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

[JeremyEbert]

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?

 

[dodo]

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.

 

[JeremyEbert]

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?

 

[dodo]

--- 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.

 

[JeremyEbert]

--- 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

 

[JeremyEbert]

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)

 

[JeremyEbert]

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?

 

[JeremyEbert]

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.

 

[dodo]

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.

 

[JeremyEbert]

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)

 

[dodo]

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

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

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

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

Yes, this one was established in post#15. Assuming you correct "x" as above.

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.

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.

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.

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,...

 

[JeremyEbert]

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

 

[JeremyEbert]

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?

 

[JeremyEbert]

Dodo, sorry for the late response, I've been a little distracted with matricies.

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

Over zealous typo. ha! 1-2/(n+1) is what I meant.

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.

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.

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.

 

[Raphie]

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).

 

[Raphie]

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

 

[JeremyEbert]

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.

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.

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.

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

 

[Raphie]

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

 

[Raphie]

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

 

[JeremyEbert]

Thank you Raphie, you have given me some amazing morsels to digest.

 

[JeremyEbert]

Also, just found that e ^ asinh(.5) = phi

 

[JeremyEbert]

I wonder if I should look at my y = sqrt((2xn)+n^2) parabolas rotated pi/2 radians?
Now y =( (x/ sqrt(2n) )^2) - n/2.
Can this be treated as a sort of parabolic function of n? Essentially the pattern I’m talking about comes from a conic section of a cone in complex exponential space. Where the growing amplitude and the circular motion create the cone. Make any sence?
http://en.wikipedia.org/wiki/Conic_section
where the cone looks similar to this:
http://i98.photobucket.com/albums/l267/alienearcandy/econic.png

 

[JeremyEbert]

something like this:
http://i98.photobucket.com/albums/l267/alienearcandy/prime-squarenewconic90.png

 

[JeremyEbert]

better yet:
http://i98.photobucket.com/albums/l267/alienearcandy/prime-squarenewconic90-3d-1.png

 

[JeremyEbert]

ball in cone:
http://www.vic.com/~syost/utk/BallInCone.html

 

[JeremyEbert]

a kind of modified shell theorem right?:
http://en.wikipedia.org/wiki/Shell_theorem
and
http://en.wikipedia.org/wiki/Inverse-square_law

 

[Raphie]

I wonder if I should look at my y = sqrt((2xn)+n^2) parabolas rotated pi/2 radians?
Now y =( (x/ sqrt(2n) )^2) - n/2.
Can this be treated as a sort of parabolic function of n? Essentially the pattern I’m talking about comes from a conic section of a cone in complex exponential space. Where the growing amplitude and the circular motion create the cone. Make any sence?

Jeremy, given the manner of observations you are reporting, I very much believe you would behoove yourself, contextually speaking, to at least begin to familiarize yourself with arithmetic functions. For instance, check out the Dirichlet Divisor function. A little research will make it manifest the relationship between this, the Riemann Hypothesis/Zeta Function and Lattice Points under a hyperbola.


Raphie

 

[dimension10]

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?

Yes. 1-1/4(Pi)r2 is related to e by the science of fluctuations.

 

[JeremyEbert]

Jeremy, given the manner of observations you are reporting, I very much believe you would behoove yourself, contextually speaking, to at least begin to familiarize yourself with arithmetic functions. For instance, check out the Dirichlet Divisor function. A little research will make it manifest the relationship between this, the Riemann Hypothesis/Zeta Function and Lattice Points under a hyperbola.


Raphie

Raphie
Thanks again! I can see where my animation highlights the divisor function and its summation by counting the lattice points that intersect.

http://www.tubeglow.com/test/Fourier.swf

I'lll have the equation here shortly.

 

[FlexGunship]

Is there a known relationship between Phi, e and pi?

Yes! When you add them together you get 7.47790847! Amazing!