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[AIS] (alternating) Iteration series: Half-iterate using the AIS?
#7
(12/10/2012, 10:38 AM)Gottfried Wrote:
(12/10/2012, 03:33 AM)mike3 Wrote: (...)
I've tried other ways of arranging the partial sums (like Cesaro-summing the positive-"degree" terms and the negative-"degree" terms separately), but it doesn't seem to help. Do I need to use Euler summation?

Hi Mike, I did not check Cesaro-summability; I just used the sumalt-procedure in Pari/GP:

Code:
fmt(200,12)   \\ set internal float precision to 200 and display digits to 12
                    \\                   (by user defined function)

tb=1.3 \\ set the exponential base in global variable
tl = log(tb) \\ set the log of the base

\\ procedure to allow sequential access to consecutive iterates via sumalt()
\\ nextx(x,0) - initializes glbx
\\ nextx(x,h) - if h>0 gives the next iterate towards the attracting fixpoint
\\ nextx(x,h) - if h<0 gives the next iterate towards the repelling fixpoint

      glx=0   \\ global x-variable
nextx(x,h=1)=if(h==0,glx=x,if(h>0,glx=exp(glx*tl)-1,glx=log(1+glx)/tl));return(glx)
    \\   nextx(1.5, 0)  \\ example call

\\ == procedure for doubly infinite alternating iteration series beginning at x
{asum(x)=local(su0,su1,su);
     su0  = sumalt(k=0,(-1)^k*nextx(x,k));
     su1  = sumalt(k=0,(-1)^k*nextx(x,-k));
     su   = su0+su1-x;
return(su); }
    \\ asum(0.6) \\ example

[update]: Even more simple: the alternating sum towards the fixpoint fp0=0 converges; but also the alternating sum towards the upper fixpoint fp1 can be made by separating the convergent sum of the and then add the half of the fixpoint (the dirichlet's eta at 0 is 0.5):
PHP Code:
sum(h=0,100,(-1)^h*(nextx(1.4,-h) -fp1 ) )  + 0.5fp1 

(take a meaningful upper limit for the sum instead of 100)

Ah... now I see, it seems I wasn't using enough terms in the Cesaro summation. But dang, this thing is close to 0. (EDIT: Ah! I see the scale in your graph... I didn't see that second scale on the right hand side -- guess I'm all OK here with this, now.)

Using the sumalt, or even better convergent summation, seems to yield the same values, so I guess it's rescued.
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Messages In This Thread
RE: Half-iterate using the infinite iteration-series? - by mike3 - 12/10/2012, 11:26 AM

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