# Real roots of partial sums of analytic function represented by infinite series :

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

• 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. Feb 4, 2020 at 9:27
• @Adam Latosiński thank you for the comment Feb 4, 2020 at 9:41

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.
• 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 $k$ Feb 4, 2020 at 12:49