• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Interesting value for W, h involving phi,Omega? Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/04/2008, 07:50 AM 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 bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 03/04/2008, 09:09 AM 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?! Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/04/2008, 09:29 AM 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 Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 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)*(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 Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 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))= 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 Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/08/2008, 10:48 AM (This post was last modified: 03/08/2008, 10:50 AM by 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. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 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: $\lim_{n\to\infty}{\frac{\theta n}{n} = 1$ 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 Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 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) Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/10/2008, 11:07 AM 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 Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 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. « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Interesting commutative hyperoperators ? tommy1729 0 1,797 02/17/2020, 11:07 PM Last Post: tommy1729 Very interesting topic Ansus 0 2,186 10/01/2019, 08:14 PM Last Post: Ansus Limit of mean of Iterations of f(x)=(ln(x);x>0,ln(-x) x<0) =-Omega constant for all x Ivars 10 22,842 03/29/2015, 08:02 PM Last Post: tommy1729 regular tetration base sqrt(2) : an interesting(?) constant 2.76432104 Gottfried 7 16,197 06/25/2013, 01:37 PM Last Post: sheldonison (MSE): Comparision of powertowers -.Possibly interesting thread in MSE Gottfried 0 3,343 05/22/2013, 07:02 AM Last Post: Gottfried Continuum sums -- a big problem and some interesting observations mike3 17 33,131 10/12/2010, 10:41 AM Last Post: Ansus Something interesting about Taylor series Ztolk 3 9,625 06/29/2010, 06:32 AM Last Post: bo198214 interesting pattern in hyper-operations Base-Acid Tetration 8 18,316 05/04/2009, 09:15 PM Last Post: BenStandeven Infinite tetration giving I*Omega costant=I*0.567143... Ivars 13 22,353 03/09/2008, 01:16 PM Last Post: bo198214 Observations on power series involving logarithmic singularities jaydfox 29 47,950 11/12/2007, 10:59 AM Last Post: bo198214

Users browsing this thread: 1 Guest(s)