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 Cardinality of Infinite tetration Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 06/17/2008, 01:02 PM (This post was last modified: 06/18/2008, 08:45 AM by Ivars.) Probably has been answered many times,but I have to ask: If cardinality of Natural numbers is Aleph_0, and their powerset P(N) has cardinality of continuum, |P(N)|= c = 2^Aleph_0 which is also the cardinality of R, then: x^x has cardinality of R^R = c^c= cardinality of power set of Reals=|P(R )|=2^c=2^2^Aleph_0 then for each step of tetration we have to add 2, so via recursion: Cardinality |x[4]1|=|P(N)|= |R|=2^Aleph_0 |x[4]n|= |P(x[4]n-1|= 2^| x[4]n-1| and generally: |x[4]n|= |P(x[4]n-1|= (2[4]n)^| x[4]1| What would be cardinality for infinite tetration , known to converge to real values or give complex values by analytic continuation? And what is the cardinality of a number obtained as a result of tetration, even for finite n? Some references here: Cardinality of continuum Another question is , what is than the cardinality of: x[4]y, x[4]I? From previous considerations, can it be (for y>1): |x[4]y|= |P(x[4]y-1|=2^? What is cardinality of I^I? And what would such cardinalities mean? Similarly the same question arises when trying to understand what is fractional or real Cartesian product of sets? May be the answer is in tetration. Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 06/17/2008, 07:59 PM (This post was last modified: 06/17/2008, 09:06 PM by Ivars.) I found interesting links generalizing Integer dimensions of sets via Euler measure: Quote:The Euler characteristic of an object is the most basic dimensionless quantity associated with it . by Tom Leinster. Here: Euler characteristic, category andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 06/17/2008, 08:52 PM Ivars Wrote:c=2^Aleph_0 This has not been proven. In fact this is the continuum hypothesis. Andrew Robbins Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 06/18/2008, 08:12 AM (This post was last modified: 06/18/2008, 08:48 PM by Ivars.) andydude Wrote:Ivars Wrote:c=2^Aleph_0 This has not been proven. In fact this is the continuum hypothesis. Andrew Robbins The question is not so much about the continuum Hypothesis which may be true, may be not, but generally about cardinality of sets produced via operations, including cardinality of Number. E.g. If we have integer number N, its cardinality is assumed |N |which is just the number of elements in set {1,2,...,N}; However, when this N is produced as a result of multiplication of 2 sets representing n1 and n2 than the resulting set is multiset, so its cardinality is greater than |N|. E.g. If we have set {0,1} and {0,1} each of them has cardinality |2|.But product: {0,1}*{0,1} = {0*0, 0*1, 1*0,1*1} = {0,0,0,1} has cardinality |4| with 0 repeated 3 times. So my question arose from looking at exponentiation: If we have a number N2 such that N2= n1^n1 than its cardinality is also |n1^n1| as a result of such operation. It corresponds to choices of n1 from n1, or, combinatorically, if we have n1 differently labeled botlles of beer, how many ways (permutations) we can arrange them in n1 boxes with repetition and when order matters. If we now calculate n1^(n1^n1) the number of boxes goes up to n1^n1, number of permutations n1^(n1^n1), which is also the cardinality of Number N3 = n1[4]3 when obtained via tetration? Unfortunately, numbers in tetration are big, but Integers that can in principle have such cardinality and are possible to partition exponentially in 2 integers are not so many (if we exclude 1 as meaningless in exponentiation): 4=2^2, 8=2^3, 9= 3^2, 16=2^4=2^2^2, 25=5^2, 27=3^3, 32= 2^5, 36=6^2, 49=7^2, 64=2^6=4^3, 81=3^4=3^2^2 ,100=10^2, 125=5^3, 128=2^7, 216=6^3, 243=3^5, 256=2^8=2^2^3, 343=7^3, 512=2^9=2^3^2, 625=5^4, 729=3^6, 1024=2^10, 1296=6^4, 2048=2^11, 2187=3^7, 2401=7^4, 3125=5^5, .. Of these, only few can be partitioned in 3 integers with repetition: 16=2^2^2, 81=3^2^2, 256=2^2^3, 512=2^3^2, 625=5^2^2, 1296=6^2^2, 2401=7^2^2, Without repetition, we get 2 ways to partition some numbers infinitely : exponential factorials n^..4^3^2 and 3_factorials 2^3^....n. May be I have missed some. So the question I put forth in initial post is about cardinality of Numbers resulting from x[4]2, x[4]n, x[4]oo. if x is integer, it is relatively simple to have a multiset as a result of exponentiation which includes all permutations . This can be used also in cases n^(n-1)^(n-2) leading to parts of exponential factorial, and n^(n+1)^(n+2) leading to parts of 3_factorial. If x is fractional, it requires interpretation of irrational cardinality, as e.g. (1/2)^(1/2)= sqrt(2) if x is irrational or transcendental or imaginary, such cardinality gets very strange. E.g. such cardinality of i^i is e^(-pi/2). The only place I have found any ideas in that direction so far is Euler characteristic of categories (and polyhedral sets) which can be imaginary, negative and fractional(see Leistner, Baez, Propp): Propp. Euler measure as generalized cardinality Ivars bo198214 Administrator Posts: 1,384 Threads: 90 Joined: Aug 2007 06/20/2008, 02:16 PM What is |x[4]2|? I was not aware that x[4]2 is a set. 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 bo198214 Administrator Posts: 1,384 Threads: 90 Joined: Aug 2007 06/20/2008, 09:18 PM Ivars Wrote: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. Whatever, but $n$ is a variable so you can regard $5^5$ as a set in this sense but not $n^n$ or $x^x$ as it depends on $n$ or $x$ respectively. Quote:Then next derivatives and their general form are easy to see: $x^{(x-n)}*x+x^{x}*(\ln(x))^n=x^{x-n+1}+x^{x}*(\ln(x))^n$ Unfortunately its not that easy, e.g: $\frac{\partial^2 x^x}{(\partial x)^2} = {{x}^{x} {\left( \ln \left( x \right) + 1 \right)}^{2} } + {x}^{x - 1}$ Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 06/21/2008, 07:27 PM bo198214 Wrote:Whatever, but $n$ is a variable so you can regard $5^5$ as a set in this sense but not $n^n$ or $x^x$ as it depends on $n$ or $x$ respectively. I have to read into this , but category theory deals with variable sets. ( I do not know if this is applicable, but seems so). There is an article in web by Prof. Bell "Abstract and variable sets in category theory". Quote:Unfortunately its not that easy, e.g: $\frac{\partial^2 x^x}{(\partial x)^2} = {{x}^{x} {\left( \ln \left( x \right) + 1 \right)}^{2} } + {x}^{x - 1}$ Thanks for correction. Anyway first differential and separated further derivatives show that there is a difference by which of x such function is differentiated, and these impact in growth speed can be separated in principle. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 07/13/2008, 08:07 PM (This post was last modified: 07/19/2008, 07:43 AM by Ivars.) As I have been reading about cardinalities a little, I was wondering what is the cardinality of number of values of complex logarithm, which are, as well known, infinite in number with a cycle +- 2pi*I, and made a short training excercise. If we have a circle with imaginary diameter I, it has a circumference I*pi (2*ln(I))=ln(i^2) If we have a circle with imaginary diameter I/2, if has circumference I*pi/2 = ln(I) Circle with imaginary diameter I/4 , circumference I*pi/4= 1/2 ln(I) = ln (sqrt(I)) Circle with imaginary diameter I/8, circumference I*pi/8=1/4*ln(I)= ln(I^(1/4)) Circle with imaginary diameter I/(2^n), circumference I*pi/(2^n)=1/2^(n-1)*ln(I)=ln(I^(1/2^(n-1)) Now if we sum circumferences of infinite number of such progression of circles, we get Pi*I*( 1+1/2+1/4+1/8+ +1/2^(n-1)+..)= 2*pi*I = 4*ln(I)= ln(I^4) = ln(1) by the sum of infinite series So we can assume that if we would have added to each I/(2^n) its infinite number of periodic values in form + - (2*pi*k)/n we would have sum of circumferences exactly Ln(1) : Ln(1) = 0, +-2**I*pi, +-4*I*pi, +- 6 *I*pi ............2*pi*I*n. n natural number on other hand we have series of logarithms of roots of I which also sums up to the same value: Ln(-1)+ ln(sqrt(I) + ln (I^(1/4)) + ln (I^(1/) + ln(I^1/16))+ .. = Ln(-1) + ln(sgrt(I)*(I^(1/4)* (I^(1/)*I^(1/16)* ....) Obviosly, also right logarithm including roots must sum to Ln(-1). , then Ln(-1)+Ln(-1) = Ln(-1*-1)= Ln(1) as above. But if we look at the structure of the roots we know that n-th root has n values, so the number of possible combinations of roots in product grows as 2^(n*(n+1)/2)) -triangular numbers, or Pascal triangle. (sgrt (I) has 2 values each of which can combine with 4 values of 4th root of I which makes totally 8 combinations which can combine with 8 values of 8th root of I so there are 2^6 = 64 combinations which combined with 16 values of 16th roots (2^4) give 2^10 = 1024 combinations which combine with 32 values of 32th root (2^5) gives 2^10*2^5= 2^15 combinations etc.). So in fact, the powers of 2 of number of combinations are triangular numbers. It could be that some roots in this multiplication cancel out so some paths become equal and the number of effective combinations is reduced? Now, we can have infinite number of ways we can construct converging or diverging values for complex logarithm as sum of circles with imaginary radiuses , and each of them, when expressed as product of roots of I, will have differerent tree structure behind them. So what is the cardinality of infinite values of complex logarithm? Ivars « Next Oldest | Next Newest »

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