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Background

We start by defining the following function $S(x)$:

$$ S(x) + S(x^2) + S(x^3) + \dots = x $$

Using the mobius inversion formula:

$$ S(x) = \sum_{r=1}^\infty \mu(r) x^r $$

We now define a nil-potent $(n+1) \times (n+1)$ matrix $\epsilon$ such that:

$$ \epsilon^{n+1} = \hat 0$$

Hence,

$$ S(\epsilon) = \sum_{r=1}^n \mu(r)\epsilon^r $$

and,

$$ S(\epsilon)+ S(\epsilon^2)+ \dots + S(\epsilon^n)= \epsilon $$

Let, us call the above equation the epsilon-n equation.

$$ \implies \frac{\partial}{\partial \epsilon} (S(\epsilon)+ S(\epsilon^2)+ \dots + S(\epsilon^n))= 1 $$

$$ \implies S'(\epsilon)+ 2 \epsilon S'(\epsilon^2)+ \dots + n \epsilon^{n-1 }S'(\epsilon^n)= 1 $$

We know $ (\epsilon^k)_{1,n+1} = 0 $ where $(\epsilon^k)_{1,n+1} $ is the n'th row and 1'st row and n'th column. Hence,

$$ \implies S'(\epsilon_{1,n+1})+ 2 \epsilon S'((\epsilon^2)_{1,n+1})+ \dots + n \epsilon^{n-1 }S'((\epsilon^n)_{1,n+1})= 1 $$

$$ \implies S'(0) = 1 $$

Now, taking the second derivative of the epsilon-n equation:

$$ \frac{\partial^2}{\partial \epsilon^2} (S(\epsilon)+ S(\epsilon^2)+ \dots + S(\epsilon^n))= 0 $$

$$ S''(\epsilon) + 2S'(\epsilon^2) + 4 \epsilon S''(\epsilon^2) + \dots = 0 $$

Again, comparing the 1'st row n'th column of each martix:

$$ S''(0) + 2 S'(0) = 0 $$

Repeating the above procedure $n$ times, we obtain a $n \times n$ matrix, $A$ :

\begin{equation*} \underbrace{\begin{pmatrix} 1&0&0&\dots&0\\ 2&1&0&\dots&0 \\ \vdots& & &\dots& \\ \end{pmatrix}}_{A} \underbrace{\begin{pmatrix} S'(0) \\ \vdots \\ S'^n (0) \\ \end{pmatrix}}_{B} = \underbrace{\begin{pmatrix} 1\\ \vdots\\ 0\\ \end{pmatrix}}_{C} \end{equation*}

Futher, we know, $S'^k(0) = \mu(k) k!$

\begin{equation*} \implies \underbrace{\begin{pmatrix} 1&0&0&\dots&0\\ 2&1&0&\dots&0 \\ \vdots& & &\dots& \\ \end{pmatrix}}_{A} \underbrace{\begin{pmatrix} \mu(1) \\ \vdots \\ \mu(n) n! \\ \end{pmatrix}}_{B} = \underbrace{\begin{pmatrix} 1\\ \vdots\\ 0\\ \end{pmatrix}}_{C} \end{equation*}

As the system of equations given by $AB=C$ is solvable, $A^{-1}$ must exist. Multiplying from the right a column matrix $D= \begin{pmatrix} 1/1! & 1/2! & 1/3! & \dots & 1/n!\end{pmatrix}$:

$$\implies BD = A^{-1} CD $$.

Taking the trace:

$$ M(n) = \sum_{r=1}^n \mu(r) = \text{Tr} (BD) = \text{Tr} (A^{-1} CD) $$

Questions

Is this correct? Is there any general formula for the general $ k \times k$ matrix of $A$ or $A^{-1}$ matrix? Is this method viable to calculate Merten's function? Does this method already exist?

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    $\begingroup$ yes it exists, it is mostly a variation on the mathworld.wolfram.com/RedhefferMatrix.html . search 'mertens matrix', 'mertens determinant' or 'riemann hypothesis matrix', or these kind of things, to get an idea of all the possible variations on those matrix formulation of the Mertens/Möbius function and of the Riemann hypothesis. $\endgroup$ – reuns Mar 20 '16 at 17:41
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    $\begingroup$ in my opinion, the main idea behind that is that the Möbius/Mertens function are very elementary functions, that can be defined from very simple arithmetic means. hence, considering only those elementary formulations, you will have many difficulties to take in account non-elementary facts such as the functional equation for $\zeta(s)$, and that's a huge problem because the generalized Riemann hypothesis (the Selberg class) tells us that the functional equation is very important for the Riemann hypothesis to be possibly true. $\endgroup$ – reuns Mar 20 '16 at 17:47
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    $\begingroup$ hence, you'll have to find a way to take in account non-elementary (complex analysis) facts in your elementary formulation of the Riemann hypothesis, which is not so easy (good luck). a possiblity, is to write an explicit formula for the Mertens function : $\displaystyle M(x) = \sum_\rho \frac{x^\rho}{\zeta'(\rho) \rho}$ (supposing all the zeros are simple, and $\rho$ are the trivial and non-trivial zeros) $\endgroup$ – reuns Mar 20 '16 at 17:48

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