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There is a well known infinite product both for $\phi(x)=\sin x$ and $\phi(x)=\cos x$. These are particular cases of the Weierstrass factorization theorem. What about $\phi(x)=a_1\cos b_1 x + a_2\cos b_2 x + a_3\cos b_3 x$, where all coefficients are real? More specifically, under what conditions on the coefficients $a_n,b_n$ do we have the simplified product $$\phi(x)=c\cdot \prod_{k=1}^\infty \Big(1-\frac{x}{\rho_k}\Big)$$

where the product is over all real and complex roots (some of them possibly multiple) ordered in the following way:

  • Roots are ordered by increasing moduli
  • Conjugate and opposite roots are grouped together

I am particularly interested in factoring these two expressions:

$$\phi_1(\sigma, t) = \sum_{n=1}^\infty (-1)^{n+1}\frac{\cos(t\log n)}{n^\sigma},\\ \phi_2(\sigma, t) = \sum_{n=1}^\infty (-1)^{n+1}\frac{\sin(t\log n)}{n^\sigma}. $$

The reason is because when and only when $\phi_1(\sigma,t)=\phi_2(\sigma,t)=0$, then $s=\sigma+it$ is a non-trivial zero of $\zeta(s)$. See here for details. I am interested to see how the roots of $\phi_1$ and $\phi_2$ are jointly distributed. According to the Riemann Hypothesis, they can never be equal unless $\sigma=\frac{1}{2}$. I am wondering if this fact is also true for other similar types of non-periodic trigonometric series, one involving cosines, and its sister involving sines.

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The product you wrote for a finite linear combination of cosine is divergent, unless you group opposite zeros. Since your function is even, the correct product is this: $$c\prod\left(1-\frac{z^2}{\rho^2}\right).$$ This follows from Hadamard's theorem. Since your function is even, you can write it as $f=g(z^2)$ where $g$ has order $1/2$, thus genus zero. Genus zero exactly means that it can be represented by the above product.

In your second question, you have an infinite sum of cosines, so it is not directly related to the first question.

Remark. I do not even see why your $\phi_1,\phi_2$ are entire functions in $t$. The series absolutely converge only when $|\Im z|<\sigma-1.$

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  • $\begingroup$ $\phi_1,\phi_2$ are not holomorphic anywhere, they don't satisfy the CR equations. $\endgroup$ – Christian Remling Jan 4 at 15:00
  • $\begingroup$ Yes that's what I said in my question: you need to group not only opposite, but also conjugate roots, just like you would do for $\cos$, or for the well-known product (not the one involving primes) for $\zeta$ in the critical strip. $\endgroup$ – Vincent Granville Jan 4 at 20:53
  • $\begingroup$ For $\phi(x)=\cos b_1 x + \cos b_2 x$ there is of course a convergent infinite product for any $b_1,b_2$ since $\phi(x)=2\cos((b_1+b_2)x/2)\cos((b_1-b_2)x/2)$. $\endgroup$ – Vincent Granville Jan 4 at 21:08
  • $\begingroup$ @Christian Remling: It did not cross my mind that he thinks of a complex variable $\sigma+it$. In my answer I treated them as functions of two complex variables $\sigma$ and $t$. $\endgroup$ – Alexandre Eremenko Jan 5 at 0:32

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