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Understanding Precision in Matlab

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Ahmad Gad
Ahmad Gad 2020 年 7 月 1 日
コメント済み: Walter Roberson 2020 年 7 月 1 日
I have a simple question. How to split the different between the two numbers into 1e6 increment (unisg linspace)?
The numbers: 5.047639467288342e+4 and 5.047639467288343e+4.
Thanks
  2 件のコメント
Stephen23
Stephen23 2020 年 7 月 1 日
"How to split the different between the two numbers into 1e6 increment (unisg linspace)?"
You can't. There are no binary floating point numbers inbetween the two values that you gave.
We can confirm this very easily by noting that their hex representations differ only by 1:
>> A = 5.047639467288342e+4;
>> B = 5.047639467288343e+4;
>> num2hex(A)
ans = 40e8a58ca12906dd
>> num2hex(B)
ans = 40e8a58ca12906de
Ahmad Gad
Ahmad Gad 2020 年 7 月 1 日
Thank you! Then, I have to consider changing the algorithm then.

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回答 (1 件)

Walter Roberson
Walter Roberson 2020 年 7 月 1 日
You cannot do that with double precision.
If you have the symbolic toolbox then:
r = linspace(sym('5.047639467288342e+4'), sym('5.047639467288343e+4'), 1e6+1);
Note that the results will be symbolic numbers with higher precision than double(), so if you were to double() them, it would be with loss of precision
  2 件のコメント
Ahmad Gad
Ahmad Gad 2020 年 7 月 1 日
The code is numeric and those values will be inserted a matrix determinant function. I think it will be time consuming.
Walter Roberson
Walter Roberson 2020 年 7 月 1 日
Determinants can have a lot of problems with cancellation. If you were to use symbolic numbers it would be better to use
r = linspace(sym(5.047639467288342e+4,'f'), sym(5.047639467288343e+4,'f'), 1e6+1);
However, symbolic determinants are slow. To put 1e6 numbers into an array you would need an array at least 1e3 x 1e3, and a proper accurate determinent of that would involve over 4 * 10^2567 terms.

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