Talk:Entire function

Is this article statement true or false?
The article says:
 * A function that is defined on the whole complex plane except for a set of poles is called a meromorphic function.

I thought a meromorphic function can be defined just on a subset of the complex plane, not on the whole complex plane minus some singularities. Oleg Alexandrov 06:01, 18 February 2005 (UTC)


 * You are right and the article is right. You are talking about a meromorphic function on the subset of the plane. The article is talking about a meromorphic function on the plane. Maybe the article should be more specific.  franklin   21:46, 30 December 2009 (UTC)

Completeness of examples
Are the only entire analytic functions polynomials in both x and exp(x)? njh 02:06, 23 June 2005 (UTC)
 * No. Any function of the form


 * $$a_0 + a_1 x+ a_2 x^2 + ... $$


 * with an infinite radius of convergence is entire function. Oleg Alexandrov 02:41, 23 June 2005 (UTC)


 * An example of an entire function that is not a polynomial in $$x$$ or $$\exp(x)$$ is $$\exp(\exp(x))$$. Another is $$\exp(x^2)$$. Eric Kvaalen (talk) 11:50, 8 February 2013 (UTC)

Properties
How would you define a function being holomorphic at ∞? —Preceding unsigned comment added by JumpDiscont (talk • contribs) 03:52, 10 November 2009 (UTC)


 * Holomorphic at infinity means that it's holomorphic in 1/z. For example, the function 1/z itself. A function cannot be holomorphic everywhere including at infinity unless it is constant. Eric Kvaalen (talk) 11:50, 8 February 2013 (UTC)

An entire function can grow faster than any given function
Thanks, user:PMajer, for the correction in this (adding a constant). But I don't think it's really necessary to give the reader such a long explanation. And the way you did it is overkill! For instance, if $$g(x)$$ is zero for $$x\le 11$$ and then jumps to $$e$$ at 12, then your method gives $$f(11)\ge 1.1^{20}\approx e^2$$.

I would just put the following (I have changed the wording a bit):

Entire functions may grow as fast as any increasing function: for any increasing function g : [0,+∞) → [0,+∞)  there exists an entire function f(z) such that f(x) > g( |x| ) for all real x. Such a function may easily be found in the form:


 * $$f(z)=c+\sum_{k=1}^{\infty}\left(\frac{z}{k}\right)^{n_k}$$,

for a constant c and a strictly increasing sequence of positive integers nk. Any such sum converges everywhere and thus defines an entire function, and if the exponents are chosen appropriately, the inequality f(x) > g( |x| ) also holds, for all real x. (The exponent n1 can be chosen so that this is true in the interval [2, 3], and then n2 can be chosen so as to satisfy the inequality in the interval [3, 4], and so on.) Furthermore, if g(x) and h(x) are continuous functions with $$g(x)<h(x)$$ for all real x, one can find an entire function f(x) such that $$g(x)<f(x)<h(x)$$ for all real x.

(You told me by e-mail that that last statement follows from the preceding, but it's not obvious so it would be good to give a reference -- unless you can give a very brief explanation.)

Eric Kvaalen (talk) 11:50, 8 February 2013 (UTC)


 * Hi, well, yes, I initially thought that too many details are not needed, but I have recently added them for you, since you seemed to be interested. Anyway, apart the triviality about the constant, I thought the initial explanation after all was clear enough. So maybe I would just go back to the previous version (without forgetting the constant c ). If I remember well The result about the order density of entire functions in the continuous is due to T.Carleman (1927; I will find the exact reference): actually I planned to add this information, as it seems  relevant enough to be included. The proof is not difficult but not trivial; a brief explanation may be nice indeed. --pm a  14:19, 11 February 2013 (UTC)


 * Further remarks:


 * 1) overkill: Actually, here the task is just to construct an entire f above the given g, only this. The given form of f allows a quick  solution (because it is automatically an entire function, and then one can  fix the constant c  and the exponents nk so that the inequality be satisfied). BUT, of course, this form do not allow a close approximation as you are asking : for instance, such an f has all its derivatives positive on the positive half-line, which means it has in any case quite a strong growth. No way to be close to g, if e.g. g is a logarithm. So the objection about the example with $$ \scriptstyle g(11)=e$$ and $$ \scriptstyle f(11)=e^2$$ looks quite naive (if you really care about this issue, we could do better using a sharper inequality than  $$ \scriptstyle \log\bigg( 1+\frac{1}{k}\bigg)\ge \frac{1}{2k},  $$  like e.g. $$ \scriptstyle \log\bigg( 1+\frac{1}{k}\bigg)\ge \frac{1}{k+1}, $$ that would give a new exponent around half as big).


 * 2) The sentence you suggest:
 *  Furthermore, if g(x) and h(x) are continuous functions with $$g(x)<h(x)$$ for all real x, one can find an entire function f(x) such that $$g(x)<f(x)<h(x)$$ for all real x 
 * is ok, but I think the stated property (Carleman's theorem) deserves more emphasis. Maybe a section of its own (e.g. about "density", after the one about "growth").


 * 3) I'll think about how to make shorter the subsequent explanation. At the moment I have cut the last passages, and left the explicit formula for c and n_k. I agree that the latter is not optimal from the point of view of getting f as smaller as possible, but it is (I think) the simpler choice , and, what matters, shows that the whole construction is really constructive. I did not like so much your  suggestion
 * ...and then n2 can be chosen so as to satisfy the inequality in the interval [3, 4], and so on,
 * because it sounds a bit misleading, in that it suggests that the choice of the nk has to be done inductively (one after the other), whereas it's somehow simpler, as each one can be fixed directly.


 * --pm a 12:28, 25 February 2013 (UTC)

Nonzero zeros?
What does that mean: "zk are the non-zero roots of f"? &mdash; Sebastian 16:27, 17 January 2018 (UTC)
 * It means that the $$z_k$$ are roots of $$f$$: $$f(z_k) = 0,$$ but that they are themselves nonzero numbers: $$z_k \neq 0.$$ –Deacon Vorbis (carbon &bull; videos) 18:37, 17 January 2018 (UTC)
 * Thanks for the good explanation! &mdash; Sebastian 00:14, 18 January 2018 (UTC)
 * But you are right. Indeed, the expression "non-zero zeros" sounds delightfully nonsensical... (That's why we like it ;) ) pm a 14:08, 28 March 2020 (UTC)