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Matlab returns different values with two equivalent functions

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Marcos Baños
Marcos Baños 2020 年 3 月 8 日
コメント済み: Marcos Baños 2020 年 3 月 10 日
Let's say I have two functions:
f = @(x) (1.01.*exp(4.*x) - 4.62.*exp(3.*x) - 3.11.*exp(2.*x) + 12.2.*exp(x) - 1.99)
F = @(x) (((1.01.*exp(x) - 4.62).*exp(x)-3.11).*exp(x) + 12.2).*exp(x) - 1.99
These two functions are equivalent but when I want to execute with MATLAB it gives me different results with the same input. (for example with 0.925)
I would like to ask why cause this situation and which one has should I trust.
Thanks in advance!
  2 件のコメント
darova
darova 2020 年 3 月 8 日
>> f(0.925)
ans =
-24.2489
>> F(0.925)
ans =
-24.2489
Works as always
Marcos Baños
Marcos Baños 2020 年 3 月 8 日
ans =
-24.2489252551835
ans =
-24.2489252551834
as you see it's not the same, and with 1.53 the difference is bigger

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採用された回答

Thiago Henrique Gomes Lobato
Thiago Henrique Gomes Lobato 2020 年 3 月 8 日
編集済み: Thiago Henrique Gomes Lobato 2020 年 3 月 8 日
The difference between then in my computer with your example is -3.197442310920451e-14 (depending of your processor may be slightely different but still extremely low). This difference is negletible and can be regarded as numerical/approximation error between taking the exp of a number and multiplying two (to four, in your case) exp's. You can reproduce it easily with an easier example:
exp(3.245*2)-exp(3.245)*exp(3.245)
ans =
-1.136868377216160e-13
The results are then actually the same and you can trust both of them with the given accuracy, which should be enough for almost any application you may think.
  6 件のコメント
Stephen23
Stephen23 2020 年 3 月 8 日
編集済み: Stephen23 2020 年 3 月 8 日
Marcos Baños
Marcos Baños 2020 年 3 月 10 日
thanks!

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