# How to test a 'float' or 'double' zero numerically?

Discussion in 'C++' started by Peng Yu, Sep 13, 2008.

1. ### Peng YuGuest

Hi,

Suppose T is 'float' or 'double'.

T x;

x < 10 * std::numeric_limits<T>::epsilon();

I can use the above comparison to test if 'x' is numerically zero. But
I'm wondering what should be a good multiplicative constant before
epsilon?

Thanks,
Peng

Peng Yu, Sep 13, 2008

2. ### Anders DalvanderGuest

I can use the above comparison to test if 'x' is numerically zero.

No, as x can also be negative.
Epsilon is the smallest value such such that 1.0 + epsilon != 1.0. You
need to scale it with the numbers to compare with. Comparing against
zero is always hard. You are probably best of with using abs(x) <
your_own_epsilon. Set your_own_epsilon to what ever you want, such as
0.00000001 perhaps.

Regards,
Anders Dalvander

Anders Dalvander, Sep 13, 2008

3. ### Peng YuGuest

Right, I meant std::abs(x).
Therefore, there is no general accept such epsilon?

Thanks,
Peng

Peng Yu, Sep 13, 2008
4. ### Juha NieminenGuest

No you can't. A value of x distinct from zero might also test as
"zero" with that.

Juha Nieminen, Sep 13, 2008
5. ### Ron AF GreveGuest

Hi,

Consider a machine where the smallest number that can be represented is
0.0001

Lets asume I have the following calculation (lets assume the 0.00005 would
be the result of some calculaton).
0.0001 -0.00005 - 0.00005
Now it is obvious that this should result in zero. However the last two
results would be zero since the machine can only have up to four digits
behind the dot. So what should be zero is actually 0.0001 so a correct value
for a multiplier for epsilon would be 0.0002. Reasoning 0.0001 < 0.0002
therefore it is zero?

Consider then the following

The same formula only we also divide by 0.0001 afterwards
( 0.0001 -0.00005 - 0.00005 ) / 0.0001 = 1 However the one actually should
be a zero therefore our first conclusion was incorrect. A correct multiplier
for epsilon should be 10001

Of course one could go on, epsilons multiplier could be anything.

Conclusion there is not a correct multiplier for epsilon. There can be one
per formula but that is probably not very practical.

Regards, Ron AF Greve

http://www.InformationSuperHighway.eu

Ron AF Greve, Sep 13, 2008
6. ### Peng YuGuest

I see. Then the problem is how to derive it for a particular formula.

Probably, I need to write down the formula and take the derivatives of
all its arguments, check how much errors there could be for each
arguments. Then I would end up with a bound of the rounding error
(epsilon is equivalent to it). Right?

Thanks,
Peng

Peng Yu, Sep 14, 2008
7. ### Rune AllnorGuest

Numerical analysis is an art in itself. There are departments
in universities which deal almost exclusively with the analysis
of numerics, which essentially boils down to error analysis.

In my field of work certain analytical solutions were formulated
in the early '50s, but a stable numerical solution wasn't found
until the early/mid '90s.

You might want to check with the math department at your local
university on how to approach whatever problem you work with.

Rune

Rune Allnor, Sep 14, 2008
8. ### Peng YuGuest

In my field of work certain analytical solutions were formulated
Would you please give some example references on this?

Thanks,
Peng

Peng Yu, Sep 14, 2008
9. ### Erik WikstrÃ¶mGuest

No, different applications requires different precision, some would
consider a variable equal to zero if it was 0.0001 from zero while
others might require 0.0000001. You have to analyse your problem to find
a value that suites you.

Erik WikstrÃ¶m, Sep 14, 2008
10. ### Rune AllnorGuest

At the risk of becoming inaccurate, as I haven't reviewed
the material in 5 years and write off the top of my head:

Around 1953-55 Tompson and Haskell proposed a method to
compute the propagation of seismic waves through layered
media. The method used terms on the form

x = (exp(y)+1)/(exp(z)+1)

where y and z were of large magnitude and 'almost equal'.
In a perfect formulation x would be very close to 1.

Since y and z are large an one uses an imperfect numerical
representation, the computation errors in the exponents
become important. So basically the terms that should
cancel didn't, and one was left with a numerically unstable
solution.

There were made several attempts to handle this (Ng and Reid
in the '70s, Henrik Schmidt in the '80), with varoius
degrees of success. And complexity. As far as I am concerned,
the problem wasn't solved until around 1993 when Sven Ivansson
came up with a numerically stable scheme.

What all these attempts had in common was that they took
the original analytical formulation and organized the terms
in various ways to avoid the complicated, large-magnitude
internal terms.

I am sure there are simuilar examples in other areas.

As for an example on error analysis, you could check out the
analysis of Horner's rule for evaluating polynomials, which
is tretaed in most intro books on numerical analysis.

Rune

Rune Allnor, Sep 14, 2008
11. ### Rune AllnorGuest

Typo correction: The problematics terms were on the form

x = exp(y)-exp(z)

where y and z are large and x is small.

Rune

Rune Allnor, Sep 14, 2008
12. ### Ron AF GreveGuest

Hi,

You could indeed do an analysis that way. Actually that kind of thing is
also done when measuring something and one has to know the error in the
measurement. Taking in account the accuracy of measuring equipment and the
kind of operation you (multiplying, addition etc) you can the tell what the
error range is (like I measured 5V +/- 0.5V.

It is a lot of work though.

Regards, Ron AF Greve

http://www.InformationSuperHighway.eu

Ron AF Greve, Sep 14, 2008
13. ### James KanzeGuest

If you want to test whether x is numerically zero, "x == 0.0" is
the only correct way.
There isn't one, since the idiom is broken (in general---there
are specific cases where it might be appropriate).

James Kanze, Sep 15, 2008
14. ### James KanzeGuest

It depends on the computation. There are a lot of contexts
where you get 0.0 exactly, and that's what you want to test for.
There are less contexts where this is true for other values (0.0
is a bit special), but they also exist.

James Kanze, Sep 15, 2008
15. ### James KanzeGuest

No. The problem is how to implement the formula so that it
gives the correct results.
Not necessarily. You need to better understand how machine
floating point works, and the mathematics which underlies it.

Think of it for a minute. If I had a system in which sin(0.0)
returned anything but 0.0 (exactly), I'd consider it defective.
For other values, this is somewhat less obvious, but 0.0 (and in
some contexts, 1.0 and -1.0) are a bit special.

James Kanze, Sep 15, 2008