# Is this Fourier like transform equal to the Riemann zeta function?

This question builds upon the answer to this question. This new question has only minor changes compared to the previous question, but the scale factor of the output from the Fourier like transform is different. I hope this is enough to ask a new question.

The question is if:

$$\Re[\zeta(1/2+It)] = +a\cdot\Re\left[\frac{\left(\frac{1}{2}+i t\right)}{\left\lceil \frac{\log (\text{N})}{\text{xres}}\right\rceil } \sum _{n=1}^{\left\lceil \frac{\log (\text{N})}{\text{xres}}\right\rceil } \left(e^{x_n}\right)^{-\frac{1}{2}+i t} \left(\text{SawtoothWave}\left[e^{x_n}\right]-1\right)\right]$$

$$\Im[\zeta(1/2+It)] = -a\cdot\Im\left[\frac{\left(\frac{1}{2}+i t\right)}{\left\lceil \frac{\log (\text{N})}{\text{xres}}\right\rceil } \sum _{n=1}^{\left\lceil \frac{\log (\text{N})}{\text{xres}}\right\rceil } \left(e^{x_n}\right)^{-\frac{1}{2}+i t} \left(\text{SawtoothWave}\left[e^{x_n}\right]+0\right)\right]$$

is equal to the Riemann zeta function on the critical line, where $a$ is some scale factor?

Edit 6.10.2012: I conjecture the scale factor $a$ to be equal to the N-th harmonic number.

Plot 1: Real and imaginary parts of the discrete Fourier like transform of exponential sawtooth waves, times some number $a$ as a scale factor.

Plot2: Real and imaginary parts of the Riemann zeta function on the critical line.

Plot3: The two plots above superimposed.

The Mathematica program for the plots is:

NN = 1000000;
xres = .002;
x = Exp[Range[0, Log[NN], xres]];
tmax = 60;
tres = .060;
Monitor[errList1 =
Table[Total[((x^(-1/2 + I t)*(SawtoothWave[x] - 1)))]*(1/2 +
I t), {t, Range[0, tmax, tres]}];, t]
Monitor[errList2 =
Table[Total[((x^(-1/2 + I t)*(SawtoothWave[x] + 0)))]*(1/2 +
I t), {t, Range[0, tmax, tres]}];, t]
g1 = ListLinePlot[
14.134725141734693790*{Re[errList1]/Length[x], -Im[errList2]/Length[x]},
DataRange -> {0, tmax}, Axes -> True, PlotRange -> {-2.6, 4},
ImageSize -> Full, Filling -> Axis];
g2 = Plot[Re[Zeta[1/2 + I t]], {t, 0, tmax}, PlotRange -> {-2.6, 4},
ImageSize -> Full];
g3 = Plot[Im[Zeta[1/2 + I t]], {t, 0, tmax}, PlotRange -> {-2.6, 4},
ImageSize -> Full];
Print["Real and imaginary parts of the discrete Fourier transform \
like plot of exponential sawtooth waves, times some number as a scale \
factor"]
Show[g1]
Print["Real and imaginary parts of the Riemann zeta function on the \
critical line"]
Show[g2, g3]
Print["The two plots above superimposed"]
Show[g1, g2, g3]


In the program the scale factor happens to be close to the first Riemann zeta zero, but this is not true since changing the parameter "NN" requires a different scale factor. I apologize if the plots are a bit oversized.

Edit 19.1.2013:

It appears that it is also possible to calculate the Zeta function with the Fast Fourier Transform as in this Mathematica program:

scale = 1000000;
xres = .00001;
x = Exp[Range[0, Log[scale], xres]];
RealPart = -Log[x]*FourierDST[(SawtoothWave[x] - 1)*x^(-1/2)];
ImaginaryPart = -Log[x]*FourierDCT[(SawtoothWave[x] + 0)*x^(-1/2)];
datapointsdisplayed = 300;
ymin = -0.012;
ymax = 0.018;
g1 = ListLinePlot[{RealPart[[1 ;; datapointsdisplayed]],
ImaginaryPart[[1 ;; datapointsdisplayed]]},
PlotRange -> {ymin, ymax}, DataRange -> {0, 68.00226987379779},
Filling -> Axis];
Show[Flatten[{g1,
Table[Graphics[{PointSize[0.013],
Point[{N[Im[ZetaZero[n]]], 0}]}], {n, 1, 16}]}],
ImageSize -> Large]


As latex this would be written something like:

$$\Re\left(\zeta \left(\frac{1}{2}+i x\right)\right)=-\log (\exp(x)) \text{FourierDST}\left[\frac{\text{SawtoothWave}[\exp(x)]-1}{\sqrt{\exp(x)}}\right]$$

$$\Im\left(\zeta \left(\frac{1}{2}+i x\right)\right)=-\log (\exp(x)) \text{FourierDCT}\left[\frac{\text{SawtoothWave}[\exp(x)]+0}{\sqrt{\exp(x)}}\right]$$

Edit 8.8.2013:

Plots of the sawtooth waves for Real and Imaginary parts:

Also, changed the +1 in the formulas to +0 after some exerimenting and consideration.

Plots of differences of the sawtooth waves for Real and Imaginary parts:

This lower plot was generated with the Mathematica code below:

scale = 1000000;
xres = .00001;
x = Exp[Range[0, Log[scale], xres]];
RealPart = (SawtoothWave[x] - 1)*
x^(-1/2); ImaginaryPart = (SawtoothWave[x] + 0)*x^(-1/2);
SumOfPartsReMinusIm = RealPart - ImaginaryPart;
SumOfPartsImMinusRe = -RealPart + ImaginaryPart;
datapointsdisplayed = 300000;
ymin = -1.5;
ymax = 1.5; g1 =
ListLinePlot[{RealPart[[1 ;; datapointsdisplayed]],
ImaginaryPart[[1 ;; datapointsdisplayed]],
SumOfPartsReMinusIm[[1 ;; datapointsdisplayed]],
SumOfPartsImMinusRe[[1 ;; datapointsdisplayed]]},
PlotRange -> {ymin, ymax},
DataRange -> {0, xres*datapointsdisplayed}, Filling -> Axis];
Show[g1, ImageSize -> Large]


https://groups.yahoo.com/neo/groups/harmonicanalysis/conversations/messages/230

Terence Tao: 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"

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@daniel I don't know. Are you saying that the expression in this question is the same as in the previous question? I thought the plus and minus 1 after the sawtooth function would change this. I frequently make mistakes, so bear with me. –  Mats Granvik Sep 27 '12 at 15:46
The expressions are different but are they different in a way that means anon's answer to your first question does not apply? –  daniel Sep 27 '12 at 17:51
@daniel I don't know. Here is the algebra I now tried to do: –  Mats Granvik Sep 27 '12 at 18:08
$$S(\omega)=\sqrt{\frac{2}{\pi}}\mathrm{Im}\left(\frac{\zeta(s)-1}{s}\right). \tag{R}$$ $$\frac{S(\omega)}{\sqrt{\frac{2}{\pi}}}=\mathrm{Im}\left(\frac{\zeta(s)-1}{s}\r‌​ight). \tag{R}$$ $$\frac{s\cdot S(\omega)}{\sqrt{\frac{2}{\pi}}}=\zeta(s)-1. \tag{R}$$ $$\frac{s\cdot S(\omega)}{\sqrt{\frac{2}{\pi}}}+1=\zeta(s). \tag{R}$$ $$\zeta(s)=\frac{s\cdot S(\omega)}{\sqrt{\frac{2}{\pi}}}+1. \tag{R}$$ where $$S(\omega)$$ is the Fourier sine transform of the exponential sawtooth function. There is the plus 1 and times the complex variable s. Is that then the same as in the current question? –  Mats Granvik Sep 27 '12 at 18:09
In A, B, and C of the answer to the previous question anon shows that the F transform you use is in fact related to the zeta function. anon refers to (6) and (5) and I don't see those anywhere... Your algebra looks correct. I'm not sure it's a separate question is all I am saying. At this point you have 4 upvotes so it's sort of moot. It's a nice question. –  daniel Sep 27 '12 at 18:48
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