🚀 go-pugleaf

In article <1qnj79$h7f@network.ucsd.edu> fisher@inls1.ucsd.edu (Yuval Fisher) writes:
>
>
>                               In the spirit of wasting (in this case 
>not so much) time and so on, I have written up a little program which does 
>pixel counting to estimate the "length" of a Cantor set. This program 
>seems to converge to the wrong length in some cases. I think this puts the 
>onus on the pixel counters of the world to explain the problems in this 
>code and why these don't occur for pixel-counting the area of M. It is, no 
>doubt, due to some oversight, misunderstanding, confusion, or sheer 
>clumsiness on my part. Or maybe not. 
>
>In this simple case, the actual measure can be estimated, and the pixel
>counting portion of the program appears to converge to the wrong result. 
>The comments at the top of the code explain what is being done and how to 
>modify the code and so on. I want to thank Jay Hill for many useful 
>comments and observations. 
>



This is an interesting approach.  But I think the problem is
not with the bit counting, but with your estimates of the "true" measure.
If the infinite product in question converges too slowly,
then taking 50,000 factors will not give you a very good
approximation.  I tried it with this:

 #define prop(i)  (1.0/(double)(4*(i+2)*(i+2)))  

An exact value can be computed for the
infinite product (i=0 to infinity) (1-prop(i)).
The value is 2/pi = 0.637; this is from the infinite
product for thesine function.  Your program yields:

  Pixel Counting gives 513 out of 1024 = 0.500976562500000000

which may not be close to 0.637, but even after 50,000 factors, 

  Estimate is 0.848835521687477002

is even worse.

-----------------------

A problem for calculus students:  How many factors in the
infinite product would be required to be sure the answer
is correct to 2 significant figures?

-- 
  Gerald A. Edgar                Internet:  edgar@mps.ohio-state.edu
  Department of Mathematics      Bitnet:    EDGAR@OHSTPY
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