Associative Property for Power Towers?

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Hertz
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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
 
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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 ?
 
jbriggs444 said:
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 ?

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?
 
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Hertz said:
Power towers ARE exponentiation, only all the exponents are the same.

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

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?
 
jbriggs444 said:
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

This is an excellent point, thanks for sharing.

Nugatory said:
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?

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

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Thanks for the help guys, I'm heading to work and I'll revisit the thread when I get back.
 
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.
 
Hertz said:
I'm sorry but I'm not sure what you mean by the left and right associative towers.

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

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