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Operation with large and small numbers

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Franklin
Franklin 2023-1-5
评论： Image Analyst 2023-1-6
1^4 + 1^4 + 23512^4 - 23512^4 = 2
but
Matlab says
1^4 + 1^4 + 23512^4 - 23512^4 = 0.
  1 个评论
Franklin
Franklin 2023-1-5
Please what is the quickest way to solve this problem? I need help with variable precision integer arithmetic.

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回答（2 个）

VBBV
VBBV 2023-1-5
(1^4 + 1^4) + (23512^4 - 23512^4)
  2 个评论
VBBV
VBBV 2023-1-5
编辑：VBBV 2023-1-5
Add operation is done first on numbers and then subtracts. To get desired result use () which shows precedence in operator actions on numbers
Franklin
Franklin 2023-1-6
Thank you.
My numbers are variable. The order in wich they are grouped in parenthesis is important. (1^4 + 23512^4) + (1^4 - 23512^4) = 0.

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Image Analyst
Image Analyst 2023-1-5
It's called truncation error. You should have learned about it in your linear algebra or numerical analysis course.
https://en.wikipedia.org/wiki/Truncation_error_(numerical_integration)
Basically it's caused by adding a gigantic number to a small number (2). There are not enough digits in 64 bits to accurately represent the 2 in the ones column of the number. The number if basically truncated and the 2 and perhaps even some other digits (like the 10's place) are lost.
To fix, use parentheses:
result = (1^4 + 1^4) + (23512^4 - 23512^4)
The first term is 2 and the second term is 0. Those numbers are in the same range and can be added with no truncation.
  2 个评论
Franklin
Franklin 2023-1-6
Thank you.
My numbers are variable. The order in wich they are grouped in parenthesis is important. (1^4 + 23512^4) + (1^4 - 23512^4) = 0.
Image Analyst
Image Analyst 2023-1-6
See the FAQ:
https://matlab.fandom.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_(or_similar)_not_equal_to_zero?
You're still bound by truncation error.
Why is the order important, and why do you have such enormous numbers added to such tiny numbers?
You might have some luck with the symbolic toolbox, but I don't have that so can't advise you on that. Call tech support about that.

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