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

• in the first cos-formula I think, the exponent at x should be $x^{2k}$ – Gottfried Helms Oct 10 '13 at 20:15
• Yes, the Taylor series of an entire function determines the function. So, yes, we can prove that all the zeros of $cos$ are real from the Taylor expansion. But you are not going to like the way I am thinking to do it because it doesn't help you for the other function. – OR. Oct 10 '13 at 20:52
• @Gottfried Thanks for spotting the typo. – mike Oct 10 '13 at 21:00
• @ABC I would really like to see it. The reason is that I think that I made some progress in finding a method to prove that all the zeros of $\cos(x)$ are real from the Taylor expansion. And this method might be applicable to other entire functions like $\xi(z)$. – mike Oct 10 '13 at 21:05
• The prologue of Rudin's 'Real and complex analysis' has the arguments. – OR. Oct 10 '13 at 21:20

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!

• Thanks a lot for the answer. But proving that zeros of sin(z) are real from its Taylor series expansion [not from the assumption that all zeros of cos(z) are real and relation sin'(z)=cos(z)] seems to be equally harder. -mike – mike Oct 10 '13 at 22:43
• @mike I agree that moving from cos to sin isn't of much help, since they are similar. Nonetheless, I think $\int \xi$ and $\xi$ are quite different. – user1337 Oct 10 '13 at 22:54
• I agree with you. LAGARIAS and MONTAGUE had a paper at arxiv on $\int \xi$. – mike Oct 10 '13 at 23:10

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 $\Im z\neq0$.

• This is A simple proof of the roots of $\cos z$ are real. But it is not what I was seeking for. Thanks anyway! – mike Aug 16 '17 at 12:15

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