A model is described where the divisor function is seen as a wave function. This divisor wave model introduces an error in the solution. Last section has some questions I am unable to answer. I cannot summarize more than written below unfortunately.

Wave divisor function: $\sigma_{0}(x)$

The integer divisor function can be described as a summation of repeating waves. Each wave filters out numbers. Divisor wave $\mathbb{X}=7$ wil filter: 7, 14, 21, 28, 35 etc. The divisor function can be described as:

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

N should be a positive even integer, only then positive pulses occur. If $N\rightarrow\infty$ discrete pulses occur on the intervals determined by: $\mathbb{X}$. This definition of the divisor function does not take 1 in account, for the conventional definition 1 should be added to the wave divisor function.

With help of Euler’s formula and the binomial theorem the function can be rewritten as:

$$ \sigma_{0}(x)=\sum_{\mathbb{X}=2}^{\infty}e^{i\left( \frac{N\pi}{\mathbb{X}}x \right)} 2^{(-N)} \sum_{k=0}^{N} \binom{N}{k} e^{-i\left( \frac{\pi}{\mathbb{X}}kx \right)} $$

The solution for the divisor function occurs when the angular component is 0 only then pulses of magnitude 1 occur. For the divisor function we can set: $e^{i\left( \frac{N\pi}{\mathbb{X}}x \right)}=1$. So the "Wave Divisor Function" becomes:

$$ \sigma_{0}(x)=\sum_{\mathbb{X}=2}^{\infty} 2^{(-N)} \sum_{k=0}^{N} \binom{N}{k} e^{-i\left( \frac{\pi}{\mathbb{X}}kx \right)} $$

The n choose k notation can be written in a trigonometric formulation.

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

$$ \Im(\sigma_{0})=-i \sum_{\mathbb{X}=2}^{\infty}\cos^{N} \left( \frac{\pi}{\mathbb{X}}x \right) \sin \left( \frac{N\pi}{\mathbb{X}}x \right) $$

This is only valid with the following criteria (found by setting above equations equal):

$$ \cos^{2} \left( \frac{N\pi}{\mathbb{X}}x \right) + \sin^{2} \left( \frac{N\pi}{\mathbb{X}}x \right)=1$$

Thus, the solution of the divisor function is only valid for integer values of $x$. The wave divisor function consists of repeating wave packages with different frequencies. The wave divisor function will have an error. It appears that the error is dependent upon the pulse width. N determines the pulse width of the individual wave packages.

enter image description here

N the pulse width definition.

The wave divisor function consists of repeating wave packages. The width of a wave package can be described as the pulse height $L$ at $\Delta x$:

$$ \cos^{N} \left( \frac{\pi}{\mathbb{X}} \Delta x \right)=L$$

So N can be calculated, we can determine $N(\mathbb{X})$ for every divisor wave. Note that N should be an even number or else negative pulses can occur. Note that the rounding to its closest even number has a randomizing effect.

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

enter image description here

For $(\mathbb{X} \rightarrow \infty)$ N can be approximated as Taylor series (see link):

$$ N(\mathbb{X}) = \frac{\log(L)}{\log \left( \cos \left( \frac {\pi}{\mathbb{X} } \Delta x \right) \right)} \approx - \frac{2 \mathbb{X}^2 \log(L)}{\pi^2 \Delta x^2} + \frac{\log(L)}{3}+ \mathcal{O} \left( \frac{1}{\mathbb{X}^2} \right)$$

Wavepulse outline.

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

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

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

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

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

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

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

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

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

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

Error of the Wave Divisor Function.

The error of the wave divisor function is majorly determined by neighbor pulses like: $\sigma(x-1)$ and $\sigma(x+1)$. The maximum error from a direct neighbor can be determined from the wave pulse outline:

$$ max(\varepsilon)=exp \left( \frac{\log(L)}{\Delta x^2} \right)$$

Error caused by $\sigma(x-m)$ and $\sigma(x+m)$ also contribute to the error. For pulses m steps away from $x$:

$$ \varepsilon(m)=exp \left( \frac{\log(L)}{\Delta x^2} m^{2} \right)$$

enter image description here

The high frequency component $HF(\mathbb{X})$ scales linear with $\mathbb{X}$ (see link for more information). This linear dependence will result in a arcsine distribution only if the divisors at e.g. $\sigma(x-1)$ and $\sigma(x+1)$ are random distributed, which is supposed true for large numbers. Note that $N(\mathbb{X})$ also is a randomizing factor while it's rounded to its closest even number.

$$ HF(\mathbb{X})= \cos \left( \frac{N\pi}{\mathbb{X}} \right) \approx \cos (constant \cdot \mathbb{X}))$$

The statistical variation of the arcsine distribution for direct neighbor pulses is determined as:

$$ Var(\mathbb{X})=\frac{1}{2} \cdot max^{2}(\varepsilon)$$

For other divisors m steps away:

$$ Var(\mathbb{X})=\frac{1}{2} \cdot \varepsilon^{2}(m)$$

The total error is summed. It appears that the error follows a random walk over an arcsine distribution. The total number of neighbor divisors determine the total variation. The total error will be the contribution of direct and neighbor pulses:

$$ Var(\sigma_{0}(x)) =\frac{1}{2} max^{2}(\varepsilon) \left( \sum_{m=1}^{\infty} \frac{\sigma_{0}(x+m) \cdot \varepsilon^{2} (m)}{max^{2}(\varepsilon)} + \sum_{m=1}^{\infty} \frac{\sigma_{0}(x-m) \cdot \varepsilon^{2} (m)}{max^{2}(\varepsilon)} \right)$$

enter image description here

Though, for narrow pulse widths it is found that the total error converges to direct neighbor divisors/pulses (see link with more information). The above relation actually takes an sort of average of the divisor count, so:

$$ Var(\sigma_{0}(x)) \approx max^{2}(\varepsilon) \cdot \overline{\sigma_{0}(x)} $$

The mean divisor growth is defined by Dirichlet. For now we do not included the error term $\mathcal{O}(x^{\Theta^{*}})$. Note that an extra $(-1)$ is added the wave divisor function is excluding divisor: 1.

$$ \overline{ D(x)} \approx \log(x) + 2 \gamma -1 -(1)$$

The error is proportional to the divisor count and the divisor count is proportional to: $\sim log(x)$ for large numbers. The standard deviation in the wave divisor function than is then proportional to:

$$ Stdev(\sigma_{0}) \approx max(\varepsilon) \cdot \sqrt{\log(x)+ 2 \gamma -2}$$

The error will grow very slowly proportional to: $\sqrt{ln(x)}$.

enter image description here


  • I assume that the error in the wave divisor function grows like a random walk. Is this correct?
  • Can for a given pulse width $L, \Delta x$ the error be estimated as a arcsine distribution?
  • Is my derivation correct that the main divisor growth and error in wave divisor function correlate?
  • When plotting the error $\varepsilon (x)$ (see Jupyter notebook) positive errors occur more often why?

More information.

Jupyter notebook:



Partial answer: Why do Positive error occur more often?

When plotting the error $\varepsilon (x) =\sigma (x)_{Wave} - \sigma(x)_{discrete}$ positive errors occur more often. The plot below shows the $\varepsilon(x)$ for 1001 pulse width settings of $L$ and $\Delta x$.

enter image description here

I found a new clue; more positive errors occur for odd $x’s$. The error for odd $x’s$ originate from the divisors of even (neigbour) numbers.

In my understanding the error of odd numbers behave not symmetrical (skewed) and divisors of even numbers symmetrical.

The error is proportional to:

$$\large \varepsilon (x) \propto \sum_{\mathbb{X}\vert (x-1)}^{} \cos(k \mathbb{X}) + \sum_{\mathbb{X}\vert (x+1)}^{} \cos(k \mathbb{X})$$

Here $\mathbb{X} \vert (x-1)$ means: $\mathbb{X}$ divides $(x-1)$. Where k is a constant and determines the pulse width of each divisor wave. Where k is a constant and determines the pulse width of each divisor wave, see above for more information.

$$\large k=-\frac{2 \log(L)}{\pi \Delta x^{2}}$$

Hopefully it's possible to answer my question. The question is rephrased and posted. Hopefully defined such that no background information is required:

Divisor Function Symmetry Neighbor Divisors

  • $\begingroup$ If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review $\endgroup$ – Matthew Daly Nov 17 '19 at 14:37
  • $\begingroup$ @MatthewDaly Is there a maximum length a post can have? If this is the case and the poster wants to add additional information, what can he do except to post this addendum as an answer? $\endgroup$ – Viktor Glombik Nov 17 '19 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.