(s\right) < h

of the ss-plane. The convergence is absolute and uniform in any closed interior strip −h+ε≤Re(s)≤h−ε-h+\varepsilon \leq \mathrm{Re}\left(s\right)\leq h-\varepsilon, ε>0\varepsilon >0; there the series defines an analytic function Φ(s)\Phi (s). The function Φ(s)\Phi (s) vanishes identically only if all the coefficients a0,a1,a2,…,an,…a_{0},a_{1},a_{2},\ldots ,a_{n},\ldots vanish. [Compute

F(u)=12πi∫a−i∞a+i∞Φ(s)e−uss2 ds,0<a<h,u>0;problem 155.] F(u) = \frac{1}{2\pi i}\int _{a-i\infty }^{a+i\infty } \Phi (s)\frac{e^{-us}}{s^{2}}\, ds, \quad 0<a<h, \quad u>0; \quad \text{problem 155}.]
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Problem 3.249

We assume that the power series

f(z)=a0+a1z+a2z2+⋯+anzn+⋯ f(z) = a_{0}+a_{1}z+a_{2}z^{2}+\cdots +a_{n}z^{n}+\cdots

converges inside the unit disk and that f(z)f(z) and all its derivatives tend to 00 as zz tends to 11 on the real axis, i.e.lim⁡z=x→1f(n)(z)=0\lim \limits_{z=x\to 1} f^{(n)}(z) = 0, n=0,1,2,…n=0,1,2,\ldots Then there are two possibilities:

  1. f(z)f(z) vanishes identically;

  2. z=1z=1 is a singular point for f(z)f(z). If the power series converges in a disk larger than the unit disk, i.e.if

lim sup⁡n→∞log⁡∣an∣n<0, \limsup _{n\to \infty } \frac{\log \left|a_{n}\right|}{n} < 0,

we are necessarily dealing with the first case. The weaker condition

lim sup⁡n→∞log⁡∣an∣n<0 \limsup _{n\to \infty } \frac{\log \left|a_{n}\right|}{\sqrt{n}} < 0

admits the same conclusion. [Construct the function Φ(s)\Phi (s) of problem 248.]

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Problem 3.250

The proposition 249 is not valid if the condition on the coefficients a0,a1,a2,…,an,…a_{0},a_{1},a_{2},\ldots ,a_{n},\ldots of f(z)f(z), namely lim sup⁡n→∞log⁡∣an∣n<0\limsup \limits_{n\to \infty }\dfrac {\log \left|a_{n}\right|}{\sqrt{n}}<0, is replaced by

lim sup⁡n→∞log⁡∣an∣nμ<0,0<μ<12. \limsup _{n\to \infty } \frac{\log \left|a_{n}\right|}{n^{\mu }} < 0, \quad 0<\mu <\frac{1}{2}.

[Put

f(z)=∫0∞e−xμcos⁡μπsin⁡(xμsin⁡μπ)e−x(1−z) dx;problems 153, and Part II, problem 222.] f(z) = \int _{0}^{\infty } e^{-x^{\mu }\cos \mu \pi }\sin (x^{\mu }\sin \mu \pi )e^{-x(1-z)}\, dx; \quad \text{problems 153, and Part II, problem 222.}]
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