# Is there a way to relate prime numbers and the fourier transform

According to what I know about Fourier transforms, any continuous periodic signal can be represented as a combination of sine and cosine functions. To me, this looks analogous to the "Fundamental theorem of arithmetic" (every integer can be represented as a unique product of primes")[The integer is the signal and the primes are the sine and cosine].

Any ideas of way to connect the Fourier transform with Prime Factorization?

(P.S. I found this paper that does something close http://arxiv.org/pdf/1410.2054v1.pdf )

• One of the references you linked to is a paper by Wolfgang Schramm. I made an entry in the OEIS about his type of GCD-Fourier calculation here: oeis.org/A231425 From it you also get the von Mangoldt function that encodes the fundamental theorem of arithmetic: oeis.org/A140256 oeis.org/A014963 – Mats Granvik Oct 21 '14 at 16:48

The question is quite broad, but here is a possible relation between Fourier transformation and prime numbers.

We know that the Riemann zeta function is defined as $$\zeta(s) = \sum_{n=1}^\infty\frac{1}{n^s},$$ for all $\Re(s)>1$.

The second thing, that the Riemann zeta function is related to prime numbers via Euler product formula.

$$\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s} = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}},$$

for all $\Re(s)>1$.

As a generalization of Riemann zeta function the Hurwith zeta function is defined as $$\zeta(s,q) = \sum_{n=0}^\infty \frac{1}{(q+n)^{s}},$$ for all $\Re(s)>1$ and $\Re(q)>0$. The Riemann zeta function is $\zeta(s,1)$.

Another way to generalize Riemann zeta function is using polylogarithm, which is defined as

$$\operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s}.$$

For $\Re(s)>1$ we have $\operatorname{Li}_s(1)=\zeta(s)$. Is is also true, that $\operatorname{Li}_s(-1)= (2^{1-s}-1)\zeta(s)$.

The sequence of $N$ complex numbers $x_0,x_1,\dots,x_{N-1}$ is transformed with discrete Fourier transformation into an $N$-periodic sequence of complex numbers with

$$X_k\ =\ \sum_{n=0}^{N-1} x_n \cdot e^{-i 2 \pi k n / N}, \quad k\in\mathbb{Z}\,.$$

Of course by using Euler's formula we could also use trigonometric functions.

Here comes the connection. The discrete Fourier transform of the Hurwitz zeta function with respect to the order $s$ is the Legendre chi function, which is defined as $$\chi_s(z) = \sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)^s}.$$ The Legendre chi function is related to polylogarithms as the following. $$\chi_s(z) = \frac{1}{2}\left[\operatorname{Li}_s (z) - \operatorname{Li}_s (-z)\right].$$

So it is also true that

$$\chi_s(1) = (1-2^{-s})\zeta(s),$$

for all $s>0$ real numbers.

So if we put this all together.

$$\prod_{p \text{ prime}} \frac{1}{1-p^{-s}} \stackrel{\text{DFT}}{\implies} (1-2^{-s})\prod_{p \text{ prime}} \frac{1}{1-p^{-s}},$$

for all $s>0$ real numbers.

I hope that my argument is right, it's just've come into my mind while reading your question. While doing the DFT at the end we've substituted $z=1$ into Legrende chi function.

There is a fundamental relationship between Fourier series and prime numbers. All prime counting functions can be expressed as infinite series of Fourier series. Please see Illustration of Fourier Series for Prime Counting Functions.

• I don't know much about Fourier series. Isn't "express[ing] as an infinite series of Fourier series" something we can do for many functions, not just just ones that count primes? – pjs36 Jan 10 '17 at 1:34
• @pjs36 Perhaps so, but my primary interest at this point in time is the investigation of Fourier series for prime counting functions, and prime counting functions and their representation as Fourier series are germane to the topic of this question. – Steven Clark Jan 10 '17 at 2:16
• @pjs36 With respect to your comment about Fourier series representations of functions other than prime counting functions, the following link defines and illustrates a general method for derivation of a Fourier series representation for $f(x)=\sum\limits_{n=1}^x a(n)$ related to the Dirichlet series $F(s)=\sum\limits_{n=1}^\infty\frac{a(n)}{n^s}$: primefourierseries.com/?page_id=7595. – Steven Clark May 14 '18 at 17:16