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Consider the following polynomial:

$$P_n(x)= \sum_{k=1}^n a_kx^k$$

We know (and can verify) that , for every $n$ , $P_n(x)$ has only real roots

Now, define

$$P(x)= \sum_{k=1}^\infty a_kx^k$$

We know that $P(x)$ analytic and entire .

Question :

Does this mean that $P(x)$ has only real roots ? If not please elaborate with examples

Note: I'm not sure this is MSE or MO question (?)

Edit: Converse of the above hypothesis isn't true

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    $\begingroup$ I can tell that the inverse isn't true. Consider $P(x) = \sin^2 x$ and $P_6(x)= x^2 - x^4/3 + 2x^6/45$. It's easy to check that $P(x)$ is entire, analytic, and has only real roots, while $P_6(x)$ has non-real roots. $\endgroup$ Feb 4, 2020 at 9:27
  • $\begingroup$ @Adam Latosiński thank you for the comment $\endgroup$
    – bambi
    Feb 4, 2020 at 9:41

1 Answer 1

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Let us assume that $P(z)$ has an root $z_0$, $\Im(z_0) \neq 0$ (without a loss of generality, we can assume $\Im(z_0)>0$). Since $P(z)$ is an entire function, this root is necessarily isolated (unless $P(z) \equiv 0$, but then $P_n(z) \equiv 0$ doesn't fullfill the assumptions). Let then $\gamma$ be a closed curve surrounding $z_0$, not surrounding any other of roots of $P(z)$, and entirely enclosed within the semiplane $\Im(z)>0$.

Then, for any $k\in\mathbb N$, $\frac{(z-z_0)^k}{P_n(z)}$ is holomorphic in the region bounded by $\gamma$, while $\frac{(z-z_0)^k}{P(z)}$ has at most a single singularity within, at $z_0$. We have

$$0 = \int_\gamma \frac{(z-z_0)^k}{P_n(z)} dz = \lim_{n\rightarrow\infty} \int_\gamma \frac{(z-z_0)^k}{P_n(z)} dz = \int_\gamma \frac{(z-z_0)^k}{P(z)} dz = 2\pi i {\rm Res}_{z_0} \frac{(z-z_0)^k}{P(z)}$$

Since this is true for any $k\in\mathbb N$, that means that the whole part of Laurent series of $\frac{1}{P(z)}$ around $z_0$ with negative powers of $(z-z_0)$ vanishes, i.e. $\frac{1}{P(z)}$ doesn't have a singularity at $z_0$. This contradicts the assumption that $P(z)$ has a root at $z_0$.

Therefore, $P(z)$ cannot have non-real roots.

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    $\begingroup$ Hurwitz theorem gives the result immediately since the Taylor series of a function converges uniformly to it on compact sets, so in particular, for any zero of $P$ there will be nearby zeroes of $P_k$ for large $k$ $\endgroup$
    – Conrad
    Feb 4, 2020 at 12:49

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