The divisor function counts the number of divisors of an integer. A model is described where the divisor function is seen as summation of repeating continuous waves. The divisor function now has a real and imaginary component. This divisor wave model introduces an error in the solution. The wave divisor function method is presented, also a description of the error is given. 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 $\mathbb{X}=7$ will filter: 7, 14, 21, 28, 35 etc. The divisor function can be described as:

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

Here from $x$ the number of divisors is determined excluding divisor $1$. $N$ should be a positive even integer; only then positive pulses occur so $N \in 2 \mathbb{N}$. If: $N \rightarrow \infty$ discrete pulses with magnitude $1$ 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 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:

$$e^{i\left( \frac{N\pi}{\mathbb{X}}x \right)}=1$$

While $N \pi$ will always be a multiple of $2 \pi$ because $N$ must be a positive even integer. 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. A wave pulse outline is modulated with a high frequency. When N increases in size the wave packages become narrower and the frequency of the signal increases. One can select a $N$ for every value of $\mathbb{X}$ such that the pulse width for all waves becomes similar.

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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 $\Delta x$:

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

From the above equation we can calculate the magnitude of $N$. The wave package width will also vary depending upon the value of $\mathbb{X}$. Thus, $N$ is a function of $\mathbb{X}$. $N(\mathbb{X})$ can derived:

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

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

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

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: $\mathbb{X} \rightarrow \infty$.

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

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

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 $x$:

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

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In between the limits the error will occur. The exact value of the error is determined by $HF(x)$. The frequency of $HF(x)$ scales almost linear with $\mathbb{X}$. For direct neighbor divisors the error can be formulated. Where $\mathbb{X}|(x-1)$ means $\mathbb{X}$ divides $(x-1)$, $k$ is a constant determined by the pulse width.

$$ \varepsilon (x) \approx max(\varepsilon) \cdot \left[ \sum_{\mathbb{X}\vert (x-1)}^{} \cos(k \mathbb{X}) + \sum_{\mathbb{X}\vert (x+1)}^{} \cos(k \mathbb{X}) \right]$$

It is assumed that for large values $x$ its divisors are randomly distributed. Also, the rounding of $N$ to its closest even integer causes a randomizing effect. It is expected that the error is picked from an arcsine distribution. The Variance in the case of an arcsine distribution can be calculated. For neighbor pulses at $(x-1)$ and $(x+1)$ the variance is:

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

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:

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

The error description is not ideal. Errors $m$ steps away can be counted duplet, like divisor of $\mathbb{X}=2$ could be counted double. Though, when the pulse width is small $\Delta x \rightarrow 0$ the error converges. The error will be determined by direct neighbor divisors. Thus, counting duplets is not the case. This relation takes a sort of mean value of the divisor count:

$$ Var(x) \approx \frac{1}{2} \cdot max^{2}(\varepsilon) \cdot (\sigma_{0}(x+1) +\sigma_{0}(x-1))$$

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

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 standard deviation in the wave divisor function than is then proportional to:

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

Simulation of the error.

For a given pulse width $L=0.5$, $\Delta x=0.2$ the divisor count can be determined. The error in the Wave Divisor can be calculated as:

$$\varepsilon (x)=\sigma_{0}(x)_{Wave}-\sigma_{0}(x)_{Discrete}$$

The error is calculated for all integers x till the number 50000 in the presented simulation. The boundaries are determined and plotted as: $3Stdev$ $(99.7 \%)$. Several observations can be made:

  • There occur more positive errors.
  • 99.606% is counted within the boundaries while 99.7 % is expected.

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  • When plotting the error $\varepsilon (x)$ positive errors occur more often why?
  • Is the error growing as a random walk over an arcsine distribution? (are divisors of large numbers randomly distributed?)

More information and references.

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

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