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

Mathematica knows that:

$$\log(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{n^{(s - 1)}}\right)\;\;\;\;\;\;\;\;\;\;\;\; (1)$$

The von Mangoldt function should then be:

$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (2)$$

The question is if the following Riemann zeta function product is equal to the Fourier transform of the

von Mangoldt function?

$$\text{Fourier Transform of } \Lambda(n) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(1/2+i \cdot t)\sum\limits_{d|n}\frac{\mu(d)}{d^{(1/2+i \cdot t-1)}} \;\;\;\;\;\;\; (3)$$

In the Fourier transform the von Mangoldt function has this form:

$$\log (\text{scale}) ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,...,\Lambda(\text{scale})$$

$$scale = 1,2,3,4,5,6,7,8,9,10,...k$$

Or as latex:

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

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

TableForm[Table[Table[If[n == 1, Log[q], MangoldtLambda[n]], {n, 1, q}], {q, 1, 12}]]

Put in pictures:

Is this Dirichlet series:

equal to this Fourier transform?

The pictures are respectively: Spectral Riemann zeta or Riemann zeta function product found below as Image 5. Fourier transform of von Mangoldt function found below as Image 6.

Edit 4 4 2014: Matrix $$T_2$$ defined below as the Dirichlet inverse of the Euler totient function as a function of the Greatest Common Divisor (GCD) of row index $$n$$ and column index $$k$$;

$$T_2(n,k) = a(GCD(n,k))$$

where:

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

starts:

$$\displaystyle T_2 = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix}$$

By the answer at Mathematics Stackexchange given by joriki, we know that the Dirichlet series in the rows and columns tend to the von Mangoldt function $$\Lambda(n)$$ and $$\Lambda(k)$$:

$$\displaystyle \begin{bmatrix} \frac{T_2(1,1)}{1 \cdot 1}&+\frac{T_2(1,2)}{1 \cdot 2}&+\frac{T_2(1,3)}{1 \cdot 3}+&\cdots&+\frac{T_2(1,k)}{1 \cdot k} \\ \frac{T_2(2,1)}{2 \cdot 1}&+\frac{T_2(2,2)}{2 \cdot 2}&+\frac{T_2(2,3)}{2 \cdot 3}+&\cdots&+\frac{T_2(2,k)}{2 \cdot k} \\ \frac{T_2(3,1)}{3 \cdot 1}&+\frac{T_2(3,2)}{3 \cdot 2}&+\frac{T_2(3,3)}{3 \cdot 3}+&\cdots&+\frac{T_2(3,k)}{3 \cdot k} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{T_2(n,1)}{n \cdot 1}&+\frac{T_2(n,2)}{n \cdot 2}&+\frac{T_2(n,3)}{n \cdot 3}+&\cdots&+\frac{T_2(n,k)}{n \cdot k} \end{bmatrix} = \begin{bmatrix} \frac{\infty}{1} \\ +\frac{\Lambda(2)}{2} \\ +\frac{\Lambda(3)}{3} \\ \vdots \\ +\frac{\Lambda(n)}{n} \end{bmatrix}$$ $$=\;\;\;\;\;\;\;\;\;\;\;$$ $$\displaystyle \begin{bmatrix} \frac{\infty}{1}&+\frac{\Lambda(2)}{2}&+\frac{\Lambda(3)}{3}+&\cdots&+\frac{\Lambda(k)}{k} \end{bmatrix}\;\;\;\;\;\;\;\;\;\;\;$$

Also by the answer below by GH from MO we know that the von Mangoldt function can written as a Dirichlet generating function when the $$s$$ goes towards the value $$1$$. Numerically for non complex values of $$s$$ the following seems to be true:

$$\displaystyle \begin{bmatrix} \frac{T_2(1,1)}{1 \cdot 1^s}&+\frac{T_2(1,2)}{1 \cdot 2^s}&+\frac{T_2(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T_2(1,k)}{1 \cdot k^s} \\ \frac{T_2(2,1)}{2 \cdot 1^s}&+\frac{T_2(2,2)}{2 \cdot 2^s}&+\frac{T_2(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T_2(2,k)}{2 \cdot k^s} \\ \frac{T_2(3,1)}{3 \cdot 1^s}&+\frac{T_2(3,2)}{3 \cdot 2^s}&+\frac{T_2(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T_2(3,k)}{3 \cdot k^s} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{T_2(n,1)}{n \cdot 1^s}&+\frac{T_2(n,2)}{n \cdot 2^s}&+\frac{T_2(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T_2(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} \end{bmatrix}$$ $$=\;\;\;\;\;\;\;\;\;\;\;$$ $$\displaystyle \begin{bmatrix} \frac{\infty}{1^s}&+\frac{\Lambda(2)}{2^s}&+\frac{\Lambda(3)}{3^s}+&\cdots&+\frac{\Lambda(k)}{k^s} \end{bmatrix}\;\;\;\;\;\;\;\;\;\;\;$$

but I don't know for sure. In order for the sum of columns sums to converge at all, the first term can not be allowed to be $$\frac{\infty}{1}$$. Therefore I set the terms in the first column equal to zero, and the matrix becomes:

$$\displaystyle \begin{bmatrix} \frac{0}{1 \cdot 1^s}&+\frac{T_2(1,2)}{1 \cdot 2^s}&+\frac{T_2(1,3)}{1 \cdot 3^s}+&\cdots&+\frac{T_2(1,k)}{1 \cdot k^s} \\ \frac{0}{2 \cdot 1^s}&+\frac{T_2(2,2)}{2 \cdot 2^s}&+\frac{T_2(2,3)}{2 \cdot 3^s}+&\cdots&+\frac{T_2(2,k)}{2 \cdot k^s} \\ \frac{0}{3 \cdot 1^s}&+\frac{T_2(3,2)}{3 \cdot 2^s}&+\frac{T_2(3,3)}{3 \cdot 3^s}+&\cdots&+\frac{T_2(3,k)}{3 \cdot k^s} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ \frac{0}{n \cdot 1^s}&+\frac{T_2(n,2)}{n \cdot 2^s}&+\frac{T_2(n,3)}{n \cdot 3^s}+&\cdots&+\frac{T_2(n,k)}{n \cdot k^s} \end{bmatrix} = \begin{bmatrix} \frac{\zeta(s)}{1}-\frac{1}{1} \\ +\frac{\zeta(s)\sum\limits_{d|2} \frac{\mu(d)}{d^{(s-1)}}}{2} -\frac{1}{2} \\ +\frac{\zeta(s)\sum\limits_{d|3} \frac{\mu(d)}{d^{(s-1)}}}{3} -\frac{1}{3} \\ \vdots \\ +\frac{\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n} -\frac{1}{n} \end{bmatrix}$$

$$=\;\;\;\;\;\;\;\;\;\;\;$$ $$\displaystyle \begin{bmatrix} \frac{\Lambda(1)}{1^s}&+\frac{\Lambda(2)}{2^s}&+\frac{\Lambda(3)}{3^s}+&\cdots&+\frac{\Lambda(k)}{k^s} \end{bmatrix}\;\;\;\;\;\;\;\;\;\;\;$$

So if this is correct then the sum of row sums and the sum of column sums would be:

$$-\frac{\zeta'(s)}{\zeta(s)}=\sum\limits_{k=1}^{\infty} \frac{\Lambda(k)}{k^s} = \sum\limits_{n=1}^{\infty} \left(\frac{1}{n}\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}-\frac{1}{n}\right)\;\;\;\;\;\;\;\;\;\;\;\;(4)$$

as in wikipedia about the von Mangoldt function.

Derivative of Riemann zeta, is this inequality true?

Edit Apr 2 2014 after GH from MO answer below:

The algorithm originating from Michael Rubinstein -> Brian Conrey -> Chris King -> Heike, with the modification that we let:

$$\Lambda(1) = H_{\text{scale}}$$ $$f(n) = H_{\text{scale}} + \sum _{n=2}^{n=scale} \Lambda (n)$$

Then

$$\text{Fourier Transform of } \Lambda(n) = g(k)$$

$$g(k) = \frac{\Im(\sum _{x=0}^{\log (\text{scale})} \left(e^x\right)^{(i k+\frac{1}{2}-1)} \left(f \left\lfloor e^x\right\rfloor -e^x\right))}{\text{length of vector x}}$$

which simplified is:

$$g(k) = \Im(\sum _{x=0}^{\log (\text{scale})} \left(e^x\right)^{(i k+\frac{1}{2}-1)} \left(f \left\lfloor e^x\right\rfloor -e^x\right))$$

according to the algorithm, where "scale" is some number that ideally should go to infinity, but for computation is set to some chosen large number.

Again the input (f[[Floor[x]]] - x) to the Fourier transform In the algorithm;

Clear[f, x]
scale = 1000;
f = ConstantArray[0, scale];
f[[1]] = N@HarmonicNumber[scale];
Monitor[Do[
f[[i]] = N@MangoldtLambda[i] + f[[i - 1]], {i, 2, scale}], i]
xres = .002;
x = Exp[Range[0, Log[scale], xres]];
tmax = 60;
tres = .015;
Monitor[errList =
Table[(f[[Floor[x]]] - x), {k, Range[0, 60, tres]}][[1 ;; 20]];, k]
ListLinePlot[errList, DataRange -> {0, 60}, PlotRange -> {-10, 15},
PlotStyle -> Blue, ImageSize -> Large]

looks like this:

From this one can conjecture that the Fourier transform of an exponential sawtooth wave will give the Riemann zeta function.

Then going back from the Riemann zeta function to the original input domain with the following two Mathematica lines:

x = Exp[Range[0, Log[scale], xres]];
a = -FourierDCT[Log[x]*FourierDST[(SawtoothWave[x] - 1)*(x)^(-1/2)]];

as part of a longer code snippet, where FourierDCT is the Fourier discrete cosine transform and FourierDST is the Fourier discrete sine transform, one gets the following plot:

Mimicing this plot I have written:

(*Mathematica 8*)
Clear[x, xx]
scale = 200;
xres = .001;
x = Exp[Range[0, Log[scale], xres]];
xx = Flatten[{0, Differences[Floor[Exp[Range[0, Log[scale], xres]]]]}];
ListLinePlot[xx*x^(-1/2), PlotRange -> {-0.1, 0.8},
ImageSize -> Large]

Although this kind of numerical mimicing does not anymore give accurate plots of the Riemann zeta function, I have based on it tried to imagine what the von Mangoldt function to be Fourier transformed, might look like with the following program:

(*Mathematica 8*)
Clear[x, xx]
scale = 200;
xres = .001;
x = Exp[Range[0, Log[scale], xres]];
xx = Flatten[{0, Differences[Floor[Exp[Range[0, Log[scale], xres]]]]}];
yy = Accumulate[xx]*xx;
zz = Table[
If[yy[[i]] == 0, 0, MangoldtLambda[yy[[i]]]], {i, 1, Length[yy]}];
ListLinePlot[zz*x^(-1/2), PlotRange -> {-0.1, 0.8},
ImageSize -> Large]

which gives this plot:

But this probably far from the correct input signal to the Fourier transform to get the similarity with the Riemann zeta function product involving the Möbius function.

To find the correct logarithmic Dirac comb like input one should probably look at the exponential sawtooth for each row in the von Mangoldt function Matrix $$T_2$$ below, see what sawtooth wave each accumulated row is, and Fourier transform it twice. Then possibly one could match it to a variant of the von Mangoldt function. A hint of what it might be is given in this plot:

Mobius function -> Dirichlet series -> Riemann zeta function product -> Fourier transform -> von Mangoldt function:

for which the program is found lower down.

In the question I ask: Is it equal to? What I mean in the third equation $$(3)$$ with the $$\sim$$ symbol is if the left hand side and the right hand side are equal to each other if you multiply one side with a scale factor $$a$$ like this

$$\Im(\sum _{x=0}^{\log (\text{scale})} \left(e^x\right)^{(i k+\frac{1}{2}-1)} \left(f \left\lfloor e^x\right\rfloor -e^x\right)) = a \cdot \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(1/2+i \cdot t)\sum\limits_{d|n} \frac{\mu(d)}{d^{(1/2+i \cdot t-1)}} \;\;\;\;\;\;\; (3)$$

Introduction:

In the case of the logarithm of $$2$$, log(2), this is equal to a table (red and green) called $$\zeta(s)$$:

$$k \mid n : 1$$ else $$0$$.

matrix multiplied by a second table (red, green and yellow) called $$(1 - 2^{1-s})$$:

$$n=k: 1$$ else if $$n=2 \cdot k: -2$$ else $$0$$

Picture of tables $$\zeta(s)$$ and $$(1 - 2^{1-s})$$ [Image 1]:

Taking the matrix product of the two tables above we get [Image 2]:

The first column then has the numerators of the alternating series.

This is as said:

$$\log(2) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - 2^{(1 - s)}\right)$$

which with a sign change in the exponent is equal to:

$$\log(2) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{2^{(s - 1)}}\right)$$

Before I get to the actual question I show the Dirichlet series for logarithm of n.

$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}... =\log(2)$$

which is the second column in:

Logarithms of $$n$$:

$$T_1 = \begin{bmatrix} 0&0&0&0&0&0&0 \\ 1&-1&1&-1&1&-1&1 \\ 1&1&-2&1&1&-2&1 \\ 1&1&1&-3&1&1&1 \\ 1&1&1&1&-4&1&1 \\ 1&1&1&1&1&-5&1 \\ 1&1&1&1&1&1&-6 \end{bmatrix}$$

It turns out that:

$$1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}... =\log(3)$$

and:

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

in general. That is same as the Mathematica known formula above.

The recurrence describing matrix $$T_1$$ is:

$$\displaystyle T_1(1,n)=0; n>1: T_1(n,k) = \sum\limits_{i=1}^{n-1} T_1(n,k-i)$$

It is natural to ask what this recurrence might do when applying it symmetrically, like this:

$$\displaystyle T_2(n,1)=1, T_2(1,k)=1, n>=k: -\sum\limits_{i=1}^{k-1} T_2(n-i,k), n

von Mangoldt function matrix:

$$\displaystyle T_2 = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix}$$

Here we then have $$\log(2)$$ in the second column and second row. Like wise $$\log(3)$$ in third row and third column.

By doing a Möbius inversion on each row or column one can conjecture that the von Mangoldt function is found as:

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

Alternatively by inspecting the numbers in the decimal digits one also finds conjecturally that:

$$\Lambda(k)=\sum\limits_{n=1}^{\infty} \frac{T_2(n,k)}{n}$$

Also, matrix $$T_2$$ is a matrix with the Greatest Common Divisor as lookup index:

$$T_2(n,k) = a(GCD(n,k))$$

where conjecturally:

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

which is also known as the Dirichlet inverse of the Euler Totient starting:

$$1,-1,-2,-1,-4,+2,-6,...$$

And as a side note we can by a result of Wolfgang Schramm calculate the Möbius function from this matrix as:

$$\displaystyle \mu(n) = \frac{1}{n} \sum\limits_{k=1}^{k=n} T_2(n,k) \cdot e^{i 2 \pi \frac{k}{n}}$$

So much for the elementary introduction.

Next we will look at the function:

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

in the complex plane where $$s=a+ib$$ is a complex number, and thereby get closer to the question,

The plot of this I call the Riemann zeta function product. With a modification of Jeffrey Stopple's

code;

Show[Graphics[
RasterArray[
Table[Hue[
Mod[3 Pi/2 +
Arg[Sum[Zeta[sigma + I t]*
Total[1/Divisors[n]^(sigma + I t - 1)*
MoebiusMu[Divisors[n]]]/n, {n, 1, 30}]],
2 Pi]/(2 Pi)], {t, -30, 30, .1}, {sigma, -30, 30, .1}]]],
AspectRatio -> Automatic]

it looks like this:

The Riemann zeta function product (30-th partial sum) [Image 3]:

Again as a sidenote we can compare with the usual Riemann zeta function:

Normal or usual zeta [Image 4]:

which is plotted in the same (complex) region as the Riemann zeta function product (-30 to +30 and -30i to +30i).

The critical strip of the Riemann zeta function product;

scale = 50; (*scale = 5000 gives the plot below*)
Print["Counting to 60"]
Monitor[g1 =
ListLinePlot[
Table[Re[
Zeta[1/2 + I*k]*
Total[Table[
Total[MoebiusMu[Divisors[n]]/Divisors[n]^(1/2 + I*k - 1)]/(n*
k), {n, 1, scale}]]], {k, 0 + 1/1000, 60, N[1/6]}],
DataRange -> {0, 60}, PlotRange -> {-0.15, 1.5}], Floor[k]]

looks like this:

The Riemann zeta function product [Image 5]:

What a coincidence then that if we take the Fourier transform of the von Mangoldt function with Heike's algorithm;

Clear[f]
scale = 100000;
f = ConstantArray[0, scale];
f[[1]] = N@HarmonicNumber[scale];
Monitor[Do[
f[[i]] = N@MangoldtLambda[i] + f[[i - 1]], {i, 2, scale}], i]
xres = .002;
xlist = Exp[Range[0, Log[scale], xres]];
tmax = 60;
tres = .015;
Monitor[errList =
Table[(xlist^(1/2 + I k - 1).(f[[Floor[xlist]]] - xlist)), {k,
Range[0, 60, tres]}];, k]
ListLinePlot[Im[errList]/Length[xlist], DataRange -> {0, 60},
PlotRange -> {-.01, .15}]

where we have set the first term in the von Mangoldt function sequence:

Table[Limit[Zeta[s]*Total[MoebiusMu[Divisors[n]]/Divisors[n]^(s - 1)], s -> 1], {n, 1, 32}]

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

equal to a Harmonic number:

$$H_{\text{scale}} ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,...,\Lambda(\text{scale})$$

and where scale is the scale is the scale of the Fourier transform matrix and the number of terms used in the von Mangoldt

function sequence, we get the same plot:

Fourier transform of von Mangoldt function [Image 6]:

This is unclear to me. But the parameter scale in the program anyways, is the value of the Harmonic Number $$H_{\text{scale}}$$.

Put simpler I have taken a matrix $$A$$:

$$A = \left( \begin{array}{cccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2^{\frac{1}{2}-i t} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 3^{\frac{1}{2}-i t} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 4^{\frac{1}{2}-i t} & 2^{\frac{1}{2}-i t} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 5^{\frac{1}{2}-i t} & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 6^{\frac{1}{2}-i t} & 3^{\frac{1}{2}-i t} & 2^{\frac{1}{2}-i t} & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 7^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 8^{\frac{1}{2}-i t} & 4^{\frac{1}{2}-i t} & 0 & 2^{\frac{1}{2}-i t} & 0 & 0 & 0 & 1 & 0 & 0 \\ 9^{\frac{1}{2}-i t} & 0 & 3^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 10^{\frac{1}{2}-i t} & 5^{\frac{1}{2}-i t} & 0 & 0 & 2^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 1 \end{array}\right)$$

calculated it's matrix inverse and multiplied by the Riemann zeta function:

$$B=\zeta(\frac{1}{2}+i t)\left( \begin{array}{cccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -2^{\frac{1}{2}-i t} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -3^{\frac{1}{2}-i t} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -2^{\frac{1}{2}-i t} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ -5^{\frac{1}{2}-i t} & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 6^{\frac{1}{2}-i t} & -3^{\frac{1}{2}-i t} & -2^{\frac{1}{2}-i t} & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ -7^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2^{\frac{1}{2}-i t} & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & -3^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 10^{\frac{1}{2}-i t} & -5^{\frac{1}{2}-i t} & 0 & 0 & -2^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 1\end{array}\right)$$

and then plotted the function that is matrix $$B$$. What I seem to get is the Riemann zeta zero spectrum as a Riemann zeta function product

resembling the Fourier transform of the von Mangoldt function where $$\Lambda(1) = H_{scale}$$

So the question is if image 5 and image 6 are equivalent, differing only by a scale factor? Or as in the title: Is the Riemann

zeta function product $$f(t)$$ (Spectral Riemann zeta):

$$f(t)=\sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(1/2+i \cdot t)\sum\limits_{d|n} \frac{\mu(d)}{d^{(1/2+i \cdot t-1)}}$$

similar or equal to the Fourier transform of:

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

times a scale factor?

In support of this claim is that if we take the Fourier transform of the Riemann zeta function product at the critical line (real part equal 1/2);

Clear[n, k, t, A, nn, B]
nn = 60
A = Table[
Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
nn}], {n, 1, nn}]; MatrixForm[A];
B = FourierDCT[
Table[Total[
1/Table[n, {n, 1, nn}]*(Total[
Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]] - 1)], {t, 1/1000,
600, N[1/6]}]];
g1 = ListLinePlot[B[[1 ;; 700]], DataRange -> {0, 60},
PlotRange -> {-7, 25}];
mm = 11.35/Log[2];
g2 = Graphics[
Table[Style[Text[n, {mm*Log[n], 5 - (-1)^n}],
FontFamily -> "Times New Roman", FontSize -> 14], {n, 1, 16}]];
Show[g1, g2, ImageSize -> Large]

we get:

Mobius function -> Dirichlet series -> The Riemann zeta function product -> Fourier transform -> von Mangoldt function:

which looks a lot like the von Mangoldt function in Heikes algorithm, in the way it is input there. Notice that the line in the

last program above

B[[1 ;; 700]]*Table[Sqrt[n], {n, 1, 700}]

is not analytically correct. A larger image of the same plot: http://i.stack.imgur.com/02A1p.jpg

The partial sums of the Riemann zeta function product:

12 first curves together or partial sums:

and a larger plot: http://i.stack.imgur.com/5LirM.jpg

This whole question has previously been asked here: https://math.stackexchange.com/questions/424530/is-this-similarity-to-the-fourier-transform-of-the-von-mangoldt-function-real

Clear[n, k, t, A, nn, h]
nn = 60;
h = 2; (*h=2 gives log 2 operator, h=3 gives log 3 operator and so on*)
A = Table[
Table[If[Mod[n, k] == 0,
If[Mod[n/k, h] == 0, 1 - h, 1]/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
nn}], {n, 1, nn}];
MatrixForm[A];
g1 = ListLinePlot[
Table[Total[
1/Table[n*t, {n, 1, nn}]*
Total[Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]]], {t, 1/1000,
nn, N[1/6]}], DataRange -> {0, nn}, PlotRange -> {-3, 7}];
mm = N[2*Pi/Log[h], 12]
g2 = Graphics[
Table[Style[Text[n*2*Pi/Log[h], {mm*n, 1}],
FontFamily -> "Times New Roman", FontSize -> 14], {n, 1, 32}]];
Show[g1, g2, ImageSize -> Large]

Matrix inverse of Riemann zeta times log 2 operator:

Clear[n, k, t, A, nn, dd]
dd = 220;
Print["Counting to ", dd]
nn = 20;
A = Table[
Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
nn}], {n, 1, nn}];
Monitor[g1 =
ListLinePlot[
Table[Total[
1/Table[n*t, {n, 1, nn}]*
Total[Transpose[
Re[Inverse[
IdentityMatrix[nn] + (Inverse[A] - IdentityMatrix[nn])*
Zeta[1/2 + I*t]]]]]], {t, 1/1000, dd, N[1/100]}],
DataRange -> {0, dd}, PlotRange -> {-7, 7}];, Floor[t]]
mm = N[2*Pi/Log[2], 20]
g2 = Graphics[
Table[Style[Text[n, {mm*n, 1}], FontFamily -> "Times New Roman",
FontSize -> 14], {n, 1, 32}]];
Show[g1, g2, ImageSize -> Large]

Matrix Inverse of (matrix inverse times zeta function (on critical line)):

The following is a relationship:

Let $$\mu(n)$$ be the Möbius function, then:

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

$$T(n,k)=a(GCD(n,k))$$

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

$$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^s} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s} \zeta(z)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^c} = \frac{\zeta(s) \cdot \zeta(c)}{\zeta(c + s - 1)}$$

which is part of the limit:

$$\frac{\zeta '(s)}{\zeta (s)}=\lim_{c\to 1} \, \left(\zeta (c)-\frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}\right)$$ $$\int \frac{\zeta '(s)}{\zeta (s)} ds = \log(\zeta(s))$$ $$\exp(\log(\zeta(s)))$$ $$=\zeta(s)$$

Plot of the Dirichlet generating function for the matrix on the critical line:

Clear[s, c, t]
c = 1 + 1/100;
Plot[Re[Zeta[1/2 + I*t]*Zeta[c]/Zeta[1/2 + I*t + c - 1]], {t, 0, 60},
PlotRange -> {-5, 105}]

Edit 9.11.2018:

(*Mathematica divergent sum start*)
nn = 500;
f[t_] = D[RiemannSiegelTheta[t], t];
g2 = Plot[
Sign[RiemannSiegelZ[t]]*
Abs[Zeta[1/2 + I*t]*
Total[Table[
Total[MoebiusMu[Divisors[n]]/Divisors[n]^(1/2 + I*t - 1)]/
n, {n, 1, nn}]]]/(f[t] + HarmonicNumber[nn]), {t, 0, 60},
PlotStyle -> Thickness[0.004], ImageSize -> Large,
PlotRange -> {-4, 4}]
(*end*)

(*Mathematica analytic continuation start*)
f[t_] = D[RiemannSiegelTheta[t], t];
epsilon = 1/40;
g3 = Monitor[
Plot[Sign[RiemannSiegelZ[t]]*
Abs[Zeta[1/2 + I*t]/Zeta[1/2 + I*t + (1 + epsilon) - 1]*
Zeta[(1 + epsilon)]]/(f[t] + 1/epsilon + EulerGamma), {t, 0,
60}, PlotStyle -> {Red, Thickness[0.004]}, ImageSize -> Large,
PlotRange -> {-4, 4}], n]

Show[{g2, g3}, ImageSize -> 1000]
(*code line: "Show[{g2, g3}, ImageSize -> 1000]" added by Roger L. Bagula*)
(*end*)

The graph above cross the x-axis at nontrivial Riemann zeta zeros.

• this has got to be the longest yet very cool MO question so far! – Suvrit Apr 3 '14 at 21:06
• I'm impressed by the effort put into this question – Julian Barathieu Mar 6 '18 at 10:13

Your first two relations follow quickly from the asymptotic behavior of the functions $s\mapsto\zeta(s)$ and $s\mapsto d^{1-s}$ as $s\to 1$. Let me demonstrate this for the second relation for $n>1$ (note that it fails for $n=1$).

For $|s-1|<1$ we have $$\zeta(s)=1/(s-1)+O(1)\quad\text{and}\quad d^{1-s}=1-(s-1)\log d+O_d((s-1)^2),$$ hence $$\sum_{d\mid n}\frac{\mu(d)}{d^{s-1}} =\sum_{d\mid n}\mu(d)-(s-1)\sum_{d\mid n}\mu(d)\log d+O_n((s-1)^2).$$ On the right hand side the first sum is zero, and the second sum is $-\Lambda(n)$, because the convolution of the Möbius function with $1$ and $\log$ at $n>1$ equals $0$ and $\Lambda(n)$, respectively. It follows that $$\sum_{d\mid n}\frac{\mu(d)}{d^{s-1}} \sim (s-1)\Lambda(n)\quad\text{as}\quad s\to 1,$$ whence $$\lim_{s\to 1}\zeta(s)\sum_{d\mid n}\frac{\mu(d)}{d^{s-1}} =\Lambda(n).$$

Now I don't understand your main question. What do you mean by the Fourier transform of $\Lambda(n)$? The Fourier transform is initially defined for sequences in $L^1(\mathbb{Z})$, and then by a limit procedure for sequences in $L^2(\mathbb{Z})$. However, $\Lambda(n)$ is not even bounded. Please define precisely the left hand side of your third display, and explain what you mean by the symbol $\sim$ there.

P.S. Similarly, I don't understand what you mean by $f(t)$. You define it by an infinite series, but I don't see that the series converges.

• I have tried to correct the definitions in the question. $f(t)$ is a function of the imaginary (is that right?) variable $t$. The convergence of $f(t)$ I don't know anything about. What I do know is that by plotting the function for nn = 10, nn = 50, nn = 300, nn = 5000 there isn't much difference in the shape and position of the graph plotted. – Mats Granvik Apr 2 '14 at 18:23
• I still don't understand what you talk about. The definition of $g(k)$ depends on some quantity "scale", so that quantity should also be part of the argument. But then we talk about finite series, so an exact match with a Riemann zeta-like function is unlikely. Also, there is no $f(t)$ until you define it properly (i.e. you need to show the convergence). I understand that you have some experimental observations, but you should turn it into a precise mathematical question (otherwise it is like asking someone's opinion about a painting or musical piece). – GH from MO Apr 2 '14 at 18:32
• I had an error in the summation index. k was meant to be n=scale. scale is just some large number. I might have other errors in here to. I will have a look tomorrow. – Mats Granvik Apr 2 '14 at 18:44

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.

• when you write "the Fourier transform of $\Lambda(1) \ldots \Lambda(k)$" you mean the Fourier (Laplace) transform of the weighted logarithmically spaced Dirac comb $\sum_{n=1}^\infty \Lambda(n) \delta(x-\ln n)$ which is $\int_{-\infty}^\infty e^{-sx} \sum_{n=1}^\infty \Lambda(n) \delta(x-\ln n) dx = \sum_{n=1}^\infty \Lambda(n) n^{-s}$, which is also the Mellin transform of the weighted (but this time linearly spaced ) Dirac comb $\sum_{n=1}^\infty \Lambda(n) \delta(x-n)$ which is $\int_0^\infty x^{-s} \sum_{n=1}^\infty \Lambda(n) \delta(x-n) dx = \sum_{n=1}^\infty \Lambda(n) n^{-s}$. – reuns Mar 6 '16 at 5:16
• and a change of variable $x= e^y$ shows that the Mellin transform of $f(x)$ is the Fourier (bilateral Laplace) transform of $f(e^x)e^{-x}$ – reuns Mar 6 '16 at 5:18

## protected by LuciaAug 17 '16 at 23:53

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