Cardinality of Infinite tetration
#6
bo198214 Wrote:What is |x[4]2|?
I was not aware that x[4]2 is a set.

Hmm. I thought it can be perceived as set by analogy to n^n which is a set of all permutations of n from n with order and repetitions.

So 2^2 (A^B) is a also multiset of all 4 distinct combinations of elements with repetition (if 2 is a set) and :

lower 2 upper 2 ( AB)
upper 2 lower 2 (BA)
lower 2 lower 2 (AA)
upper 2 upper 2 (BB)

etc.

Now, if x is a set eg defined by whatever is proper e.g ({........|next real after x}) then my question was can

x^x=x[4]2 represent the set of all combinations of lower and upper x so its cardinality will be|x[4]2|= R^R. I know R^R is a cardinality of all functions from reals to reals, but also if we look at permutations which go over all reals smaller than x than the set of all such permutations should have cardinality R^R. Is that impossibleSad

I was also looking at partial derivatives of x^y to see the difference between upper and lower arguments ( and then look at x^x by replacing y with x):

I was wondering if function \( z=x^y \) or \( z=u^v \) does not lead to mere insights about tetration in general and \( x[4]2=x^x \) in particular. \( x^x \) is \( x^y \) with \( y=x \).

If we look at partial derivatives of \( x^y \), they exibit different dependance of change in function value from each of arguments.

If we keep \( y=const \) than :

\( \frac{\partial x^y}{\partial x}=x^{(x-1)}*y \)

If we keep \( x=const \) than:

\( \frac{\partial x^y}{\partial y}=x^{y}*\ln(x) \)

Derivative of \( x^x \) is \( x^{x}*(\ln(x)+1) \). This can be obtained by replacing \( y=x \) and summing both derivatives:

\( x^{(x-1)}*x+x^{x}*\ln(x)=x^{x}*(\ln(x)+1) \)

But we can also maintain 2 parts separately, each one relating to different x-es:

\( x^{(x-1)}*x+x^{x}*\ln(x)=x^{x}+x^{x}*\ln(x) \)

Then next derivatives and their general form are easy to see:

\( x^{(x-n)}*x+x^{x}*(\ln(x))^n=x^{x-n+1}+x^{x}*(\ln(x))^n \)

We can name the First term of the 2 summands as derivative by TOP x and second as derivative by BOTTOM x.

Obviously, both partial derivatives by TOP x and bottom x are infinitely differentiable but follows different patterns. To me it seems there is difference between both x so I thought they can be looked at as somewhat independent sets and their exponentiation will besides giving 1 real value also produce an infinite set with cardinality R^R.

Or have I made some mistake very early again?

Ivars
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Messages In This Thread
Cardinality of Infinite tetration - by Ivars - 06/17/2008, 01:02 PM
RE: Cardinality of Infinite tetration - by Ivars - 06/17/2008, 07:59 PM
RE: Cardinality of Infinite tetration - by Ivars - 06/18/2008, 08:12 AM
RE: Cardinality of Infinite tetration - by Ivars - 06/20/2008, 04:28 PM
RE: Cardinality of Infinite tetration - by Ivars - 06/21/2008, 07:27 PM
RE: Cardinality of Infinite tetration - by Ivars - 07/13/2008, 08:07 PM

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