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Re: 0^0=1
Posted:
Apr 22, 2012 11:33 AM
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On Apr 22, 5:06 pm, Pubkeybreaker <pubkeybrea...@aol.com> wrote:
> On Apr 22, 9:34 am, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
>
> > Pubkeybreaker wrote:
>
> > > On Apr 22, 2:22 am, Dave <dave_and_da...@juno.com> wrote:
>
> > > > Whether x^a is complex or undefined when x < 0 and a is irrational depends on whether you say that sqrt(-1) is undefined or i, doesn't it?
>
> > > No, it doesn't.
>
> > Well, I'm not so sure. If "you say" that sqrt(-1) is undefined, then
> > that implies that reals only are being considered; and in that case, if
> > x^a were (non-real) complex, then it would be undefined.
>
> x^a is undefined EVEN WHEN sqrt(-1) IS defined.
>
> (hint: x^a for a irrational has infinitely many possible values)
Well, I'm not sure: since we can put x^a = e^{a Log(x)} , with Log =
complex logarithm (Log Z = log|Z| + i*arg(z), with log = the usual
real logarithm function) it's all a matter of chosing some branch of
this function (many times "taking away" the non-positive real axis and
t), and thus we have a nice function: for irrational a we'd have that
its argument is 0 (or another integer multiple of 2Pi*i, if another
branch is chosen).
Tonio
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