Does any one how to prove that every entire algebraic function is a polynomial? I'm under the impression that this can be achieved by showing that an algebraic function grows no faster than a polynomial.
Algebraic functions are polynomials?
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1Not every entire function is a polynomial. For example, consider $e^z$. Could you be a bit more specific with your question? – 2012-02-02
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1The exponential function is entire and it is not a polynomial. – 2012-02-02
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1Do you mean every entire algebraic function is a polynomial? – 2012-02-02
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1Based on the title, I believe the question is how to prove that every *algebraic* entire function is a polynomial. – 2012-02-02
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0yes I'm sorry I meant every entire algebraic function is a polynomial. – 2012-02-02
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0I don't see how growth rate is going to distinguish between polynomials and other algebraic functions. $f(z)=\sqrt{z^2+1}$ is algebraic but not polynomial and has the same growth rate as the polynomial $g(z)=z$. – 2012-02-02
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0@Gerry: I guess Ricky Maher's idea is that if it can be shown that algebraic functions have growth rate dominated by a polynomial, and that every nonpolynomial entire functions does not, then the problem would be solved. Since $\sqrt{z^2+1}$ is not entire it wouldn't cause a problem with this approach. – 2012-02-02
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0@Jonas, so, the idea is to show that growth rate distinguishes between polynomials and other entire functions, more precisely, that nonpolynomial entire functions of necessity grow faster than any polynomial. Well, that should work - if it's true - and if you can prove it. – 2012-02-02
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0Yes, suppose $f(z)$ is an entire function with polynomial growth, i.e. there are constants $C$ and $N$ such that $|f(z)| < C |z|^N$ for all sufficiently large $|z|$. Then $f(z)/z^N$ has a removable singularity at $\infty$, so $f(z)$ has at most a pole there, and so $f(z)$ is meromorphic on the Riemann sphere. But then $f(z)$ is a rational function, and since it has no poles in $\mathbb C$ it is a polynomial. – 2012-02-02
2 Answers
I assume you mean every entire algebraic function is a polynomial. Suppose $g$ is an entire algebraic function of order $n$. Thus there are polynomials $c_j(z)$, $j=0 \ldots, n$ with $c_n$ not identically $0$ such that $\sum_{j=0}^n c_j(z) g(z)^j = 0$. For all but finitely many complex numbers $w$, $\sum_{j=0}^n c_j(z) w^j$ is not identically $0$, and so there are at most finitely many $z$ for which $\sum_{j=0}^n c_j(z) w^j = 0$: those are the only $z$ for which we can have $g(z) = w$. Thus for all but finitely many $w$, $g(z)$ takes the value $w$ only finitely many times. But an entire function that is not a polynomial has an essential singularity at $\infty$, and by the Great Picard Theorem it takes all but one value infinitely many times.
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0For what it is worth, I have a survey by V. I. Arnold that mentions the function $z(a,b,c)$ that satisfies $$z^7 + a z^3 + b z^2 + c z + 1 = 0$$ and says in a footnote that it is "an entire algebraic function (without poles)" and asks whether it can be expressed as a superposition of such functions in two variables. – 2012-02-02
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0@Will, I noticed that one, too. I convinced myself that in the context of that piece, "entire algebraic function" doesn't mean "function that is both entire and algebraic," but something else. – 2012-02-02
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0@GerryMyerson, thanks, I decided not to leave that as a comment to the OP partly because I was uncertain what was going on. Technical matter, it seems we can reply to a comment with merely an @ sign and a first name? Because the software did tell me about your comment, in the "responses" flag in my personal profile page. Hmmm, if I look at your profile, some things are not visible to me, such as your "response" flag, so I cannot quickly experiment with this. – 2012-02-02
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0@Wil, my understanding is that the first three letters of the name suffice. – 2012-02-02
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0@Ger, it worked. Thanks. – 2012-02-02
Hopefully not, because $\: \operatorname{exp} : \mathbb{C} \to \mathbb{C} \:$ defined by $\:\:\:\: f(z) \:\: = \:\: \displaystyle\sum_{n=0}^{\infty} \: \left(\frac1{n!} \cdot \left(z^n\right)\right)$
is an entire function that is not a polynomial.
See http://en.wikipedia.org/wiki/Exponential_function#Complex_plane.
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1See the title; I believe the functions are also supposed to be algebraic. – 2012-02-02
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0so far i have the following: 1. If (1/abs value z)(abs value f) holds then f is a polynomial. So assume theorem not true... – 2012-02-02
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0@RickyMaher: "$\frac{1}{|z|}|f|$" is not a statement. What do you mean by "$\frac{1}{|z|}|f|$ holds"? – 2012-02-02