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Suppose that I have an analytic function $f(z)=\sum_{n=0}^\infty a_n z^n$ which converges on some disk around the origin.

For a particular function I encountered, I wished to prove that every coefficient, $a_n$, is non-negative.

I am wondering what complex analytic methods exist to detect negative coefficients if all my coefficients are real. What nice ways are there to check if all of the coefficients of my power series are the same sign?

In more generality, are there methods which detect whether eventually all of the coefficients are of the same sign? (That is, whether or not there exists $N$ such that for all $n,m>N$, $a_n$ and $a_m$ will be the same sign)

I am really interested in any, and many, thoughts on this problem. What strategies could possibly work?

Thanks!

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    @user8268: Yes, for this problem they are real. I edited to correct it. Although that raises another question: Is there a way to detect whether or not all of the coefficients of my power series are real?2011-05-07
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    Would induction work? $f(0)=a_0$. Then differentiate to get $a_1$ and so on.2011-05-07
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    @Eric Naslund: (a trivial remark:) reality of coefficients is equivalent to reality of $f(z)$ for $z$ real, or to $f(\bar{z})=\overline{f(z)}$ - but I'm not sure what you imagine you know about $f$2011-05-07
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    You've got to give more detail. It would be surprising if there were tools for this that would work in every situation. For illustration, I have published a paper in which the difference of two series with integer coefficients is shown to have non-negative coefficients. So, the provenance of your series is really the whole story here.2011-05-07
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    @Will Jagy: This is a good point, perhaps I should of elaborated more in the statement. I of course don't expect to find way that works in every case. My question is moreso about what general approaches can be used to solve such problems. What strategies _could_ work?2011-05-07
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    Reminds me... if $f(z)$ has "oscillatory" behavior, that precludes the possibility that all the $a_n$ be nonnegative, no?2011-05-08

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