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

Asked by Monkeymengmeng on 25 Jun 2018
Latest activity Commented on by Monkeymengmeng on 26 Jun 2018
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.

  2 Comments

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.
thanks so much.Stephen.

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2 Answers

Answer by Ameer Hamza
on 25 Jun 2018
Edited by Ameer Hamza
on 25 Jun 2018
 Accepted Answer

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.

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Answer by Walter Roberson
on 25 Jun 2018

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