Posts: 366
Threads: 26
Joined: Oct 2007
02/01/2008, 08:38 PM
(This post was last modified: 02/03/2008, 01:47 PM by Ivars.)
One more interesting value is h(i*(e^(pi/2))) = h(I*(I^(1/I)).
And it is: h(i*i^(1/i) = h(i^((1/i)+1)) = h( i^(1+i)/i)) = 0,213934198848366+I*0,213934198848366
So that form is a+i*a, and Arg is pi/4, so in exponential form:
h( i^(1+i)/i))= +0,30254864546678200*e^(I*pi/4).
May be I have mixed some sign for imaginary part.
Ivars
Posts: 366
Threads: 26
Joined: Oct 2007
02/10/2008, 09:38 AM
(This post was last modified: 03/02/2008, 09:12 PM by Ivars.)
We can bring this a little further by taking x=phi=1,61803399...Golden Mean and using the property that 1/phi = phi1
BTW interesting to note that multiplication of true and inverse self roots of phi (This is not generally true for other numbers):
(phi^(1/phi))*((1/phi)^phi))= 1/phi = phi1= h((phi)^(1/phi)) = 0,61803399 . One more interesting selfroot is:
(i^(1/i))*((1/i)^i))= (i^(1/i))^2=(e^(pi/2))^2=e^(pi)=23,14069263
so:
h((I/phi)^(phi/I)) = h(I*(1/phi)^(phi/I))=h((I*(phi1))^(phi/I)) = phi/I = I*phi=((1/phi)+1)*I = ((I/phi) +I)
If we multiply/divide by 2, since I/2=sin (I*ln(phi)) we get:
h((I/phi)^(phi/I))) = (phi/2)/(I/2) = (phi/2)/(sin (I*ln(phi)))
Also:
h((1/phi)^phi) = h(phi1)^phi) = phi = 1,61803399
Posts: 366
Threads: 26
Joined: Oct 2007
02/20/2008, 12:25 AM
(This post was last modified: 02/20/2008, 12:30 PM by Ivars.)
I am not sure if this is true, but seems rather close:
W(ln2/(2*sqrt2)=W(ln4/(4*4^(1/4)))= W(ln2/(2^(3/2))=W(ln4/(4^(5/4)))= W(ln(2^(1/(2^(3/2))))= ln2/2=ln4/4
than
W(ln2/2)=W(W(ln(2^(1/(2^(3/2))))= ln2
But rather it seems just quite close approximation.
Yes,it is just that: Difference is 2,87556E10
Ivars
Ivars
Posts: 366
Threads: 26
Joined: Oct 2007
02/21/2008, 11:35 PM
(This post was last modified: 02/29/2008, 11:30 AM by Ivars.)
To finalize one even more interesting finding:
Here is probably well known feature, anyway interesting value of W:
W( ln( ((e^I)/I) ^ (I/(e^I))) = ((pi/2)1)*I = lnI  I=lnIln(e^I)=ln(I*e^I)
W(ln(((I*e^I)^(1/(I*e^I) = ((pi/2)1)*I = (1pi/2)*I = lnI+I= lnI+ln(e^I)= ln((e^I)/I)
correspondingly:
h( ((e^I)/I) ^ (I/(e^I))) = I* e^I = sin1I*cos1
h((I*(e^I)^(1/I*(e^I))= I*e^I = sin(1)+I*cos1
and also :
I*e^I= e^((pi/2)I+I)= e^(1pi/2)*I
I*e^I=e^((pi/2)*II)=e^(pi/21)*I
are rather specific rotations in complex plane.
Ivars
Posts: 366
Threads: 26
Joined: Oct 2007
02/25/2008, 01:13 PM
(This post was last modified: 02/28/2008, 10:27 AM by Ivars.)
Omega constant =0.5671432904097838729999686622 is defined by :
Omega*(e^Omega)=1 and
ln(Omega)=Omega
So selfroot of Omega:
Omega^(1/Omega) = e^ln(Omega^(1/Omega) = e^((1/Omega)*(ln(Omega))=e^((1/Omega)*(Omega)) =e^(1)=1/e=0,367879441
And:
Omega^Omega = e^ln(Omega^Omega) = e^(Omega*ln (Omega) = e^(Omega^2)=0,724950783
(Omega^(Omega))=(1/Omega)^(Omega)=e^ln(Omega^(Omega))=e^Omega*ln(Omega)=e^(Omega^2)=1,379403986
Infinite tetration of selfroot of Omega:
h(Omega^(1/Omega))=h(1/e) = W(ln(1/e)/(ln(1/e))= W(1)/1=Omega=0,56714329=ln(Omega),
Square superroot of (Omega^1/Omega) :
ssrt(Omega^(1/Omega) = ln(1/e)/W(ln(1/e))= 1/W(1)= 1/(0.318131505204764 + 1.337235701430689*I) = 0.16837688705553+0.707755195958823*I.
W(1) = 0.318131505204764 + 1.337235701430689*I is one of two conjugate values. Jaydfox :"ith iteration of natural exponentiation of base e has two primary fixed points at 0.318131505204764 + 1.337235701430689*I." in
Imaginary iterates of ....
Also:
h((1/Omega)^Omega))= h( e^(Omega^2)) = W(Omega^2)/(Omega)^2 = Omega/Omega^2= 1/Omega=1/ln(Omega)
Generally, h( e^(Omega^n)= 1/(Omega^(n1)) ; n>1
h((Omega^2)^(1/Omega^2))= Omega^2=0,321651512
h(Omega^3)^(1/Omega^3))= Omega^3=0,182422497 etc .Generally:
h(Omega^n)^(1/Omega^n))= Omega^n if n>=1.
We can compare this to h(i^(1/i)) = h((1/i)^i))=  i = i^3=1/i.
Posts: 366
Threads: 26
Joined: Oct 2007
02/25/2008, 02:01 PM
(This post was last modified: 02/29/2008, 11:57 AM by Ivars.)
This is obvious imaginary selfroot of Omega, but might prove useful, linking I and Omega and e:
Omega^(I/Omega)=e^I=1/(e^I)=0.540302306I*0.841470985=cos1I*sin1
(Omega^(I/Omega))^pi/2=(e^(I*pi/2))=i
(Omega^(I/Omega))^pi=(e^(I*pi))=1
(Omega^(I/Omega))^2*pi=e^(I*pi) = 1
Basically, if we have Omega, there is no need for e as it can be always defined and found as 1/self root of Omega. Opposite is not so easy.
That means base 1/e is Omega^(1/Omega), base 1/e^I is Omega^(I/Omega), similarly formed as base i^1/i = e^(pi/2) (and all inverses).
This begs a question is Omega itself an interesting base, or 1/Omega,or I*Omega, or I/Omega.
For exapmple, log omega (I) = (I/Omega)*pi/2
To get rid of e more, i will use formula i derived earlier in this thread, putting x=1/Omega.
W((1/Omega)*(pi/2)  I*(1/Omega)*ln((1/Omega)) ) = ln(1/Omega) I*(pi/2) for x>1;
(W(pi/(2*Omega)I) = Omega(I*pi/2)
Then:
h(((e^(pi/2))^(1/Omega))*(e^I)) = (Omega+(I*(pi/2))/(pi/(2*Omega)+I) and, since
e= Omega^(1/Omega) , e^(I) =( Omega^(I/Omega))
h(Omega^(pi/2)*Omega^ (I/Omega)) = h((Omega^((Omega*pi/2I)/Omega))= (Omega+(I*(pi/2))/(pi/(2*Omega)+I)
The same without pi/2 in h:
h(Omega^(pi/2)*Omega^ (I/Omega)) = h((Omega^((1I)/Omega))*(I^(1/I))=(Omega+(I*(pi/2))/(pi/(2*Omega)+I)
This value is:
h(((e^(pi/2))^(1/Omega))*(e^I)) =(Omega+(I*(pi/2))/(pi/(2*Omega)+I)= Omega*I=I*0,56714329..
The other branch of W should give I*Omega?
This was changed from previous mistaken result, due to wrong signs.
Posts: 366
Threads: 26
Joined: Oct 2007
02/26/2008, 03:57 PM
(This post was last modified: 11/17/2008, 02:04 PM by Ivars.)
Previous result was:
h(Omega^(pi/2)*Omega^ (I/Omega)) = h((Omega^((1I)/Omega))*(I^(1/I))=(Omega+(I*(pi/2))/(pi/(2*Omega)+I)
h(((e^(pi/2))^(1/Omega))*(e^I)) =Omega*I=I*0,56714329
Previosly here was damped harmonic oscillator. This time it has dissappeared, but will reappear later as undamped if W is real:
Let us take the formula:
W(x*(pi/2)  I*x*ln(x) ) = ln(x)  I*(pi/2) for x>1
And substitute (1/W(y)) = x>1 ,so y<e for the time being. Than:
W((1/W(y))*(pi/2) I* (1/W(y))*ln(1/W(y))) = ln (1/W(y))I*pi/2
further, I only will write W instead of W (y):
W( (1/W)*(pi/2) + I *(lnW)/W)=lnW  I*pi/2
this means ln (arg h ) = (1/W)*pi/2I*(lnW)/W and
h( e^((pi/(2*W))*e^(I*lnW/W)) = (lnW+I*pi/2)/ ((1/W)*(pi/2)  I *(lnW)/W) which leads to :
h( e^((pi/(2*W))*e^(I*lnW/W)) = I*(W+ W*pi^2/(4* (lnW)^2))/ ((pi^2/(4*(lnW)^2))+1)
by denoting
m = 1/((pi^2/4*(lnW)^2+1)
k= (W^2*pi^2)/(4*(lnW)^2)*(1/((pi^2/4*(lnW)^2+1))=(W^2*pi^2)/(4*(lnW)^2)*m
k/m =(W^2*pi^2)/(4*(lnW)^2)
The result above represents linear response function of undamped harmonic oscillator with equation:
y''+k/m*y=0
h(e^((pi/(2*W))*e^(lnW/W) = Z= I*(W*mk/W) which has imaginary resonance frequency (for W real and >1):
w rez =sqrt(k/m) = I*W*pi/(2*lnW) = I*pi/2*(log base W of (z)1)
because from definition of W function, lnz=lnW+W; W=lnzlnW; W/lnW = lnz/lnW1 = log base W of z1
f rez = (I/4)*(W/lnW)
So when y=1, W(1) = omega, w rez = (I *Omega*pi)/(2 *ln Omega) =  I*pi/2, f=I/4  consistent with results obtained earlier. For other y, W(y), ln W(y) , k, m, w rez, Zm etc. will be different.
If W= I , w rez =sqrt((I^2*pi^2)/(4*(I*pi/2)^2)) = sqrt( 1) = + I, f= (+I/2*pi)
As we can see, both m and K are functions of "frequency" W, so this seems to be rather strange system.
Impedance module Zm =((W+ W*pi^2/(4* (lnW)^2))/ ((pi^2/(4*(lnW)^2))+1))
Phase arctan((W+ W*pi^2/(4* (lnW)^2))/ ((pi^2/(4*(lnW)^2))+1)).
Output of this system if driven by Fo*cos(Wt):
y=Fo/[W*((W W*pi^2/(4* (lnW)^2))/ ((pi^2/(4*(lnW)^2))+1))]*sin(Wtarctan(Z/I))
I hope this is more correct than previous attempt.
Interesting? What would be oscillating in number world?
Little more play with rezonance frequency:
w rez =sqrt(k/m) = I*W*pi/(2*lnW) = I/2*pi*W/ln(W)
Since I/2= sinh (ln(phi)) where phi is golden ratio:
w rez = sinh(ln(phi) * pi* (W/lnW)
If y= 1, W(1) = Omega, ln(W(1))=  Omega and W/lnW=1
w rez= sinh (ln(phi))* pi =  I*pi/2
Ivars
Posts: 1,386
Threads: 90
Joined: Aug 2007
Ivars Wrote:Interestingly, this can be transformed into form corresponding to linear response function impedance of a damped harmonic oscillator;
I dont understand, how can a constant correspond to a function?
Posts: 366
Threads: 26
Joined: Oct 2007
bo198214 Wrote:Ivars Wrote:Interestingly, this can be transformed into form corresponding to linear response function impedance of a damped harmonic oscillator;
I dont understand, how can a constant correspond to a function?
Neither do I  yet. One idea is that this constant is a result of some process we have no idea about so it is in fact variable or function of some other mathematical parameter e.g y may be continuos dimensionality of mathematical hyperoperation (from 0zeration to [infinity] , t  dimensionality or hypervolume of hyperdimensional mathematical structure achieved via infinite tetration (from 1 to infinity ) or vice versa?.
That is just a very wild guess at too early stage. I am looking for such interdependence, may be in wrong place with wrong approach...
Ivars
Posts: 366
Threads: 26
Joined: Oct 2007
02/28/2008, 02:04 PM
(This post was last modified: 02/28/2008, 03:15 PM by Ivars.)
If we work with this:
W(x*(pi/2)  I*x*ln(x) ) = ln(x)  I/pi/2) for x>1
by substituting y=(pi/2) I*ln(x)), y> pi/2 we get lnx = I*y+I*pi/2 and x= e^ln(x) = I*e^(I*y)
W((I*y)*e^(I*y)) = Iy ; y>pi/2
then infinite tetration of any complex number that is a self root of e^(I*y) is :
h(e^(I*y)^e(I*y)) = (e^(I*y)) this works for y>=pi/2, at least.
Its no big help, since to find y anyway Lambert function is needed, but if we just vary y, we will get argument and h for each y.
since e^I= Om^(I/Om), where Om=Omega constant, 0,567143... it is also true that:
h((Om^((I*y/Om))^((Om)^(I*y/Om)))) =h((Om^(I*y*Om))^(I*y/Om1))= Om^((I*y)/Om))
We can see that result is purely imaginary with values +i only if y= pi/2, 3pi/2, 5pi/2 etc. Than means that Gottfrieds spider graphs values + i oscilate at:
e^I*n*pi/2^e^(I*n*pi/2) for n>=1 and odd or:
h(i^(1/i)) = h(4,81407738096535) = e^(I*pi/2)= I
h(i^3^(1/i^3)) =h(4,81407738096535) = e^(3pi/2) = I
h(i^5^(1/i^5) =h(4,81407738096535) = e^(5pi/2) = I
h((i^7)^(1/i^7) = h(4,81407738096535)= e^(7pi/2) = I etc.
Purely Real values of h happen at y= pi/2*k, where k = 2,4,6....
h(e^(Ipi)^(e^(Ipi))) = h( 1) = e^Ipi= 1
h(e^(2Ipi)^(e^(2I*pi))=h(1) = e ^2pi= 1
h(e^(3Ipi)^(e^(3I*pi))=h(1) = e^3pi = 1
h(e^(4Ipi)^(e^(4*pi))= h(1) = e^4pi= 1 etc.
It is interesting is there a formula for y<pi/2 which means argument for h with fractions of I.
I hope there were not too many mistakes, but if there were, please help me to find them.
Ivars
