x^(x^(x^(x^x^...)))....)))) = 2
Sujit
What is solution to
x^(x^(x^(x^x^...)))....)))) = 2?
( That is x raised to x raised to x ..infinite times... = 2)
One immediate thing comes to mind is:
it is same as x^(the whole thing which is 2) =2
so x = sqrt 2
but if we consider,
x^(x^(x^(x^x^...)))....)))) = 4
still we get (if apply same logic)
x=sqrt 2
but then x^(x^(x^(x^x^...)))....)))) is 2 or 4, if x=sqrt 2?
Surely if x=sqrt2 this series should blow up..and can't remain finite.....
Then what is the way to find x?
Thanks,
---
Regards,
Sujit P Gujar.
IISc Bangalore.
web: http://people.csa.iisc.ernet.in/sujit
Sujit
But right now I don't have proof...
Surely,
I will like to have experts' opinion.
Thanks,
---
Regards,
Sujit P Gujar.
IISc Bangalore.
Pubkeybreaker
> Hello all,
>
> What is solution to
> x^(x^(x^(x^x^...)))....)))) = 2?
> ( That is x raised to x raised to x ..infinite times... = 2)
>
> One immediate thing comes to mind is:
> it is same as x^(the whole thing which is 2) =2
> so x = sqrt 2
Here is a hint:
For what positive real number z is z^z maximal?
David W. Cantrell
> Hello all,
>
> What is solution to
> x^(x^(x^(x^x^...)))....)))) = 2?
> ( That is x raised to x raised to x ..infinite times... = 2)
>
> One immediate thing comes to mind is:
> it is same as x^(the whole thing which is 2) =2
> so x = sqrt 2
That is correct.
> but if we consider,
> x^(x^(x^(x^x^...)))....)))) = 4
> still we get (if apply same logic)
> x=sqrt 2
> but then x^(x^(x^(x^x^...)))....)))) is 2 or 4, if x=sqrt 2?
It's only 2.
I think there is a result due to Euler that the equation
x^(x^(x^(x^x^...)))....)))) = c has a real solution iff 1/e <= c <= e.
David
G. A. Edgar
> > but then x^(x^(x^(x^x^...)))....)))) is 2 or 4, if x=sqrt 2?
>
> It's only 2.
>
> I think there is a result due to Euler that the equation
> x^(x^(x^(x^x^...)))....)))) = c has a real solution iff 1/e <= c <= e.
Perhaps it is better to say "positive" not "real"...
If x = -1, would you say that x^(x^(x^(x^x^...)))....)))) converges?
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
David W. Cantrell
> > > but then x^(x^(x^(x^x^...)))....)))) is 2 or 4, if x=sqrt 2?
> >
> > It's only 2.
> >
> > I think there is a result due to Euler that the equation
> > x^(x^(x^(x^x^...)))....)))) = c has a real solution iff 1/e <= c <= e.
>
> Perhaps it is better to say "positive" not "real"...
> If x = -1, would you say that x^(x^(x^(x^x^...)))....)))) converges?
Indeed! I might even make the verb past tense:
If x = -1, x^(x^(x^(x^x^...)))....)))) has already converged. ;-)
David
Denis Feldmann
Euler, I dont know, but it is reasonably easy to prove. Except it is
first *essential* to define x^(x^(x^(x^x^...)))....)))) as the limit of
the sequence u_0=x, u_1=x^x,..., u_(n+1)=x^(u_n) ; then classical tests
of convergence of this sequence (here using the mean value theorem) give
the result above
Wlodzimierz Holsztynski (wlod)
>
> Here is a hint:
>
> For what positive real number z is z^z maximal?
Pardon me?
Here is a hint:
z = 1000 does not work,
nor z = 1000000, nor...
Wlod
Jonathan Hoyle
>
> What is solution to
> x^(x^(x^(x^x^...)))....)))) = 2?
> ( That is x raised to x raised to x ..infinite times... = 2)
>
> One immediate thing comes to mind is:
> it is same as x^(the whole thing which is 2) =2
> so x = sqrt 2
>
> but if we consider,
> x^(x^(x^(x^x^...)))....)))) = 4
> still we get (if apply same logic)
> x=sqrt 2
> but then x^(x^(x^(x^x^...)))....)))) is 2 or 4, if x=sqrt 2?
> Surely if x=sqrt2 this series should blow up..and can't remain finite.....
> Then what is the way to find x?
For any x^x^x^... = z, then it must be true that x^z = z. Let us
assume that x,z real with x,z>0. If x^z=z, then x=z^(1/z). Since 0 <
z^(1/z) < ~1.4446... for all z, then we see that x^x^x^... cannot
converge for x > ~1.4446...
This does not prove that x^x^... will converge everywhere on
(0,1.4446...), but if it did converge anywhere in this interval, it
must converge to one of the two values which solves x=z^(1/z).
Gottfried Helms
> On May 16, 7:20 am, Sujit <sujit.gu...@gmail.com> wrote:
>> Hello all,
>>
>> What is solution to
>> x^(x^(x^(x^x^...)))....)))) = 2?
>> ( That is x raised to x raised to x ..infinite times... = 2)
>
I want to add (just) an opinion on notation and convention.
One cannot evaluate this -
x^(x^(...)) = 2
because there is no initial value to start with the
evaluation.
I think you (I'm adressing the OP here) should write
...^(x^(x)) = 2
and then
...^(x^(x)) = y
x^y = y
Now you may search for x such that x,x^x,x^x^x,...^x^x^x
converges and find a range for x, where this converges.
(Euler-range for x):
1/e^e < x < e^(1/e)
or
-1*e < ln(x) < e^-1
Insert this and you find
x = 2^(1/2) = sqrt(2)
which is in the (Euler-) range.
Now you say, but the same x can be used to express:
...^(x^(x)) = 4
because
x^4 = 4
x = 4^(1/4)
But how would you introduce a function (of x), which provides
two values for the same x? Functions are *defined* to provide
exactly one value for each x (if at all) .
And also: you may try any value for x from the convergent
Euler-range and approximate- you'll never arrive at 4.
It's the same as with x^2 = y
Formulated as a *function* x=f(y)
x = f(y)= y^(1/2)
you have only one solution x for each y (for y=2 we have x = 1.414... )
As a function in x
y = f(x) = x^2
we have then one value for y for each x (where different x may give
the same value for y, for x0=-sqrt(2), x1=+sqrt(2) y0=y1=2 )
Using the above problem it is
x = f(y) = y^(1/y)
By the definition of a function we have one solution x for each y
(which may be arrived by different y, namely x=sqrt(2) for y=2 and y=4).
If we want to understand this as *function* of x
y = f(x) = lim ...^(x^(x^x))
then by definition we can have only one value of y for any x by the
definition of a function.
(Again: many x may give the same value for y - but this was not
asked in the OP)
If we want to understand y=...^(x^(x^x)) as *inverse* f^-1(y) of
x = f(y) = y^(1/y), with a *set of solutions* in y we have to
introduce a new convention, different to that of functions.
For the inverse of the exponential-function, y = f(x) = exp(x),
we have the same problem, we find not only two (as in the
sqrt-example) but even infinitely many (complex) values x giving
the same y:
we have a set of values of x mapping to the same y : {x0,x1,x2,...} -> y
and a notion for an inverse of this mapping y-> {x0,x1,x2,...} needs
then the introduction of another concept; often simply formulated as
"multivalued function".
I find this an unlucky formulation, and think, there should
be a better name for this. Maybe this name was chosen, because
for y=exp(x) the set of {x0,x1,x2,...} has a simple structure:
xk = x0 + k*2*pi*I , k=0..inf
(But that's only speculation, don't really know the exact whereabouts
of the introduction of the name "multivalued function" besides Euler's
treatise of the log-function, where he dealt with this)
---
Hmm, my own argumentation still doesn't look very straight, as I read
this second time before posting, sorry; I seem to need some more practice
with this discussion myself...
Gottfried
--
---
Gottfried Helms, Kassel
Pubkeybreaker
> > Here is a hint:
>
> > For what positive real number z is z^z maximal?
>
> Pardon me?
Of course that should have been z^(1/z)
Sujit
---
Regards,
Sujit P Gujar.
IISc Bangalore.
Web : http://people.csa.iisc.ernet.in/sujit
> On May 16, 3:36�pm, "Wlodzimierz Holsztynski (wlod)"
Tim Little
> What is solution to
> x^(x^(x^(x^x^...)))....)))) = 2?
> ( That is x raised to x raised to x ..infinite times... = 2)
As written it isn't very well defined. I'd formalise it a bit more by
defining the tetration operator ^^ in terms of repeated
exponentiation, and asking for a solution in x to lim(n->oo) x^^n = 2.
> One immediate thing comes to mind is: it is same as x^(the whole
> thing which is 2) =2 so x = sqrt 2
Yes, that is a solution to the equation.
> but if we consider, x^(x^(x^(x^x^...)))....)))) = 4 still we get (if
> apply same logic) x=sqrt 2
That isn't a solution if you define the problem in terms of limits.
The unique value of lim(n->oo) sqrt(2)^^n is 2. There is no solution
to lim(n->oo) x^^n = 4.
The idea of finding a limit of by looking at fixed points is not bad.
For a continuous function, the limit will be a fixed point: but the
converse is not necessarily true.
> Surely if x=sqrt2 this series should blow up..and can't remain finite.....
If y < 2 then sqrt(2)^y < 2, so the series can't blow up.
One interesting question you might like to work on: how large can x be
without the series blowing up?
- Tim