Number of precision digits( sum(sum(A)) not equal to sum(A:)

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Monkeymengmeng 2018-6-25

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评论： Monkeymengmeng 2018-6-26

Hey guys. I am new to Matlab. And I need to compute a two-dimension matrix(65 x 65). called matrix A. sth like:

-0.043209482621010  -0.043209460507728  -0.043209425775628  -0.043209372126255
-0.043209488597448  -0.043209470284528  -0.043209441521445  -0.043209397092189
-0.043209517396291  -0.043209517396291  -0.043209517396291  -0.043209517396291
-0.043209590936714  -0.043209637700394  -0.043209711149500  -0.043209824603580
....

But when I do the operation

sum(sum(A)) it computes:1.847411112976260e-13

and the operation

sum(A(:)) it computes:5.652422974122828e-14

shouldn't they be the same consequence? I am confusing with that. Maybe it is about the precision digits. Can anyone help me?Thanks so much.

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Stephen23 2018-6-25

在 MATLAB Online 中打开

This is just accumulated floating point error, and is easy to demonstrate:

>> M = [+1e7,-1e7;1e-7,-1e-7]
M =
  1.0000e+007  -1.0000e+007
  1.0000e-007  -1.0000e-007
>> sum(sum(M))
ans = 0
>> sum(M(:))
ans =  0.00000000058284

This is exactly the same as every other question asked on this forum regarding floating point numbers: different operations on floating point numbers can (and usually do) produce different output values. Summing some values, then summing those sums is clearly a different set of operations than rearranging and then summing.

Monkeymengmeng 2018-6-26

thanks so much.Stephen.

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Answer 1

Ameer Hamza 2018-6-25

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编辑：Ameer Hamza 2018-6-25

It might be happening because of finite precision of floating point numbers, Also, calling the sum() twice seems to be amplifying the errors caused by the finite precision. Since you are dealing with very small numbers, you might want to use arbitrary precision arithmetic to get accurate answers.