# Tetration Forum

Full Version: Interesting value for W, h involving phi,Omega?
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Pages: 1 2 3 4
I added asymtotic values of negative infinite of base e heptation and [9] to the sum I mentioned before:

Sum [5] =1/1,85035452902718^8-1/(2*1,8503545290271^7+1/(3*1,8503545290271^6-1/(4*1,8503545290271^5+1/(5*1,8503545290271^4=0,007297583=1/137,0316766

From Andrew's graph, I found the values to be roughly e[7]-infinity = -3,751 and e[9]-infinity = -5,693.

Then I put them in the same sum, obtaining:

Sum [7] =1/3,751^8-1/(2*3,751)^7+1/(3*3,751)^6-1/(4*3,751)^5+1/(5*3,751)^4=3,20285E-05

Sum[9] = 1/5,693^8-1/(2*5,693)^7+1/(3*5,693)^6-1/(4*5,693)^5+1/(5*5,693)^4=3,15992E-05

Then I made Sum [5,7,9] = Sum[5]-Sum[7]+Sum[9] = 0,007297583-3,20285E-05+3,15992E-05=0,0072971534=1/137,039738252

So after this, approximation of alpha =0.07297352570(5) got even better, as I expected, but of course I do not know the exact values of e[7]-infinity and e[9]-infinity and more.

Then we could see how does the sum Sum[5]-Sum[7]+Sum[9]-Sum[11]+Sum[13]-Sum[15]+.........converge.

Ivars
Ivars Wrote:From Andrew's graph, I found the values to be roughly e[7]-infinity = -3,751 and e[9]-infinity = -5,693.

Can you be a bit more detailed how you obtained those values?!
Well, his graph shows them. That is what I understood. as asymptotes going to -x , while asymptoes of even negative ntations are going to -y, and are -2,-4,-6 etc on x.

From the graph, just knowing that e[5]-infinity is -1,85035452902718 as calculated by jaydfox in the beginning of this thread.

I magnified the Anderw's graph a little bit, and hoped his coordinates is linear in picture so axis do not change scale, so distance from negative x axis would give values in proportion to distance from x axis to e[5]-infinity, which is known an also present on the graph as first negative asymptote in the direction in -x.

Since the graph only shows values at x=-10 of course I may be wrong in hoping the proportion holds to infinity but for the alpha approximation it is enough to have just 2 -3 decimal signs to see the trend.

That is why I asked for exact values so I do not need this guesswork, but did not get them.

Ivars
I Need to add few more formulae and check before we can explain(? ) oscillations related to Omega and W(1):

Omega^(1/(I*Omega) = e^I

Omega^(-1/(I*Omega)=e^-I

sin (z) = (-I/2)* (Omega^(z/(I*Omega))-Omega^(-z/(I*Omega)))
cos (z) = (1/2)*((Omega^(z/(I*Omega))+Omega^(-z/(I*Omega)))

So if so if z has simple form (p/q)*I*Omega,

sin((p/q)*I*Omega)=(-I/2)* (Omega^(p/q)-Omega^-(p/q))

if p=q=1,

sin((I*Omega)=(-I/2)*(Omega-1/Omega) = (-I/2)*1,19607954..=-I*0,59803977..
Cos(I*Omega)=(1/2)*(Omega+1/Omega) =(1/2)*2.330366124=1,161830623...

Also

(I*Omega)^(1/Omega) =-0.342726848178+I*0.13369214926..

Module ((I*Omega)^(1/Omega)) = 1/e = Omega^(-1/Omega)

An Interesting complex number with module 1/e.

The angle between these 2 formula values is 2,1988261.. rad =125,983.. degrees.

(1/(I*Omega))^Omega = 0.86728..- I* 1.07264..

Module ((1/(I*Omega))^Omega ) = Omega^Omega

Arg ((1/(I*Omega))^Omega ) = atan(-0.808545..)= -0.67993..rad = -38. 957 degrees

Ivars
Interestingly, if we take

z=I*ln(phi)= I* ln(1.6180399..) and

z =I*log omega (phi)= I* (ln(1.6180399..)/ln(Omega))= -0.8484829....,

using:

sin (z) = (-I/2)* (Omega^(z/(I*Omega))-Omega^(-z/(I*Omega))),
cos (z) = (1/2)*((Omega^(z/(I*Omega))+Omega^(-z/(I*Omega)))

sin(I*log omega (phi)= (-I/2)
cos (I*log omega (phi)) = (1/2) *(sqrt(5))= phi-1/2=1.6180399-0.5=1.1180399

but (I/2)=sin(I*ln(phi), so

sin(I*ln(phi)*sin(I*logomega (phi)) = 1/4
sin(I*ln(phi)+sin(I*logomega (phi)) =0
sin(I*ln(phi)/sin(I*logomega (phi)) =-1
sin(I*ln(phi)-sin(I*logomega (phi)) =-I

(sin(I*ln(phi))^sin(I*logomega (phi)) =(sin(I*logomega (phi)))^sin(I*ln(phi)) = e^(pi/2)

So far so good.

Ivars
Consider circle map (Arnold map) :

$\theta'= \theta+\Omega-{\frac{K}{2*\pi}}*sin(2\pi\theta)$

Let $\Omega=0.567143..$

And $K= {\frac{\pi}{2*\Omega}..$

then map becomes:

$\theta'= \theta+\Omega-{\frac{1}{4*\Omega}}*sin(2\pi\theta)$

I did 1800 iterations for $\theta->= \theta'$ starting from $\theta=0$ with 50 digit accuracy ( This was my first try) and the resulting conjecture is:

$lim (n->infinity) {\frac{\theta n}{n} = 1$

monotonically from below, no oscillations. So the resulting angle is 1 rad again. I was expecting it.
To evaluate previous conjecture analytically, we need few more tools:

$\lim_{n\to\infty}{\frac{\theta n}{n} = 1$

Let us put $\theta(0) = 0$

$\theta (1)= 0+\Omega-0=\Omega$

$\theta(2)= \theta(1)+\Omega-{\frac{1}{4*\Omega}}*sin(2\pi\theta(1)) = \Omega+\Omega+\frac{1}{4*\Omega}*sin(2\pi\Omega)$

To evaluate $sin(2\pi\Omega)$ we may use formula derived earlier:

sin (z) = (-I/2)* (Omega^(z/(I*Omega))-Omega^(-z/(I*Omega))),

so sin(2pi*Omega) = (-I/2)*(Omega^(2pi*Omega/(I*Omega))-Omega^(-2pi*Omega/(I*Omega))) = (-I/2)*((Omega^(-I*2pi)-(Omega^(+I*2pi))

But $\Omega^{-I*2\pi}= -0.9123233..-I*0.4094706...$

module $\Omega^{-I*2\pi} =1$
Arg $\Omega^{-I*2\pi} =0.41287 rad = 24.17158.. degrees$
$\Omega^{I*2\pi} = -0.9123233..+I*0.4094706...$

module $\Omega^{I*2\pi} =1$

So $sin(2\pi\Omega)= {\frac{-I}{2}}*-I*0.8189142..= - 0.4094706.$

$\frac{1}{4*\Omega}*sin(2\pi\Omega)= -0.180497$

$\theta(2) = \Omega+\Omega+0.180497= 1.31478..$

Which coincides with numerical calculation.
Ivars
Since

z=W(z)*e^W(z)

ln(z)=lnW(z)+W(z)

ln(W(z))=ln(z)-W(z)

h(W(z)^(1/(W(z))=W(z)

and

ln(W(z)^(1/W(z))= (1/W(z))*ln(W(z)= ln(z)-W(z)/W(z)=ln(z)/lnW(z)-1

In base W(z) that would be just log base W(z) (z)-1

ln (W(z)^(1/W(z))= W(z)/ln(W(z))= ln(z)/lnW(z)-1 = log base W(z) (z)-1

then

h(W(z)^(1/W(z))= -W(-(log base W(z) (z))+1) /((log base W(z) (z))-1))

So now there is a continuous (?) base for logarithms that gives infinite tetration result, if tetrated number is representable as self root of W function.

Superroot: Ssroot(W(z)^(1/W(z)) = ((log base W(z) (z))-1)) /W((log base W(z) (z))-1)
We can apply easily this to values of x^(1/x) if x is integer or unitary fraction.

If x=odd negative unit fraction, 1/n where n= odd, than:

(-1/n)^(-n) = 1/((-1/n)^n)= -n^n

If x=even negative unit fraction 1/m, where m=even:

(-1/m)^(-m)=1/((-1/m)^m)=m^m

so

h((-1/n)^(-n))= - 1/n if n odd, in this case argument in h is negative;
h((-1/m)^(-m))= -1/m if m even in this case argument in h is positive;

Examples:

h((-1/3)^(-3))=h(1/((-1/3)^3))= h(1/(-1/27)) =h(-27) = -1/3

h((-1/2)^(-2))=h(1/((-1/2)^2)))=h(1/(1/4))=h(4)= -1/2 ????

Inverting this:

h(-3^(-1/3))=h(1/((-3^(1/3)) ? there are 3 cubic roots of -3, 3^(1/3)*(cubic roots of -1).

3^(1/3)*(e^(ip/3), e^(-ip/3), -1)

All of them has the same value for h, namely h(-3^(-1/3))=h(1/((-3^(1/3)) =-3

h(-2^(-1/2))=h(1/((-2^(1/2))) / There are 2 square roots of -2 . 2^(1/2)*(-i; +i)

Both have the same value for h((-2^(-1/2))=-2.

I must be making some stupid mistake here.Please let me know so I do not continue

Ivars
I was studying the graph of selfroot of Lambert function:

W(x)^(1/W(x)).

It has maximum value at x=(e*(e^e)) and is e^(1/e) ; so

W(e*(e^e)) = e

Numerically,

W(41,1935556747..)^(1/(W(41,1935556747)= 1,444667861

I multplied e*(e^e)* Omega constant =

41,193556747..*0,567143...=23,36263675...

On other hand, I took logarithm of (e*(e^e))

ln (e*(e^e)) = 1+e = 3,718281828.....

I multiplied it with Pi :

3,718281828.....* 3,141592..= 11,68132688

And I multiplied this with 2:

11,68132688..*2 = 23,362653...

So:

pi = approx((e*(e^e)*Omega)/(2*(e+1)))

Since e=Omega^(-1/Omega), its just an approximation containing 2 and Omega.

This approximation seems to be good for 5 decimals. I wonder why and can it be improved.
Pages: 1 2 3 4