Operations by matrix disease: 1.000.. - 1 is not equal to zero

Question

Leonardo Coccia 2021-6-14

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/856070-operations-by-matrix-disease-1-000-1-is-not-equal-to-zero

回答： Leonardo Coccia 2021-6-15

Goodmorning everyone, I'm having some trouble with my CLA code. The product delta = P*beta - m_origin must be equal to [0; 0; 0] because P*beta is equal to [1; 2; 3] and m_orign is [1; 2; 3] but the difference between them, in my code, is'nt equal to [0; 0; 0] but to [2.2204e-16; 0; 0]. I've just checked if the first element of the arrays is approximated but this is'nt the case. Thansk to everyone.

n_asset = 3;
C = [1, 0.5, 0; 0.5, 1, 0; 0, 0, 1];
A = ones(n_asset,1)';
P = [C, A'];
m_orig = [1; 2; 3];
beta = [-1.428571428571429; 0.571428571428571; 0.857142857142857; 2.142857142857143];    
delta = P*beta-m_orig 

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Scott MacKenzie 2021-6-14

编辑：Scott MacKenzie 2021-6-14

It is not true that

P*beta is equal to [1; 2; 3]

Here's a little modification I made to your code to illustrate what is going on:

format longg

x = P*beta

y = m_orig

delta = x-y

Output:

x =

0.999999999999999

2

3

y =

1

2

3

delta =

-8.88178419700125e-16

0

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Steven Lord 2021-6-14

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/856070-operations-by-matrix-disease-1-000-1-is-not-equal-to-zero#answer_724690

在 MATLAB Online 中打开

Welcome to the world of floating-point arithmetic.

Please try this little experiment. Find something to write with and something to write on (ideally compatible things; pencil and paper not pencil and whiteboard.)

Step 1: Using long division (like you learned in school) divide 1 by 7. Call the result x. You are allowed to write as many decimal places of the result as you want, but only those you explicitly write can be used in step 2. No indicating a set of repeated digits to get "an infinite" number of places.

Step 2: Multiply x by 7. Call the result y.

In exact arithmetic we know (1/7)*7 is exactly 1. But x is not one seventh. It is slightly smaller than one seventh because you rounded off one seventh to fit it into x. Therefore y will not be 1, it will be slightly smaller than 1.

The numbers in your beta variable are not exactly multiples of one seventh.

beta = [-1.428571428571429; 0.571428571428571; 0.857142857142857; 2.142857142857143];
q = 7*beta
q = 4×1
  -10.0000
    4.0000
    6.0000
   15.0000
format longg
beta - [-10; 4; 6; 15]/7 % not all 0's
ans = 4×1
     -4.44089209850063e-16
     -4.44089209850063e-16
     -1.11022302462516e-16
                         0

Those differences between the elements of beta and multiples of one seventh are small, but they have an impact on the calculations performed using beta.