Maybe this is obvious, but it comes to my mind now. I was thinking about the zeros of $\sin(z).$ Imagine, we think in an analytic function on $\mathbb{C}$ with one zero in $0$ and all the other zeros very far away. So we can take for example $P_R(z)\doteq \sin\left (\frac{\pi z}{R}\right)=\sum_{k=0}^{\infty} \frac{(-\pi/R)^k}{(2k+1)!} z^k=\pi \frac{z}{R}\prod_{k=1}^{\infty}\left(1-\frac{z^2}{(kR)^2}\right),$ where $R>0.$ Suppose a vector $\epsilon=(\epsilon_0,\ldots,\epsilon_N)$ where $\epsilon_i\in [-\delta,\delta]$ and define the perturbation $P_{R,N,\epsilon}(z)\doteq \sum_{k=0}^{N} \left(\frac{(-\pi/R)^k+\epsilon_k}{(2k+1)!}\right) z^k.$ It is intuitively obvious that if $1>>\delta>0,$ then $z\in \{z\neq 0:P_{R,N,\epsilon}(z)=0\}\Rightarrow |z|>f(R,N),$ where $f(R,N)$ is increasing in $R$ for each $N.$ Moreover, it seems obvious also that there exists an increasing function $g:\mathbb{R}\to\mathbb{N}$ such that $f(R,g(R))\to \infty$ as $R\to \infty.$ However, is it true?
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1What is the significance of making the $\epsilon_j$ random variables? Are you really asking if these claims hold with prob $1$ ? – Christian Remling Jun 09 '14 at 18:46
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You are right. I changed it. – user39115 Jun 09 '14 at 18:53
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2Also, your perturbation doesn't look small at all, in a relative sense ($\delta\gg (\pi/R)^k$ eventually). – Christian Remling Jun 09 '14 at 18:57
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That's true. I still having the question in this case, i.e. when $\epsilon_i\in [-\delta^i,\delta^i]$, that should be simpler. – user39115 Jun 09 '14 at 19:01
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This is not a direct and explicit answer.
You may check papers by Varga and Carpenter etc "Zeros of partial sums of $\cos(z)$ and $\sin(z)$ I(2000), II(2001), III(2010)" and by Kappert (1996) "On the zeros of the partial sums of $\cos(z)$ and $\sin(z)$" and references therein.
The general take-home message I got is that the a lot of zeros of the normalized partial sums of the Taylor series for $\sin(x)$ and $\cos(x)$ are clustered around the Szego curves. The other zeros are on the real axis between $-e^{-1}$ and $+e^{-1}$. So I suspect that the small perturbation of such series that you considered will also have real and finite zeros.
mike
mike
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Thanks, I m since recently aware of this as I saw in the answers to http://mathoverflow.net/questions/4329/roots-of-truncations-of-ex-1 . Obviously there are finitely many zeros, I think that in most of the cases they are not real, but I want to know how far they can be from the one of the unperturbed series. – user39115 Jun 10 '14 at 16:28
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