raising negative number to a power

>> c=-1.5^2
c=-2.2500
>> c=-1.5*-1.5
c=2.2500

The latter result is correct. The former is incorrect. How to raise negative numbers to a power that is internally consistent and correct in Matlab.

Answer 1

[Callum Clarke]

Just put your negative values in brackets:

(-1.5)^2 = 2.2500

In your case Matlab is applying the sign afterwards, as it sees it as a standard operator and is following the usual operation order by calculating the power before subtraction.

[William Perkins]

Thank you. Not intuitive, especially given the syntax for multiplication, but, alas, just one of those things and easy enough to work with.

[John D'Errico]

But it is completely intuitive, completely logical, and completely consistent, in the sense of the order of operations.

Suppose you wrote the expression

1+2^3

How would you expect that to be interpreted? I'd expect to get 9 as a result. Thus, MATLAB knows that an exponentiation is done BEFORE addition and subtraction. So power is first, then multiplication and division, and only then does MATLAB do addition and subtraction.

What you do NOT want to see is MATLAB evaluate the addition first, so getting the number 3, then raise that to the third power. So you would NOT want to get 27 as a result. That would be absurd.

Similarly,

1 - 2^3

should yield -7. It should not generate -1 as a result.

Ok, so then what do you expect to see from

-2^4

Remember, - is still the minus operator. It gets applied only AFTER exponentiation, which has a higher order of precedence.

Thus, -2^4 should generate -16 as a result, as it does.

-2^4
ans =
   -16

If you want to enforce the minus operator to happen first, then use parens.

(-2)^4
ans =
    16

Finally, you need to appreciate that

-2

and

- 2

are the same number. In both cases, it is a unary minus applied to the number 2.

So, completely logical.

[Jason]

Hello,

I typed a rather simple expression: (-8)^(1/3)

but the answer I got was 1.0000 + 1.7321i, instead of a nice -2. Any ideas...?

Thank you, Jason

[Steven Lord]

There are three values that when raised to the third power give -8. [Yes, the real parts of y are not all exactly -8 and y has a non-zero imaginary part. Such is floating point arithmetic.]

format longg
x = roots([1 0 0 8]) % roots of x^3+8 = 0 or x^3 = -8

x = 
                         -2 +                     0i
                          1 +      1.73205080756888i
                          1 -      1.73205080756888i

y = x.^3

y = 
                         -8 +                     0i
          -7.99999999999999 +  4.44089209850063e-16i
          -7.99999999999999 -  4.44089209850063e-16i

Alternately if you know that you want the real root, use nthroot.

z1 = nthroot(-8, 3)

z1 = 
    -2

w1 = z1.^3

w1 = 
    -8