# Can the Riemann hypothesis be relaxed to say that this matrix A consists of square roots?

I realize that asking this question is like presenting to a patent attorney a wheel-less skateboard while asking to patent a hoverboard.

Anyways. Lagarias version of the Riemann hypothesis sets a bound on the sum of divisors:

$$\sigma(n) \le H_n + e^{H_n} \ln H_n$$

where $H_n$ is a harmonic number.

The von Mangoldt function matrix, as I call it, can be generated from the matrix product of two matrices:

$$A = \left( \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 1 & \sqrt{2} & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & \sqrt{3} & 0 & 0 & 0 & 0 \\ 1 & \sqrt{2} & 0 & \sqrt{4} & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & \sqrt{5} & 0 & 0 \\ 1 & \sqrt{2} & \sqrt{3} & 0 & 0 & \sqrt{6} & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & \sqrt{7} \\ \vdots&&&&&&&\ddots \end{array} \right)$$

which is equal to $A(n,k)=\sqrt{k}$ if $k$ divides $n$, else $A(n,k)=0$

The matrix inverse of $A$ is by its terms essentially equal to the matrix:

$$B = \left( \begin{array}{ccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 & \cdots \\ 0 & -\sqrt{2} & 0 & -\sqrt{2} & 0 & -\sqrt{2} & 0 \\ 0 & 0 & -\sqrt{3} & 0 & 0 & -\sqrt{3} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -\sqrt{5} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \sqrt{6} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -\sqrt{7} \\ \vdots&&&&&&&\ddots \end{array} \right)$$

which is equal to $B(n,k)=\mu(n)\sqrt{n}$ if $n$ divides $k$, else $A(n,k)=0$

where $\mu(n)$ is the Möbius function

defined by:

$$\mu(n)=\begin{cases} (-1)^{\omega(n)}=(-1)^{\Omega(n)} &\text{if }\; \omega(n) = \Omega(n)\\ 0&\text{if }\;\omega(n) \ne \Omega(n).\end{cases}$$

• $\mu(n) = 1$ if $n$ is a square-free positive integer with an even number of prime factors.
• $\mu(n) = -1$ if $n$ is a square-free positive integer with an odd number of prime factors.
• $\mu(n) = 0$ if $n$ has a squared prime factor.

The von Mangoldt function matrix is then the matrix product $A$ times $B$:

$$T = A.B = \left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$

And the von Mangoldt function is then:

$$\Lambda(n) = \sum\limits_{k=1}^{k=\infty}\frac{T(n,k)}{k}$$

as proven by joriki here.

or as the Dirichlet generating functions of the columns as proven here by GH from MO at Mathoverflow:

$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$

Here comes the hoverboard / wheel-less skateboard:

Since according to the explicit formula, the von Mangoldt function is a sum of logarithmic square root waves as follows:

$$\sum_{n=1}^{n=k} \Lambda(n) = \Re\left(-\sum _{j=1}^{\infty} \left(\frac{x^{1-\rho _j}}{1-\rho _j}+\frac{x^{\rho _j}}{\rho _j}\right)-\frac{1}{2} \log \left(1-\frac{1}{x^2}\right)+x-\log (2 \pi )\right)$$

or as a Mathematica one-liner from Alex Kontorovich web page:

Plot[Re[X - Log[2 Pi] - Log[1 - 1/X^2]/2 -
Sum[X^(N[ZetaZero[j]])/(N[ZetaZero[j]]) +
X^(1 - N[ZetaZero[j]])/(1 - N[ZetaZero[j]]), {j, 1, 30}]], {X,
1.1, 30}]


Can the Riemann hypothesis be relaxed/be made precise to say that the so called von Mangoldt function matrix $T$ is a matrix product $T=A.B$ as in the example above?

(*Matrix T Mathematica  8*)
nn = 32;
A = Table[
Table[If[Mod[n, k] == 0, k^(ZetaZero[k]), 0], {k, 1, nn}], {n, 1,
nn}];
B = Table[
Table[If[Mod[k, n] == 0, MoebiusMu[n]*n^(ZetaZero[-n]), 0], {k, 1,
nn}], {n, 1, nn}];
MatrixForm[T=N[A.B]]


It appears to work for any complex number sequence in the exponents as long as the sum of the two matrices $A$ and $B$'s respective real parts is equal to 1, and the imaginary parts are each others negatives. In other words a condition that applies to any two complex number sequences of that form, of which the zeta zeros are a subset, so no progress.

To demonstrate this I have made this variant of the program above:

(*Matrix T Mathematica 8 start*)nn = 32;
a = Table[RandomComplex[], {n, 1, 32}]
A = Table[
Table[If[Mod[n, k] == 0, k^(a[[k]]), 0], {k, 1, nn}], {n, 1, nn}];
B = Table[
Table[If[Mod[k, n] == 0, MoebiusMu[n]*n^(1 - a[[n]]), 0], {k, 1,
nn}], {n, 1, nn}];
MatrixForm[T = Chop[N[A.B]]]
(*end*)


which produces matrix $T$. This probably has to do with the elementary fact that:

$$n=n^{a} n^{1-a}$$

$n$ is here a substitute for the terms in matrix $T$.

So for some arbitrary complex number sequence $a$ like for example:

$$a=0.771518+0.640552I,0.192739+0.923147I,0.931096+0.758704I,...$$

or the non-trivial Riemann zeta zeros:

$$a=0.5 + 14.1347 I, 0.5 + 21.022 I, 0.5 + 25.0109 I,...$$ we have in general the matrices:

$$A = \left( \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 1 & 2^{a_2} & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 3^{a_3} & 0 & 0 & 0 & 0 \\ 1 & 2^{a_2} & 0 & 4^{a_4} & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 5^{a_5} & 0 & 0 \\ 1 & 2^{a_2} & 3^{a_3} & 0 & 0 & 6^{a_6} & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 7^{a_7} \\ \vdots&&&&&&&\ddots \end{array} \right)$$

$$B = \left( \begin{array}{ccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 & \cdots \\ 0 & -2^{1-a_2} & 0 & -2^{1-a_2} & 0 & -2^{1-a_2} & 0 \\ 0 & 0 & -3^{1-a_3} & 0 & 0 & -3^{1-a_3} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -5^{1-a_5} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 6^{1-a_6} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -7^{1-a_7} \\ \vdots&&&&&&&\ddots \end{array} \right)$$

Which have the same property as the earlier matrices $A$ and $B$ producing matrix $T$ as the matrix product:

$$T = A.B = \left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$

• there are many matrix formulation of the Riemann hypothesis, for example some bounds on the spectral radius or the eigenvectors of some matrix defined from number-theoritic functions. but, your description of $B$ and $T$ is confused and the rest after is even more. so give more details, write down your formulas, as when you say "Since the von Mangoldt function is a sum of logarithmic square root waves" can you define precisely what it means ? – reuns Jan 24 '16 at 12:28
• Thank you for the comment. I have tried to improve the question now as you suggested. – Mats Granvik Jan 24 '16 at 13:01
• $\sum_{k=1}^\infty \frac{T(1,k)}{k}$ diverges, how do you prove that for $n \ne 1$ : $\Lambda(n) = \sum_{k=1}^\infty \frac{T(n,k)}{k}$ ? and everybody knows what is $\mu(n)$, but nobody understand what you mean with " the von Mangoldt function is created from square roots as described above". – reuns Jan 24 '16 at 14:06
• you understand nearly nothing of your question, even the proof that $\Lambda(n) = \sum_{k=1}^\infty \frac{1}{k} \sum_{d | gcd(n,k)} d \mu(d)$ you don't understand it, so ... work ! – reuns Jan 24 '16 at 14:19
• I know that $\sum_{k=1}^\infty \frac{T(1,k)}{k}$, but when taking the Fourier transform of the von Mangoldt function defined this way, I believe that a good choice is to truncate it: $\sum_{k=1}^{k=n} \frac{T(1,k)}{k}$ with $n$ equal to the number of von Mangoldt function terms used in the Fourier transform. This because then the spectra like plot will be zero at locations of zeta zeros. Or at least it looks like the plot is zero at zeta zeros. – Mats Granvik Jan 24 '16 at 14:21