when I was in a bookshop yesterday, I came across a book about chaos and
dynamics in which I found an interesting piece. It was about an article
that
was posted to USENET (I don't know if it was this group) in which the
following matter was discussed.
It appears that in the point 0.75+0i+epsilon (and also 0.25+0i+epsilon)

with epsilon being a fixed value  the number of iterations needed to
make
sure the point goes to infinity is quite interesting. See below (I hope I
remember this correctly) ...
epsilon  iterations needed

0.1  3
0.01  31
0.001  314
0.0001  3141
0.00001  31415
.  .
.  .
The number of iterations needed appears to be some power of 10 times Pi,
with the decimals cut off.
Does anybody know more about this ? Does anybody know the original poster
?
Any feedback would be absolutely bananas :)
Bye,
Paul

me...@cs.utwente.nl
> It appears that in the point 0.75+0i+epsilon (and also
> 0.25+0i+epsilon) with epsilon being a fixed value  the number of
> iterations needed to make sure the point goes to infinity is quite
> interesting. See below (I hope I remember this correctly) ...
> epsilon  iterations needed
> 
> 0.1  3
> 0.01  31
> 0.001  314
My initial reaction was that this was nonsense, since the point never
reaches infinity. And to "make sure" it will approach infinity, where
c is near 0.25, all that's necessary is that epsilon be more than zero.
But then I tried it for 0.25, seeing how many iterations it takes for
Z to exceed two. The results, in double precision C (about 15 decimal
digits), were suggestive. So I tried it again in MUMPS which has a
higher precision (about 20 decimal digits). Here were the results:
.1 9
.01 31
.001 98
.0001 313
.00001 992
.000001 3141
.0000001 9934
.00000001 31415
.000000001 99345
.0000000001 314158
.00000000001 993458
.000000000001 3141592
.0000000000001 9934590
.00000000000001 31416004
(The first number is epsilon. The second number is the number of
iterations it takes for Z = Z*Z + (0.25+epsilon) to exceed 2.0.)
Maybe those pilike numbers are just a coincidence. But then again,
maybe not. The abberant last number can be explained away as due to
precision loss.
The value "2.0" isn't important. I would have gotten almost identical
numbers had I chosen 1.0, pi, or a trillion. Once Z leaves the vicinity
of 0.5, it takes off like an alien spaceship in _Independence Day_.
Here's the one line MUMPS program I used:
s e=1 f s e=e/10,z=0,c=0.25+e x "f j=1:1 q:z>2 s z=z*z+c" w e," ",j,!
> Does anybody know the original poster?
I did a Dejanews search, and found several articles on the topic that
were just over a year old. Some of them referred to an even earlier
article, supposedly from 1992. I can forward the 1995 articles if
you like.
I haven't looked at 0.75. But doesn't it converge on both (real)
sides of that number? I thought all real numbers from 2.0 to 0.25
were in the Mandelbrot set.

Keith Lynch, k...@access.digex.net
http://www.access.digex.net/~kfl/
That would be Dave Boll, 24 April 1991,
"Pi & the Mandelbrot Set", posted in sci.math.
> I did a Dejanews search, and found several articles on the topic that
> were just over a year old. Some of them referred to an even earlier
> article, supposedly from 1992. I can forward the 1995 articles if
> you like.
>
> I haven't looked at 0.75. But doesn't it converge on both (real)
> sides of that number? I thought all real numbers from 2.0 to 0.25
> were in the Mandelbrot set.
Correct. You need to approach from above (or below) the real axis.

Gerald A. Edgar ed...@math.ohiostate.edu
Didn't the above mentioned book reproduce the original posting? I'll
have to check my copy at home.
>I haven't looked at 0.75. But doesn't it converge on both (real)
>sides of that number? I thought all real numbers from 2.0 to 0.25
>were in the Mandelbrot set.
Use z = (0.75+i*eps).
Also try z = (1.25+i*eps), z = 2  eps.
>
>Keith Lynch, k...@access.digex.net
>http://www.access.digex.net/~kfl/
Jay

int main(){float g,s,f,r,i;char*_="!/,;<:!lnb/bh`r/ylqbAmmhI/S/x`K\n";int m,u,
e=0;_[32]++;for(;e<3919;){u=(256*(s=(r=.0325*(m=e%80)2)*r+(i=.047*(e/80)1.128
)*i)96)*s+32*r<3?25:16+32*r+16*s<1?31:0;if(u==(s=f=0))do g=s*sf*f+r;while((f=
2*s*f+i)*f+(s=g)*g<4&&++u<27);putchar(_[++e>3840&&m<25?31m:m>78?32:u]^1);}}