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Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)
#21
Gottfried Wrote:[quote=Ivars]
[quote=Gottfried]
Hmm, for 1+I-1-I+1+I-1-I+.... I get 1/(1-I) = 1/2 + 1/2I by the formula just invoking Pari/GP on that, and also 1/2 + 1/2I by conventional Euler-summation.
0 + Ix +0x^2 - Ix^3 + 0 x^4 + I x^5 .... = I-I+I-I+..= Ix *(1/(1+x^2)) -> 1/2 I
or
I - Ix + Ix^2 - Ix^3 .... = I*(1/(1+x)) -> 1/2 I

Gottfried

That seems logical. Thanks, I forgot the trick with generating series where x->1. Instead, and BTW, why can't it go to I as well (x->I) ? Certainly putting x=I will also lead to some sort of summation, albeit slightly different generating series. E.g.

0 + x -0x^2 + x^3 - 0 x^4 + x^5 .... = I-I+I-I+.. =x* (1/(1-x^2)) ->x=I-> I*(1/(1+1))= 1/2* I

But also I/2= sin(I*ln(GoldenMean)) = sin(I*ln(1+0.618033..)= sin (I*( 1+0.618+0.618^2/2+0.618^3/3+0.618^4/4+...+0.618^n/n) = sin (I+I*0.618+I/2*0.618^2+I/3*0.618^3+..+I/n*0.618^n+..) -> plus extend sinus in series.

See here Formula 26.

while I try to understand increasing height series, what would be Your take on harmonic imaginary series Sum : I/n? Or easier case, I/n^2?

I was similarly(?) to You wandering that perhaps we need infinite product power series to evaluate tetration ( because of its speed)?

so b[4]n= b^b^b^b... =(a1*b)^1*(a2*b)^2*(a3* b)^3*(a4* b)^4*...*(an*b)^n = ?= c1*b^1+c^2*b^b+c^3*b^b^b+..?

You probably have a more clear view on that.

Ivars
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Messages In This Thread
RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/26/2008, 07:43 AM

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