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 Cardinality of Infinite tetration Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 06/20/2008, 04:28 PM 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 impossible 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 « Next Oldest | Next Newest »

 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 andydude - 06/17/2008, 08:52 PM RE: Cardinality of Infinite tetration - by Ivars - 06/18/2008, 08:12 AM RE: Cardinality of Infinite tetration - by bo198214 - 06/20/2008, 02:16 PM RE: Cardinality of Infinite tetration - by Ivars - 06/20/2008, 04:28 PM RE: Cardinality of Infinite tetration - by bo198214 - 06/20/2008, 09:18 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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