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Re: Re: A problem of numerical precision
*To*: mathgroup at smc.vnet.net
*Subject*: [mg52650] Re: [mg52642] Re: [mg52610] 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> <200412040907.EAA13400@smc.vnet.net> <200412050708.CAA26788@smc.vnet.net>
*Reply-to*: drbob at bigfoot.com
*Sender*: owner-wri-mathgroup at wolfram.com
You used exact versions of A and B in all your trials, where the OP (I think) and I (definitely) defined A and B based on the approximation to b.
I believe Peter Pein did the opposite thing -- he used the lower-precision version of A and B when using the higher-precision b to define it[[1]].
So we all got different results, just as we could expect to do.
Short of using exact arithmetic throughout, I think the fixed-precision method is authoritative. The result was indistinguishable from the machine-precision values, for a very good reason -- errors don't actually accumulate in this calculation, as Mathematica's precision arithmetic assumes they do. Using fixed-precision OR machine-precision sidesteps that arithmetic and prevents the test in f from going wrong.
If we reach an eigenvector corresponding to A's first eigenvalue, we should STAY there (unless the test in f comes True). If we're anywhere else, A.{x,y} tends to diminish error, not increase it. Even B.{x,y} decreases error for many inputs, and the term isn't usually added anyway. If it isn't added when it shouldn't be, no problem!
Hence numerical instability comes almost exclusively from the test in f, and that's affected far less by accumulated error than by artificially diminished precision.
Bobby
On Sun, 5 Dec 2004 02:08:38 -0500 (EST), Daniel Lichtblau <danl at wolfram.com> wrote:
> Guofeng Zhang wrote:
>>
>> 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
>>
>
>
> You are seeing the effect of plotting "zeros" of negative accuracy.
>
> Try it as below.
>
> delta = 1/100;
> a = 9/10;
> b = -3/10*Sqrt[2];
> A = {{1,0}, {b,a}};
> B = {{-1,1}, {-b,b}};
> f2[v_,x_] := UnitStep[Abs[v-x]-delta]
> M = 3000;
>
> itm = Table[0, {i,M}, {j,2}];
> itm[[1]] = N[{-delta/2, (a+b)*(-delta/2)}, MachinePrecision];
> Do[itm[[i+1]] = A.itm[[i]] + f2[itm[[i,1]],itm[[i,2]]]*B.itm[[i]],
> {i,1,M-1}]
>
> it200 = Table[0, {i,M}, {j,2}];
> it200[[1]] = N[{-delta/2, (a+b)*(-delta/2)}, 200];
> Do[it200[[i+1]] = A.it200[[i]] + f2[it200[[i,1]],it200[[i,2]]]*B.it200[[i]],
> {i,1,M-1}]
>
> it500 = Table[0, {i,M}, {j,2}];
> it500[[1]] = N[{-delta/2, (a+b)*(-delta/2)}, 500];
> Do[it500[[i+1]] = A.it500[[i]] + f2[it500[[i,1]],it500[[i,2]]]*B.it500[[i]],
> {i,1,M-1}]
>
> (These could also be computed with FoldList).
>
> Now look at the last elements, or take the Precision, of it500. You will
> find that fewer than 13 digits are correct. At lower initial precision
> you get what amount to indeterminate values (except we have some
> information bounding their magnitudes). Using machine arithmetic there
> is no way to tell from the result whether it is "correct" or just how
> far off it might be. It turns out that they are just fine in comparison
> to it500. So plots of itm and it500 are reliable.
>
> The last 1000+ values for it200 are a different matter. They all appear
> to be zero. For example:
>
> In[156]:= it200[[1888]] == {0,0}
> Out[156]= True
>
> A closer inspection reveals that they have negative accuracy. The
> 1888th, for example, could be anywhere up to around 262 digits in
> magnitude (and it gets worse as we move further along in it200).
>
> In[157]:= InputForm[it200[[1888]]]
> Out[157]//InputForm= {0``-262.03379085102995, 0``-261.84220176123233}
>
> So they plot as zeros, but that's not a very reliable indication of the
> actual values.
>
>
> Daniel Lichtblau
> Wolfram Research
>
>
>
>
>
--
DrBob at bigfoot.com
www.eclecticdreams.net
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