What is the Visual Prime Pattern Based on Trig and Harmonics?

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That's awesome. I wish I could have thought of that. I wonder what kinds of applications this could be used in.
 
galoisjr said:
That's awesome. I wish I could have thought of that. I wonder what kinds of applications this could be used in.

Thank you so much. Its been hard geeting any feedback from this on here. I'm 100% self taught so its hard to get points across when you're not formally trianed. I have the equations behind the visuals but I think there is a way to use collision detection to efficiently determin when concentric circles intersect with evenly spaced parallel verticle lines from which you can decifer every square root, with prime square roots only occurring on the first parabola. Do you see what I'm talking about?
 
wow, amazing article
 
Hi, Jeremy,
I just have one question. Suppose you replace all parabolas by straight lines. That is, no sqrt; the first parabola becomes a line with slope 1 (y=x), and the other parabolas would be replaced by lines with slopes 2,3,4,... (the lines y=2x, y=3x, y=4x, ...). As you draw horizontal lines passing through the marks on your first line (the one with slope 1), would that horizontal still intersect none of the marks on other lines only at prime numbers of the first line?
 
Dodo said:
Hi, Jeremy,
I just have one question. Suppose you replace all parabolas by straight lines. That is, no sqrt; the first parabola becomes a line with slope 1 (y=x), and the other parabolas would be replaced by lines with slopes 2,3,4,... (the lines y=2x, y=3x, y=4x, ...). As you draw horizontal lines passing through the marks on your first line (the one with slope 1), would that horizontal still intersect none of the marks on other lines only at prime numbers of the first line?
Dodo,
Yes, all primes P would only intersect on y=1x and y=Px with composites intersecting on their divisors but you loose your relation to the Fourier series and the unit circle which I think are very important.
 
I also find it interesting that the first parabola has a vertex of 1/2.
 
Right; your parabolas do not pass through the origin, instead they have been shifted so that the parabola representing the multiples of n passes through the point in the first parabola that represents the integer n. (This way, the horizontal lines will only intersect true multiples of n, clearing up other instances of n itself.)

A similar thing can be done by shifting the lines I mentioned before; the line with slope n would pass not through the origin, but through the point (n,n) on the first line. Attached is a drawing.

In fact, graphs of any monotonic curve (x^2, x^3, exp x, ln x, ...) would also produce the primes in the same manner (namely, in the manner of http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes" ).

Edit: My bad, x^2 is not, overall, monotonic. I was referring to curves that are increasingly monotonic on the first quadrant; that is, for x>0, whenever y>x you have f(y)>f(x), so that the vertical ordering of the points is preserved.
 

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  • #10
it seems to me that preservation of order would be intrinsic to any effective sieve. Correct me if I’m wrong but, I don’t think the function I’m using to generate my sieve is necessarily monotonic, although the results can be viewed that way.

where d={1,2,…,x} , z=1-2d/x, n=x/x-2d and y= sqrt((x-d)*d)

tan(acos(z))*n = y (concentric circles intersection with vertical lines)

and factors of y when d=1 where q={1,2,…,y}

y-q^2/2q = 0 mod (1/2) (horizontal intersection of y with vertical lines)
 
  • #11
This seems like the right place to post this question...

I have been extremely curious about the square roots of prime numbers ever since I had a dream that seemed to indicate there was some sort of characteristics of the resulting irrational numbers. This may not make any sense (as it was a dream, but try to follow what I'm asking), but there was a feeling that the square roots of smaller prime numbers exhibited more "chaotic" behavior in their decimal expansion than larger primes.

If that made no sense at all, I'm simply trying to find some research into the properties of the square roots of prime numbers. I can't seem to find anything on the internet, but if anyone knows of a paper or a link etc I'd appreciate it.
 
  • #12
Well, there is a sqrt(). Do an experiment: change all your sqrt() to log() in your Flash code, just like that, and then tell me if anything significant has changed. Even better: change all the calls to sqrt() to some function defined by you, thefun(); there you can play with returning sqrt(), log(), or whatever.

I've been skimming through your code, and I'm wondering where are you introducing the tan(acos(z)) part, because I can't find it.

srfriggen: you may want to start a new thread with your question. Personally I don't have an answer, but someone else may.
 
  • #13
Dodo said:
Well, there is a sqrt(). Do an experiment: change all your sqrt() to log() in your Flash code, just like that, and then tell me if anything significant has changed. Even better: change all the calls to sqrt() to some function defined by you, thefun(); there you can play with returning sqrt(), log(), or whatever.

I've been skimming through your code, and I'm wondering where are you introducing the tan(acos(z)) part, because I can't find it.

srfriggen: you may want to start a new thread with your question. Personally I don't have an answer, but someone else may.

Dodo,
I understand the point you are making but the sqrt() is essential in my equation because it perfectly defines the Moiré pattern created by concentric circles and parallell lines. All other functions will miss the intersections of this pattern. My inquiry into this pattern came from an article I read here:
http://www.egge.net/~savory/maths9.htm
harmonics:
http://en.wikipedia.org/wiki/File:Moodswingerscale.svg
the unit circle:
http://upload.wikimedia.org/wikiped...r.svg/1000px-Unit_circle_angles_color.svg.png
and the inverse square law:
http://www.splung.com/cosmology/images/magnitude/inversesquare.jpg
 
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  • #14
Dodo said:
I've been skimming through your code, and I'm wondering where are you introducing the tan(acos(z)) part, because I can't find it.

You won't find the tan(acos(z)) part in my code but I mimic its output.
 
  • #15
Well, where I was heading to, is that primes are produced because of the sieving process, which in turn comes from the vertical order of the points; and this is not really related to the intersection with the circles.

Leaving the primes apart, you seem interested in the coincidence of the paraboles and the circles, precisely at the lines projected out of the unit circle. I wrote some notes in a PDF that may help with the trigonometry of the situation, and with the reason why the intersections occur precisely at roots of consecutive integers, if that's what you're ultimately asking. The notes also show why that tan(acos(...)) formula is not really right.
 

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  • #16
Dodo,
Thank you so much for your notes. They made perfect sense to me. I see now that sin(angle) keeps my secant line inside the unit circle with a height of Py = d+1/2 * sin(angle) and my tan(angle) is outside the unit circle with a height of Py = d+1/d+1-2 * tan(angle).


I also see your point about the vertical order of the points. In fact I have a excel spreadsheet with this exact table on it from when I started down this path years ago.

01 02 03 04 05 06 07 08 09 10 11
02[04]06 08 10 12 14 16 18 20
03 06[09]12 15 18 21 24 27
04 08 12[16]20 24 28 32
05 10 15 20[25]30 35
06 12 18 24 30[36]
07 14 21 28 35
08 16 24 32
09 18 27
10 20
11

This ordering is key because it shows the congruence of squares exposing Fermat’s factorization method which is the basis for the quadratic sieve and the general number field sieve. For example look at 36:

36 – 1^2 = 35
36 – 2^2 = 32
36 – 3^2 = 27
36 – 4^2 = 20
36 – 5^2 = 11

I find it more than a coincidence that the simple pattern of parallel lines intersecting with concentric circles produces this ordering exactly showing that primes only have a congruence of square( (P-1)/2)^2 to square ((P+1)/2)^2.

As to your comment that “the sinusoid is a pretty artifact used ONLY to split the diameter on the unit circle”, I have to disagree. Fundamental frequency division produces harmonics. The sinusoid shown is the harmonic produced by dividing the unit circle or fundamental frequency. The intersection of these divisions on the unit circle directly mark the deformation points of the fundamental frequency’s sinusoid when you “mix” the two frequencies (fundamental + harmonic), hence the my comment on the link to the Fourier series and harmonic analysis.
 
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  • #18
  • #19
JeremyEbert said:
The intersection of these divisions on the unit circle directly mark the deformation points of the fundamental frequency’s sinusoid when you “mix” the two frequencies (fundamental + harmonic), hence the my comment on the link to the Fourier series and harmonic analysis.
To be clear, I'm talking about the orthogonal projection onto the time axis as regards sin with the "directly mark" part here: "intersection of these divisions on the unit circle directly mark the deformation points of the fundamental frequency’s sinusoid"
 
  • #20
Sorry, Jeremy, but I really don't understand what do you mean. Which is the "time axis" for you, the horizontal axis? What are "deformation points"? If you mean the intersection of the sinusoid with the horizontal diameter of the unit circle, anything I can see is that the diameter is being split in equal parts; I fail to understand where do you see a Fourier series, given that no sinusoids are being added together, or when, for the only sinusoid in sight, the amplitude seems to play no role at all. Is there a calculation involving the sinusoid in one iteration and the sinusoid in the next iteration, and if so, precisely what calculation?
 
  • #21
Dodo is basically asking to see your equations if you have any. Then everyone can see for themselves their form and what they do.
 
  • #22
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  • #24
Hi, Jeremy,
surely you realize that, in those sites that you cite, sinusoids are being added together. A formula that looks something like this is used,
f(x) = a1 sin(x) + a2 sin(2x) + a3 sin(3x) + ...
where the a1,a2,a3 are the amplitudes (the ones controlled by different slides on those pages).
This is what I fail to see in your drawing, where the sinusoid just stands alone in the middle of the unit circle, and that is why I made the remark about it being used only to split a segment in equal parts.
 
  • #25
Dodo said:
Hi, Jeremy,
surely you realize that, in those sites that you cite, sinusoids are being added together. A formula that looks something like this is used,
f(x) = a1 sin(x) + a2 sin(2x) + a3 sin(3x) + ...
where the a1,a2,a3 are the amplitudes (the ones controlled by different slides on those pages).
This is what I fail to see in your drawing, where the sinusoid just stands alone in the middle of the unit circle, and that is why I made the remark about it being used only to split a segment in equal parts.
Oh yes, I definitely understand that and I know my animation does not show the mixing of the sinusoids, it just shows one at a time. What I intend to show is how FFT can be used to identify prime harmonics. A prime number harmonic will only have energy at its frequency and its fundamental (1) across the spectrum, whereas a composite number harmonic will have energy at all its factors across the spectrum. ex: a 1/4 or 4th harmonic of a fundamental frequency will have energy in the 1/2 or 2nd harmonic. Make any sense?
 
  • #26
Now, that makes more sense. Go ahead.
 
  • #27
Dodo said:
Now, that makes more sense. Go ahead.
Will do. I'm working on the translation to complex exponentials. Its something new for me.
 
  • #28
So let me see if I understand the Fourier series in complex notation using Euler's formula e^iwt. The imaginary part is the sinusoids frequency and polarization. A positive imaginary part depicting a left-hand (counter-clockwise) polarization with a negative value depicting a right-hand (clock-wise) polarization. The real part of the complex exponent depicts a change in amplitude over t (the period), a positive value for growth and a negative value for decay. Is this correct?
 
  • #29
interesting side note.

golden ratio

phi = 1.6180339887 = sqrt(5)/2 + 1/2

1/4 + sqrt(5)/2 + 1/4 = Phi

sqrt(5)/2 + 1/4 = 1.3680339887498948482045868343656

http://www.forexlive.com/165384/all/not-sure-whats-up-there-at-1-3680-but-it-must-be-big
 
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  • #30
golden ratio

phi/2 = 0.80901699437494742410229341718282

acos(0.80901699437494742410229341718282) = 36
 
  • #31
JeremyEbert said:
So let me see if I understand the Fourier series in complex notation using Euler's formula e^iwt. The imaginary part is the sinusoids frequency and polarization. A positive imaginary part depicting a left-hand (counter-clockwise) polarization with a negative value depicting a right-hand (clock-wise) polarization. The real part of the complex exponent depicts a change in amplitude over t (the period), a positive value for growth and a negative value for decay. Is this correct?

And the sinusoid is both the sine and cosine added together equaling e^iwt, so at any point in 4D space (3D + time) on the complex plane, you have frequency, polarization, amplitude and phase encoded into the complex exponent. savvy?
 
  • #32
Hi, Jeremy,
here I am abstaining myself, because you're asking for the physical meaning of quantities - maybe these questions would be more appropriate on the Physics / Classical Physics forum, or maybe someone else can add something.

What I can say is that e^iwt represents a point on the unit circle (radius=1) in the complex plane. The parameter "w" can be thought of as "angular frequency" (radians per unit of t). The whole expression has a complex value, with a real part = cos(wt) and an imaginary part = sin(wt). Thus, when this complex number (a point in the plane moving along a circle) is projected on the real axis, the projected point on the axis oscillates, and the same for the projection on the imaginary axis. All this doesn't go further than the 2-D plane.

The expression does not speak of "amplitude" either, since the point doesn't move away from the unit circle (and, in any case, you'd need to refer to the amplitude of the sinusoidal movement of the projections on the axes; the complex point itself does not describe a sinusoid, it just goes in a circle as t varies). For adding something like amplitude, you'd need to scale the equation, for example by multiplying it by a constant, as in c.e^iwt; here "c" is now the new circle radius, and the projections on the axes are now sinusoids with a different amplitude than before. Or this "c" could be a function of t, instead of a just a constant.

I'm aware that you have seen images of an helix (the same circle moving forward in a third dimension). but this equation alone does not express that. Again, I suspect a forum dealing with electromagnetism may have better answers; I wouldn't know how "polarization" fits in here, for example.
 
  • #33
Dodo said:
Hi, Jeremy,
here I am abstaining myself, because you're asking for the physical meaning of quantities - maybe these questions would be more appropriate on the Physics / Classical Physics forum, or maybe someone else can add something.

What I can say is that e^iwt represents a point on the unit circle (radius=1) in the complex plane. The parameter "w" can be thought of as "angular frequency" (radians per unit of t). The whole expression has a complex value, with a real part = cos(wt) and an imaginary part = sin(wt). Thus, when this complex number (a point in the plane moving along a circle) is projected on the real axis, the projected point on the axis oscillates, and the same for the projection on the imaginary axis. All this doesn't go further than the 2-D plane.

The expression does not speak of "amplitude" either, since the point doesn't move away from the unit circle (and, in any case, you'd need to refer to the amplitude of the sinusoidal movement of the projections on the axes; the complex point itself does not describe a sinusoid, it just goes in a circle as t varies). For adding something like amplitude, you'd need to scale the equation, for example by multiplying it by a constant, as in c.e^iwt; here "c" is now the new circle radius, and the projections on the axes are now sinusoids with a different amplitude than before. Or this "c" could be a function of t, instead of a just a constant.

I'm aware that you have seen images of an helix (the same circle moving forward in a third dimension). but this equation alone does not express that. Again, I suspect a forum dealing with electromagnetism may have better answers; I wouldn't know how "polarization" fits in here, for example.

it seems like there is 2 parts to e, the imaginary and real. Re e^0 = 1 = Amplitude , I am e^iwt = Rotation
This is what I've been using as a reference:
http://demonstrations.wolfram.com/TheComplexExponential/
It seems like it expresses all of the things I mentioned. What else is needed?
 
  • #34
--- I'm sorry, I am overcomplicating things. Give me a moment to rewrite this.

Ok. Now that I see the demo you link to, the equation used looks like z = e^bt, where both "z" and "b" are complex numbers. (The demo uses the greek letter "lambda", I'm using the letter "b" to save me some exotic typing.)

Suppose that this complex number "b" has real part = "a" and imaginary part = "w", so that b = a + i.w ; now the equation looks like

z = e^bt = e^(a+iw)t = (e^a) . (e^iwt)

The first factor, e^a, is a constant, representing the amplitude (of the sinusoids projected upon the axes); or the radius of the circle, in the complex plane. The second factor, e^iwt, is as described in my previous post, when I took the simplifying idea of thinking "w" to be just a real number.

- - - - - - -

Another thing that may help is this (though probably it is obvious to you already): the equation represents either motion on the 2-D plane (that is, two dimensions x,y plus time), or just a point in 3-D (the three dimensions represented by the three variables x,y,t). If you make a graph of y versus x, you'll see a circle; if, instead, you graph x versus t, or graph y versus t, then you see a sinusoid.
 
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  • #35
Dodo said:
--- I'm sorry, I am overcomplicating things. Give me a moment to rewrite this.

Ok. Now that I see the demo you link to, the equation used looks like z = e^bt, where both "z" and "b" are complex numbers. (The demo uses the greek letter "lambda", I'm using the letter "b" to save me some exotic typing.)

Suppose that this complex number "b" has real part = "a" and imaginary part = "w", so that b = a + i.w ; now the equation looks like

z = e^bt = e^(a+iw)t = (e^a) . (e^iwt)

The first factor, e^a, is a constant, representing the amplitude (of the sinusoids projected upon the axes); or the radius of the circle, in the complex plane. The second factor, e^iwt, is as described in my previous post, when I took the simplifying idea of thinking "w" to be just a real number.

- - - - - - -

Another thing that may help is this (though probably it is obvious to you already): the equation represents either motion on the 2-D plane (that is, two dimensions x,y plus time), or just a point in 3-D (the three dimensions represented by the three variables x,y,t). If you make a graph of y versus x, you'll see a circle; if, instead, you graph x versus t, or graph y versus t, then you see a sinusoid.

yes perfect! that helps! I'm finding e so much in my equation. the ((n-1)/(n+1))^(-n/2) ~ e part has really got me preoccupied
 
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  • #36
JeremyEbert said:
yes perfect! that helps! I'm finding e so much in my equation. the ((n-1)/(n+1))^(-n/2) ~ e part has really got me preoccupied
Just for clarification the (n-1)/(n+1) part is the x = 1-(2/d+1) part of the equation.
so its also 1-(2/d+1)^(-d/2)~e
and the rest would be
sin(acos(1-(2/d+1)) * (d+1)/2 = sqrt(d)
 
  • #37
JeremyEbert said:
Just for clarification the (n-1)/(n+1) part is the x = 1-(2/d+1) part of the equation.
so its also 1-(2/d+1)^(-d/2)~e
and the rest would be
sin(acos(1-(2/d+1)) * (d+1)/2 = sqrt(d)

So I guess that means:

z = e^bt = e^(a+iw)t = (e^a) . (e^iwt) ~ (e^a) . (e^i(-1/d/2))

Or basicaly wt~(-1/d/2) right ?

Would my vertical and horizontal lines then be Gaussian integers?
 
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  • #38
also I think this means in Euler's formula e^ix = cos(x) + i sin(x)
in my equation:
x=acos(1-(2/d+1))

where 1/d = the harmonic series.
 
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  • #39
Honest, I can't make head or tails from this.

Defining the variables you're using would help a lot.

"d", last time I understood something, was an integer, going 1, 2, 3, ...
"x" .. I'm no longer sure.
Who is n, and how (n-1)/(n+1) = 1-(2/d-1) ?

And I don't really know what is this symbol ~ , or what are you trying to say with it. But begin by defining the variables that you're using. "x" is the distance from here to here. Simple things like that.

---

Here is something else that may also help you being understood. I'm noticing that you have troubles using parenthesis in formulas. Here is some advice.

If you see a formula like
\frac {a+b} {c+d}
it is incorrect to write it as a+b/c+d, because people will understand
a+\frac b c + d
You have programmed computers, so you know that the division symbol has more "precedence" than the sum. Keeping these things in mind will help everyone else understand.
 
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  • #40
Ha! Sorry, I got a little carried away. I'll start over. To be clear I'm using in radians not degrees.

n = 1,2,3,..., infinity

(n-1)/(n+1) = 1-(2/(n-1))

x = 1-(2/(n-1))

sqrt(n) = sin(acos(x)) / 2 * (n+1)

x^(-n/2) quickly converges to the constant e

t=acos(x)

Eulers formula e^it=cos(t) + i sin(t)

cos(t) = x = (n-1)/(n+1) = 1-(2/(n-1))

edit------e^(-1/(n/2)) quickly converges to x

edit-------

also

d=1,2,3,...(n-1)

redfine x as

x = 1-((2/(n-1)*d))

if n is prime then d can equal 1 and (n-1)
 
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  • #41
JeremyEbert said:
(n-1)/(n+1) = 1-(2/(n-1))

x = 1-(2/(n-1))
I believe you mean 1-2/(n+1), with a "+" sign.

JeremyEbert said:
sqrt(n) = sin(acos(x)) / 2 * (n+1)
Yes, this one was established in post#15. Assuming you correct "x" as above.

JeremyEbert said:
x^(-n/2) quickly converges to the constant e
Now, you have to be careful when trying things out on the computer. This may converge to e, or to something close to e but not e. At this moment I'm not entirely sure, but go on.

JeremyEbert said:
t=acos(x)

Eulers formula e^it=cos(t) + i sin(t)

cos(t) = x = (n-1)/(n+1) = 1-(2/(n-1))
Ok, as long as 1-2/(n-1) is changed to 1-2/(n+1).
But now you should go somewhere with these definitions; personally, I don't see where you're going to, or why the introduction of a complex variable is needed.

JeremyEbert said:
d=1,2,3,...(n-1)

redfine x as

x = 1-((2/(n-1)*d))
So be it, but again I don't see where you're heading to.

JeremyEbert said:
if n is prime then d can equal 1 and (n-1)
And now I'm lost. In which way n begin prime is a restriction to "d"? You said d was another free variable going 1,2,3,...
 
  • #42
Sorry for the delay. I’ve been working on the next piece to this and got distracted by an interesting vector of this equation.

Sorry to jump around here but I think this is where I need some more help explaining the big picture.

I noticed that the parabolas created by this pattern, follow this equation per quadrant on the x,y grid.

Where n = 1,2,3,…infinity
Quadrant 1: y = sqrt((2xn)+n^2) = the square root of integer multiples of n

Question, if I use complex numbers, x + iy will I get both quadrants of the parabola? (1&4 for positive n and 2&3 for negative n)

The interesting vector I noticed is at 45 degrees or pi/4 radians.
The sqrt((2xn)+n^2) parabolas intersect that vector at sqrt(((1+sqrt(2))*n)^2).
I have attached an image demonstrating this.

The thing I find most interesting about these intersections is this:

q=((1+sqrt(2))*n)^2
u=((n^2)*6) – q
(q*u)^(1/4) = n

Also interesting side note:

q=((1+sqrt(2))*n)^2
u=((n^2)*12) – 2q
(2q*u)^(1/2) = 2(n^2) = maximum number of electrons an atom's nth electron shell can accommodate
 

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  • #43
I noticed something else this weekend while working with the complex exponentials demonstration here:
http://demonstrations.wolfram.com/TheComplexExponential/

if you set the equation to e^(1+3.1415 i)t you get this:

http://i98.photobucket.com/albums/l267/alienearcandy/e.png

which looks a lot like the golden ratio here:

http://i98.photobucket.com/albums/l267/alienearcandy/phi.png

in fact if you overlay them you can see they are very close:

http://i98.photobucket.com/albums/l267/alienearcandy/e-phi.png

Is there a known relationship between Phi, e and pi?
 
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  • #44
Dodo, sorry for the late response, I've been a little distracted with matricies.

Dodo said:
I believe you mean 1-2/(n+1), with a "+" sign.
Over zealous typo. ha! 1-2/(n+1) is what I meant.
Dodo said:
Now, you have to be careful when trying things out on the computer. This may converge to e, or to something close to e but not e. At this moment I'm not entirely sure, but go on.
Youre right. It is just an assumption bassed on a small subset of data.
Dodo said:
But now you should go somewhere with these definitions; personally, I don't see where you're going to, or why the introduction of a complex variable is needed.
I'm not entirely sure yet either, It’s just an idea or theory that I'm trying to prove. I can visualize it; I'm just trying to depict it mathematically. I really appreciate your feedback. It helps greatly.
Dodo said:
And now I'm lost. In which way n begin prime is a restriction to "d"? You said d was another free variable going 1,2,3,...
I'm going to come at this from a different perspective... give me a few.
 
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  • #45
JeremyEbert said:
Is there a known relationship between Phi, e and pi?

5 arccos (Golden Ratio/2) = pi
2 cos (pi/5) = Golden Ratio

Or, alternatively, to better express the symmetrical relationship in terms that requires nought but a single sign change and an exchange of variables...

(7+3)/2 cos^-1 (phi/((7-3)/2)) = pi
(7- 3)/2 cos^1 (pi/((7+3)/2)) = phi

Ceiling [(sqrt e)^(n-2)] is a Fibonacci Forgery for n = 1-->10 (stemming from the fact that phi^2 = 2.618033988... and e = 2.718281828...)

e is associated with compound growth, while the Golden Ratio is associated with optimal reception and transmission of information. More another time, particularly in relation to all 3 mathematical constants and the binomial theorem. (e.g. Both the Eulerian and Stirling Triangles give row sums equal to n!. As such, if you sum the inverses of both these triangles by row sum, you will eventually, at the limit = infinity, converge to e)

Also, Jeremy, note the following relationship:

zeta (n) / 2^(n-1) gives lower bounds on the density of lattice sphere packings in n-dimensions
Kissing Number
http://mathworld.wolfram.com/KissingNumber.html

2^n, of course, is the row sum of Pascal's Triangle [SUM C (n,k) for n = row number and -1 < k < n+1 ]. If, instead, you take the sum of squares of row entries you get the Central Binomial Coefficients, which can be related (in tandem with the powers of 2) to both pi and, by extension, the summed volume of 2n-dimensional spheres.

- RF

P.S. One interesting note in relationship to pi vs. phi. pi is transcendental, but not phi, which is "only" irrational (but according to some, the most irrational of the irrationals).
 
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  • #46
Also consider, Jeremy, the following relationship:

Let x = Divisors of 12 = 1,2,3,4,6,12

x + 1 = 2, 3, 4, 5, 7, 13 --> {4 U Unique Prime Divisors of the Leech Lattice} --> {4 U First Five Mersenne Prime Exponents}

Let y = Divisors of Divisors of 12 = 1, 2, 2, 3, 4, 6

y - 1 = 0, 1, 1, 2, 3, 5 --> {Fibonacci_n (modulo 6)}

The point being that Dodo may be on the right track in cautioning you not to over-extrapolate regarding the apparent Golden Ratio relationship you have come across.

Which is not to say you are wrong or that, even if you are, you might not be on to something important:

1, 2, 3, 4, 6 are the only positive integer solutions to 2*cos (2*pi/n) is in N. This is the formula that is used in the short proof of the Crystallographic Restriction Theorem.

- RF

P.S. Also note that for K_n a Maximal Lattice Packing for Dimension n...

1*2 = x_1 * x_2 = 2 = K_1
2*3 = x_2 * x_3 = 6 = K_2
3*4 = x_3 * x_4 = 12 = K_3
4*6 = x_4 * x_5 = 24 = K_4
6*12 = x_5 * x_6 = 72 = K_6
 
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  • #47
Raphie said:
5 arccos (Golden Ratio/2) = pi
2 cos (pi/5) = Golden Ratio

Or, alternatively, to better express the symmetrical relationship in terms that requires nought but a single sign change and an exchange of variables...

(7+3)/2 cos^-1 (phi/((7-3)/2)) = pi
(7- 3)/2 cos^1 (pi/((7+3)/2)) = phi
I posted arcos(Phi/2) = 36 deg earlier so I guess I knew this although I really like the symmetry you have shown.
Raphie said:
Ceiling [(sqrt e)^(n-2)] is a Fibonacci Forgery for n = 1-->10 (stemming from the fact that phi^2 = 2.618033988... and e = 2.718281828...)
Hence the similarity of the images I assume.
Raphie said:
e is associated with compound growth, while the Golden Ratio is associated with optimal reception and transmission of information. More another time, particularly in relation to all 3 mathematical constants and the binomial theorem. (e.g. Both the Eulerian and Stirling Triangles give row sums equal to n!. As such, if you sum the inverses of both these triangles by row sum, you will eventually, at the limit = infinity, converge to e)

Also, Jeremy, note the following relationship:

zeta (n) / 2^(n-1) gives lower bounds on the density of lattice sphere packings in n-dimensions
Kissing Number
http://mathworld.wolfram.com/KissingNumber.html

2^n, of course, is the row sum of Pascal's Triangle [SUM C (n,k) for n = row number and -1 < k < n+1 ]. If, instead, you take the sum of squares of row entries you get the Central Binomial Coefficients, which can be related (in tandem with the powers of 2) to both pi and, by extension, the summed volume of 2n-dimensional spheres.

- RF

P.S. One interesting note in relationship to pi vs. phi. pi is transcendental, but not phi, which is "only" irrational (but according to some, the most irrational of the irrationals).
I’m reading your post here:
https://www.physicsforums.com/showthread.php?p=3212052#post3212052
fascinating to say the least.
JeremyEbert said:
The interesting vector I noticed is at 45 degrees or pi/4 radians.
The sqrt((2xn)+n^2) parabolas intersect that vector at sqrt(((1+sqrt(2))*n)^2).
I have attached an image demonstrating this.

The thing I find most interesting about these intersections is this:

q=((1+sqrt(2))*n)^2
u=((n^2)*6) – q
(q*u)^(1/4) = n
Evidently the 1+SQRT(2) part is known as the Silver Ratio:
http://en.wikipedia.org/wiki/Silver_ratio
which is the also limiting ratio of consecutive Pell numbers
 
  • #48
JeremyEbert said:
Evidently the 1+SQRT(2) part is known as the Silver Ratio:
http://en.wikipedia.org/wiki/Silver_ratio
which is the also limiting ratio of consecutive Pell numbers

Indeed. Which, as we (now) know, is also not unrelated to the Sophie Germain Triangular Numbers. I believe there to be some manner of as yet undiscovered linkage between those and Sophie Germain Primes, the first three of which are 2, 3, 5 = (5-1)/2, (7-1)/2, (11-1)/2, for 5, 7 & 11 the first 3 safe primes associated with the Ramanujan Congruences (aka "0-Dimensional Ono Primes")

Pell Numbers are resursively constructed thusly:

1A + 2B = C

Whereas Fibonacci Numbers are recursively constructed thusly:

1A + 1B = C

So, in spite of my caution, I would not, if I were you, assume as a given that the (provisional) Golden Ratio relationship you came across is just a chimera.

- RF
 
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  • #49
JeremyEbert said:
I posted arcos(Phi/2) = 36 deg earlier so I guess I knew this although I really like the symmetry you have shown.

One of the benefits of geting the two formulas "talking together in the same language" is that it then becomes a simple matter to construct a single generating formula uniting both constants, phi and pi:

For...
(7- 3)/2 cos^1 (pi/((7+3)/2)) = phi
(7+3)/2 cos^-1 (phi/((7-3)/2)) = pi

Let..
b = (n) (mod 2)
a = (n + 1) (mod 2)

Then...
(7- 3(-1)^b))/2 cos^((-1)^b) ((phi^a*pi^b)/((7+3(-1)^a))/2)) = phi^b*pi^a

And then you can generate other formulas from those same variables...

(7- 3(-1)^b))/2 cos^((-1)^b) ((phi^a*pi^b)/((7+3(-1)^a))/2))
*
(7- 3(-1)^a))/2 cos^((-1)^a) ((phi^b*pi^a)/((7+3(-1)^b))/2))
= phi*pi

The phi*pi product is known as the "Biwabik Sum" and it relates to the set of all odd numbers, of which all primes, excluding 2, are a subset. See...

Pi, Phi and Fibonacci Numbers
http://goldennumber.net/pi-phi-fibonacci.htm

- RF
 
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  • #50
Thank you Raphie, you have given me some amazing morsels to digest.
 

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