# Infinite products for linear combinations of sines or cosines

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.

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.
Remark. I do not even see why your $$\phi_1,\phi_2$$ are entire functions in $$t$$. The series absolutely converge only when $$|\Im z|<\sigma-1.$$
• $\phi_1,\phi_2$ are not holomorphic anywhere, they don't satisfy the CR equations. Commented Jan 4, 2021 at 15:00
• 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. Commented Jan 4, 2021 at 20:53
• 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 $\phi(x)=2\cos((b_1+b_2)x/2)\cos((b_1-b_2)x/2)$. Commented Jan 4, 2021 at 21:08
• @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 $t$. Commented Jan 5, 2021 at 0:32