Logo by DsyneGrafix - Contribute your own Logo!

END OF AN ERA, FRACTALFORUMS.COM IS CONTINUED ON FRACTALFORUMS.ORG

it was a great time but no longer maintainable by c.Kleinhuis contact him for any data retrieval,
thanks and see you perhaps in 10 years again

this forum will stay online for reference
News: Support us via Flattr FLATTR Link
 
*
Welcome, Guest. Please login or register. July 23, 2018, 08:06:02 AM


Login with username, password and session length


The All New FractalForums is now in Public Beta Testing! Visit FractalForums.org and check it out!


Pages: [1]   Go Down
  Print  
Share this topic on DiggShare this topic on FacebookShare this topic on GoogleShare this topic on RedditShare this topic on StumbleUponShare this topic on Twitter
Author Topic: exponential fractal, logarithm fractal, sinus fractal  (Read 3110 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 888 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
Pages: [1]   Go Down
  Print  
 
Jump to:  

Related Topics
Subject Started by Replies Views Last post
Fractal Explorer - fast fractal generator for Android Smartphones / Mobile Devices Black 6 5090 Last post November 29, 2010, 10:18:20 AM
by Cyclops
Where do the Fractal Science Kit users congregate for fractal posting. Fractal Science Kit wmauzey 4 2513 Last post February 27, 2012, 12:39:58 AM
by wmauzey
Raw Fractal Cosmic Friend ( 100% pure fractal ) Wildstyle milo 0 1013 Last post May 07, 2012, 02:08:58 PM
by milo
Raw Fractal Cosmic Child ( 100% pure fractal ) Wildstyle milo 0 865 Last post May 07, 2012, 02:11:40 PM
by milo
Shells, Fractal Hair - fractal number lines.. Images Showcase (Rate My Fractal) Eric B 0 1008 Last post October 20, 2012, 05:47:13 PM
by Eric B

Powered by MySQL Powered by PHP Powered by SMF 1.1.21 | SMF © 2015, Simple Machines

Valid XHTML 1.0! Valid CSS! Dilber MC Theme by HarzeM
Page created in 0.158 seconds with 27 queries. (Pretty URLs adds 0.011s, 2q)