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Re: Searching list for closest match to p

  • To: mathgroup at smc.vnet.net
  • Subject: [mg79531] Re: Searching list for closest match to p
  • From: Peter Pein <petsie at dordos.net>
  • Date: Sat, 28 Jul 2007 05:42:55 -0400 (EDT)
  • References: <f81m5q$ll1$1@smc.vnet.net>

chuck009 schrieb:
> I'm working on an interesting theorem in Complex Analysis called Jentzsch's Theorem in which the zeros of the partial sums of a Taylor series for an analytic function in the unit disk, all converge to values on the unit disk.  So I choose a point on the unit disk, p=Cos[pi/3]+iSin[pi/3], calculate the normal series for f[x]=Log[1+x] for n ranging from 1 to 100, calculate the zeros, then for each polynomial, search the zeros for the one closest to the point.  Here's my code.  I feel the Table part is messy with the First[First[Position... construct in it.  Can anyone recommend a more concise way of searching the zero lists and finding the one closest to p3?
> 
> Thanks,
> 
> 
> p3color = Red; 
> p3 = Cos[Pi/3] + I*Sin[Pi/3];
>  
> p3mintable = 
>    Table[zlist = x /. N[Solve[Normal[Series[Log[1 + x], {x, 0, nval}]] == 0], 
>         6]; minz = zlist[[First[First[Position[mins = (Abs[#1 - p3] & ) /@ 
>             zlist, Min[mins]]]]]], {nval, 1, 100}]; 
> 
> p3vals = ({Re[#1], Im[#1]} & ) /@ p3mintable; 
> 
> lp3 = ListPlot[p3vals, PlotRange -> {{-1.4, 1.4}, {-1.4, 1.4}}, 
>     AspectRatio -> 1]; 
> 
> Show[{lp3, Graphics[{p3color, PointSize[0.03], Point[{Re[p3], Im[p3]}]}], 
>    Graphics[Circle[{0, 0}, 1]]}]
> 
Dear Chuck,

maybe you formulated inexact.

I do not have Jentzsch's theorem at hand, but it would intuitively make
more sense to me if: Let f be analytic *on C*. Then the zeros of the
partial sums of the Taylor series of f all converge to values on the
unit *circle*.

f(0)=0 doesn't seem to matter.

 I didn't try to prove this (too lazy) but I've got an counterexample
for your version:
f(x)=x/(x-2)^2 is analytic on the unit disc, but:
p3color=Red;
circleColor=Blue;
p3=Exp[I*Pi/3];
f[x_]=x/(x-2)^2;
s100=SeriesCoefficient[Series[f[x],{x,0,100}],#]x^#&/@Range[100];

p3vals=Through[{Re,Im}[#]]&@@@
      Table[zlist=x/.NSolve[Total[Take[s100,nval]]\[Equal]0];
        Pick[zlist,#,Min[#]]&[Abs[zlist-p3]],{nval,100}];

lp3=ListPlot[p3vals,PlotRange\[Rule]All,AspectRatio\[Rule]Automatic,

Epilog\[Rule]{circleColor,Circle[{0,0},1,{0,Pi/2}],p3color,PointSize[.03],
        Point[Through[{Re,Im}[p3]]]}]

shows that the zeroes do not converge to a z0 from the boundary of the
unit disk.
And constructing an easy case of a removable discontinuity with
f[x_]=x/(Exp[x] - 1) shows, that the p3vals are all equal to zero.
Using f[x_]=(x-Pi)/(Exp[x]-Exp[Pi]) to have a removable discontinuity
outside D while being anlytic on D leads to Union[p3val] = {0} too.

I have had a look at the net for this theorem (my books don't know it)
but found only consequences. If you've got this in a common format (*.ps
*.txt *.pdf or whatever) please send a copy to my email.

Thanks in advance,
Peter


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