MATLAB: Is it ok to have rounding errors when rounding to integers

bugMATLABround

Run the following:

round(100000,-5)

In my opinion it should return 100000 exactly, but it returns 100000+eps(100000). Do you think this is a bug, or is it an expected round off error because of floating point arithmetic?

Best Answer

Does appear to be a "quality of implementation" issue. Hadn't ever used it w/ the negative optional argument but looks like does have some artifacts, not clear quite where this one would come from. Since it's builtin function can't see implementation but that's what it looks like happens...altho for this input the "deadahead"

>> fix(100000/1E5)*1E5
ans =
    100000
>>

works so not sure what must be going on internally.

Looks to me to be worth a bug report.

A workaround I'm sure you've already thought of is to wrap the call in another round...

>> round(round(100000,-5))
ans =
    100000
>>

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MATLAB: What does the following MATLAB code do

It takes ‘x’, raises it to the ‘n’-th power, adds ‘delta’ to it, then rounds that result toward zero, divides that result by ‘10^n’ and prints the result.

The intent appears to be to round ‘x’ to 7 decimal places.

You can do that easily with the current version of the round function:

r = round(x, 7)

that will produce the same result.

MATLAB: Matlab floating point precision.

Never trust the least significant bits of a floating point number, at least unless you know enough about the extent that you can trust them. But in general, don't.

A double precision floating point number carries roughly 16 digits, actually 52 binary bits of precision.

What happens when you add two numbers that are of different magnitude?

X0 = 1/3
X0 =
         0.333333333333333

So X0 is 1/3. Not exactly because we cannot represent the fraction 1/3 exactly in binary. But that is not really pertinent here.

sprintf('%0.55f',X0)
ans =
    '0.3333333333333333148296162562473909929394721984863281250'

Suppose we add and subtract delta?

delta = 100000.1
delta =
                  100000.1
X1 = X0 + delta - delta
X1 =
       0.333333333328483

What happened? We added a number that was more than 1e5 times as large as X0. Stored in a double, that means we lose the bottom 5 digits of the sum. Subtracting delta off again, does not help, because we already lost that information. Once information is lost, it cannot be magically regained.

Best Answer

Related Solutions

MATLAB: What does the following MATLAB code do

MATLAB: Matlab floating point precision.

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