# Audio Analysis Divisor Function

#### OOOVincentOOO

Hello,

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.:cold: Last attempt to get some input.

Best regards,

Vincent

(note the described method are excluding 1 as an divisor, solution should actually be +1)

1 person

#### idontknow

I dont see it reasonable te get banned for posting advanced math.

1 person

#### OOOVincentOOO

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 definition 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.

https://mybinder.org/v2/gh/oooVincentooo/Shared/master?filepath=Wave%20Divisor%20Function%20Audio.ipynb

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 other's Fourier complement.

Best regards,

Vince

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#### OOOVincentOOO

Hello,

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 :devil: leave me alone.

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

I only did validation in the Jupyter notebook.

Thank you,

Vince

#### OOOVincentOOO

Properties Frequency Spectrum (uncertainty principle).

Properties Frequency Spectrum (uncertainty principle).

The following got me puzzled the last weeks:

When the pulsewidth in the time domain gets smaller the frequency spectrum of the divisors tend to be identified more clear. From Fourier transform properties one would expect the frequency bandwidth to become wider as the time domain pulse gets narrow the: Uncertainty principle.

The spectrum of the wave divisor function seems to behave opposite to the uncertainty principle. Below the z-score of the wave divisor spectra is calculated. The z-score describes the behavior and the uncertainty principle is remained.

Time domain $f(x)$:

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

The pulsewidth in the time domain is determined by: $L$ pulseheight at position $\Delta x$. In the equations described later we will vary the pulsewidth in the time domain. Onward we set $L=0.5$ as an constant and the time domain pulsewidth is varied by reducing $\Delta x \rightarrow 0$.

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

$$\large b(\mathbb{X}) = \frac{N}{\mathbb{X}}\pi \approx - \frac{2 \space \log(L)}{\pi \space \Delta x^{2}} \mathbb{X} = constant \cdot \mathbb{X}$$

Frequency domain $\hat{f} (\xi)$:

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

The frequency pulses can be seen as normal distributions. The standard deviation of a pulse in the frequency domain is proportional to:

$$\large Stdev(\hat{\sigma}_{0}(\xi)) \propto \sqrt{-a}$$

The minimal frequency distance between two neighbor pulses is:

$$\large \Delta \xi = b(\mathbb{X}+1)-b(\mathbb{X})=b(1)$$

The z-score between two neighbor frequency pulses then is:

$$\large Z \propto \frac{b(1)}{\sqrt{-a}} \propto \frac{1}{\Delta x}$$

When the time domain pulse gets narrow $\Delta x \rightarrow 0$ the $z-score$ in the frequency domain gets bigger. Thus the individual pulses in the frequency domain become better identified. One can say that the pulsewidth in frequency domain $\sqrt{-a}$ grows more slowly then the frequency difference between two neighbor divisors $b$.

https://mybinder.org/v2/gh/oooVincentooo/Shared/master?filepath=Wave%20Divisor%20Function%20Audio.ipynb

New Question:
In the first post 3 equations where given for the wave divisor function (in time domain as to say).

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

After more thinking I still think :spin: $1)$ is a sort of frequency spectrum of $2)$ and $3)$ is this correct? Has someone seen these 3 equations in maybe another form or application?

Best regards,

Vince

#### idontknow

Try to express X$$\displaystyle =\phi(\sigma_{0}) \;$$ for $$\displaystyle \sigma_{0} (X)=1$$.
And you have the prime-function .
If you can do so, post it.
$$\displaystyle \phi ( \sigma )$$ can generate the primes.
Series Reversion -- from Wolfram MathWorld

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1 person

#### OOOVincentOOO

Try to express X$$\displaystyle =\phi(\sigma_{0}) \;$$ for $$\displaystyle \sigma_{0} (X)=1$$.
And you have the prime-function .
If you can do so, post it.
$$\displaystyle \phi ( \sigma )$$ can generate the primes.
Series Reversion -- from Wolfram MathWorld
Looks cool. Bit complicated for my knowhow. But another view and more doors to open!

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1 person

#### OOOVincentOOO

Audio Wave Divisor Function

Original video/audio did not have full spectrum of all divisors till 1000. New video does.

Video divisors with time pulse width: dx=0.15, L=0.5.

Audio from Re: wave divisor function with:

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

And:

$$\large N(\mathbb{X})= \frac{\log (L)}{\log \left( \cos \left( \frac {\pi}{\mathbb{X} } \Delta x \right)\right)} \quad N \in 2 \mathbb{N}$$

Do not know where audible patterns come from. I think once brain tries to find structure.

There is an audible repeating pattern: low to high -> high to low sort of chromatic.

I would like to determine where that pattern comes from any suggestions how to identify source?

Gr,

Vincent

1 person

#### OOOVincentOOO

"Wave Divisor Function"

My topic the: "Wave Divisor Function" reaches it's end.

All I am capable of is done for now. The mostly virtual/imaginary feedback on the forum helped me to focus my thoughts.

I posted a summary on stacksexchange hoping someone will check my hobby homework and questions there or here. :unsure:

https://math.stackexchange.com/q/3427431/650339

Best regards and thank you for your patience with another fool who thinks different,

Vincent Preemen

1 person

#### ISP

Have you tried the Physics Forums or StackExchange? edit: I just read your above post. Hope this helps anyways.

https://www.physicsforums.com/

https://math.stackexchange.com/

I usually start my threads here because I go through a lot of drafts and typos (my eyes ain't what they used to be and it's legal for me to smoke weed, so I'm usually sipping a cocktail and high as hell when posting here...).

I then go on to Physics forums because they are lot smarter there than they are here.

I finally go on to StackExchanges were the big boys play.

All of the work I have done recently has made it through the Physics forums and SE, but I still don't have an answer, so there may not be one for my question(s). Maybe there is for you though. Good luck!