x^x

[Paul]

f(x) = x^x
Given this function, defined, let's say for all real numbers, is there any way to tell when x is rational versus irrational for integer values of f(x)?
e.g.
x^x = 4
x = 2
x^x = 27
x = 3
x^x = 3
x = 1.825455054...

Thanks!

[Tom Mattson]

You mean other that simply taking the xth root of x?

[Paul]

Do you mean the xth root of f(x)? Or am I misunderstanding? And yes, I meant other than simply looking at specific argument and function values.

[Tom Mattson]

Do you mean the xth root of f(x)?


Whoops, I sure do. :redface:


And yes, I meant other than simply looking at specific argument and function values.

I'm not sure off the top of my head, but let me play with it.

[abertram28]

3^3 doesnt equal, so x=3 is irrational. sorry to nitpick.

interesting problem though. i bet there isnt a way besides applying the interger set for x and assuming all others will be irrational. is there any value for f(x) that results in a non interger rational number? if so, that negates using the interger set to find the y.

[Paul]

Thanks, I corrected the typo. And I guess that would put us on the path to a solution. Essentially, given that x is not an element of Z (the integer set), is x^x \in Z possible?

[Nexus[Free-DC]]

Okay, suppose x=a/b, where a,b are integers. Assume that the fraction is reduced, ie. gcd(a,b)=1. Then x^x= (a/b)^(a/b)=(\frac{a^a}{b^a})^\frac{1}{b}

But gcd(a^a,b^a)=1, and therefore (\frac{a^a}{b^a})^\frac{1}{b} is irrational. It follows that if x^x is an integer, then either x is an integer or transcendental.

[abertram28]

so there are also no solutions that are non intergers but rational, like 1/3 and 1/4?
i understand that x^x is an interger if x is an interger.

that tells us that the set of intergers for x gives us the solution set of y? there are no fractional xs for interger ys? sorry im dumb. i want to learn though!

[StonedPanda]

I was wondering the same thing about taking the logarithm or ln of a function.

[Zurtex]

I was interested in this quite some time ago. I finally found an article saying there is no known way of rearranging the equation:

y=x^x

In to some function of y in terms of x. But I did find an iterative formulae so you could approximate to as much accuracy as you wanted if you had y and wanted to know x. I'll see if I can find it again.