[Guest]

Something I found rather fascinating. For which value(s) of x does the following function converge (and provide a proof).

f(x) = x^xxx...

[Guest]

EDIT: Sorry, wrong answer, I didn't know what converge meant...

Edited June 13, 2010 by Anza Power

[Guest]

Something I found rather fascinating. For which value(s) of x does the following function converge (and provide a proof).

f(x) = x^xxx...

for any value less than 1 and greater than 0 it will converge.

and for negative values of x it should be less than 1.

[Guest]

for any value less than 1 and greater than 0 it will converge.

and for negative values of x it should be less than 1.

Wouldn't it also converge for x = 0?

[Guest]

x to x to the ...

Edited June 13, 2010 by bmm914

[Guest]

X would converge on the intervals(-∞<X≤-1] and [0≤X≤1]. If -∞<X<1, then denominator under 1 becomes infinitely large and ratio converges to zero. If X=-1, X=0, or X=1 then the answer is always 1. If 0<X<1, then problem will converge to zero since the power is positive and less than 1. If -1<X<0, or 1<X, then problem goes off to infinity.

[bonanova]

For starters, check out a for f(x) = 2.

You can remove the bottom-most number without changing the value [if it exists] of the [infinite] stack.

Thus f(x) = x ^f(x).

So for example if x= sqrt(2), can f(x) be both 2 and 4?.

[Guest]

We can simplify the function by stating x = x^x. Thus x is f(x) as stated in the question. The only solutions to that equation are 1, 0 and -1.

[Guest]

Hmmmm.....I think this will work.

The two answers which will definitely work are 1 and -1.

Numbers less than -1 will alternate between huge and tiny numbers as you keep raising to the next power so they will not work.

For the numbers between -1 and 0, only rational numbers which have odd denominators (in reduced form) will work and converge to -1. All other numbers will yield non real results.

Numbers between 0 and 1 will all work because a positive fraction in this range raised to itself will become smaller and smaller and so will be come 0 in the limit. So, you'll get a number to the 0th power which will be 1.

Any number greater that 1 will get huge quickly and so will go to infinity and thus won't work.

[Guest]

We can simplify the function by stating x = x^x. Thus x is f(x) as stated in the question. The only solutions to that equation are 1, 0 and -1.

This is wrong. Please disregard.

[Guest]

For x < -1, f(x) = Impossible

For x = -1, f(x) = -1

For -1 < x < 0, f(x) = Impossible

For x = 0, f(x) = 0 or 1

For 0 < x < 1, f(x) = x

For x = 1, f(x) = 1

For x > 1, f(x) = Infinite

The function f(x) converge for x = {-1, ]0,1]}

(0 is not included because f(0) converges to multiple points...)

That's what I think.

[Guest]

No correct answers so far.

Most people are saying that x<1. However bonanova has shown that x can be sqrt(2) and still converge. Funny how you get 2 and 4 as solutions. Try it on a calculator and it actually converges to 2.

[bonanova]

Let y = f(x) = x^xxx...

f(x) exists only for certain values of x.

When f(x) exists, clearly y = x^y.

This leads to the nice solution x = g(y) = y^1/y

The function g begins small, rises to a maximum of x = e^1/e at y = e, then tends to x=1 for large y.

For small values of y, g diverges into two stable solutions.

Close inspection shows divergence for y < .065988 ... = 1/(e^e)

where x = 0.3678 ... = 1/e.

Solutions exist for in the region y in [.3678..., 2.718...] = [1/e, e].

With corresponding values of x of [.0659..., 1.444...] = [1/e^e, e^1/e].

Proof [not merely graphical inspection] is given and

[Guest]

Let y = f(x) = x^xxx...

f(x) exists only for certain values of x.

When f(x) exists, clearly y = x^y.

This leads to the nice solution x = g(y) = y^1/y

The function g begins small, rises to a maximum of x = e^1/e at y = e, then tends to x=1 for large y.

For small values of y, g diverges into two stable solutions.

Close inspection shows divergence for y < .065988 ... = 1/(e^e)

where x = 0.3678 ... = 1/e.

Solutions exist for in the region y in [.3678..., 2.718...] = [1/e, e].

With corresponding values of x of [.0659..., 1.444...] = [1/e^e, e^1/e].

Proof [not merely graphical inspection] is given and

Nice presentation.

f(x) converges if and only if e^-e <= x <= e^1/e

(approximately between 0.0659 and 1.4446) as shown by Leonhard Euler (1783)

and Ferdinand G. M. Eisenstein (1844).