Interesting value for W, h involving phi,Omega? - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Interesting value for W, h involving phi,Omega? (/showthread.php?tid=110) Pages: 1 2 3 4 RE: Interesting value for W, h involving phi? - Ivars - 02/01/2008 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 RE: Interesting value for W, h involving phi? - Ivars - 02/10/2008 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 RE: Interesting value for W, h involving phi? - Ivars - 02/20/2008 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 RE: Interesting value for W, h involving phi, Omega? - Ivars - 02/21/2008 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 RE: Interesting value for W, h involving Omega,phi? - Ivars - 02/25/2008 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. RE: Interesting value for W, h involving Omega, phi? - Ivars - 02/25/2008 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. RE: Interesting value for W, h involving phi? - Ivars - 02/26/2008 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 y1): 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 RE: Interesting value for W, h involving phi? - bo198214 - 02/26/2008 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? RE: Interesting value for W, h involving phi? - Ivars - 02/26/2008 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 RE: Interesting value for W, h involving phi? - Ivars - 02/28/2008 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