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 A new set of numbers is necessary to extend tetration to real exponents. marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 03/11/2015, 07:56 PM I will make a different definition of the elemental functions to extend it to tetration. By starting with zero, and taking the successor, is possible to obtain the entire set of natural numbers, but no other type of number, like a rational, or a complex. Once addition and his inverse function, subtraction is considered, is possible to “discover” the set of the Integers, because to solve any equation is necessary to use negative numbers. Product forces to use rational numbers, and exponentiation needs to discover the complex set. So, is natural to suspect that tetration requires an more general set of numbers to solve any equation, and I think that it is the root of the difficulty in extending tetration to real exponents. Is possible to define a product a.b this non traditional way: Which means that zero is added to a, b times. I would call zero (the identity element of addition) the implicit base of product. It is trivial that a.0=0 It is also possible to define exponentiation a^b as multiplication of 1 (the identity element of product) by a, b times: By definition, it is trivial that a°=1 It leads to define a^-b=c as the number c which In other words, It also leads to define as the number c which For that reason, I would define tetration on a subtle different way than the custom: Tetration of a, to exponent b, and implicit base z is defined as Ttr(a,b,z) this way: When z = 1, it may be omitted and written Ttr(a,b,1) = Ttr(a,b) It is immediate that Ttr(a,b,0) = Ttr(a,b-1) also, So, is not necessary to make a distinction between left and right parenting notation. Trivially, Ttr(a,0) = 1, because no number modifies 1 by being raised to 1 To find the value of tetration for any real number b, in Ttr(a,b), is necessary to know the meaning for any negative integer b, and for any number equal to 1/b (b integer). So, let’s start from Ttr(a,0) = Ttr(a,-b+b) = Ttr(a,b,Ttr(a,-b)) = Ttr(a,b,c) = 1 or So, for a natural number b, For a simple case, let’s take c = Ttr(a,-1), then it should be By taking logarithm: c . ln(a) = 0, so, if c is a complex number, it can only be zero, for any real a≠1. That is inconvenient, because impedes to extend the concept of tetration to negative integer exponents. But we should remember that each new level of arithmetic operation needed new, more general set of numbers to solve any equation, so is necessary to ask if to extend tetration to real exponents, is necessary to use a more general set of numbers, were a power of a number different to zero may be zero. So, I argue that to understand tetration is essential to use sets of numbers with non trivial roots of zero. Such roots of zero may be nilpotent numbers, or any set with non trivial Idempotent elements, which would have non trivial zero divisors. There are many sets with non trivial zero divisors. The question is which one would be the more useful one. Are the Split-complex numbers? something from hypercomplex? It should have the complex numbers as a subset. marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 03/12/2015, 06:04 PM (This post was last modified: 03/13/2015, 03:46 PM by marraco.) (03/11/2015, 07:56 PM)marraco Wrote: For a simple case, let’s take c = Ttr(a,-1), then it should be By taking logarithm: c . ln(a) = 0, so, if c is a complex number, it can only be zero, for any real a≠1. Actually, Still n=0 is one solution, so it requires a new set of numbers to solve for c, but for n≠0: and with the same reasoning this is just logarithm of 1 in base a iterated b times Why the iterated logarithm of 1? because 1 was the implicit base. More generally: Is easy to verify that for any pair of integers n,m For example: (when taking power with m≠0, m represent m roots or more, and not all the roots are valid solutions to a^^-1, for example y=-1^(2n/(1+2m)), m produces 1+2m roots, and not all those roots give y=1) This shows that all the rational numbers are solution for 1^^(-1), but we know that all the real numbers r have the property 1^r=1, so: -Either there are more solutions than given by the logarithm function, or -Any real number can be expressed as a rational number. Yet, we know that transcendental numbers are not equal to any rational number, but maybe this is saying that since numbers like PI can be approximated as much as desired, then pi is also equal to a rational number, as 1,999999... is equal to 2. By the way, it is interesting that ln(x)=Ttr(e,-1,x) ,and e(x)=Ttr(e,1,x) marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 03/12/2015, 10:58 PM (This post was last modified: 03/12/2015, 11:00 PM by marraco.) Curiously, IF the infinite tower for negative integer exponent converges, then it converges to the same value than the positive one. I don't know if it converges for negative exponents, and if it does, probably c is not unique. marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 03/12/2015, 11:10 PM (This post was last modified: 03/12/2015, 11:34 PM by marraco.) This paper (how do I insert a link here? the icons don't work) Henryk Trappmann, Dmitrii Kouznetsov, Methods for real analytic tetration https://bbuseruploads.s3.amazonaws.com/b...ain.pdf%22 That paper argues that Ttr(a,-1) = 0, and Ttr(a,-2) = ln(0) = -∞ , so it limits the domain of tetration exponents to (-2,∞) But it does not account for multiple valued results for Ttr(a,-1), which give meaningful values for any negative integer (-∞,∞). As I argued, Ttr(a,-1) = i.n.2.pi/ln(a) So, there are infinite values, and for each n, Ttr(a,-2) also gives infinite vales: ∞², and Ttr(a,-3) haves ∞³ answers sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 03/14/2015, 10:37 PM (This post was last modified: 03/14/2015, 10:38 PM by sheldonison.) (03/12/2015, 11:10 PM)marraco Wrote: This paper (how do I insert a link here? the icons don't work) Henryk Trappmann, Dmitrii Kouznetsov, Methods for real analytic tetration https://bbuseruploads.s3.amazonaws.com/b...ain.pdf%22 That paper argues that Ttr(a,-1) = 0, and Ttr(a,-2) = ln(0) = -∞ , so it limits the domain of tetration exponents to (-2,∞) But it does not account for multiple valued results for Ttr(a,-1), which give meaningful values for any negative integer (-∞,∞). As I argued, Ttr(a,-1) = i.n.2.pi/ln(a) So, there are infinite values, and for each n, Ttr(a,-2) also gives infinite vales: ∞², and Ttr(a,-3) haves ∞³ answers The analytic tetration function referenced in the link has a logarithmic singularity at -2. Take any real number z which is greater than -2. Winding around the singularity counter clockwise, between -2 and -3, and then returning to z, you wind up with $\text{Tet}_e(z)+2\pi i$. Winding around the singularity clockwise, between -2 and -3, and returning to z, then you have $\text{Tet}_e(z)-2\pi i$. This is well known. To understand the behavior of $\text{Tet}_e(z)$, one has to understand the analytic function and it's Riemann surface. - Sheldon marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 03/19/2015, 12:47 AM (This post was last modified: 03/19/2015, 02:57 AM by marraco.) $a =\, ^{1}{a} \, =\, ^{\frac{n}{n}}{a} \,=\, ^{\frac{1}{n_n}+\frac{1}{n_{n-1}}+...+\frac{1}{n_1}}{a} \,=\, ^{n}({^{\frac{1}{n}}{a}})$ So, we can define the n$^{^{^{th}}}$ root of "a" this way $c \,=\, ({^{\frac{1}{n}}{a}}) \,\Leftrightarrow\, ^nc\,=\, a$ Tetration exponent product is non commutative. In general: $^n(^{\frac{1}{m}}a) \,\neq\, ^{\frac{1}{m}}(^na)$ The inequality stands as long as m ≠ n, so if a product is defined to operate on tetration exponents, it is not the common product, or it do not operates on the complex field, or results are multivalued to preserve the equality. sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 03/19/2015, 09:59 PM (03/19/2015, 12:47 AM)marraco Wrote: .... Tetration exponent product is non commutative. In general: $^n(^{\frac{1}{m}}a) \,\neq\, ^{\frac{1}{m}}(^na)$ I would also add that for analytic tetration in general: $^n(^{\frac{1}{n}}a) \,\neq a$ - Sheldon marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 03/19/2015, 10:45 PM (This post was last modified: 03/19/2015, 10:53 PM by marraco.) (03/19/2015, 09:59 PM)sheldonison Wrote: (03/19/2015, 12:47 AM)marraco Wrote: .... Tetration exponent product is non commutative. In general: $^n(^{\frac{1}{m}}a) \,\neq\, ^{\frac{1}{m}}(^na)$ I would also add that for analytic tetration in general: $^n(^{\frac{1}{n}}a) \,\neq a$ I defined $^n(^{\frac{1}{n}}a) \, = a$ as ttr($(^{\frac{1}{n}}a)$ ,n,1)=a were $(^{\frac{1}{n}}a)$ may be multivalued. It has some problems, like $\lim_{n \rightarrow \infty} (^{\frac{1}{n}}a) = a^{\frac {1}{a}}$, which has "a" solutions, (for a E N), against °a=1 by definition. So, it looks like this definition of $^{\frac{1}{m}}a$ is wrong all the way. Still, there should be some, non integer number t, such $c \,=\, ({^t{a}}) \,\Leftrightarrow\, ^nc\,=\, a$ « Next Oldest | Next Newest »

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