Services & ResourcesWolfram ForumsMathGroup Archive

[Date Index] [Thread Index] [Author Index]

Re: Re: A problem of numerical precision

• To: mathgroup at smc.vnet.net
• Subject: [mg52649] Re: [mg52637] Re: A problem of numerical precision
• From: DrBob <drbob at bigfoot.com>
• Date: Tue, 7 Dec 2004 04:09:34 -0500 (EST)
• References: <200412030815.DAA25377@smc.vnet.net> <cos0ps$dhq$1@smc.vnet.net> <200412050708.CAA26762@smc.vnet.net>
• Reply-to: drbob at bigfoot.com
• Sender: owner-wri-mathgroup at wolfram.com

I'm using 5.1 and got very different answers. Here's the second case:

delta = 1/100;
a = 9/10;
b = -3*1.4142135623730950``200/10;
A = {{1, 0}, {b, a}};
B = {{-1, 1}, {-b, b}};

f[v_, x_] := If[Abs[(10^20)*v - (10^20)*x] > (10^20)*delta, 1, 0];

M = 3000;
it = Table[0, {i, M}, {j, 2}];
it[[1]] = {-delta/2, (a + b)*(-delta/2)};

For[i = 1, i < M, it[[i + 1]] = A.it[[i]] +
         f[it[[i, 1]], it[[i, 2]]]*B.it[[i]]; i++];
temp = Take[it, -1000];
ListPlot[temp];

(a single point, where the other plot was a starburst)

You do get the same result for both if you do this the second time:

b = -3*1.4142135623730950``200/10;
M = 3000;
it = Table[0, {i, M}, {j, 2}];
it[[1]] = {-delta/2, (a + b)*(-delta/2)};

For[i = 1, i < M, it[[i + 1]] = A.it[[i]] + f[it[[i,
     1]], it[[i, 2]]]*B.it[[i]]; i++];
temp = Take[it, -1000];
ListPlot[temp];

A and B didn't get reset, so it's a very different series.

Bobby

On Sun, 5 Dec 2004 02:08:18 -0500 (EST), Steve Luttrell <steve_usenet at _removemefirst_luttrell.org.uk> wrote:

> In version 5.1 (Windows XP) I get exactly the same result in both cases.
>
> Steve Luttrell
>
> "Guofeng Zhang" <guofengzhang at gmail.com> wrote in message
> news:cos0ps$dhq$1 at smc.vnet.net...
>>
>> Hi,
>>
>> I met one problem when I do some iteration: By increasing numerical
>> precision, the results are so different from the original one!  I
>> don't know what went wrong, and hope to get some answers.
>>
>> The code is
>>
>> delta = 1/100;
>> a = 9/10;
>> b = -3*1.4142135623730950/10;
>> A = { {1,0}, {b,a} };
>> B= { {-1,1}, {-b,b} };
>>
>> f[v_,x_] := If[ Abs[ (10^20)*v-(10^20)*x]>(10^20)*delta, 1, 0 ];
>>
>> M = 3000;
>> it = Table[0, {i,M}, {j,2} ];
>> it[ [1] ] = { -delta/2, (a+b)*(-delta/2) };
>>
>> For[ i=1, i<M, it[ [i+1]]  = A.it[[i]]+f[ it[ [i,1] ], it[ [i,2] ]
>> ]*B.it[ [i] ]; i++ ];
>> temp = Take[it, -1000];
>>
>> ListPlot[ temp ];
>>
>> By evaluating this, I got an oscillating orbit. I got the same using
>> another system. However, if I increase the numerical precision by using
>> b = -3*1.4142135623730950``200/10;
>> to substitute the original b, the trajectory would converge to a fixed
>> point instead of  wandering around!
>>
>> I don't know why. I am hoping for suggestions. Thanks a lot.
>>
>> Guofeng
>>
>> --
>> Guofeng Zhang
>> PhD student
>> Dept. of Mathematical and Statistical Sciences
>> University of Alberta
>> Edmonton AB
>> Canada T6G 2G1
>> www.ece.ualberta.ca/~gfzhang
>>
>
>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net

• References:
  • Re: A problem of numerical precision
    • From: "Steve Luttrell" <steve_usenet@_removemefirst_luttrell.org.uk>