# Wave Divisor Function

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#### OOOVincentOOO

Dear Math Forum,

For quite some time (years), I have been playing around with the divisor function (counting the number of divisors for a given integer). I created a 10 slide summary in the attached presentation.

Wave Divisor Function:

I have been able to describe the divisor function with the help of waves (periodic functions). This wave description (now Re and Im) introduces an error in the (Re) solution. However, this error seems proportional to the mean divisor count.

With the described method, it might (maybe) be possible to numerically refine/determine the non-leading terms of the Divisor Summation

Every time I am working on this subject, I have the feeling this is an original way to look at the divisor function (as waves). I make up the idea that the discrete math can be expressed as waves and vice versa (analogue as quantum mechanics).

My wish and hope a mathematician has a look on the attached summary of my findings.

(My skills are too limited to continue any further.)

Vince

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#### OOOVincentOOO

Hello,

My conclusion of the wave divisor function:

A method is described to express the divisor function as a summation of waves function. This wave formulation results in a real and imaginary divisor solution. The divisor count in the wave representation will have and error. The variance of the error is estimated to be proportional to the mean divisor count.

The variance of the error in the wave divisor model can be determined for unlimited pulse width settings. If the error in the divisor model is truly proportional to the mean divisor count this method might be useful refining the non leading terms of the mean divisor function.

Though, my math skills are to limited to come to an prove.

I was not happy how I justified the: n choose k is similar with the trigonometric notation. I adapted the document on small issues (concept / conclusion remains the same):

Greetings,

Vince

#### OOOVincentOOO

Hello,

I was hoping someone could give some feedback on the "Wave Divisor Function" see previous post! All home brew concept. I think is rather cool. This method/concept has many strange properties like:

- Error growth Involves Brownian motion / Random Walk.
- "Wave divisor function" limit results in divisor function.
- The error growth is very small Sqrt(ln(x)).
- At infinity the error actually stops growing.
- The pulses for primes at infinity will fluctuate rapidly between -1 and +1.
- Etc Etc Etc

I do not understand allot of math I can find on number theory. I am just naÃ¯ve and liked to puzzle on the subject. Maybe you find (QA):

a) It's Nonsence.
b) Described with poor math.
c) Do not bother me!
d) Leave fantasy boy in bubble.
e) Does not have any value.
f) I have enough other thing to do.
g) Nonsense: trying to describe discrete math with waves like...
h) This concept already exists (in a similar form).

At least you can make some video's with the Wave Divisor Function in Im-Re space . Homebrew in excel. Note that the divisor count is -1, add 1 offset to the Re solution to find conventional divisor count.

Video 1 (large scale):

Video 2 (more interesting: zoomed in on origin):

Best regards,

Vince

#### idontknow

Post a simple example of your theory, then let's compare it with the known divisor function. Last edited by a moderator:

#### OOOVincentOOO

Good question! I need some more time to think of a simple example. It is still a concept in my opinion.

One way of describing/summarizing:

I started off with:

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

Where x is the number from which the number of divisor count is determined. N should be a positive even integer to obtain positive pulses. N is defined in such a way that the pulse widths are similar for each divisor wave Xw (see presentation first posts). To keep it simple, see it as a large number. When N is infinite, we obtain pulses of height 1 at different frequencies (like: 2, 4, 6, 8, 10â€¦. or: 7, 14, 21, 28â€¦.). The summation of waves Xw is to infinity, but not required.

Basically, I came up with two possible representations of the divisor function (sigma 0). These follow naturally from cos^N. Now there is an Re and Im solution, the Re solution is:

$$\displaystyle \sigma_0(x)=\sum_{X_w=2}^{(\infty)}2^{-N}\sum_{k=0}^{N} \dbinom{N}{k}\cos \left( \frac{2{\pi}k}{X_w}x\right) =\sum_{X_w=2}^{(\infty)}\cos^N \left( \frac{\pi}{X_w}x \right) \cos \left( \frac{N \pi}{X_w}x \right)$$

This representation of the divisor function will introduce an error. The error grows like a Random walk / Brownian motion. The error is likely proportional to the mean divisor count (Dirichlet). My skills are too limited to come to a real proof (see presentation in second post on forum). Empirically, it seems to fit. No idea with non-leading terms Dirichlet (cause and effect).

Variance of Error:

$$\displaystyle \epsilon(x)_{Var}=\max(x) \space (\ln(x)+2\gamma-1)$$

Max error is determined by the pulse width (see presentation previous post).

I could not help it. But basically, the function can also be written as other sigma functions (is that called generalization? :unsure . But I have not determined the error contributions in these cases.

$$\displaystyle \sigma_i(x)=\sum_{X_w=2}^{(\infty)} X_w^i \space 2^{-N}\sum_{k=0}^{N} \dbinom{N}{k}\cos \left( \frac{2{\pi}k}{X_w}x\right) =\sum_{X_w=2}^{(\infty)} X_w^i \space \cos^N \left( \frac{\pi}{X_w}x \right) \cos \left( \frac{N \pi}{X_w}x \right)$$

In the meanwhile, I'll continue to think of a simple example. I was thinking to write out some of the series you obtain for each wave. But they grow quite rapidly in numbers. This method is not the most efficient way to determine the divisors :help:. If you love big numbers, this is a nice method!

(I am stubborn, in my definition 1 is excluded the real sigma solution is plus 1 )

Greetings,

Vince

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#### idontknow

I understand what is being computed but not much on the derivation.
Just compare the error growth of your function with the known divisor function.

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#### OOOVincentOOO

Hello,

Thank you for your reply. I had to go back on my notes from a couple of years back. Phew, I think this was the method (Euler formula and Binomial Theorem) hope I did not make too many type errors.

$$\displaystyle \cos ^N \left( a \right)= \left( \frac{e^{ia}+e^{-ia}}{2} \right)^N= \left( e^{-\ln(2)+ia}+e^{-\ln(2)-ia} \right)^N=\sum_{k=0}^{N}\dbinom{N}{k} \left( e^{(-\ln(2)+ia)(N-k)}e^{ (-\ln(2)-ia)(k)} \right)=...$$

$$\displaystyle ... \left( \frac{1}{2} \right) ^{N} \sum_{k=0}^{N}\dbinom{N}{k} e^{(ia)(N-2k)}= \left( \frac{1}{2} \right) ^{N} \space e^{iaN} \sum_{k=0}^{N}\dbinom{N}{k} e^{-i2ak}$$

The trick is that e^iaN has a supporting role only. For the divisor function we are only interested in the Re solution (a=0) so I defined: e^iaN=1. That is the method how I got my concept. The original pulses of the cos^N are now modulated with a high frequency see:

â€¦â€¦.
About the error and your comment indeed it is better to express in terms of sigma alone. The variance (0.5) of the error comes from the high frequency distribution (arc sine). For large numbers and narrow pulse width the variance of the error would approach:

$$\displaystyle \epsilon(x)_{Var}=Var(\epsilon) \space Max(\epsilon) \space \left( \sigma_{0}(x-1)+\sigma_{0}(x+1) \right)$$

$$\displaystyle \epsilon(x)_{Var}=Max(\epsilon) \space \frac{1}{2} \left( \sigma_{0}(x-1)+\sigma_{0}(x+1) \right)$$

I hope I made some thing more clear.

Gr,

Vince

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#### OOOVincentOOO

Hello,

In my posted method for the "Wave Divisor Function", I encountered several limit values. I have used Wolfram Alpha to find the solution. That was satisfying until now.

I would like to learn more about the following limit. Is there anyone who can explain this limit in simple language? I am no math pro and verified it by making a parametric plot see:

Solution that fits my needs perfectly (hope I did not make typing errors :spin: I type formula manually! (I just observed that spinning man arms move like arc sine distribution!)):

$$\displaystyle \lim_{{}X_W\rightarrow \infty } \frac {\ln(L). \ln \left( cos \left( \frac {\pi}{X_{W}} \right) \right)}{ \ln \left( \cos \left( \frac {\pi \space \delta x}{X_{W}} \right) \right)}= \frac {\ln(L)}{\delta x^{2}}$$

Wolfram's satisfying answer (slight variations in numerator alter the solution!):

https://www.wolframalpha.com/input/?i=limit+ln(L)*ln(cos(pi%2FX))%2Fln(cos(pi*Delta%2FX))+as+X-%3Einfinity

Thank you,

Vince

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#### OOOVincentOOO

Hello,

Maybe my question was unclear. Asking a good question is difficult. Starting point:

$$\displaystyle f(x)=\cos ^N \left( \frac{x}{X_w} \pi \right)$$

Where $$\displaystyle X_w$$ is an integer and determines the frequency.

N is defined by: $$\displaystyle L$$ and $$\displaystyle \Delta x$$. They determine the pulsewidth of cos^N, the pulse height $$\displaystyle f(x)=L$$ at width $$\displaystyle x=\Delta x$$.

$$\displaystyle N=\frac {\ln(L)}{\ln \left( \cos \left( \frac {\space \Delta x }{X_{W} } \pi \right) \right)}$$

One of the limits of interest is this.

$$\displaystyle g(x)= \lim_{X_W\rightarrow \infty}\cos ^N \left( \frac{x}{X_w} \pi \right) = L^{ \frac {x^2}{\Delta x^2}} =\exp \left( \frac { \ln(L) }{\Delta x^2} x^2 \right)$$

Solution:

https://www.wolframalpha.com/input/?i=lim+(cos(pi*n%2Fx))%5E(ln(L)%2Fln(cos(pi*delta%2Fx)))+for+x+to+inf

This has the similar outcome as the one posted before. The previous one was on coordinate x=1 (and the logarithm of the solution) with more intuitive notation from function analysis.

Instead of a periodic function we obtain a single pulse/distribution around the origin. The pulses from cos^N will eventually follow this limit for large Xw.

Is there someone who can explain this limit for me? How is such a solution determined? Thank you in advance, The are many more questions about the project that I have started and do not know where to begin .

Vince

(I just learned how to type a capital $$\displaystyle \delta$$ and $$\displaystyle \Delta$$. In my previous post, I used small delta. Just to add more confusion ).

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#### idontknow

About the limit: If I figure out a better solution, I will post it later.

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