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 Pages:    Go Down       Author Topic: exponential fractal, logarithm fractal, sinus fractal  (Read 4798 times) Description: new math operators coming soon 0 Members and 1 Guest are viewing this topic.
hgjf2
Fractal Phenom      Posts: 456 « on: December 14, 2013, 08:55:21 AM »

If we imaginate a math function whick have a property as f:C->C a complex function but give f(z+1)=P(f(z)) whatever z<-C , where P(z) is a complex polynom Logged
hgjf2
Fractal Phenom      Posts: 456 « Reply #1 on: December 14, 2013, 09:12:31 AM »

The exponential fractal:
let P=z^2+1; let  f(z+1)=[f(z)]^1+1;
I can make a range: f(0)=0; f(1)=1; f(2)=2; f(3)=5; f(4)=26; f(5) = 677; f(6) = 458330; f(7) = 210066388901; to infinity.
Then f(-1) = +-i ;  f(-2) = sqrt(1+-i) wher i =sqrt(-1). But f(-oo) would have roots placed at Julia set {z|P(P(...P(z)...))<>oo,P(z)=z^2+1} whick is disconnected.
But how give f(0,5) or f(i) ?
This must definetly a new type of complex function: the exponential fractal ef[P]:C->C with ef[P](z+1)=P(ef[P](z)) \/z<-C.
So we can definetly ef[z^2+1](2) as lim{n->oo}sqrt[sqrt[sqrt[...[sqrt[2^(2^n)]-1]...]]] where (sqrt) is by (n) times. then ef[z^2+1](2)
give 1,69379309 whick a irrational number. So ef[z^2+1](3) =3,868935032 and ef[z^2+1](4) = 15,96865828 etc.
Then for approximate from interpolate ef[z^2+1](2,5) , the function ef(z) must be comparate with f(z) = 2^(2^z) because
ef[z^2](z) = 2^(2^z) but 2^(sqrt(2)) = 2,665144143 whick is a difficult irrational number.

More explains and researches coming soon Logged
hgjf2
Fractal Phenom      Posts: 456 « Reply #2 on: December 14, 2013, 09:14:43 AM »

The logarithm fractal.
If giveing ef[P](z) as exponential fractal.
So the logarithm fractal lof[P](z) is inverse of exponential fractal : let v = lof[P](z) then z=ef[P](v). Logged
hgjf2
Fractal Phenom      Posts: 456 « Reply #3 on: December 14, 2013, 09:27:03 AM »

The sinus fractal:
Can we define a complex function like sinus, if e^(iz) is a periodical function because is a complex helix e^(iz) = e^(iz+2ki{pi})
e^(in) is placed on a complex circle {a+bi|a^2+b^2=1} when n<-R (real number).
But let we imaginate a complex function f:C->C where the values would be placed on a Julia set and a border of Julia set if this Julia fractal would be
connected (example {z|P(P(P...(z)...)))<>oo,P(z)=z^2-1}) as soon as have f(z)=[f(z/2)]^2-1 when z=(sqrt(5)+1)/2 whick is inside border these Julia set. Then can we definithly f(z) as cosinus fractal cosf(z) or cf(z). Only that the periodicity this function would chose we because a Julia set don't have length. We know that a length of fractal give infinity, if length of koch snowflake border give lim{n->+oo}(4/3)^n.
If we chose at periodicity of sinus fractal and cosinus fractal as 1: cosf(z) = cosf(z+k) whatever (k) is integer. Defineing a sinus fractal sinf(z) or sf(z) as
cosf((1/4)-z) like as at classic sinus.

More explain and researches coming soon. Logged
hgjf2
Fractal Phenom      Posts: 456 « Reply #4 on: December 14, 2013, 11:23:46 PM »

The sinus fractal:
Can we define a complex function like sinus, if e^(iz) is a periodical function because is a complex helix e^(iz) = e^(iz+2ki{pi})
e^(in) is placed on a complex circle {a+bi|a^2+b^2=1} when n<-R (real number).
But let we imaginate a complex function f:C->C where the values would be placed on a Julia set and a border of Julia set if this Julia fractal would be
connected (example {z|P(P(P...(z)...)))<>oo,P(z)=z^2-1}) as soon as have f(z)=[f(z/2)]^2-1 when z=(sqrt(5)+1)/2 whick is inside border these Julia set. Then can we definithly f(z) as cosinus fractal cosf(z) or cf(z). Only that the periodicity this function would chose we because a Julia set don't have length. We know that a length of fractal give infinity, if length of koch snowflake border give lim{n->+oo}(4/3)^n.
If we chose at periodicity of sinus fractal and cosinus fractal as 1: cosf(z) = cosf(z+k) whatever (k) is integer. Defineing a sinus fractal sinf(z) or sf(z) as
cosf((1/4)-z) like as at classic sinus.

More explain and researches coming soon.

Through exist a posible relation between exponential fractal and sinus fractal.
cosf[P](exp(z)) can be ef[P](z+mi) unde (m) is a real number,  because cosf[P](2*exp(z)) = P(cosf[P](exp(z)) and ln(2*exp(z)) = z+ln(2) a real increment with something. Logged
hgjf2
Fractal Phenom      Posts: 456 « Reply #5 on: December 14, 2013, 11:27:45 PM »

The sinus fractal:
Can we define a complex function like sinus, if e^(iz) is a periodical function because is a complex helix e^(iz) = e^(iz+2ki{pi})
e^(in) is placed on a complex circle {a+bi|a^2+b^2=1} when n<-R (real number).
But let we imaginate a complex function f:C->C where the values would be placed on a Julia set and a border of Julia set if this Julia fractal would be
connected (example {z|P(P(P...(z)...)))<>oo,P(z)=z^2-1}) as soon as have f(z)=[f(z/2)]^2-1 when z=(sqrt(5)+1)/2 whick is inside border these Julia set. Then can we definithly f(z) as cosinus fractal cosf(z) or cf(z). Only that the periodicity this function would chose we because a Julia set don't have length. We know that a length of fractal give infinity, if length of koch snowflake border give lim{n->+oo}(4/3)^n.
If we chose at periodicity of sinus fractal and cosinus fractal as 1: cosf(z) = cosf(z+k) whatever (k) is integer. Defineing a sinus fractal sinf(z) or sf(z) as
cosf((1/4)-z) like as at classic sinus.

More explain and researches coming soon.

The values of cosinus fractal can be follow along Julia set's biomorph stripes.
A graph explain coming soon, now I checking picture for can post here for pass filters and posting rules , else may be reject with posting error. Logged
hgjf2
Fractal Phenom      Posts: 456 « Reply #6 on: December 15, 2013, 10:13:31 PM »

cosin fractal diagram:  julia_FF_.JPG (12.34 KB, 300x300 - viewed 1454 times.) Logged
SamTiba
Safarist      Posts: 83 « Reply #7 on: January 20, 2017, 04:01:12 AM »

There's some time that passed by since the initial posts, but this is the actual sine-fractal (Julia-Set of sin(x)):  Logged

Some of my images: Pinterest
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