Can we prove that all zeros of entire function cos(x) are real from the Taylor series expansion of cos(x)? Q1: Can we prove that all zeros of cos(x) are real from the following Taylor series expansion of cos(x)?
$$ \cos(x) = \sum_{n=0}^\infty \frac{(-1)^k}{(2k)!}x^{2k} $$
The Riemann $\xi(z)$ function is an entire function related to the Riemann $\zeta(s)$ function ($s=1/2+iz$) via (Titchmarsh, p16):
$$ \xi(z) = \frac{1}{2}s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s) $$
The functional equation is given by: 
$$ \xi(z)=\xi(-z)$$
$\xi(z)$ function can be expressed as a Taylor series ($b_k>0$):
$$ \xi(z) = \sum_{n=0}^\infty \frac{(-1)^k}{(2k)!}b_{k}z^{2k} $$
Q2: Can we prove that all zeros of an entire function,  like $\xi(z)$, are real from the Taylor series expansion of $\xi(z)$?
Any references are appreciated.
-mike
 A: Well, I think I can show it, but the idea requires knowing in advance that $\sin x$ has only real zeros: 
We will use the following proposition ,which is given as an exercise in Ahlfors' text:

Show that if $f(z)$ is of genus $0$ or $1$ with real zeros, and if $f(z)$ is real
  for real z, then all zeros of $f'(z)$ are real.    Hint: Consider $\text{ Im} \frac{f'(z)}{f(z)}$.

Integrating the Taylor series of the cosine gives the Taylor series of the sine:
$$\sin(z)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} z^{2n+1} $$
Since the coefficients are real, we see that the sine function is real for real arguments. 
Using the formula 
$$\rho=\limsup_{n\to\infty}\frac{n\ln n}{-\ln|a_n|} $$
for the order of the entire function $\sum a_n z^n$, we can see that $\sin(z)$ has order $\rho=1$, and according to Hadamard's factorization theorem we find that its genus is $\leq 1$.

In order to apply this on your example, you should ask whether $\xi(z)$ has an antiderivative with genus $\leq 1$, which vanishes exclusively on the real axis. (the real coefficients give the third condition automatically).
Hope this helps!
A: For $z=x+i y$, one verifies that
$$ |\cos z|^{2}=\cos^{2} x+\sinh^{2} y,$$
by using the addition formula for cosine and Euler's formula (possible to deduce from the power series expansions of the involved functions). Consequently, one has the inequality
$$|\cos z|^{2}\geq\sinh^{2} y.$$
Since $\sinh y=(e^{y}-e^{-y})/2=0$ iff $y=0$, the above inequality immediately implies that $\cos z$ cannot vanish for $\Im z\neq0$.
A: I found out the answer from a paper on the web, but I can not find it anymore.  So I will write down from what I remembered and also filled in the steps that were omitted.
First we rewrite the expression as
$$ \cos(x) =\lim_{n\to\infty}g_n(x/(2n)) $$
$$ g_n(x/(2n)):=\sum_{k=0}^n (-1)^k \binom{2n}{2k}\frac{x^{2k}}{(2n)^{2k}} $$
The function $g_n(x)$ is called the Jensen polynomial associated with entire function $\cos(x)$.
This is possible because
$$A(n,k):=\binom{2n}{2k} \frac{(2k)!}{(2n)^{2k}} =\frac{(2n)_{2k}}{(2n)^{2k}}=\frac{(2n)}{(2n)}\frac{(2n-1)}{(2n)}...\frac{(2n-2k+1)}{(2n)}$$
So
$$\lim_{n\to\infty} A(n,k)=1$$
Using Mathematica 7.0 we found out that
$$g_n(x/(2n))=\frac{1}{2}\left(1+\frac{ix}{2n}\right)^{2n}+\frac{1}{2}\left(1+\frac{-ix}{2n}\right)^{2n}$$
It is interesting to see that
$$\lim_{n\to\infty}g_n(x/(2n))=\frac{1}{2}\lim_{n\to\infty}\left(\left(1+\frac{ix}{2n}\right)^{2n}+\left(1-\frac{ix}{2n}\right)^{2n}\right)=\frac{1}{2}(e^{ix}+e^{-ix})=\cos(x)$$
Let $\omega_{k}$ and $-\omega_{k}$ with $n=1,2,...,n $ be the $2n$ roots of $y^{2n}=-1$, they are given by:
$$\omega_{k}=\exp\left({i\pi}\frac{2k+1}{2n}\right)$$
then roots of $g_n(x/(2n))$ are given by:
$$x_k=-(2in)\frac{\omega_{k}-1}{\omega_{k}+1}=2n\tan\left(\frac{\pi(2k+1)}{4n}\right)$$
$$x_{n+k}=-(2in)\frac{-\omega_{k}-1}{-\omega_{k}+1}=-2n\cot\left(\frac{\pi(2k+1)}{4n}\right)$$
When $n\to\infty$ the first $n$ zeros survived (remained finite)
$$\lim_{n\to\infty}x_k=\lim_{n\to\infty}2n\tan\left(\frac{\pi(2k+1)}{4n}\right)=\frac{\pi}{2}(2k+1)$$
the last $n$ zeros are pushed to $-\infty$
$$\lim_{n\to\infty}x_{n+k}=\lim_{n\to\infty}(-2n)\tan\left(\frac{\pi}{2}-\frac{\pi(2k+1)}{4n}\right)=\lim_{n\to\infty}(-2n)\tan\left(\frac{\pi}{2}\right)=-\infty$$
-mike
A: 
I found out the answer from a paper on the web

Varadarajan 2006 Euler through time - a new look at old themes p. 76 ff
An objection to Euler's solution of the Basel problem was that it assumed that sin z had no other roots than the obvious real ones. To answer this, Euler proved the product formula for sinh(z)/z. 
He first wrote exp(z) as (1+z/i)^i where "i" denotes an infinite number, and next found the roots of (1+z/n)^n - (1-z/n)^n by grouping in pairs the roots of unity for finite "n".
Tannery's theorem for infinite products with a parameter is needed to justify his final limit argument n->i.
This is where the binomial coefficients come from in the Jensen polynomials for entire functions (also discovered by Laguerre in 1880, before Jensen in 1911).
J. Gélinas
