05/25/2008, 10:12 PM
Ivars Wrote: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.Gottfried Wrote:Here are some plots of that tri-furcation, see uncommented list below.
I find the bi-,tri- and multifurcation an interesting subject, as we ask: can we assign an individual value if the iteration oscillates/is furcated - since this is somehow related to the partial evaluation of non-convergent oscillating series, to which we assign a value anyway.
Yes, that is an interesting idea, that these seemingly convergent iterations are actually divergent but get the value in the same way like e.g. series 1-1+1-1..........= 1/2. What then, one would assign to the point \( e^{-\pi/2} \) which when iterated with \( z=I \) on top oscillates between \( +I \) and \( - I \) ?
From complex geometric series , would we have to assign value to that Iteration by analogy with divergent (?) sum:
\( 1/(1+I)=I{^0}+I{^1}-I{^2}+I{^3}-I{^4}+..............=1+I+1-I-1... \)
\( {1/(1+I)} =1/2-I/2 \) whose module is \( {\sqrt2/2} \) and argument \( -\pi/4 \), so value would be:
\( {(\sqrt2/2)}*e^{-I*\pi/4} \)
No, the sum is not the same as \( I-I+I-I... \). First I have to generate such sum where only odd powers of I are present.
Ivars
For the second, to clarify the series you must assign an indexing-scheme, for instance powers of x.
Did you mean:
0 + Ix +0x^2 - Ix^3 + 0 x^4 + I x^5 .... = Ix *(1/(1+x^2)) -> 1/2 I
or
I - Ix + Ix^2 - Ix^3 .... = I*(1/(1+x)) -> 1/2 I
The limits should evaluate this way then (good to see in in the view of matrix-operations on formal powerseries, then the index is always clear under any transformation... ) Fortunately, they come out to be the same, but such equality may not be concluded only by the non-indexed (perhaps compressed by eliminating intermediate zeros) notation.
Gottfried
Gottfried Helms, Kassel

