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Interesting value for W, h involving phi,Omega?
#11
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
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#12
We can bring this a little further by taking x=phi=1,61803399...Golden Mean and using the property that 1/phi = phi-1

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 = phi-1= 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*(phi-1))^(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(phi-1)^phi) = phi = 1,61803399
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#13
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 -2,87556E-10

Ivars

Ivars
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#14
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=lnI-ln(e^-I)=ln(I*e^I)

W(-ln(((I*e^-I)^(1/(I*e^-I) = -((pi/2)-1)*I = (1-pi/2)*I = -lnI+I= -lnI+ln(e^I)= ln((e^I)/I)


correspondingly:

h( ((e^I)/I) ^ (I/(e^I))) = -I* e^I = sin1-I*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^(1-pi/2)*I
I*e^-I=e^((pi/2)*I-I)=e^(pi/2-1)*I


are rather specific rotations in complex plane.

Ivars
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#15
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^(n-1)) ; 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.
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#16
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)=0.540302306-I*0.841470985=cos1-I*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/2-I)/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^((1-I)/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.
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#17
Previous result was:

h(Omega^(pi/2)*Omega^ (-I/Omega)) = h((Omega^((1-I)/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/2-I*(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*m-k/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=lnz-lnW; W/lnW = lnz/lnW-1 = log base W of z-1

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(Wt-arctan(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
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#18
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?
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#19
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 0-zeration 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
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#20
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/Om-1))= 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
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