[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 \( \pm (x_{-h} - fp1) \) 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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