# Can we Relate Radius of Convergence of Taylor Series and Asymptotic Rate of Growth?

I still need to be disabused of the belief that there is some simple connection between the finiteness of the radius of convergence and the asymptotic rate of growth.

1. Can we develop any specific criteria based on the asymptotic growth rate to determine when the radius of convergence is finite or infinite?

2. Does $f$ analytic with $f=\mathscr{O}(1)$ imply that it has infinite radius of convergence?

3. Does $f$ analytic with $f=\omega(\ln(x)) \land f=\mathscr{O}(e^x)$ imply that $f$ has infinite radius of convergence?

4. (Converse of 2. and 3.) Do the logarithm and exponential represent lower and upper bounds, respectively, for rates of growth which an analytic function diverging at infinity may have such that its radius of convergence is infinite?

And thus all other analytic functions with infinite radius of convergence are $\mathscr{O}(1)$, i.e. bounded?

I recognize that the answers to the second through fourth questions are almost certainly no, but to what extent, if any, are they almost true? It seems like some not-so-much-weaker version of them might be true. The motivation for these conjectures/these questions is probably best illustrated through some examples:

1. The Gamma Function $\Gamma$ - A super-exponential function with finite radius of convergence.

Taking logarithms, I believe that the Taylor expansion of the logarithm of the gamma function implies that $\ln \Gamma = \Omega(\ln(x))$, whereas obviously $\ln e^x = \mathscr{O}(1)$.

2. The Logarithm - A sub-polynomial function with finite radius of convergence

Note that $\ln(x)=o(x^{\epsilon})$ for any $\epsilon > 0$. Moreover, $\int_1^{\infty} \frac{1}{t^{1+\epsilon}} \text{d}t <\infty$ for any $\epsilon >0$, whereas $\ln(x)=\int_1^x \frac{1}{t} \text{d}t$ diverges as $x \to \infty$. Hence the logarithm seems to represent some sort of fundamental "border" function for the polynomials, which obviously all have infinite radii of convergence and are actually exactly the class of all functions with terminating (finite) Taylor series.

3. Sine and Cosine - Two functions with infinite radii of convergence which are $\Theta(1)$.

4. Polynomials - Infinite radii of convergence, terminating Taylor series, and $\omega(\ln(x))$ and $o(e^x)$.

5. Other Trigonometric Functions ($\tan,\cot,\sec,\csc$) - Finite Radii of Convergence, and satisfying $\omega(e^x)$ as $x \to \frac{\pi}{2}$ or $x \to \pi$.

6. Bessel Functions of the First or Second Kind - these functions are bounded and analytic and (coincidentally?) also have infinite radius of convergence.

7. A Possible Counterexample?

As @Wojowu suggested, consider the function $\Xi(x):=\displaystyle \sum \frac{x^n}{(n!)^2}$

Does $\Xi(x)$ have finite or infinite radius of convergence? Does $\Xi(x)=o(e^x)$ or $\Xi(x)=\Theta(e^x)$? (I am assuming that $\Xi(x)=\mathscr{O}(e^x)$, but I could be wrong.)

Background:

This is a follow-up to a question I asked previously: Is Every (Real) Analytic Function (with Non-Degenerate MacLaurin Series) Asymptotically Greater Than any Polynomial?

The asymptotic rate of growth of an entire (everywhere analytic function) will be determined both by which of its terms vanish as well as the signs of the terms.

However, as comparing the examples of $\Gamma, \sin,$ and $\ln$ show, no simple criteria based solely on the number of vanishing terms nor on their signs is enough to make definitive conclusions about asymptotic rate of growth (although there is almost certainly some complicated connection).

• I believe the answer to question #1 is simply "no." Take any entire function which is real on the real axis, then multiply it by $1/(1+x^2)$. The resulting function will have a finite radius of convergence and will be within $\Theta(x^{-2})$ of the original function as $x \to \pm \infty$. If I recall correctly it's always possible to find a real entire function which has any given asymptotic growth behavior, so this argument can produce a real analytic function with any asymptotic growth but whose power series has a finite radius of convergence. (...) – Antonio Vargas May 8 '16 at 18:04
• (...) I hope someone will chime in to correct or confirm my suspicion. – Antonio Vargas May 8 '16 at 18:04
• I like that argument -- it's sounds like a good way to generate counterexamples. Do you remember where you might have learned that it's always possible to find a real entire function which has any given asymptotic growth behavior? This is what confuses me, because if there is no limit, then why should the gamma function not be entire? I hypothesized that analytic functions must just be the closure of the polynomials under "appropriately bounded infinite sums" and that therefore there must be some "maximal" element to this set (in terms of growth) - I hypothesized the exponential -- but only – Chill2Macht May 9 '16 at 2:12
• I learned it here on MSE, actually, but I can't seem to find the thread. I recall that the proof used the Hadamard Factorization Theorem in some way. I also learned that there is no "fastest growing" entire function, basically due to the same argument. Sorry I can't be more help. – Antonio Vargas May 9 '16 at 2:40
• Re Question 7 (this is really too many for one post, btw...) $\Xi(x) = \Theta (x^{-1/4} e^{2\sqrt{x}})$ as $x \to \infty$. – Antonio Vargas May 9 '16 at 2:44

For an arbitrary sequence $a_n$, there is an entire function $f$ which interpolates the sequence: $f(n) = a_n$ for all $n \in \mathbb N$. This follows from Mittag-Leffler's theorem and the Weierstrass factorization theorem. See also this paper of I.M. Sheffer

• Can we say the same for analytic but not entire functions (i.e. finite radius of convergence)? – Chill2Macht May 9 '16 at 4:42
• Also this means then that there is an entire function which interpolates the factorials? This makes me question why the Gamma function is the "natural choice" then (I read the reason once, but I need to find what it was again) – Chill2Macht May 9 '16 at 4:48
• 1) Of course. For example, if you want poles at $\pm i$, take an entire function that interpolates $a_n (1 + n^2)$ and divide by $1 + z^2$. – Robert Israel May 9 '16 at 5:08
• 2) Perhaps you were referring to the Bohr-Mollerup theorem? – Robert Israel May 9 '16 at 5:11
• Yes thank you! Both of these are really helpful again! I really appreciate it. – Chill2Macht May 9 '16 at 14:27

@AntonioVargas I think I found what you were referring to:

https://en.wikipedia.org/wiki/Entire_function#Growth

The relevant conclusion from the article is:

Entire functions may grow as fast as any increasing function: for any increasing function $g: [0, \infty) \to [0, \infty)$ there exists an entire function $f(x)$ such that $f(x) > g(x)$ for all $x \ge 0$.

An example of such a function is: $$f(x) = g(2) + \displaystyle\sum\limits_{k=1}^{\infty} \left( \frac{z}{k} \right)^{2 \lceil k \ln g(k+2) \rceil}$$

As a result, 4. is clearly false. I would also suspect more strongly than before as a result that 1.-3. are false, but I will have to do more research for more rigorous proof.

Further Reading: which I have not yet done but aim to do in the future

Boas, Ralph P. - 1954 Entire Functions. Academic Press. OCLC 847696
Levin, B. Ya. - 1996 Lectures on Entire Functions. American Mathematical Society.