Re: Old Question

[thompson(at)orpheus.nascom.nasa.gov (William Thompson)]

Ben Tupper <pemaquidriver@tidewater.net> writes:



>Jacques Basson wrote:

>> Hi all
>>
>> Sorry, this has got to be an old question, but I can't seem to locate
>> the answer.  What is the way around the following problem?
>>
>> IDL> a = -1
>> IDL> print, -1^(1./3)
>>      -1.00000
>> IDL> print, a^(1./3)
>>           NaN
>> % Program caused arithmetic error: Floating illegal operand
>>
>> Thanks
>> Jacques

Um, I don't know if you realize this, but the only reason that the first
example DOESN'T fail is because

	IDL> print, -1^(1./3)

is equivalent to

	IDL> print, -( 1^(1./3))

In other words, the minus sign doesn't come into play until after the
exponentiation is applied.  If you typed in

	IDL> print, (-1)^(1./3)

you'd get the NaN result, with the error message, as above.  Note that you can
test this by typing in

	IDL> print, -1^2, -(1^2), (-1)^2
	      -1      -1       1

Of course, as is implied by the answer below, the proper way to evaluate your
example would be

	IDL> a = -1
	IDL> print, a^complex(1./3, 0)
	(     0.500000,     0.866025)

William Thompson


>Hello,

>I now know why it happens.  In the documentation I see...

>Exponentiation

>The caret (^) is the exponentiation operator. A^B is equal to A raised to
>the B power.

>* If A is a real number and B is of integer type, repeated multiplication
>is applied.
>* If A is real and B is real (non-integer), the formula A^B = e^(B ln A)
>is evaluated.
>* If A is complex and B is real, the formula A^B = (re^(iq))^B = r^B *
>(cosBq + isinBq) (where r is the real part of A and iq is the imaginary
>part) is evaluated.

>* B is complex, the formula A^B = e^(B ln A) is evaluated. If A is
>also complex, the natural logarithm is computed to be ln(A) = ln(re^(iq))
>= ln(r) + iq (where r is the real part of A and iq is the imaginary
>part).
>* A^0 is defined as 1.

>Your example falls into the second type of operation.   I don't know how
>to get around that but would like to know also.

>Ben

>--
>Ben Tupper
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