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.
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:
$$ N(\mathbb{X})= \frac{\log (L)}{\log \left( \cos \left( \frac {\pi}{\mathbb{X} } \Delta x \right)\right)} \quad N \in 2 \mathbb{N} $$
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 $x$:
$$ \varepsilon(m)=exp \left( \frac{\log(L)}{\Delta x^2} m^{2} \right)$$
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:
$$ Var(x) =\frac{1}{2} max^{2}(\varepsilon) \left( \sum_{m=1}^{\infty} \frac{\sigma_{0}(x+m) \cdot \varepsilon^{2} (m)}{max^{2}(\varepsilon)} + \sum_{m=1}^{\infty} \frac{\sigma_{0}(x-m) \cdot \varepsilon^{2} (m)}{max^{2}(\varepsilon)} \right)$$
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.
Questions.
- 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:
pdf:
https://drive.google.com/open?id=1Etu4vOfjsnbaysk_UR6HIA9R7EDybH-n
