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October 9th, 2019, 10:44 AM   #1
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Audio Analysis Divisor Function


The divisor function can be written as a summation of repeating pulses with a frequency. It can be represented with the functions below:

$$1) \space \sigma_{0}(x)=\sum_{\mathbb{X}=2}^{\infty} 2^{(-N)} \sum_{k=0}^{N} \binom{N}{k} e^{-i\left( \frac{\pi}{\mathbb{X}}kx \right)} $$
$$2) \space \Re(\sigma_{0}(x))=\sum_{\mathbb{X}=2}^{\infty} \cos^{N} \left( \frac{\pi}{\mathbb{X}}x \right) \cos \left( \frac{N\pi}{\mathbb{X}}x \right) $$
$$3) \space \Im(\sigma_{0}(x))=-i \sum_{\mathbb{X}=2}^{\infty}\cos^{N} \left( \frac{\pi}{\mathbb{X}}x \right) \sin \left( \frac{N\pi}{\mathbb{X}}x \right) $$

$N (\mathbb{X})$ is chosen in a way that all pulses have the same pulse width. Note that $N$ should be an even integer.

The solution has been calculated up until x=1000 and the corresponding audio signal has been created (more info in description youtube):

I like to study more on the subject. Questions:
  • How to analyse the frequency domain? Can the functions be transformed in frequency domain?
  • Is function $1)$ already a sort of frequency domain of $2)$ and $3)$?
Hoped for some input. Hopefully I don't get banned for another question. Last attempt to get some input.

Best regards,


(note the described method are excluding 1 as an divisor, solution should actually be +1)
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October 10th, 2019, 04:00 AM   #2
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I dont see it reasonable te get banned for posting advanced math.
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idontknow is offline  
October 13th, 2019, 04:01 AM   #3
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Fourier Transform Wave Divisor Function

Fourier Transform Wave Divisor Function.

The wave divisor function consists of a pulse outline modulated with a high frequency component. The real solution of the wave divisor function is:

$$\large \Re(\sigma_{0})=\sum_{\mathbb{X}=2}^{\infty}\cos^{ N} \left( \frac{\pi}{\mathbb{X}}x \right) \cos \left( \frac{N\pi}{\mathbb{X}}x \right) $$

N is determined by the pulse width of $cos^{N}$ and calculated with ($L$ pulseheight at position $\Delta x$). N should be an positive even integer to obtain positive pulses only:

$$\large N(\mathbb{X}) \approx \lim_{\mathbb{X} \rightarrow \infty} \frac{\log(L)}{\log \left( \cos \left( \frac {\pi}{\mathbb{X} } \Delta x \right) \right)} = - \frac{2 \mathbb{X}^2 \log(L)}{\pi^2 \Delta x^2} $$

The first term $cos^N$ can also be simplified, this is the pulse outline. The pulse outline forms a bell shaped distribution arround the origin for $\mathbb{X} \rightarrow \infty$:

$$\large O(x)=\lim_{\mathbb{X} \rightarrow \infty}\cos^{N} \left( \frac{\pi}{\mathbb{X}}x \right)= e^{a x^{2}}$$

$$\large a=\frac{\log(L) \space}{\Delta x^{2}}=constant$$

The high frequency component $HF(\mathbb{X})$ scales linear with $\mathbb{X}$ (see link for more information) for: $\mathbb{X} \rightarrow \infty$.

$$\large HF(\mathbb{X})= \cos \left( \frac{N\pi}{\mathbb{X}} x \right) \approx \cos (b x)$$

$$\large b(\mathbb{X}) = \frac{N}{\mathbb{X}}\pi \approx \alpha \mathbb{X} = constant \cdot \mathbb{X}$$

So for $\mathbb{X} \rightarrow \infty$ the wave divisor function becomes:

$$\large \Re(\sigma_{0})\rightarrow \sum_{\mathbb{X}=2}^{\infty}e^{a x^{2}} \cos (b x) $$

The wave divisor at infinity can be Fourier transformed in the frequency domain. The following Fourier transform definitation was used:

$$\large \hat{f}(\xi)=\int_{-\infty}^{\infty}f(x) \space e^{-2 \pi ix \xi} \space dx$$

With help of Wolfram Alpha the Fourier transform is determined (see link below). The frequency spectra of an individual divisor wave will consist of a bell shape mirrored in the y-axis.

$$\large \hat{\sigma}_{0}(\xi)= \frac{\sqrt{\pi}}{2 \sqrt{-a}} \left( e^{(b-2 \pi \xi)^{2} /4a} + e^{(b+2 \pi \xi)^{2} /4a} \right) $$

Every number will have at least one divisor wave. Because of the linearity properties of the Fourier transform we can sum the spectra to obtain the complete spectra of a number. The simulation below shows the time domain wave and the frequency spectra. Also the wave has been transposed to an audible signal.

I think I have answered my own question from the first post.

My assumption in original post is false: trigonometric and n choose k notation are not each others Fourier complement.

Best regards,


Last edited by OOOVincentOOO; October 13th, 2019 at 04:05 AM.
OOOVincentOOO is offline  
October 14th, 2019, 10:11 AM   #4
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Someone notified me on a mistake.


This would be an better notation:

$$\large N(\mathbb{X}) = \frac{\log(L)}{\log \left( \cos \left( \frac {\pi}{\mathbb{X} } \Delta x \right) \right)} \approx - \frac{2 \mathbb{X}^2 \log(L)}{\pi^2 \Delta x^2} \space (\mathbb{X} \rightarrow \infty)$$

I am non math pro, I hope the mad(th) demons leave me alone.

Previous post, are the last three formulas of the Fourier transform correct?

I only did validation in the Jupyter notebook.

Thank you,

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