A new math function inspired from Julia's fractal

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Offline hgjf2

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« on: February 15, 2018, 09:29:20 PM »
A new math function inspired from the Julia's fractals theory under construction I working in progress and can creating a new math branch . It is named ef[P](z) where
 (ef) mean exponential fractal where P is the polynom of own Julia Set and z<-C a complex number.
 Has those properties: ef[P](z)=P(ef[P](z-1)) for whatever complex number z.
This function is good for who is curious how to solving a functional equations like "f(f(z))=z^2-1 what gives f(z)?"
 Moment ago I gave some values: ef[z^2](z)=a^(2^z); ef[z^3](z)=a^(3^z); ef[z^2-2](z)=2cos(aTT(2^z));
ef[z^3-3z^2+3](z)=-1+2cos(aTT(3^z)) due cos(2x)=2cos(x)^2-1 and cos(3x)=4cos(x)^3-2cos(x); lim[x->-oo](ef[P](x)) contain all numbers from the border of the Julia set JP(z)={z|P(P...P(z)...)not infinity}.
Other properties are: ef[P(z)](z)+z0=ef[P(z-z0)+z0](z);Q(ef[P(z)](z)=ef(Q(P(Q(z)))](z) where Q is inverted Q :
Q(Q(z))=Q(Q(z))=z, whatever P,Q the complex functions.
The inversion of ef(z) is lof[P](z) with the egalities ef[P](lof[P](z))=lof[P](ef[P](z))=z whatever z,P. I taken the term "lof" that mean logarithm fractal.
Also ef[z](z)=a; ef[z+k](z)=a+kz and ef[kz](z)=a*k^z.  Moment ago I founded term "a" like a integrall constant C used and at the differential equations. The function ef[P](z) has complex period : existing v<-R for whick ef[P](z)=ef[P](z+vi) if P is polynom unliniar (kind 2 and above)
More properties of this function coming soon, I still working to find new properties.
If defineing the function efL[P](z)=ef[P](ln(z)) then efL[P](z) is a periodical function like cos(z) and prod[k<-Z](cos(z+kTTi) if P is polynom, due efL[P](z)= P(efL[P](z/e)) has route on the border of the Julia set owned by the polynom P, if z is included in the line whick contains all periods.

Example: If having f[z^2-1](0)=f(2TT)= (1+sqrt(5))/2=z0; f(TT)=-(1+sqrt(5))/2=-z0; f(TT/2)=sqrt(1+z0)=P(z0); where P(z)=
z^2-1 and P is the inverse; f(TT/4)=sqrt(1+sqrt(1+z0))=P(P(z0)); f(2TT/3)=(1-sqrt(5))/2=z1;P(P(z1))=z1; f(TT/3)=sqrt(1+z1)=P(z1);  f(2TT/7)=z2 (a solution of equation degree 6 whick maybe without solutions with radicals) from P(P(P(z2)))=z2 etc due f[z^2-1](z)=f(z/2)^2-1 where f[z^2-1](z)=ef[z^2-1](log[2](z))=efL[z^2-1](e^(log[2](z))).
The determining of this f(i) and more formulas coming soon...
« Last Edit: July 01, 2018, 02:07:06 PM by hgjf2, Reason: completing »

Offline hgjf2

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« Reply #1 on: April 08, 2018, 08:47:23 PM »
The new complex function ef[P](z) can be developing as infinity sum:
An example:
ef[z^2-1](log[2]ln(x))=x+(1/2x)+(1/8(x^3))-(1/16(x^5))+(11/128(x^7))-(9/256(x^9))+-... whick seem to be chaotical sum, but not due can be determining with a reccursive formula:
(x+(1/2x)+(1/8(x^3))-(1/16(x^5))+(11/128(x^7))-(9/256(x^9))+-...)^2=x^2+1+(1/2(x^2))+(1/8(x^6))-(1/16(x^10))+(11/128(x^14))-(9/256(x^18))+-...  exacthly the infinity sum for sqrt(1+sqrt(1+...+sqrt(1+x^(2^n))...)) with n-1 paranthesyses yet (n) sqrt-s.
The terms of this sums for deepdelving can be determining with algorithmic languages like MATHLAB or VISUAL BASIC or HTML.
If typing f(x)=ef[z^2-1](log[2]ln(x)) then f(x)=f(sqrt(x))^2 -1.
Then can solving the equation f(f(x))=x^2-1 with numerical method much better than interpolations like Lagrange or Birkoff.
That g(x)=x+(1/2x)+(1/8(x^3))-(1/16(x^5))+(11/128(x^7))-(9/256(x^9))+-... and g(x^2)=x^2+(1/2(x^2))+(1/8(x^6))-(1/16(x^10))+(11/128(x^14))-(9/256(x^18))+-... = g(x)^2-1 then f(g(x)) = g(x^(sqrt(2)))=g(x^(sqrt(2)))=x^(sqrt(2))+(1/2*(x^(sqrt(2))))+
(1/8*(x^(3*sqrt(2))))-(1/16*(x^(5*sqrt(2))))+(11/128*(x^(7*sqrt(2))))-(9/256*(x^(9*sqrt(2))))+-... if f(f(x)=x^2-1, then defineing t=g(x) where f(t)=g(x^(sqrt(2))) and x=g(t) then f(t)=g((g(t))^(sqrt(2))) => f(x)=g((g(x))^(sqrt(2))), and this inverse of g(x) named g(x) can be rendered numerical creating numerical tables like those from the eingineering normatives.  q.e.d.
This exercices can revoluting the math analysis.
The theory deserve to be published in doctoral thesis.
This my theory has allready copyright of my research thesis at math institute
« Last Edit: July 15, 2018, 09:56:05 AM by hgjf2, Reason: notice added »

Offline hgjf2

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« Reply #2 on: October 27, 2018, 10:57:17 AM »
A posible formula for area of Julia set founded on MATH.STACKEXCHANGE.COM based by Milnor's Dynamics in one complex variable of area expressed as a series based on Gronwall's area theorem:

If takes a function f(v)=v(1+a2/(v^2)+a4/(v^4)+a6/(v^6)+...) where f(v^2)=[f(v)]^2+c like in this example:
ef[z^2-1](log[2]ln(v))=v+(1/2v)+(1/8(v^3))-(1/16(v^5))+(11/128(v^7))-(9/256(v^9))+-... [case c=-1]  where here a2=1/2;a4=1/8;a6=-1/16;a8=11/128;a10=-9/256; etc.
The formula for area of Julia set would be S=pi*(1-|a2|^2-|a4|^2-|a6|^2-...) . I yet not sure if this formula is right formula or approximate formula for Area of Julia set. I will investigating and researching about this.
« Last Edit: October 27, 2018, 11:16:52 AM by hgjf2 »


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