Given a non-integral real $\alpha$, is there an entire (see http://en.wikipedia.org/wiki/Entire_function) function $h(x)$ such that $x^{-\alpha}h(x)\longrightarrow 1$ for $x\rightarrow+\infty$ (with $x$ real non-negative)?

Clearly, such a function if it exists is not unique since $h(x)+e^{-x}$ and similar functions work also.

  • $\begingroup$ What is the question? $\endgroup$ – Per Alexandersson May 28 '10 at 9:05
  • $\begingroup$ Existence, yes or no, and if yes, an example, of an entire function $h$ such that $x^{-\alpha}h(x)\rightarrow 1$ for $x\rightarrow+\infty$ with $x$ real. $\endgroup$ – Roland Bacher May 28 '10 at 9:20
  • $\begingroup$ Looking at the example of $1/\Gamma(1+x)\sim (e/x)^x(2\pi x)^{-1/2}$ as $x\to+\infty$, I would say "yes" but definitely a maitre in complex analysis is wanted. :) $\endgroup$ – Wadim Zudilin May 28 '10 at 10:11
  • $\begingroup$ Casorati-Weierstrass: If $f$ has an essential singularity at $a$, then the image under $f$ of any punctured disk around $a$ is dense in $\mathbb C$. Use for $a=\infty$. Take an entire function $f(z)$ and consider the entire $g(z)=zf(z)^2$; there exists a direction $\lambda$ along which $g(z)\to C\ne0$ as $z\to\infty$. Then $f(x)=c_0\sqrt{x}g(x/\lambda)$ will give an example with $\alpha=-1/2$. $\endgroup$ – Wadim Zudilin May 28 '10 at 10:28

Start with an entire function $f$ such that $f(x)=1/x + O(1/x^2)$ for $x>0$, $x\rightarrow\infty$. For example $f(z)= (1-e^{-z})/z$.

Let F be some primitive for $f$: $F(z)=\int_1^z f(s)ds$.

We have $F(x)= ln(x)+C+O(1/x)$, with C some constant ($ \ C=\int_1^\infty \ (f(x)-{1\over x})\ dx$ ).

Then consider $h(x)=exp(\alpha F(x)-\alpha C)$.

We get ${h(x)\over x^\alpha} = exp(O(1/x))\rightarrow 1$.

  • $\begingroup$ Yes, it works. $\endgroup$ – Wadim Zudilin May 28 '10 at 10:33
  • $\begingroup$ It works up to a constant (which is easily dealt with) picked up by the integral. $\endgroup$ – Roland Bacher May 28 '10 at 15:20

As a matter of fact, real entire functions (that is, entire functions that map the real line into itself, or equivalently, functions represented by a power series centered in 0, with real coefficients and radius of convergence infinite) are dense in $C^0({\mathbb R}, \mathbb{R})$ in the sense of the order, that is:

Theorem (T.Carleman, 1927). For any two continuous real valued functions f < g there exists a real entire function $\phi$ in between:

$f(x)<\phi(x) < g(x)$ for all $x\in\mathbb{R}$.

So in particular, an entire function may be asymptotic to any continuous real function, and also, it may grow as fast as any continuous function.

  • $\begingroup$ There seems to be an issue with the latex and markdown. The last line of Pietro's box should read: $f(x)<\phi(x)<g(x)$ for all $x\in\mathbb{R}$. $\endgroup$ – j.c. May 28 '10 at 18:20
  • $\begingroup$ @pietro: do you happen to have a reference that theorem (either a modern presentation or a reference to which paper of Carleman)? The internet suggests that it is his 1927 paper Sur un theoreme de Weierstrass in Arkivfor matematik, astronomi och fyski [sorry about lack of accents] but the journal is not available in my library so I'd like to be sure before I make the librarian go get me a copy. $\endgroup$ – Willie Wong Oct 20 '15 at 14:17
  • $\begingroup$ @Willie: yes that is Carleman paper. The proper keyword for a search is "Carleman approximation". For instance this paper : projecteuclid.org/euclid.mmj/1031710533 $\endgroup$ – Pietro Majer Oct 22 '15 at 7:35

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