Revisions to Is this Riemann zeta function product equal to the Fourier transform of the von Mangoldt function?

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edited Apr 13, 2017 at 12:19

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This is not a complete answer but I want show that there is nothing mysterious going on here.

We want to prove that:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

The Dirichlet inverse of the Euler totient function is

$$a(n)=\sum\limits_{d|n} \mu(d)d$$

Construct the matrix $$T(n,k)=a(GCD(n,k))$$

which starts:

$$\displaystyle T = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix} $$

where GCD is the Greatest Common Divisor of row index $n$ and column index $k$.

joriki showed showed that the von Mangoldt function is $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

Then add this quote by Terence Tao from here, that I don't completely understand but I do almost see why it should be true:

Quote:" The Fourier transform in this context becomes (essentially) the Mellin transform, which is particularly important in analytic number theory. (For instance, the Riemann zeta function is essentially the Mellin transform of the Dirac comb on the natural numbers"

Now let us return to the matrix $T$:

First the von Mangoldt is expanded as:

$$\displaystyle \begin{bmatrix} +1/1&+1/1&+1/1&+1/1&+1/1&+1/1&+1/1&\cdots \\ +1/2&-1/2&+1/2&-1/2&+1/2&-1/2&+1/2 \\ +1/3&+1/3&-2/3&+1/3&+1/3&-2/3&+1/3 \\ +1/4&-1/4&+1/4&-1/4&+1/4&-1/4&+1/4 \\ +1/5&+1/5&+1/5&+1/5&-4/5&+1/5&+1/5 \\ +1/6&-1/6&-2/6&-1/6&+1/6&+2/6&+1/6 \\ +1/7&+1/7&+1/7&+1/7&+1/7&+1/7&-6/7 \\ \vdots&&&&&&&\ddots \end{bmatrix}$$

Edit: 24.1.2016. From here on the variables $n$ and $k$ should be permutated but I don't know how to fix the rest of this answer right now.

Summing the columns first is equivalent to what was said earlier: $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

where:

$$\Lambda(n) = \begin{cases} \infty & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

or as a sequence:

$$\infty ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,...,\Lambda(\infty)$$

And now based on the quote above let us say that: $$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) = \sum\limits_{n=1}^{n=k}\frac{\Lambda(n)}{n^s}$$

Expanding this into matrix form we have the matrix:

$$\displaystyle \begin{bmatrix} \frac{T(1,1)}{1 \cdot 1^s}&+\frac{T(1,2)}{1 \cdot 2^s}&+\frac{T(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T(1,k)}{1 \cdot k^s} \\ \frac{T(2,1)}{2 \cdot 1^s}&+\frac{T(2,2)}{2 \cdot 2^s}&+\frac{T(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T(2,k)}{2 \cdot k^s} \\ \frac{T(3,1)}{3 \cdot 1^s}&+\frac{T(3,2)}{3 \cdot 2^s}&+\frac{T(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T(3,k)}{3 \cdot k^s} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{T(n,1)}{n \cdot 1^s}&+\frac{T(n,2)}{n \cdot 2^s}&+\frac{T(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} \end{bmatrix}$$

On the right hand side we see that it sums to the right hand side of what we set out to prove, namely:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

Things that remain unclear are: What factor should the left hand side be multiplied with in order to have the same magnitude as the right hand side? And why in the Fourier transform does the first term of the von Mangoldt function appear to be $\log q$?

$$\Lambda(n) = \begin{cases} \log q & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

$$n=1,2,3,4,5,...q$$

As a heuristic, $$\Lambda(n) = \log q \;\;\;\; \text{if }n=1$$ probably has to do with that in the Fourier transform $q$ terms of $\Lambda$ are used and the first column in square matrix $T(1..q,1..q)$ sums to a Harmonic number.

This is not a complete answer but I want show that there is nothing mysterious going on here.

We want to prove that:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

The Dirichlet inverse of the Euler totient function is

$$a(n)=\sum\limits_{d|n} \mu(d)d$$

Construct the matrix $$T(n,k)=a(GCD(n,k))$$

which starts:

$$\displaystyle T = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix} $$

where GCD is the Greatest Common Divisor of row index $n$ and column index $k$.

joriki showed that the von Mangoldt function is $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

Then add this quote by Terence Tao from here, that I don't completely understand but I do almost see why it should be true:

Quote:" The Fourier transform in this context becomes (essentially) the Mellin transform, which is particularly important in analytic number theory. (For instance, the Riemann zeta function is essentially the Mellin transform of the Dirac comb on the natural numbers"

Now let us return to the matrix $T$:

First the von Mangoldt is expanded as:

$$\displaystyle \begin{bmatrix} +1/1&+1/1&+1/1&+1/1&+1/1&+1/1&+1/1&\cdots \\ +1/2&-1/2&+1/2&-1/2&+1/2&-1/2&+1/2 \\ +1/3&+1/3&-2/3&+1/3&+1/3&-2/3&+1/3 \\ +1/4&-1/4&+1/4&-1/4&+1/4&-1/4&+1/4 \\ +1/5&+1/5&+1/5&+1/5&-4/5&+1/5&+1/5 \\ +1/6&-1/6&-2/6&-1/6&+1/6&+2/6&+1/6 \\ +1/7&+1/7&+1/7&+1/7&+1/7&+1/7&-6/7 \\ \vdots&&&&&&&\ddots \end{bmatrix}$$

Edit: 24.1.2016. From here on the variables $n$ and $k$ should be permutated but I don't know how to fix the rest of this answer right now.

Summing the columns first is equivalent to what was said earlier: $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

where:

$$\Lambda(n) = \begin{cases} \infty & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

or as a sequence:

$$\infty ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,...,\Lambda(\infty)$$

And now based on the quote above let us say that: $$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) = \sum\limits_{n=1}^{n=k}\frac{\Lambda(n)}{n^s}$$

Expanding this into matrix form we have the matrix:

$$\displaystyle \begin{bmatrix} \frac{T(1,1)}{1 \cdot 1^s}&+\frac{T(1,2)}{1 \cdot 2^s}&+\frac{T(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T(1,k)}{1 \cdot k^s} \\ \frac{T(2,1)}{2 \cdot 1^s}&+\frac{T(2,2)}{2 \cdot 2^s}&+\frac{T(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T(2,k)}{2 \cdot k^s} \\ \frac{T(3,1)}{3 \cdot 1^s}&+\frac{T(3,2)}{3 \cdot 2^s}&+\frac{T(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T(3,k)}{3 \cdot k^s} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{T(n,1)}{n \cdot 1^s}&+\frac{T(n,2)}{n \cdot 2^s}&+\frac{T(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} \end{bmatrix}$$

On the right hand side we see that it sums to the right hand side of what we set out to prove, namely:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

Things that remain unclear are: What factor should the left hand side be multiplied with in order to have the same magnitude as the right hand side? And why in the Fourier transform does the first term of the von Mangoldt function appear to be $\log q$?

$$\Lambda(n) = \begin{cases} \log q & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

$$n=1,2,3,4,5,...q$$

As a heuristic, $$\Lambda(n) = \log q \;\;\;\; \text{if }n=1$$ probably has to do with that in the Fourier transform $q$ terms of $\Lambda$ are used and the first column in square matrix $T(1..q,1..q)$ sums to a Harmonic number.

This is not a complete answer but I want show that there is nothing mysterious going on here.

We want to prove that:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

The Dirichlet inverse of the Euler totient function is

$$a(n)=\sum\limits_{d|n} \mu(d)d$$

Construct the matrix $$T(n,k)=a(GCD(n,k))$$

which starts:

$$\displaystyle T = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix} $$

where GCD is the Greatest Common Divisor of row index $n$ and column index $k$.

joriki showed that the von Mangoldt function is $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

Then add this quote by Terence Tao from here, that I don't completely understand but I do almost see why it should be true:

Quote:" The Fourier transform in this context becomes (essentially) the Mellin transform, which is particularly important in analytic number theory. (For instance, the Riemann zeta function is essentially the Mellin transform of the Dirac comb on the natural numbers"

Now let us return to the matrix $T$:

First the von Mangoldt is expanded as:

$$\displaystyle \begin{bmatrix} +1/1&+1/1&+1/1&+1/1&+1/1&+1/1&+1/1&\cdots \\ +1/2&-1/2&+1/2&-1/2&+1/2&-1/2&+1/2 \\ +1/3&+1/3&-2/3&+1/3&+1/3&-2/3&+1/3 \\ +1/4&-1/4&+1/4&-1/4&+1/4&-1/4&+1/4 \\ +1/5&+1/5&+1/5&+1/5&-4/5&+1/5&+1/5 \\ +1/6&-1/6&-2/6&-1/6&+1/6&+2/6&+1/6 \\ +1/7&+1/7&+1/7&+1/7&+1/7&+1/7&-6/7 \\ \vdots&&&&&&&\ddots \end{bmatrix}$$

Edit: 24.1.2016. From here on the variables $n$ and $k$ should be permutated but I don't know how to fix the rest of this answer right now.

Summing the columns first is equivalent to what was said earlier: $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

where:

$$\Lambda(n) = \begin{cases} \infty & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

or as a sequence:

$$\infty ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,...,\Lambda(\infty)$$

And now based on the quote above let us say that: $$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) = \sum\limits_{n=1}^{n=k}\frac{\Lambda(n)}{n^s}$$

Expanding this into matrix form we have the matrix:

$$\displaystyle \begin{bmatrix} \frac{T(1,1)}{1 \cdot 1^s}&+\frac{T(1,2)}{1 \cdot 2^s}&+\frac{T(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T(1,k)}{1 \cdot k^s} \\ \frac{T(2,1)}{2 \cdot 1^s}&+\frac{T(2,2)}{2 \cdot 2^s}&+\frac{T(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T(2,k)}{2 \cdot k^s} \\ \frac{T(3,1)}{3 \cdot 1^s}&+\frac{T(3,2)}{3 \cdot 2^s}&+\frac{T(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T(3,k)}{3 \cdot k^s} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{T(n,1)}{n \cdot 1^s}&+\frac{T(n,2)}{n \cdot 2^s}&+\frac{T(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} \end{bmatrix}$$

On the right hand side we see that it sums to the right hand side of what we set out to prove, namely:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

Things that remain unclear are: What factor should the left hand side be multiplied with in order to have the same magnitude as the right hand side? And why in the Fourier transform does the first term of the von Mangoldt function appear to be $\log q$?

$$\Lambda(n) = \begin{cases} \log q & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

$$n=1,2,3,4,5,...q$$

As a heuristic, $$\Lambda(n) = \log q \;\;\;\; \text{if }n=1$$ probably has to do with that in the Fourier transform $q$ terms of $\Lambda$ are used and the first column in square matrix $T(1..q,1..q)$ sums to a Harmonic number.

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edited Jan 24, 2016 at 14:48

Mats Granvik

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This is not a complete answer but I want show that there is nothing mysterious going on here.

We want to prove that:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

The Dirichlet inverse of the Euler totient function is

$$a(n)=\sum\limits_{d|n} \mu(d)d$$

Construct the matrix $$T(n,k)=a(GCD(n,k))$$

which starts:

$$\displaystyle T = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix} $$

where GCD is the Greatest Common Divisor of row index $n$ and column index $k$.

joriki showed that the von Mangoldt function is $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

Then add this quote by Terence Tao from here, that I don't completely understand but I do almost see why it should be true:

Quote:" The Fourier transform in this context becomes (essentially) the Mellin transform, which is particularly important in analytic number theory. (For instance, the Riemann zeta function is essentially the Mellin transform of the Dirac comb on the natural numbers"

Now let us return to the matrix $T$:

First the von Mangoldt is expanded as:

$$\displaystyle \begin{bmatrix} +1/1&+1/1&+1/1&+1/1&+1/1&+1/1&+1/1&\cdots \\ +1/2&-1/2&+1/2&-1/2&+1/2&-1/2&+1/2 \\ +1/3&+1/3&-2/3&+1/3&+1/3&-2/3&+1/3 \\ +1/4&-1/4&+1/4&-1/4&+1/4&-1/4&+1/4 \\ +1/5&+1/5&+1/5&+1/5&-4/5&+1/5&+1/5 \\ +1/6&-1/6&-2/6&-1/6&+1/6&+2/6&+1/6 \\ +1/7&+1/7&+1/7&+1/7&+1/7&+1/7&-6/7 \\ \vdots&&&&&&&\ddots \end{bmatrix}$$

Edit: 24.1.2016. From here on the variables $n$ and $k$ should be permutated but I don't know how to fix the rest of this answer right now.

Summing the columns first is equivalent to what was said earlier: $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

where:

$$\Lambda(n) = \begin{cases} \infty & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

or as a sequence:

$$\infty ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,...,\Lambda(\infty)$$

And now based on the quote above let us say that: $$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) = \sum\limits_{n=1}^{n=k}\frac{\Lambda(n)}{n^s}$$

Expanding this into matrix form we have the matrix:

$$\displaystyle \begin{bmatrix} \frac{T(1,1)}{1 \cdot 1^s}&+\frac{T(1,2)}{1 \cdot 2^s}&+\frac{T(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T(1,k)}{1 \cdot k^s} \\ \frac{T(2,1)}{2 \cdot 1^s}&+\frac{T(2,2)}{2 \cdot 2^s}&+\frac{T(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T(2,k)}{2 \cdot k^s} \\ \frac{T(3,1)}{3 \cdot 1^s}&+\frac{T(3,2)}{3 \cdot 2^s}&+\frac{T(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T(3,k)}{3 \cdot k^s} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{T(n,1)}{n \cdot 1^s}&+\frac{T(n,2)}{n \cdot 2^s}&+\frac{T(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} \end{bmatrix}$$

On the right hand side we see that it sums to the right hand side of what we set out to prove, namely:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

Things that remain unclear are: What factor should the left hand side be multiplied with in order to have the same magnitude as the right hand side? And why in the Fourier transform does the first term of the von Mangoldt function appear to be $\log q$?

$$\Lambda(n) = \begin{cases} \log q & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

$$n=1,2,3,4,5,...q$$

As a heuristic, $$\Lambda(n) = \log q \;\;\;\; \text{if }n=1$$ probably has to do with that in the Fourier transform $q$ terms of $\Lambda$ are used and the first column in square matrix $T(1..q,1..q)$ sums to a Harmonic number.

This is not a complete answer but I want show that there is nothing mysterious going on here.

We want to prove that:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

The Dirichlet inverse of the Euler totient function is

$$a(n)=\sum\limits_{d|n} \mu(d)d$$

Construct the matrix $$T(n,k)=a(GCD(n,k))$$

which starts:

$$\displaystyle T = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix} $$

where GCD is the Greatest Common Divisor of row index $n$ and column index $k$.

joriki showed that the von Mangoldt function is $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

Then add this quote by Terence Tao from here, that I don't completely understand but I do almost see why it should be true:

Quote:" The Fourier transform in this context becomes (essentially) the Mellin transform, which is particularly important in analytic number theory. (For instance, the Riemann zeta function is essentially the Mellin transform of the Dirac comb on the natural numbers"

Now let us return to the matrix $T$:

First the von Mangoldt is expanded as:

$$\displaystyle \begin{bmatrix} +1/1&+1/1&+1/1&+1/1&+1/1&+1/1&+1/1&\cdots \\ +1/2&-1/2&+1/2&-1/2&+1/2&-1/2&+1/2 \\ +1/3&+1/3&-2/3&+1/3&+1/3&-2/3&+1/3 \\ +1/4&-1/4&+1/4&-1/4&+1/4&-1/4&+1/4 \\ +1/5&+1/5&+1/5&+1/5&-4/5&+1/5&+1/5 \\ +1/6&-1/6&-2/6&-1/6&+1/6&+2/6&+1/6 \\ +1/7&+1/7&+1/7&+1/7&+1/7&+1/7&-6/7 \\ \vdots&&&&&&&\ddots \end{bmatrix}$$

Summing the columns first is equivalent to what was said earlier: $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

where:

$$\Lambda(n) = \begin{cases} \infty & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

or as a sequence:

$$\infty ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,...,\Lambda(\infty)$$

And now based on the quote above let us say that: $$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) = \sum\limits_{n=1}^{n=k}\frac{\Lambda(n)}{n^s}$$

Expanding this into matrix form we have the matrix:

$$\displaystyle \begin{bmatrix} \frac{T(1,1)}{1 \cdot 1^s}&+\frac{T(1,2)}{1 \cdot 2^s}&+\frac{T(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T(1,k)}{1 \cdot k^s} \\ \frac{T(2,1)}{2 \cdot 1^s}&+\frac{T(2,2)}{2 \cdot 2^s}&+\frac{T(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T(2,k)}{2 \cdot k^s} \\ \frac{T(3,1)}{3 \cdot 1^s}&+\frac{T(3,2)}{3 \cdot 2^s}&+\frac{T(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T(3,k)}{3 \cdot k^s} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{T(n,1)}{n \cdot 1^s}&+\frac{T(n,2)}{n \cdot 2^s}&+\frac{T(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} \end{bmatrix}$$

On the right hand side we see that it sums to the right hand side of what we set out to prove, namely:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

Things that remain unclear are: What factor should the left hand side be multiplied with in order to have the same magnitude as the right hand side? And why in the Fourier transform does the first term of the von Mangoldt function appear to be $\log q$?

$$\Lambda(n) = \begin{cases} \log q & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

$$n=1,2,3,4,5,...q$$

As a heuristic, $$\Lambda(n) = \log q \;\;\;\; \text{if }n=1$$ probably has to do with that in the Fourier transform $q$ terms of $\Lambda$ are used and the first column in square matrix $T(1..q,1..q)$ sums to a Harmonic number.

This is not a complete answer but I want show that there is nothing mysterious going on here.

We want to prove that:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

The Dirichlet inverse of the Euler totient function is

$$a(n)=\sum\limits_{d|n} \mu(d)d$$

Construct the matrix $$T(n,k)=a(GCD(n,k))$$

which starts:

$$\displaystyle T = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix} $$

where GCD is the Greatest Common Divisor of row index $n$ and column index $k$.

joriki showed that the von Mangoldt function is $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

Then add this quote by Terence Tao from here, that I don't completely understand but I do almost see why it should be true:

Quote:" The Fourier transform in this context becomes (essentially) the Mellin transform, which is particularly important in analytic number theory. (For instance, the Riemann zeta function is essentially the Mellin transform of the Dirac comb on the natural numbers"

Now let us return to the matrix $T$:

First the von Mangoldt is expanded as:

$$\displaystyle \begin{bmatrix} +1/1&+1/1&+1/1&+1/1&+1/1&+1/1&+1/1&\cdots \\ +1/2&-1/2&+1/2&-1/2&+1/2&-1/2&+1/2 \\ +1/3&+1/3&-2/3&+1/3&+1/3&-2/3&+1/3 \\ +1/4&-1/4&+1/4&-1/4&+1/4&-1/4&+1/4 \\ +1/5&+1/5&+1/5&+1/5&-4/5&+1/5&+1/5 \\ +1/6&-1/6&-2/6&-1/6&+1/6&+2/6&+1/6 \\ +1/7&+1/7&+1/7&+1/7&+1/7&+1/7&-6/7 \\ \vdots&&&&&&&\ddots \end{bmatrix}$$

Edit: 24.1.2016. From here on the variables $n$ and $k$ should be permutated but I don't know how to fix the rest of this answer right now.

Summing the columns first is equivalent to what was said earlier: $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

where:

$$\Lambda(n) = \begin{cases} \infty & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

or as a sequence:

$$\infty ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,...,\Lambda(\infty)$$

And now based on the quote above let us say that: $$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) = \sum\limits_{n=1}^{n=k}\frac{\Lambda(n)}{n^s}$$

Expanding this into matrix form we have the matrix:

$$\displaystyle \begin{bmatrix} \frac{T(1,1)}{1 \cdot 1^s}&+\frac{T(1,2)}{1 \cdot 2^s}&+\frac{T(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T(1,k)}{1 \cdot k^s} \\ \frac{T(2,1)}{2 \cdot 1^s}&+\frac{T(2,2)}{2 \cdot 2^s}&+\frac{T(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T(2,k)}{2 \cdot k^s} \\ \frac{T(3,1)}{3 \cdot 1^s}&+\frac{T(3,2)}{3 \cdot 2^s}&+\frac{T(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T(3,k)}{3 \cdot k^s} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{T(n,1)}{n \cdot 1^s}&+\frac{T(n,2)}{n \cdot 2^s}&+\frac{T(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} \end{bmatrix}$$

On the right hand side we see that it sums to the right hand side of what we set out to prove, namely:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

Things that remain unclear are: What factor should the left hand side be multiplied with in order to have the same magnitude as the right hand side? And why in the Fourier transform does the first term of the von Mangoldt function appear to be $\log q$?

$$\Lambda(n) = \begin{cases} \log q & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

$$n=1,2,3,4,5,...q$$

As a heuristic, $$\Lambda(n) = \log q \;\;\;\; \text{if }n=1$$ probably has to do with that in the Fourier transform $q$ terms of $\Lambda$ are used and the first column in square matrix $T(1..q,1..q)$ sums to a Harmonic number.

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created Jul 31, 2015 at 13:08

Mats Granvik

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This is not a complete answer but I want show that there is nothing mysterious going on here.

We want to prove that:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

The Dirichlet inverse of the Euler totient function is

$$a(n)=\sum\limits_{d|n} \mu(d)d$$

Construct the matrix $$T(n,k)=a(GCD(n,k))$$

which starts:

$$\displaystyle T = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix} $$

where GCD is the Greatest Common Divisor of row index $n$ and column index $k$.

joriki showed that the von Mangoldt function is $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

Then add this quote by Terence Tao from here, that I don't completely understand but I do almost see why it should be true:

Quote:" The Fourier transform in this context becomes (essentially) the Mellin transform, which is particularly important in analytic number theory. (For instance, the Riemann zeta function is essentially the Mellin transform of the Dirac comb on the natural numbers"

Now let us return to the matrix $T$:

First the von Mangoldt is expanded as:

$$\displaystyle \begin{bmatrix} +1/1&+1/1&+1/1&+1/1&+1/1&+1/1&+1/1&\cdots \\ +1/2&-1/2&+1/2&-1/2&+1/2&-1/2&+1/2 \\ +1/3&+1/3&-2/3&+1/3&+1/3&-2/3&+1/3 \\ +1/4&-1/4&+1/4&-1/4&+1/4&-1/4&+1/4 \\ +1/5&+1/5&+1/5&+1/5&-4/5&+1/5&+1/5 \\ +1/6&-1/6&-2/6&-1/6&+1/6&+2/6&+1/6 \\ +1/7&+1/7&+1/7&+1/7&+1/7&+1/7&-6/7 \\ \vdots&&&&&&&\ddots \end{bmatrix}$$

Summing the columns first is equivalent to what was said earlier: $$\Lambda(n)=\sum\limits_{k=1}^{k=\infty} \frac{T(n,k)}{k}$$

where:

$$\Lambda(n) = \begin{cases} \infty & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

or as a sequence:

$$\infty ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,...,\Lambda(\infty)$$

And now based on the quote above let us say that: $$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) = \sum\limits_{n=1}^{n=k}\frac{\Lambda(n)}{n^s}$$

Expanding this into matrix form we have the matrix:

$$\displaystyle \begin{bmatrix} \frac{T(1,1)}{1 \cdot 1^s}&+\frac{T(1,2)}{1 \cdot 2^s}&+\frac{T(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T(1,k)}{1 \cdot k^s} \\ \frac{T(2,1)}{2 \cdot 1^s}&+\frac{T(2,2)}{2 \cdot 2^s}&+\frac{T(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T(2,k)}{2 \cdot k^s} \\ \frac{T(3,1)}{3 \cdot 1^s}&+\frac{T(3,2)}{3 \cdot 2^s}&+\frac{T(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T(3,k)}{3 \cdot k^s} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{T(n,1)}{n \cdot 1^s}&+\frac{T(n,2)}{n \cdot 2^s}&+\frac{T(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} \end{bmatrix}$$

On the right hand side we see that it sums to the right hand side of what we set out to prove, namely:

$$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$

Things that remain unclear are: What factor should the left hand side be multiplied with in order to have the same magnitude as the right hand side? And why in the Fourier transform does the first term of the von Mangoldt function appear to be $\log q$?

$$\Lambda(n) = \begin{cases} \log q & \text{if }n=1, \\\log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$

$$n=1,2,3,4,5,...q$$

As a heuristic, $$\Lambda(n) = \log q \;\;\;\; \text{if }n=1$$ probably has to do with that in the Fourier transform $q$ terms of $\Lambda$ are used and the first column in square matrix $T(1..q,1..q)$ sums to a Harmonic number.

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occurred Jul 31, 2015 at 13:08

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