For what x is x^x real?
Thaddeus J. Evert
Dear all,
Having contributed a little in the way of answers to this group, I have a
question:
For what values of x is x^x defined under the real numbers?
I've already determined x^x is real for all
1) positive real numbers
2) negative rational numbers such that if x = p/q (where p/q is in reduced
form) then q must be *odd*.
But what about negative irrational numbers or when x = 0 ? I vaguely
recall some discussion about the latter (whether 0^0 is defined) but
didn't follow the thread. Any ideas would be appreciated.
Thanks in advance,
Thaddeus
\O/ _| \ / |_ \0/
| __\0 \_ | _/ 0/__ |
/ \ | \ |0 /0\ 0| / | / \
"The time you enjoy wasting is not wasted time.
- Bertrand Russell
math...@my-dejanews.com
In article <Pine.SOL.3.91.98061...@lonestar.jpl.utsa.edu>,
"Thaddeus J. Evert" <tev...@lonestar.jpl.utsa.edu> wrote:
>
> Dear all,
>
> Having contributed a little in the way of answers to this group, I have a
> question:
>
> For what values of x is x^x defined under the real numbers?
>
> I've already determined x^x is real for all
> 1) positive real numbers
> 2) negative rational numbers such that if x = p/q (where p/q is in reduced
> form) then q must be *odd*.
>
> But what about negative irrational numbers or when x = 0 ? I vaguely
> recall some discussion about the latter (whether 0^0 is defined) but
> didn't follow the thread. Any ideas would be appreciated.
>
> Thanks in advance,
> Thaddeus
Except for the rationals (=/= 0) which you've identified that's it
since x^x = e^xln(x). For 0 this is undefined
and for negative numbers ln(x) had infinitely many complex values.
For the rationals with odd denominator one of those is real. No others
(negatives) have a real value.
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
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Stan Brown
Someone claiming to be tev...@lonestar.jpl.utsa.edu (Thaddeus J. Evert)
wrote in <Pine.SOL.3.91.98061...@lonestar.jpl.utsa.edu>:
>For what values of x is x^x defined under the real numbers?
i^i is real, too. (Well, it has an infinite number of values, but one of
them is real.)
--
Stan Brown, Oak Road Systems, Cleveland, Ohio, USA
http://www.concentric.net/%7eBrownsta/
My reply address is correct as is. The courtesy of providing a correct
reply address is more important to me than time spent deleting spam.
Earle Jones
Thad: I'm very happy to see that people are still playing with y = x^x. I
used to give this to my students to keep them awake at night.
I think you have found all of the real solutions.
x^x approaches 0 from the right as x approaches 0. That's easy to show.
There is a minimum between x = 0 and x = 1, which can be found by
logarithmic differentiation.
Now: Here's a problem for you. There is a maximum real value of x^x in
the *left* half plane. Can you find it?
earle
--
In article <Pine.SOL.3.91.98061...@lonestar.jpl.utsa.edu>,
"Thaddeus J. Evert" <tev...@lonestar.jpl.utsa.edu> wrote:
>Dear all,
>
>Having contributed a little in the way of answers to this group, I have a
>question:
>
>For what values of x is x^x defined under the real numbers?
>
>I've already determined x^x is real for all
>1) positive real numbers
>2) negative rational numbers such that if x = p/q (where p/q is in reduced
>form) then q must be *odd*.
>
>But what about negative irrational numbers or when x = 0 ? I vaguely
>recall some discussion about the latter (whether 0^0 is defined) but
>didn't follow the thread. Any ideas would be appreciated.
>
>Thanks in advance,
>Thaddeus
>
>\O/ _| \ / |_ \0/
> | __\0 \_ | _/ 0/__ |
>/ \ | \ |0 /0\ 0| / | / \
>
>"The time you enjoy wasting is not wasted time.
> - Bertrand Russell
__
__/\_\
/\_\/_/
\/_/\_\ earle
\/_/ jones
These three cubes are to remind you that
3^3 + 4^3 + 5^3 = 6^3
John Galt
math...@my-dejanews.com wrote in message
<6mb8sl$gpu$1...@nnrp1.dejanews.com>...
>In article <Pine.SOL.3.91.98061...@lonestar.jpl.utsa.edu>,
> "Thaddeus J. Evert" <tev...@lonestar.jpl.utsa.edu> wrote:
>>
>> Dear all,
>>
>> Having contributed a little in the way of answers to this group, I have a
>> question:
>>
>> For what values of x is x^x defined under the real numbers?
>>
>> I've already determined x^x is real for all
>> 1) positive real numbers
>> 2) negative rational numbers such that if x = p/q (where p/q is in
reduced
>> form) then q must be *odd*.
>>
>> But what about negative irrational numbers or when x = 0 ? I vaguely
>> recall some discussion about the latter (whether 0^0 is defined) but
>> didn't follow the thread. Any ideas would be appreciated.
>>
>> Thanks in advance,
>> Thaddeus
> Except for the rationals (=/= 0) which you've identified that's it
>since x^x = e^xln(x). For 0 this is undefined
>and for negative numbers ln(x) had infinitely many complex values.
>For the rationals with odd denominator one of those is real. No others
>(negatives) have a real value.
>
>-----== Posted via Deja News, The Leader in Internet Discussion ==-----
>http://www.dejanews.com/ Now offering spam-free web-based newsreading
Correct me if I am wrong, but I believe that i^i is defined, i being the
root of -1.
dau...@neo.lrun.com
> In article <Pine.SOL.3.91.98061...@lonestar.jpl.utsa.edu>,
> "Thaddeus J. Evert" <tev...@lonestar.jpl.utsa.edu> wrote:
> >
> > Dear all,
> >
> > Having contributed a little in the way of answers to this group, I have a
> > question:
> >
> > For what values of x is x^x defined under the real numbers?
> >
> > I've already determined x^x is real for all
> > 1) positive real numbers
> > 2) negative rational numbers such that if x = p/q (where p/q is in reduced
> > form) then q must be *odd*.
> >
> > But what about negative irrational numbers or when x = 0 ? I vaguely
> > recall some discussion about the latter (whether 0^0 is defined) but
> > didn't follow the thread. Any ideas would be appreciated.
> >
> > Thanks in advance,
> > Thaddeus
> ******
> Except for the rationals (=/= 0) which you've identified that's it
> since x^x = e^xln(x). For 0 this is undefined
--------------------------------------------
Just a fine point here: ln 0 is undefined, 0^0 is indeterminate.
dauvil
--------------------------------------------
Michael Abbott
> "Thaddeus J. Evert" <tev...@lonestar.jpl.utsa.edu> wrote:
>>
>> Dear all,
>>
>> Having contributed a little in the way of answers to this group, I have a
>> question:
>>
>> For what values of x is x^x defined under the real numbers?
>>
>> recall some discussion about the latter (whether 0^0 is defined) but
>> didn't follow the thread. Any ideas would be appreciated.
You are asking about a function f(x)=x^x -- this can be given a value at
x=0, because x^x tends to 1 as x tends to 0 from above.
A rather vague proof goes like this:
f(x) = exp(x ln x).
Substitute u = 1/x
f(x) = exp(-(ln u)/u)
We are interested in
lim{x->0+} f(x) = lim{u->+infinity} exp(-(ln u)/u)
Now it is pretty obvious that (ln u)/u tends to 0 as u goes to +infinity,
and thus the limit for f(0+) is exp(0)=1
HBr...@my-dejanews.com
> Correct me if I am wrong, but I believe that i^i is defined, i being the
> root of -1.
Quite defined, and, in fact, real!
Since e^xi equals to cos x plus i sin x,
It follows that e^(pi/2)i equals to cos (pi/2) plus i sin (pi/2)= 0 +1i= i
So i^i is equal to (e^(pi/2)i)^i which is equal to, by exponential properties,
(e^pi*i*i/2), or e^(-pi/2), which is the reciprocal of the square root of e to
the pith power.
math...@my-dejanews.com
In article <6mhv6n$13l$1...@nnrp1.dejanews.com>,
HBr...@my-dejanews.com wrote:
>
>
> > Correct me if I am wrong, but I believe that i^i is defined, i being the
> > root of -1.
>
> Quite defined, and, in fact, real!
>
> Since e^xi equals to cos x plus i sin x,
> It follows that e^(pi/2)i equals to cos (pi/2) plus i sin (pi/2)= 0 +1i= i
>
> So i^i is equal to (e^(pi/2)i)^i which is equal to, by exponential properties,
>
> (e^pi*i*i/2), or e^(-pi/2), which is the reciprocal of the square root of e to
> the pith power.
Be careful with generalizations from the reals to the complex plane!!!
i^i has infinitely many values (all of which are real!!!)