Associative Property for Power Towers?

Hertz

A power tower (x^^n) is a variable raised to the power of itself n amount of times.

x^^4 = x^x^x^x
x^^3 = x^x^x
x^^2 = x^x
x^^1 = x

I was wondering if an associative property for power towers exists.

Does x^(x^x) equal the same thing as (x^x)^x? Is x^(x^^n) equal to x^^(n + 1)?

If anybody could prove that the order of the exponents doesn't matter if the exponents are the same, that would be great, but an intuitive reasoning would be great also :)

e-
WHOA!! I didn't realize I posted this in the physics forum. If anybody would be able to move it to the general math discussion forum that would be great

jbriggs444

You begin by talking about power towers. But then you ask a question about ordinary exponentiation.

Does (x^y)^z = x^(y^z)?

well, does (2^2)^8 = 2^(2^8) ?
does 4^8 = 2^256 ?
does 65536 = 1.16 x 10^77 ?

Hertz

Power towers ARE exponentiation, only all the exponents are the same.

Does (2^2)^2 = 2^(2^2)? You tell me.

-e
You do have a good point though. In order for there to be an associative property for power towers there must be a commutative property of exponentiation.

3^27 = 7625597484987
27^3 = 19683

Evidently there is no commutative property for exponentiation.

Hmm.. New question then.. How is a power tower defined?

x^(x^(x^x))
or
((x^x)^x)^x?

jbriggs444

Fair point.

So let's try another example

Is (3^3)^3 equal to 3^(3^3)?
Is 9^3 equal to 3^27?
Is 243 equal to 7625597484987?

jbriggs444

What _is_ true is that (x^x)^x is equal to x^(x*x).
When x=2, it is true that x^x = x*x.
But when x=3 it is not true that x^x = x*x

Nugatory

Two is a special case, because (x^x)^x = x^(x^2)
But in general, exponentiation is not associative.

An interesting follow-up question: For what if any non-negative values of x will the left-associative tower converge? How about the right-associative one?

Hertz

This is an excellent point, thanks for sharing.

I'm sorry but I'm not sure what you mean by the left and right associative towers.

--
Thanks for the help guys, I'm heading to work and I'll revisit the thread when I get back.

jbriggs444

If I'm onto what you're about then...

1^^n = 1 regardless of x and is associative both ways.

0^^n = 1 for even n and 0 for odd n if you go with right associativity and don't mind getting into a flame war over the definition of 0^0.

0^^n = 1 for all n if you go with left associativity and don't mind the flame war.

There is a voice in my head trying to yell that there is a solution to x^x = x that also makes the tower converge.

But I'm feeling a bit out of my depth now.

Nugatory

The right-associative tower is:
(x^(x^(x^(x^.......))))

The left-associative tower is:
((...((x^x)^x)^x)^x)....