Which is the machine precision of Matlab?

60 次查看（过去 30 天）
MRC
MRC 2014-2-17
编辑： John D'Errico 2014-2-18
Hi all, I have the following problem: I want to find the argmax (not necessarily unique) of a function f(alpha,beta). Under some assumptions on f it is possible to obtain the argmax set following this procedure:
1) compute F=max f(alpha,beta) wrto alpha and beta
2) for each alpha compute H(alpha)=max f(alpha,beta) with respect to beta
3) pick alpha s.t. H(alpha)=F;
4) pick corresponding beta from step 2)
I'm trying to perform this procedure in Matlab; my question is: assuming that the maximization problems have been correctly solved, how much the result given by Matlab is precise? i.e. in step 3) I should pick alpha s.t. H(alpha)~= F+...? or I should round the numbers to which digit?
I have attached F and H(alpha).
Thank you very much!
  1 个评论
John D'Errico
John D'Errico 2014-2-18
编辑：John D'Errico 2014-2-18
I'm intrigued here. At one point you state you know nothing of numerical analysis. At another point, you state that "under some assumptions on f, it is possible to obtain the argmax".
So which is it? You know nothing, or a fair amount?
By the way, as optimization schemes go, I'm not terribly impressed with this one.
This almost feels like your homework assignment, given the statement above, as if you were fed that line.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Iain
Iain 2014-2-18
Matlab stores and operates on doubles. Doubles have 15-ish significant figures of precision.
If you keep your numbers all roughly the same, you'll manage to get 15ish accurate digits on your answer. If you don't, then, to demonstrate try:
zero = 10^17 + 1 - 10^17
one = 10^17 - 10^17 + 1
You and I both know that the answer to both equations is 1. Double format numbers get that wrong. If you do that sum in unsigned 64 bit integers, it gets it right.
zero = uint64(10^17) + uint64(1) - uint64(10^17)
one = uint64(10^17) - uint64(10^17) + uint64(1)
unsigned 64 bit integers have a precision of "1", and a maximum value of 2^64 - 1 (approx 4*10^18), to which you may apply an offset and scaling factor
  2 个评论
MRC
MRC 2014-2-18
Thank you very much Iain, just to be sure:
(1) " _If you keep your numbers all roughly the same, you'll manage to get 15ish accurate digits on your answer. If you don't, then, ... "_you mean that if from the very first step of my code I round all results, say,to 17 digits after the decimal point, then my final result will be accurate until the 15th digit after the decimal point?
(2) How can I say to Matlab to " keep numbers all roughly the same "? Does Matlab do it by default?
Thanks!
Iain
Iain 2014-2-18
1) No. I mean that matlab rounds every double format number to the nearest number it can represent. This allows 15ish significant figures, significant figures are NOT decimal places. 156965468456231 has 15 significant figures, just like 0.000156965468456231 has 15 significant figures.
2) Matlab does not do that automatically. You need to understand your scenario well to establish where your numerical issues will occur.

请先登录，再进行评论。

更多回答（1 个）

dpb
dpb 2014-2-18
编辑：dpb 2014-2-18
Default in Matlab is double which is stored with 53 bits of precision. This converts to roughly 15-16 decimal digits of precision (log10(2^53)-->15.95).
  5 个评论
显示 3更早的评论
John D'Errico
John D'Errico 2014-2-18
I think dpb was asleep here. A double is stored with 53 BITS of precision, not 53 DIGITS as was stated at one point.
dpb
dpb 2014-2-18
Ooops...typo indeed, edited/corrected. Thanks for pointing it out...

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 MATLAB 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by