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October 9th, 2019, 10:44 AM  #1 
Member Joined: Dec 2014 From: Netherlands Posts: 34 Thanks: 5 Math Focus: hobby  Audio Analysis Divisor Function
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): https://www.youtube.com/watch?v=5oWcu_Qjdn0 I like to study more on the subject. Questions:
Best regards, Vincent (note the described method are excluding 1 as an divisor, solution should actually be +1) 
October 10th, 2019, 04:00 AM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 
I dont see it reasonable te get banned for posting advanced math. 
October 13th, 2019, 04:01 AM  #3 
Member Joined: Dec 2014 From: Netherlands Posts: 34 Thanks: 5 Math Focus: hobby  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 yaxis. $$\large \hat{\sigma}_{0}(\xi)= \frac{\sqrt{\pi}}{2 \sqrt{a}} \left( e^{(b2 \pi \xi)^{2} /4a} + e^{(b+2 \pi \xi)^{2} /4a} \right) $$ 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/oooVincen...%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 others Fourier complement. Best regards, Vince Last edited by OOOVincentOOO; October 13th, 2019 at 04:05 AM. 
October 14th, 2019, 10:11 AM  #4 
Member Joined: Dec 2014 From: Netherlands Posts: 34 Thanks: 5 Math Focus: hobby 
Hello, Someone notified me on a mistake. Mistake.jpg 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. Question: Previous post, are the last three formulas of the Fourier transform correct? I only did validation in the Jupyter notebook. Thank you, Vince 

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analysis, audio, divisor, function, wave 
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