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Author Topic: the paradox of complex function by elliptical integral  (Read 1112 times)
Description: complex strangennes
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hgjf2
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Posts: 456


« on: May 13, 2017, 10:34:53 AM »

We known if want to make Julia fractals by f(z)=tg(z)+c, we can approximating with
...*(z-5PI/2)*(z-3PI/2)*(z-PI/2)*(z+PI/2)*(z+3PI/1)*(z+5PI/2)*...+c.
But at the elliptical integrals whick are be byperiodical functions f(z)=integral(sqrt(z^3-z))dz whick is f(z)=f(z)+2k*N+2l*Ni where
N=integral[0;1](sqrt(z^3-z))dz and k,l are integers. If we want to approximating g(z) the invert of F(z) that F(g(z))=z with
...*(z+(-3-3i)N)*(z+(-3-i)N)*(z+(-3+i)N)*(z+(-3+3i)N)*(z+(-1-3i)N)*(z+(-1-i)N)*(z+(-1+i)N)*(z+(-1+3i)N)*
(z+(1-3i)N)*(z+(1-i)N)*(z+(1+i)N)*(z+(1+3i)N)*(z+(3-3i)N)*(z+(3-i)N)*(z+(3+i)N)*(z+(3+3i)N)*...+c for to obtaining a Julia set on C# or on C++ on and ULTRAFRACTAL, then the complex function will be few different of this current elliptical integral function due the complex function will have degree
n^2 and if the function g(z) will have points of inversion transformation (having vertical asymptots) whick isn't allowed at the complex polynomial functions, and F(Z)=integral(sqrt(sin(z)))dz.
Also g(z) using the Scwartz-Cristoffel mapping : details on http://mathfaculty.fullerton.edu/mathews/c2003/SchwarzChristoffelMod.html
Also the approximating will keep the radial stalks like as at my image "Persan Kaleidoscope" posted in topic "Polyhedrons and more polyhedrons" and at my DEVIANTART


* SchwarzChristoffelMod_gr_9.gif (7.92 KB, 648x175 - viewed 131 times.)
« Last Edit: May 13, 2017, 10:42:58 AM by hgjf2, Reason: math correcting » Logged
hgjf2
Fractal Phenom
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Posts: 456


« Reply #1 on: September 03, 2017, 11:25:59 AM »

I founded another paradox at this function product.
If rendering f(z) = lim[n->oo]((prod[k=1..2n][l=1..2n](z+(-1+2k-2n)+(-1+2l-2n)i)/(prod[k=1..2n][l=1..2n]((-1+2k-2n)+(-1+2l-2n)i)) obtaining a complex function but non - periodical due f(0)=1 and f(2)=f(-2)=f(2i)=f(-2i)=exp(TT/2) and f(4)=f(-4)=f(4i)=f(-4i)=exp(2TT) and
f(2+2i)=f(2-2i)=f(-2-2i)=f(-2+2i)=-exp(TT) and f(1+i)=f(1-i)=f(-1+i)=f(-1-i)=0 and f(1)=f(-1)=f(i)=f(-i)=integral[-1;1]((1/x)arctg(1/x))dx whick is elliptical integral. If trying another approximating function
 f(z) = lim[n->oo]((prod[k=1..6n][l=1..2n](z+(-1+6k-6n)+(-1+2l-2n)i)/(prod[k=1..6n][l=1..2n]((-1+6k-6n)+(-1+2l-2n)i)) obtaining another complex function non - periodical with f(0)=1 and f(2)=f(-2)=exp(2arctg3) and f(2i)=f(-2i)=exp(TT-2arctg3) and  f(4)=f(-4)=exp(8arctg3) and
f(4i)=f(-4i)=exp(4TT-8arctg3) and f(2+2i)=f(2-2i)=f(-2-2i)=f(-2+2i)=-exp(TT) and f(1+i)=f(1-i)=f(-1+i)=f(-1-i)=0 and
f(1)=f(-1)=integral[-3;3]((1/x)arctg(1/x))dx and f(i)=f(-i)=integral[-1/3;1/3]((1/x)arctg(1/x))dx ,
exacthly how exists the paradox of the infinite sum 1-(1/2)+(1/3)-(1/4)+(1/5)-(1/6)+-... at whick if swaping the terms may not give ln(2):
1+(1/3)-(1/2)+(1/5)+(1/7)-(1/4)+(1/9)+(1/11)-(1/6)++-... gives (3/2)ln(2) not ln(2).
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