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02/20/2008, 05:43 PM
(This post was last modified: 02/20/2008, 05:45 PM by bo198214.)
andydude Wrote:If we choose to hold \( a [N1] b = a [N] (\text{hy}N\text{log}_a(b) + 1) \) to be true for all N, then the natural consequence is that \( a [0] b = b + 1 \) for all a ... .
If we choose to hold \( a [N1] a = a [N] 2 \) true for all N, then we get \( a [0] b = b + 2 \) for a == b. Both cannot be true, so we must make a choice.
Hm, the first one (demanding that a[N1](a[N]b)=a[N](b+1) for all real b and integer N) seems to be the pure approach.
Interestingly the derivation of a previous operation seems not to depend on an intial condition like a[N]1=a (which is not true for N=1, but for all N>1).
But we just could conclude from a[N1](a[N]b)=a[N](b+1) that a[0]b=b+1. And we can further conclude from
a[1](a[0]b)=a[0](b+1)
that
a[1](b+1)=a[0](b+1)
hence
a[1]b=a[0]b=b+1 moreover by induction that a[N]b=b+1 for all N<0.
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andydude Wrote:If we choose to hold \( a [N1] a = a [N] 2 \) true for all N, then we get \( a [0] b = b + 2 \) for a == b. Both cannot be true, so we must make a choice.
My suggestion is to keep: a[s1]a = a[s]2, like:
a^a = a[3]a = a[4]2 = a#2
a*a = a[2]a = a[3]2 = a^2
a+a = a[1]a = a[2]2 = a*2
a°a = a[0]a = a[1]2 = a+2
GFR
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bo198214 Wrote:..........
the bracketing must be to the right. So
ao(ao(aoa))=a+4
ao(aoa)=a+3
aoa=a+2
a = a+1 ??? Hi, Henryk!
As a matter of fact, your last line is not a criticism to the poor new zeration "baby", but it is a general "problem" of all the elements of the hierarchy, including a very old and respectable operation. Let me try to show it as follows, starting from the behaviour of the " right unit element u" of each hyperop.
a[4]u = a#u = a > u = 1, as we have just correctly shown,
a[3]u = a^u = a > u = 1, as we know from ... centuries;
a[2]u = a*u = a > u = 1, as we knew ... even before;
a[1]u = a+u = a > u = 0, bingo !, Deviation from the "rule"! However, nobody complained, until now.
a[0]u = a°u = a > u = oo rebingo! But, here, we must discuss a little bit.
As a first observation, we may see that, in case of a general hyperoperation, i.e.:
a[s]u = a > u = 1 for s > 1; u = 0 for s = 1; and u = oo for s = 0.
The suggestion to remain cool is coming from the fact that the "rule" u = 1 is violated also by the old very respectable addition. So, we see that, taking this into consideration, we are not allowed to write
neither a ° 1 = a , nor a + 1 = a.
But, this should not be a reason for doubting of zeration.
Concerning the " left unit element v" (and showing it in the inverse ranking order), we have something even more peculiar:
v[0]a = v°a = a > v = oo, and .. we shall discuss;
v[1]a = v+a = a > v = 0, OK, as above;
v[2]a = v*a = a > v = 1, idem, as above;
v[3]a = v^a = a > v = [a/]rt a = selfrt a.
Well, we know, selrt a it's not an "element", because it is changing with a. For rank 4 (tetra), we should have something similar to:
v[4]a = v#a = a > v = [a/]srt a = sselfrt a.
A deep analysis of this left/right behaviour is strongly needed.
bo198214 Wrote:Can you btw show me, how you derived the commutativity of zeration? Via the analysis of homomorphisms (see KAR). For example, for the calculations of the square superroot, the squqreroot and the half of a variable z we may use the following iterative formulas:
x = ssqrt z ≈ sqrt(p * [p/] rt z) , with x > p
x = sqrt z ≈ ( p + z/p) / 2 , with x > p
x = z / 2 ≈ (p ° (z – p)) – 2 , with x > p, and p integer.
By a change of variables (xp ↔ p), we can assume that the following relation is also verified:
x = z / 2 ≈ ((z – p) ° p) – 2
and that, therefore, zeration must be commutative. But, here, KAR might probably add something else.
In conclusion, we came again to:
a ° b = a + 1 if a > b
a ° b = b + 1 if a < b
a ° b = a + 2 = b + 2 if a = b.
Gianfranco
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02/21/2008, 02:04 PM
(This post was last modified: 02/21/2008, 02:06 PM by bo198214.)
GFR Wrote:The suggestion to remain cool is coming from the fact that the "rule" u = 1 is violated also by the old very respectable addition. So, we see that, taking this into consideration, we are not allowed to write
neither a ° 1 = a , nor a + 1 = a.
But, this should not be a reason for doubting of zeration.
No, its rather a reason for omitting this violated rule about the initial condition a[N]1=a. As was already shown if we omit this condition and rather focus on the quite venerable b[N1](b[N]x)=b[N](x+1) mother rule of hyper operations, then it directly follows that a[0]x=x+1 (and also even a[N]x=x+1 for all N<0).
Which (for me!) makes a lot of sense.
We all know that the successor operation n>n+1 is the mother operation in the Peaono style definition of natural numbers and operations on them.
And the rule a[N]a=a[N+1]2 falls for me also into the category of initial condition.
Quote:bo198214 Wrote:Can you btw show me, how you derived the commutativity of zeration?
Via the analysis of homomorphisms (see KAR).
But we have already seen that commutativity does not follow from a[0]a=a+2 and a[0](a+x)=a+x+1 (x>1) (which includes a[0](a[0]a)=a+3, etc). As I gave a counter example of an operation that satisfies both equations but which is not commutative.
This is another weak point of chosing the condition a[0]a=a+2 for defining zeration: it does not uniquely determine zeration (while it does in andydude's first path b[N1](b[N]x)=b[N](x+1)).
Quote:In conclusion, we came again to:
a ° b = a + 1 if a > b
a ° b = b + 1 if a < b
a ° b = a + 2 = b + 2 if a = b.
but it does not really follow from anything but is rather one consideration of andydude's second path.
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I think names should be given to these hyperoperations to make easier intuitive understanding.
As I understand it, positive zeration is counting, or enumeration, of infinite number of objects, while negative is counting of different objects (also infinite in number).
1 is infinite summation, or summation where order of terms does not matter as they are presumed to be infinite.
Halvation then is summation of any 2 enumerated , thirdation  any 3 etc. There is an imaginary part to these operations which corresponds to the number of ways these sums can be made from infinite numbers we have enumerated.
I would say there is a direct correspondance between imaginary part of 1/n ation and combinatorics.
The same applies to fractional operators between 1 and 2  summation and multiplication.
They represent multiplication of finite number of objects. etc.
between 23  exponentation of finite number of objects and related combinatorics expressed by?
between 34  tetration of finite number of objects and related combinatorics expressed by?
I know I am not very clear about this, but for me it makes sense to establish a clear link between operations possible in mathematics and operations defined as hyperoperationsthat would help intuitively to predict their properties and by that clarify the operations used in mathematics in general.
Have You had any ideas about this?
Ivars
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This isn't really related to zeration per se, but some time ago I searched for an operator ♦ over which addition is distributive (i.e., a+(b♦c) = (a+b)♦(a+c)). (I guess my motivation was that just as exponentiation distributes over multiplication, and multiplication distributes over addition, addition should also distribute over an operation that may be regarded to be, in some sense, "before" addition.)
Eventually, I came upon the definition: \( a\diamond b = \ln(e^a + e^b) \). As you can verify, \( a+(b\diamond c) = \ln\ e^a + \ln(e^b + e^c) = \ln(e^a(e^b + e^c)) = \ln(e^{a+b}+e^{a+c})=(a+b)\diamond(a+c) \).
This operator is associative and also commutative, and is defined for all positive reals. If you admit complex numbers, it is defined for negative reals as well (although it will be complicated by the question of which branch of logarithm should be chosen). The hypothetical identity element is ∞ (because \( \lim_{x\rightarrow\infty}e^x=0 \), which then reduces a♦(∞) to a; also, ln 0 = ∞ by the dual argument). Interestingly enough, Rubtsov's zeration operation also features an identity at ∞. This property is in fact, a consequence of the distributive law: if addition distributes over any operation #, then the identity e of the # operation must satisfy e+x=e for all x. (The derivation of this is left as an exercise for the reader). This can be seen in exponentiation as well: the multiplicative identity 1 acts as a "zero" for exponentiation: 1^x = 1 for all x.
The only problem is that my operator does not satisfy a♦a♦...♦a (n times) = a+n. The closest one can get is to switch from base e to base 2, in which case this operation satisfies a♦a♦...♦a (2^n times) = a+n.
But anyway, the whole point of this overlylong post is that, in going from addition to zeration, something has to go. Rubtsov's zeration is nonassociative and also discontinuous at one point. My operation is both associative and continuous over the positive reals, but then its n'th iteration is not equal to addition by n, which property Rubtsov's zeration does satisfy. So, in deciding which operation comes "before" addition, it's a question of which properties we want to keep, 'cos we can't have them all.
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02/21/2008, 10:18 PM
(This post was last modified: 02/21/2008, 10:20 PM by bo198214.)
quickfur Wrote:This isn't really related to zeration per se, but some time ago I searched for an operator ♦ over which addition is distributive (i.e., a+(b♦c) = (a+b)♦(a+c)). (I guess my motivation was that just as exponentiation distributes over multiplication, and multiplication distributes over addition, addition should also distribute over an operation that may be regarded to be, in some sense, "before" addition.) This distribution law does not fit into the distribution laws for exponentation and multiplication
a[3](b+c)=(a[3]b)[2](a[3]c)
a[2](b+c)=(a[2]b)[1](a[2]c)
The natural continuation would be:
a[1](b+c)=(a[1]b)[0](a[1]c)
And one can prove that there is only the trivial operation [0]:
Let c=0 then
a+b=(a+b)[0]a
which means that x[0]y=x.
What you are referring to is another kind of hierarchy differently from the hyper operations hierarchy. It is the hierarchy of operation {n} built by the law:
a{n+1}(b{n}c)=(a{n+1}b) {n} (a{n+1}c)
a{1}b=a+b
as opposed to a hierarchy of operations [[n]]:
a[[n+1]](b+c)=(a[[n+1]]b)[[n]](a[[n+1]]c)
a[[1]]=a+b
which is not even satisfyable on the natural numbers (however possible with binary trees).
one can easily verify that one solution to the {n} hierarchy is
\( a\{n\}b=\exp^{\circ n1}\left(\ln^{\circ n1}(a)+\ln^{\circ n1}(b)\right) \)
which for \( n=0 \) matches exactly your:
Quote:\( a\diamond b = \ln(e^a + e^b) \).
And all these operations {n} are associative and commutative.
However I never heard something interesting about this hierarchy, perhaps because the domain of definition with the operations {n} and {n+1} (any n) is just isomorphic to the reals with + and *.
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Wah, I forgot to welcome my dear friend (hopefully he agrees with this designation ) quickfur  or sporadically also known as slowpaw  on this forum!
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bo198214 Wrote:Wah, I forgot to welcome my dear friend (hopefully he agrees with this designation ) quickfur  or sporadically also known as slowpaw  on this forum! Yes, I agree with this designation. I feel quite honored, in fact.
Anyway, with regards to the {n} hierarchy, you're right, a class of solutions is given by \( a\{n\}b=\exp^{\circ n1}\left(\ln^{\circ n1}(a)+\ln^{\circ n1}(b)\right) \).
It does have some interesting attributes, although probably only in very specialized fields. I've had somebody tell me once that they actually contemplated the same operations and found it useful in computer graphics: this class of operations can be used for image compositing where the order of composition doesn't matter, and the pixel value must remain within a specified range.
Another interesting observation (at least, interesting to me) is that the converse construction, \( a\oslash b = e^{\ln\ a + \ln\ b} = ab \). That is to say, in stepping from addition to the diamond operator, if we "stepped in reverse", we get to multiplication. In other words, the diamond operator is in some vague sense the "same distance" from addition as addition is from multiplication. But then again, this is just because we're remapping the field structure (+,*) to (♦,+), so we're not really discovering any new things.
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quickfur Wrote:This isn't really related to zeration per se, but some time ago I searched for an operator ♦ over which addition is distributive (i.e., a+(b♦c) = (a+b)♦(a+c)). (I guess my motivation was that just as exponentiation distributes over multiplication, and multiplication distributes over addition, addition should also distribute over an operation that may be regarded to be, in some sense, "before" addition.)
Eventually, I came upon the definition: \( a\diamond b = \ln(e^a + e^b) \). As you can verify, \( a+(b\diamond c) = \ln\ e^a + \ln(e^b + e^c) = \ln(e^a(e^b + e^c)) = \ln(e^{a+b}+e^{a+c})=(a+b)\diamond(a+c) \).
Yes, you are right, that it is possible to construct this operator. It is a classical homomorphism of addition with using function exp(x). I researched in 90th years application of this operator and of some others for upgrade numerical solution systems of the differential equations. Good the outcome was with a homomorphism on the basis function ln(x).
However, this operator (a homomorphism of addition) is not Zeration. Moreover, application Zeration has allowed to achieve more simple algorithms of a homomorphism methods numerical solution of the differential equations and their systems. These researches are available in my monography (in Russian).
bo198214 Wrote:Can you btw show me, how you derived the commutativity of zeration?
Thanks for a problem on commutability Zeration.
Originally, I would like to note, that violation of logic is correctly noted:
\( \;\; a\;[4]\;1\; = {}^1a = a \)
\( \;\;\;\; a^1 = a \)
\( \;\;\;\; a \cdot 1 = a \)
\( \;\;\;\; a + 1 \ne a \)
For the last case we have \( a + 0 = a \). If to accept existence Zeration this fact means violation of a rule of construction of operation of a following rank \( n = 1 \).
Zeration it has been created at learning Tetration in 1987.
I developed algorithm evaluation superradical of the radical similar to an evaluation:
\( \sqrt x \approx \frac{{\frac{x}{a} + a}}{2} \) \( \Rightarrow \) \( ssqrt(x) \approx \sqrt {\log _a \left( x \right) \cdot a} \)
After a successful evaluation of the superradical on this algorithm, I have decided to calculate x/2:
\( \frac{x}{2} \approx \left( {x  a{\rm{?}}a} \right)  2 \) \( \Rightarrow \) \( \frac{x}{2} \approx \left( {x  a \circ a} \right)  2 \)
Here it is necessary to apply operation by a level below addition.
At definition there were problems:
1. \( a + 1 \ne a \). It means, that for Zeartion we have violation in typical representation of number 1 as "units"
2. How to install or refute a commutability?
3. How to solve the equation: \( a \circ x = a,\;\;\;\;x = ? \)
The hint on a problem 2 and 3 has Tetration.
For this purpose we shall recollect a known limit:
\( \lim\limits_{n \to \infty } \left( {{}^n\left( {\sqrt[a]{a}} \right)} \right) = a,\;\;\;a \le e,\;\;\sqrt[a]{a} > e^{  e} \)
From it follows, that the superradical from \( a \) superpower aspiring infinity has solution as \( \sqrt[a]{a} \).
Similar association can be observed, reducing a rank of operation:
\( \lim\limits_{n \to \infty } \left( {\sqrt[n]{a}} \right) = \frac{a}{a} = 1 \)
\( \lim\limits_{n \to \infty } \left( {\frac{a}{n}} \right) = a  a = 0 \)
Here it is necessary to pay attention, that at a limit there is an operation of type "radical", in summary type "radical", but on a rank below. As there is "paradox" in operation "addition" on number 1, and on number of repetitions 2 and more  is not present, it is necessary to assume, that on infinite number of repetitions on operation "addition" there will be no paradox. We shall write:
\( \lim\limits_{n \to \infty } \left( {a  n} \right) = a\Delta a = \left( {  \infty } \right) \), where "\( \Delta \)"expected inverse Zeration operation of type "radical".
Further, from the given legitimacies it is possible to install, that Zeration it is commutative. For this purpose we shall consider two functions at various values \( a \ge 2 \):
\( \;\;\;\sqrt[a]{a} = {\rm{var}}\;\;\;\; \) and \( \;\;\;\;\log _a \left( a \right) = 1 = {\rm{const}} \) .
For operations "division" and "subtraction":
\( \;\;\;\frac{a}{a} = 1 = {\rm{const}}\;\;\;\; \) and \( \;\;\;\;a  a = 0 = {\rm{const}} \).
From the reduced example, follows, that inverses operations of type "log" at identical values of argument have constant value. Inverse operation of type "radical" has variable values at a modification of identical arguments and is equal to a constant in case of coincidence with operation of type "log", that is commutabilities.
From a limit \( \lim \limits_{n \to \infty } \left( {a  n} \right) = a\Delta a = \left( {  \infty } \right) \) follows, that inverse operation of type "radical" for Zeration has unique solution for identical arguments, that is Zeration is commutative.
If Zeration it is commutative \( a\Delta a = \left( {  \infty } \right)\;\; \Rightarrow \;\left( {  \infty } \right) \circ a = a \).
bo198214 Wrote:No, thats not a strict consequence. The following is a strict consequence:
We start with the axiom
ao(ao(ao......))) = a+n for integer n>1 (and a being real or complex)
then we conclude
ao(a+n)=a+n+1 You write n> 1 and safely expand "n" up to complex? You can seriously make matching of any complex number with 1?
KAR
P.s. I bring an apology, for quality of computer translation on English.
