Posts: 22
Threads: 1
Joined: Feb 2008
bo198214 Wrote: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) I forgot to note, what I had in mind here is \( (ab)^c = a^{c}b^c \) and \( (a+b)c = ac + bc \). Of course, exponentiation/multiplication has other properties that I didn't take into account, which is not a surprise, since they impose other constraints on the resulting operation and ♦ would not qualify. But I was just using ♦ as an example that depending on what properties are desired, one can derive different operations that qualify under the generic label of "before addition".
Posts: 22
Threads: 1
Joined: Feb 2008
KAR Wrote:quickfur Wrote:[...]
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). Я согласен с Вами. My operation is not zeration, even if it does exhibit some similar properties, such has having identity element as \( \infty \). As I noted, my operator does not satisfy the defining property of the operations hierarchy, namely, that the n'th iteration of it should correspond to addition by n. Zeration as you have defined does have this property, which is good.
Quote:[...]
If Zeration it is commutative \( a\Delta a = \left( {  \infty } \right)\;\; \Rightarrow \;\left( {  \infty } \right) \circ a = a \).
Yes, it seems that any operation below addition, where some form of iteration reduces to addition, must have \( \infty \) as its identity element. As I've noted, any operation # which qualifies as being "before addition" must have as identity something that behaves like a multiplicative zero with respect to addition. \( \infty \) qualifies, because, at least intuitively speaking, \( (\infty)+x = \infty \).
The interesting thing about this, is that if we then construct inverse elements w.r.t. to #, then we must admit new numbers that lie "before" \( \infty \). This seems quite reminiscient of how constructing the inverse of addition created the negative numbers, the inverse of multiplication created the rational numbers, and the (radical) inverse of exponentiation created the real numbers (due to such constructs as \( \sqrt{2} \)). Combining the (radical) inverse of exponentiation with the negative numbers gave us the complex numbers. One can hardly wonder that the inverse of zeration would yield new numbers too. (It makes one wonder if the inverse of tetration would also create new numbers... I suspect it must've come up in this forum before, right?)
Quote:[...]
P.s. I bring an apology, for quality of computer translation on English.
Нет ничего. As long as it gets the message across... (Now it's my turn to apologize for my limited Russian.)
Posts: 174
Threads: 4
Joined: Aug 2007
quickfur Wrote:One can hardly wonder that the inverse of zeration would yield new numbers too. (It makes one wonder if the inverse of tetration would also create new numbers... I suspect it must've come up in this forum before, right?)
We didn't discuss, but we (KAR & myself) think that it must be so!
GFR
Posts: 1,630
Threads: 106
Joined: Aug 2007
KAR Wrote: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.
You mean if we have an operation * and we have two inverses / and \, i.e. a=(a*b)/b=(a/b)*b and b=a\(a*b)=a*(a\b), which are equal, i.e. x/y=y\x, then * must be commutative?
Hm, then (b*a)/b = b\(b*a) = a and hence b*a=a*b, so thats correct.
(Though you didnt even show radical equals log, but rather that the radical type inversion function is constant. That does not necessarily imply that also the radical type equals the log type.)
So what you gave is not a derivation, I admit I also love to speculate and think in analogies. However we are here in the mathematical field. Derivations, strict consequences and proofs are the things that count. So after we got some ideas of how things could possibly work, we are not bound to stay at the state of speculation (or to just postulate things without a possibility to verify, as the philosophers do) but we can go to the core and make things manifest: derive things, proving propositions, etc. And that would be what I am really interested in.
Quote: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?
Why should I extend this condition to complex n?
I just wanted to show that from the by you given conditions, i.e. aoa=a+2 and ao(aoa)=a+3, etc, from which follows ao(a+n)=a+n+1 for n>1, even if we enhance these conditions, by letting n be real, the by you given zeration is not a consequence but rather an example that satisfies these conditions.
However that the condition does not contain a complex but only real n, does not mean that zeration itself can not be defined on the complex numbers (though you did not define it on the complex numbers).
This would be *my* definition for *complex* \( a,b \):
\( a\circ b = \begin{cases}b+1 & : \Re(b)>\Re(a)+1\\a+2&: \Re(b)\le \Re(a)+1\end{cases} \)
Let us verify the conditions:
\( a\circ a = a+2 \) because \( \Re(a)\le \Re(a)+1 \),
let \( x>1 \) then \( a\circ(a+x)=a+x+1 \) because \( \Re(a+x)>\Re(a)+1 \)
and for natural \( x \) the other conditions follow \( a\circ (a\circ a)=a\circ(a+2)=a+3 \), etc.
Posts: 366
Threads: 26
Joined: Oct 2007
02/24/2008, 02:50 PM
(This post was last modified: 02/24/2008, 03:00 PM by Ivars.)
GFR Wrote:quickfur Wrote:One can hardly wonder that the inverse of zeration would yield new numbers too. (It makes one wonder if the inverse of tetration would also create new numbers... I suspect it must've come up in this forum before, right?)
We didn't discuss, but we (KAR & myself) think that it must be so!
GFR
Inverse zeration has to produce the same "numbers" that operate with operation [infinity] and which are, basically, unenumerable numbers. That is the trick needed to go around Godel if you begin everything with those "numbers". In "esoteric" sense, as bo uses it, it is perfect chaos. But it is just a ļegitimate mathematical need to close hyperoperationsunenumerable "numbers".
Or 1 number.
Ivars
Posts: 510
Threads: 44
Joined: Aug 2007
bo198214 Wrote: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).
If you'll notice in my version of the FAQ pages 1112, (based on this post) I use a piecewise function with initial conditions for converting hyperN => hyper(N+1) and I use a single definition for hyperN => hyper(N1) which depends on the fact that you don't need an initial condition. It is very interesting indeed.
Andrew Robbins
Posts: 1,630
Threads: 106
Joined: Aug 2007
03/06/2008, 04:09 PM
(This post was last modified: 03/06/2008, 04:09 PM by bo198214.)
andydude Wrote:If you'll notice in my version of the FAQ pages 1112, (based on this post) I use a piecewise function with initial conditions for converting hyperN => hyper(N+1) and I use a single definition for hyperN => hyper(N1) which depends on the fact that you don't need an initial condition. It is very interesting indeed.
That sounds to me like a request for us two to sit down and damn merge the FAQ, is it?
Posts: 1,630
Threads: 106
Joined: Aug 2007
GFR Wrote:The problem, as you also correctly said, is to define an operation (ONE hyperop) that would be the unique operation, exactly fitting in the hyperops hierarchy, at rank 0.
Only to summarize and clarify the current situation:
If we agree on the law a[n+1](b+1)=a[n](a[n+1]b) for all hyperoperations [n] for integer n and agree that a[1]b=a+b then it stringently follows (without assumptions about initial values) that a[0]b=b+1 and it also stringently follows for all hypo operations that a[n]b=b+1. (This was shown in this thread by Andrew and me.)
I would call this "exactly fitting in the hyper operations hierarchy".
If we otherwise dont accept the above law, then there are of course multiple possibilities to define an operation [0] and if we impose the conditions
a[0]a=a+2
a[0](a+n)=a+n+1 for n>1
the operation [0] is still is not uniquely defined not even on the integer numbers a.
One example of an operation obeying these rules is Konstantin's zeration and another example was given by me in this thread. Till now there was no set of equations presented for which Konstantin's zeration is the only solution.
Posts: 174
Threads: 4
Joined: Aug 2007
Your observations are clear. I know that Konstantin is preparing a comment on the various questions concerning zeration, both in the framework of this Forum and of the WRI NKS Forum (News and Announcements), accompanied by a short pdf "situation report". All this will come soon.
I hope. I mean ... I am sure!
Posts: 6
Threads: 1
Joined: Mar 2008
03/25/2008, 08:28 AM
(This post was last modified: 03/26/2008, 12:57 AM by James Knight.)
Ok you guys are getting closer and closer to what I have defined as zeration.... Anyway I have scribbled a paper that I was going to formally type up but I want to share!!! Anyway, you don't have to believe anything that I put but please have an open mind! BTW I have been researching since September...
Preamble
I want to distinguish operation types (so we're on the same page)
1st type : Regular Operations (zeration, addition, multiplication,tetration etc.)
2nd Type : Left Inverse Operations (Root Type)
3rd type : Right Inverse Operations (Log Type)
First I want to define a special function N(n). This function will give the neutral value of a particular level/rank 'n'. This will be useful to determine values of the Zeration Operation.
A Neutral Value (I think it has been called the "right unit element u" or something like an element incidence, anyway) is the value which is "between" the Right Inverse Operation and the Regular Operation.
Values
Rank n N (n)
1 0
2 1
3 1
Ok so you want proof.... here!
+1 +1 +1
1 0 1 2
Notice the differences are 1... Watch when they change to 2
+2 +2 +2 +2
2 0 2 4 6
Notice that as we go LEFT to Right >>>> (some people can be dislexic!!)
We are ADDING "2"
Notice as we move from RIGHT to Left <<<<<<<<<
We are subtracting "2"
Notice that 0 is the "MIDDLE" number between the POSTIVE and the NEGATIVE quantities in both scenarios.
EXAMPLE TWO
x2 x2 x2 x2
1/4 1/2 1 2 4
Again we can see that the Neutral Value is situated in the "MIDDLE" between the NICE whole numbers and the UGLY fractions.
EXAMPLE THREE
2^ 2^ 2^ 2^
0 1 2 4 16
Notice going left to right >>> we are applying a base of two (ie. if we have 'n' then we are finding "2^n")
and right to left <<< we are taking log base two
1 really isn't in the middle but we can see that because we are iterating an operation we are finding the value of the operation of rank one higher.
Examples
eg3) 2#1 2#0 2#1 2#2 2#3
0 1 2 4 16
eg2) 2^(2) 2^1 2^0 2^1 2^2
1/4 1/2 1 2 4
eg1 2(1) 2(0) 2(1) 2(2)
2 0 2 4
If you have noticed the pattern it is...
N(n) = x[n+1]0
where x is allowed to be any value
Ok Also I would like to lay down the LAW (sorry about these poor jokes... I am tired...)
The Minus One Law (and the Plus One Law)

So if we want to find N[n] value then we need an equation.
It is known that going Left to Right the Sequence for any Level Becomes
x[n]P x[n]P (remember the +2 x2 2^ etc?)
x[n+1](1) x[n+1]( 0) x[n+1] (1) etc...
where P is the Previous Term
Left to Right is the the Pluse One LAw
x[n](x[n+1](b)) = x[n+1](b+1)
VOILA!
OK the Minus one Law is Reversed <<<<< Right to Left
First I must define a notation
Let x [®n] (y) be the Right Inverse of the Regular Operation Rank n.
(BTW I have developed my own simple notation.. but computer eghh...)
MINUS ONE LAW
(x[n+1]b) [®n] (x) = x [n+1] (b1)
Also (n = n1)
(x[n]b) [®n1] (x) = x [n] (b1)
OK enough LAWS
Zeration
Rearanging the Minus One Law we get
x [n1] (x[n](b1)) = x [n] (b)
Substituting n = 1 and y = x+b 1
Definition 1
x [0] y = y + 1
This is where controversy and contradiction begins... Remember keep an open mind.
From this definition we can prove that zeration is NOT Commutative!!!
(In my mind, it didn't make sense that zeration was commutatave and not associative)
PROOf
x [0] y = y+1
y [0] x = x + 1
y+`1 = x+1 only when y=x therefore NOT commutative
Again it's NOT Associative
a [0](b[0]c) = c+2
(a[0]b)[0]c = c+1
c+1 = c+2 (only at NON FINTE VALUES)THerefore NOT ASSOCIATIVE!
OK this is the good stuff now!
Because zeration is not commutative it has two inverses: a left and a right.
The left inverse of zeration is defined as deltation. Basically they are hyperreal infinite or infinitesimal .
A vertical line x=a can be made by the equation y = x Δ (a1)
A "Black Graph" can be made by the equation y = x Δ (x1)
A "White Graph" can be made by the equation y = x Δ x + c
where c is everything but 1
Ok that leaves the exciting right inverse!!!
Since last fall, I had my doubts over the commutativity of zeration as well as the discontinuity. I have spent numerous hours redoing laws and being frustrated. Ok now I would like to present to you
Knightation or Nitation (struggling on what to call it...)
Knightation is the Right Inverse of Zeration.
The Operator J is used to refer to Knightation (it's supposed to look like an ear lobe idea)
if x o y = z then
y = z J x
Definition 2
x J y = x  1
Knightation and Zeration are two Parallel Lines basically
Also I looked into ranks lower than 0.
It's fairly easy you just have to use the rearranged minus one formula.
Basically Zeration is what you get but!!!
The Neutral Value changes to 1. This however doen'st affect anything...
Ok now I want to look back at the previous definitions of zeration
x o x o x o ... x = x + n
n
This doesn't work when n = 1
LS = x RS = x + 1
There has to be consistency otherwise nothing means anything.
x o x = x + 2 is one of the previous fundamental definitions.
However, if I use Knightation on both sides I get
x = x+1 which is not good.
I think the Neutral Values are to blame for this mishap.
notice that the neutral values are one one less than what they usually are. Also notice '2' is after '1'.
Well I hope you have gained something from this or have been entertained by my random jokes. I hope to actually post the paper I am working on currently.
Also I am a computer programmer and I am going to soon start a program that will compute and graph hyperoperations. Anyway, I am sooooo happy right now because I got accepted to the University of Waterloo!! Soo tired!
Talk to you later,
James
