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I added asymtotic values of negative infinite of base e heptation and [9] to the sum I mentioned before:
Sum [5] =1/1,85035452902718^81/(2*1,8503545290271 ^7+1/(3*1,8503545290271 ^61/(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] =1/3,751^81/(2*3,751)^7+1/(3*3,751)^61/(4*3,751)^5+1/(5*3,751)^4=3,20285E05
Sum[9] = 1/5,693^81/(2*5,693)^7+1/(3*5,693)^61/(4*5,693)^5+1/(5*5,693)^4=3,15992E05
Then I made Sum [5,7,9] = Sum[5]Sum[7]+Sum[9] = 0,0072975833,20285E05+3,15992E05=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
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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?!
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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
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03/06/2008, 09:53 PM
(This post was last modified: 03/07/2008, 12:41 PM by 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)*(Omega1/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
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03/06/2008, 10:20 PM
(This post was last modified: 03/07/2008, 12:58 PM by 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))= phi1/2=1.61803990.5=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
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03/08/2008, 10:48 AM
(This post was last modified: 03/08/2008, 10:50 AM by Ivars.)
Consider circle map (Arnold map) :
Let
And
then map becomes:
I did 1800 iterations for starting from with 50 digit accuracy ( This was my first try) and the resulting conjecture is:
monotonically from below, no oscillations. So the resulting angle is 1 rad again. I was expecting it.
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03/08/2008, 11:20 AM
(This post was last modified: 03/09/2008, 08:02 PM by Ivars.)
To evaluate previous conjecture analytically, we need few more tools:
Let us put
To evaluate 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
module
Arg
module
So
Which coincides with numerical calculation.
Ivars
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03/09/2008, 08:48 AM
(This post was last modified: 03/09/2008, 12:32 PM by 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)
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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
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03/14/2008, 09:55 PM
(This post was last modified: 03/15/2008, 08:16 AM by 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.
