+- Tetration Forum (https://tetrationforum.org)
+-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1)
+--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3)
+--- Thread: Interesting value for W, h involving phi,Omega? (/showthread.php?tid=110)
RE: Interesting value for W, h involving phi,Omega? - Ivars - 03/04/2008
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=1/137,0316766From 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
RE: Interesting value for W, h involving phi,Omega? - bo198214 - 03/04/2008
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?!
RE: Interesting value for W, h involving phi,Omega? - Ivars - 03/04/2008
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
RE: Interesting value for W, h involving phi,Omega? - Ivars - 03/06/2008
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...
This corresponds to angle -0,64105..rad= -36,7297..grad
Also
(I*Omega)^(1/Omega) =-0.342726848178+I*0.13369214926..
Module ((I*Omega)^(1/Omega)) = 1/e = Omega^(-1/Omega)
Arg ((I*Omega)^(1/Omega)) = atan(-2,5632)=-1,198826..rad = -68,6876759..grad
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
RE: Interesting value for W, h involving phi,Omega? - Ivars - 03/06/2008
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
RE: Interesting value for W, h involving phi,Omega? - Ivars - 03/08/2008
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.
RE: Interesting value for W, h involving phi,Omega? - Ivars - 03/08/2008
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
RE: Interesting value for W, h involving phi,Omega? - Ivars - 03/09/2008
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)
RE: Interesting value for W, h involving phi,Omega? - Ivars - 03/10/2008
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
RE: Interesting value for W, h involving phi,Omega? - Ivars - 03/14/2008
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.