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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.

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    Not 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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    The exponential function is entire and it is not a polynomial.2012-02-02
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    Do you mean every entire algebraic function is a polynomial?2012-02-02
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    Based 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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    yes I'm sorry I meant every entire algebraic function is a polynomial.2012-02-02
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    I 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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    @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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    @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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    Yes, 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

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