Suppose $f(x)=\sum_{|\alpha|=0}^{\infty}a_{\alpha} x^{\alpha}$ for all $x\in\mathbb{R}^n$. Moreover we know a priori that $f$ is an algebraic function. Is $f$ necessarily a polynomial?If not what are typical counterexamples?
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$\begingroup$ What do you mean by algebraic? Algebraic over the ring of polynomial functions on $\mathbb R^n$? $\endgroup$– John PardonCommented Dec 7, 2014 at 3:49
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$\begingroup$ Algebraic function in this sense. $\endgroup$– BigMCommented Dec 7, 2014 at 4:10
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3$\begingroup$ I think if you have convergence for all $x$ in $\mathbb R^n$, you will have bounds on the coefficients that would make it absolutely convergent for all $x$ in $\mathbb C^n$. An algebraic function will have at most polynomial growth, so Cauchy's theorem should imply that high degree coefficients vanish. $\endgroup$– Lev BorisovCommented Dec 7, 2014 at 4:15
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You wrote a series. If you mean that it converges for all $x\in R^n$ than this is an entire function that is the series also converges for $x\in C^n$. An entire algebraic function is a polynomial, indeed. This follows from the growth estimate and Cauchy's inequalities, for example.
answered Dec 7, 2014 at 5:26
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$\begingroup$ I cannot agree with you more.corrected it.and thanks for the answer. $\endgroup$– BigMCommented Dec 7, 2014 at 5:34