Re: [newbie] unexpected behaviour for x^x

[Jean Dubois]

2014-12-12 14:14 GMT+01:00 fbarbuto <address@hidden>:
> Your question, and some answers you've got, baffled me in many ways.
>
> Mathematically speaking, -0.001^-0.001 (or -0.001^(-0.001) thereof)  should
> not be compared to 0.001^0.001, because while the latter is a real number,
> the former operation yields a /complex/ number.  That is easy to see if you
> apply logs to both expressions, because since y = x^c (suppose "x" and "c"
> as real numbers), log10(y) = c*log10(x) and hence "y" can also be computer
> as y = 10**(c*log10(x)).  Thus,
>
> log10(0.001^0.001) = 0.001*log10(0.001) = 0.001*(-3) = -0.003  ===> Real
> number
>
> whereas
>
> log10(-0.001^(-0.001)) = -0.001*log(-0.001) ===> Complex number, as the log
> of a negative number does not exist in the realm of the Reals.
>
> Therefore your comparison makes no much sense at all.
>
> From Octave's standpoint I got some surprising results.  Whereas
> -0.001^(-0.001) does indeed yield -1.0069, the same does not happen when you
> create a variable x = -0.001 and perform x^x, as someone cleverly suggested:
>
>>> -0.001^-0.001
> ans = -1.00693166885180
>
>>> x = -0.001
> x = -0.00100000000000000
>>> x^x
> ans =  1.00692669984727590 - 0.00316336393000065i
>
> That is the result I would have expected to see (a complex number), which is
> consistent with:
>
>>> y = -0.001*log10(-0.001)
> y =  0.00300000000000000 - 0.00136437635384184i
>>> 10**y
> ans =  1.00692669984727590 - 0.00316336393000065i
>
> As a cherry on the top of the cake, I have just "discovered" that the old
> and much-loved (by me) Fortran/Python exponentiation syntax, "**", is also
> valid in Octave:
>
>>> 0.001**0.001
> ans =  0.993116048420934
>
> (I never fully liked the caret as an exponentiation symbol).
>
> Regards,
>
> Fausto
>
>
>
>
> --
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> http://octave.1599824.n4.nabble.com/newbie-unexpected-behaviour-for-x-x-tp4667762p4667772.html
> Sent from the Octave - General mailing list archive at Nabble.com.
>
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Thanks for going deeper into this matter
Concluding:
For x real: lim_{x-->0+} x^x=1
However for x real: lim_{x-->0-} x^x is non-existing, even though
numerically calculating lim_{x-->0-} x^x might suggest you get a
complex number
for which the real part goes to one, and the imaginary part goes to zero

jean