In deference to Halloween, I thought of posting a short note on a couple of Matlab numerical gotchas. They are not really undocumented – in some cases they are specifically documented and in other cases can be inferred from the documentation. What they share in common is that for some reason they are not widely known. Their importance comes from the fact that I often encounter them in Matlab code, and they often do not immediately cause syntax error, but rather more difficult-to-diagnose incorrect results. So I thought it would be worthwhile to formally document them here:

^[1]

• Inverse-trig of “invalid” values ^[2]
• Numeric (in)equalities ^[3]
• Cube root returns a complex number ^[4]
• Usage of “+” and “-” in arrays ^[5]

1. Inverse-trig of “invalid” values

This one came to me from Baruch Karlin:
In a program using math functions, it is easy to forget to limit the values of interim computations to the boundaries expected in the physical world. One specific example is that while we normally think of inverse trig functions (e.g., asin and acos) as returning the angles corresponding to the specified trig function result, in fact such inverse-trig functions are defined mathematically ^[6] such that their range extends past the “standard” domain (-1 to +1 in the case of asin and acos):

>> value = some_computation();
>> result = asin(value)   %value=10 in this example
result =
       1.5707963267949 -  2.99322284612638i

As the snippet shows, it is sometimes difficult to immediately see that the value variable could receive an inappropriate value, resulting in a complex result. Assuming this is undesirable, we could clamp the value:

value = some_computation();
value = min(max(real(value),-1),1);   %value is clamped to [-1,+1]
result = asin(value);   % result is assured to be real

2. Numeric (in)equalities

This is an oldie, in this case well predating Matlab, going back all the way to the early days of floating-point arithmetic. It is probably the most widely-known and yet most widely-encountered gotcha in any programming language since the days of yore. It is featured in the semi-official Matlab FAQ ^[7], which explains the reason that:

>> 1 == (3 -1.1 -0.9)
ans =
     0    %=false!
>> 2 -1.1 -0.9
ans =
    -1.11022302462516e-016

The workaround is to use eps whenever comparison of non-integer numeric values is needed, and indirect comparison (e.g., < or >) rather than direct ones (== or ~=).
The reader in referred to the above-mentioned FAQ, or to one of the following resources:

• Official Matlab documentation on floating-point accuracy and pitfalls ^[8]
• Loren’s blog post on this ^[9]
• Cleve Moler’s newsletter article ^[10]
• Wikipedia on FP numerical inaccuracies ^[11]
• Floating-point computation pitfalls article ^[12]
• What every computer scientist should know about floating-point arithmetic article ^[13]

3. Cube root returns a complex number

This one is also an oldie from the semi-official Matlab FAQ ^[7] (a bit paraphrased):
Since 8^1/3=2, we naturally expect to find that (-8)^1/3=-2. Unfortunately we find in Matlab that:

>> (-8)^(1/3)
ans =
                1 +      1.73205080756888i

In the same way there are two solutions (plus and minus) for the square root of a positive real number, there are multiple solutions for roots of negative (and complex) numbers. If you express a number as A*eiθ, its cube root takes the form (A^1/3)*e^i(θ+2kπ)/3, for k=0:2. Since -8 is 8e^iπ, θ=π or -π. Therefore, the cube roots of -8 are 2e^iπ/3, 2e^-iπ/3, and 2e^iπ. The last one simplifies to the expected real value of -2.
So we see that the cube root (and in general, the n-th root) always returns 3 roots (or in general, n roots), and Matlab chooses one of them, which may not necessarily be non-complex. As in the first gotcha, the resulting complex value is correct in the mathematical sense, but could well lead to erroneous results downstream in the Matlab code.
Matlab always returns the first solution counter-clockwise from the positive real axis. Armed with this knowledge, we can compute all or some particular root. For instance, if you want the negative real cube root, simply take the cube root of the absolute value of the number, and negate it. For a different wording and more information, see MathWorks Tech Solution 1-15M1N ^[14].
Joe Sababa suggested a method to find all the roots at once using roots on the polynomial, effectively solving x3+8=0 for x:

>> P=[1 0 0 8]; roots(P)  % solve x^3 + 8 = 0
ans =
                         -2
          0.999999999999999 +      1.73205080756888i
          0.999999999999999 -      1.73205080756888i

(note the FP inaccuracy, discussed in the previous gotcha)
An alternate method to obtain -2 as a cube root of -8 is to use the nthroot function:

>> x = nthroot(-8, 3)
x =
    -2

4. Usage of “+” and “-” in arrays

Usage of “+” and “-” operators is normally intuitive. Matlab knows when we refer to the unary operation and when to the binary one. But it gets somewhat less intuitive in numeric arrays:

>> [1 -2 3 4 5]  %unary minus
ans =
     1    -2     3     4     5
>> [1 - 2 3 4 5]  % binary minus
ans =
    -1     3     4     5
>> [1 +2 3 4 5]  % unary plus
ans =
     1     2     3     4     5
>> [1 + 2 3 4 5]  % binary plus
ans =
     3     3     4     5

As can be seen, the extra space between the operator and the number caused Matlab to change its treatment of the operator. This also applies to programmatic values such as: [value1 -value2 ...].

^[15]

for columnIdx = 1 : 5
   cellAddress = ['a' +columnIdx-1 '3'];  % should be, for example: 'c3'
   Excel.Range(cellAddress).Value = someValue;
end

Can you spot why Excel croaked when the Matlab program ran?
The workaround here is to either ensure that unary operators are directly adjacent to the values, or separate the values with a comma (or a semi-colon ; ) to prevent any misunderstanding by Matlab:

>> [1, - 2, 3, 4, 5]
ans =
     1    -2     3     4     5
>> [1; - 2; 3; 4; 5]
ans =
     1
    -2
     3
     4
     5

9 Comments To "Matlab numerical gotchas"

#1 Comment By Walter Roberson on Facebook On October 30, 2013 @ 14:08

#2 Comment By Chandrakanth On October 31, 2013 @ 08:58

 
m>=3  % is fine.
m>= 3 % notice the space before 3. This is fine as well.
m >= 3 % notice space before 3 and after m. This too is fine.
m >=3 % space only after m. This complains with an error.

#3 Comment By Wesley Brodsky On November 1, 2013 @ 10:28

#4 Comment By the cyclist On November 1, 2013 @ 13:56

#5 Comment By Guillaume On November 6, 2013 @ 02:24

#6 Comment By Yair Altman On November 6, 2013 @ 12:39

#7 Comment By Daniel E. Shub On November 11, 2013 @ 05:06

#8 Comment By Malcolm Lidierth On November 18, 2013 @ 09:04

#9 Comment By Sami Varjo On January 9, 2014 @ 06:35

 
>> myUint8Var = uint8(0);
>> newVector = [myUint8Var 510.14 45.24 1.2]
newVector =
    0  255   45    1

>> array = zeros(3,4,'double');
>> array(1,:) = [myUint8Var 510.14 45.24 1.2]
array =
0   255    45     1
0     0     0     0
0     0     0     0