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## Re: [newbie] unexpected behaviour for x^x

**From**: |
Julien Bect |

**Subject**: |
Re: [newbie] unexpected behaviour for x^x |

**Date**: |
Fri, 12 Dec 2014 13:40:16 +0100 |

**User-agent**: |
Mozilla/5.0 (X11; Linux i686; rv:31.0) Gecko/20100101 Thunderbird/31.3.0 |

Le 12/12/2014 13:16, Jean Dubois a écrit :

I was surprised to found out that when I calculated x^x with octave, using
floats which approach zero I get two different results depending on the
direction
in which zero is approached, this is what I mean:
octave:1> 0.001^0.001
ans = 0.99312
octave:2> -0.001^-0.001
ans = -1.0069

Hello Jean,
First thing, be careful with operator precedence:
octave:1> -0.001^-0.001== - (0.001 ^ (- 0.001))
ans = 1

`So the minus sign in front of your second expression gives the minus
``sign in front of your second result, nothing surprising.
`
Now, without the minus sign, I get
octave:1> format short
octave:2> 0.001^0.001
ans = 0.99312
octave:3> 0.001^-0.001
ans = 1.0069
which is the correct result, because
octave:4> 1 / (0.001^-0.001)
ans = 0.99312

`Last thing, x^x with a negative x will give you a complex result, but
``still converging to 1:
`
octave:5> (-0.001) ^ (-0.001)
ans = 1.00692669984728 - 0.00316336393000065i
octave:6> (-1e-10) ^ (-1e-10)
ans = 1.00000000230259 - 3.14159266082358e-10i
I hope this helps.