On the existence of rational operators

[JmsNxn]

I was wondering if anybody here has anything to say about this. I'd love to know if there are related topics.

Consider the following definition:

r:log(x) is a superfunction of log(x); where r is the iteration count.
Therefore
2:log(x) = log(log(x))
etc etc...
r:log(y:log(x)) = r+y:log(x)
and therefore:
-1:log(x) = b ^ x
-2:log(x) = b ^ (b ^ x)
so on and so forth. r:f(x) is taken to be the superfunction of f(x) aswell.
0:f(x) = x

We must first observe tetration and its connections to the superfunction of log(x).
if b {0} x = b + x and b {1} x = b * x and b {2} x = b ^ x
b {3} x will denote tetration. (Do not be fooled by the number 3).

by definition (logs base b)
log(b {3} x) = b {3} (x-1)
and therefore, connected to superfunctions:
r:log(b {3} x) = b {3} (x-r)
this holds for b > 0, b =/= 1; r < x; x > -1; b, r, x E R

As you can see rational iterations of the logarithm are defined by rational tetration. There is still no clear concensus on the evaluation of rational tetration, however, I hope to further argue the model which states over domain [-1, 0];
-1 <= f <= 0
b {3} f = f + 1

Now comes the area of my paper where one must open their minds. m,n > 0 E R
Consider:
log(m* n) = log(m) + log(n)
and
log(m ^ n) = log(m) * n
and
2:log(m ^ n) = 2:log(m) + log(n)
One can see that logarithms work to lower the operator magnitude across any operator lower than {2}
The assertion I make is that taking rational iterations of logarithms gives us rational operators.

Or:

0 <= q <= 1
q:log(m {1} n) = q:log(m) {1-q} q:log(n)
q:log(m {2} n) = q:log(m) {2-q} n

These operators would have the following property; if S(q) returns the identity of any operator:
m {q} S(q) = m
q:log(m) + q:log(S(q)) = q:log(m)
therefore
q:log(S(q)) = 0
which is true regardless of logarithm base.
m {1+q} S(1+q) = m
q:log(m) * S(1+q) = q:log(m)
therefore:
S(1+q) = 1
and which in general becomes all operators greater than or equal to one have identity one.


Operators less than or equal to one are commutative:
m {q} n = n {q} m
q:log(m) + q:log(n) = q:log(n) + q:log(m)

m {1+q} n =/= n {1+q} m
q:log(m) * n =/= q:log(n) * m

Operators less than or equal to one are associative:
m {q} (l {q} n) = l {q} (m {q} n)

Rational operators preserve the law of recursion found in natural operators.
m {1 + q} 2 = m {q} m
q:log(m) * 2 = q:log(m) + q:log(m)

therefore:
m {1+q} n = m {q} m {q} m ... {q} m n amount of times
m {2+q} n is therefore defined recursively.

if k;log(x) is the inverse of any function b {k} x:
0;log(x) = x - b
1;log(x) = x/b
2;log(x) = log_b(x)
1+q;log(b {2+q} x) = b {2+q} (x-1)
or more generally:
r: (1+q);log(b {2+q} x) = b {2+q} (x-r)
r:q;log(b {1+q} x) = b {1+q} (x-r)

Rational operators are not distributive over addition, however, as multiplication is to exponentiation and as exponentiation is to multiplication [i]{q} is to {1+q} as {1+q} is to {q}
therefore:
(m {q} n) {1 + q} l = (m {1+q} l) {q} (n {1+q} l)

(m {1+q} n) {q} m = m {1+q} (n+1)


m {1+q} 0 = S(q)
since
q:log(m) * 0 = q:log(S(q)) = 0

Now comes the difficult task of evaluating these new found operators. With our knowledge that:
r:log(b {3} x) = b {3} (x-r)
r:log(b {3} x) = b {3} (slog(b {3} x) - r)
and therefore:
r:log(m) = b {3} (slog(m) - r)

Where slog(x) is the inverse function of tetration.

now, since:
-r:log(r:log(x)) = x
m {q} n = -q:log(q:log(m) + q:log(n))
m {q} n = b {3} (slog( (b {3} (slog(m) - q)) + (b {3} (slog(n) - q))) + q)

m {1+q} n = -q:log(q:log(m) * n)
m {1+q} n = b {3} (slog((b {3} (slog(m)-q)) * n) + q)


Now, further observing the identity function:
since:
q:log(S(q)) = 0

b {3} (slog(S(q)) -q) = 0

slog(S(q)) - q = -1

slog(S(q)) = q - 1

S(q) = b {3} (q-1)

And now if the critical strip of tetration is defined as:
-1 <= f <= 0

b {3} f = f + 1

S(q) = q

and therefore:
m {q} q = m
Further notes:
Consider the function
A(x) = m {x} n

Which is a generalization of the Ackerman function, extending it to domain real.

Interesting results are:
A(x) = 2 {x} 2 peaks at A(1 - 1/ln(2)) = 4.248828844

A(x) = 2 {x} 2 is periodic with period one, and therefore has a fourier series.

Results found using the following derivatives:
(b {3} x)' = ln(b)^floor(x) * [E(k=0, floor(x)) b {3} (x - k)]
(slog(x))' = (ln(b)^floor(slog(x)) * [E(k=0, floor(slog(x))) k:log(x)])^(-1)

Where [E(k=0, n) f(k)] is an Euler product.


Edit:

Also if

(x {q} y) }q{ y = x

or if }q{ is rational division and subtraction.

x {1+q} -1 = q }q{ x

Which is a special case of a more general formula

x {1+q} e^ji = q (e^ji){q} x
if (-1){q} = }q{

(x (e^ji){q} y) (e^ji)}q{ y = x
(e^ji)}q{ = (e^(j+pi)i){q}

Not much is really known about artificial operators. They are created by multiplying any natural operator with a complex coefficient of magnitude 1. }2{ is roots. }3{ is super roots


EDIT 2:
Also, one should note that

0.5:log(0.5:log(x)) = b {3} (slog(b {3} (slog(x) - 0.5)) - 0.5) = log(x)

Which is probably my main argument for the extension of tetration that I use.

Also, if one doesn't like this extension: rational operators are an independent discovery consistent with any rational tetration. However, if domain [-1, 0] is not universal for each base rational operators become dependent on a logarithm base.

Edit 3:

Actually, I see now that there is another method of evaluating rational iterations of the logarithm function.
as long as:
-q:log(q:log(x)) = x; this should maintain consistency.

Actually nvm this last part, my rational iteration model is symmetric to the other method.

Edit 4:

Here is a graph of x {0} 3 transforming into x {1.8} 3, counting up by .2
window screen is (xmin = 0, xmax=50, ymin=0, ymax=50)


The fact that it's squiggly bewilders me and leaves me in awe.
I believe it has something to do with the extension of tetration I use...

[bo198214]

I would like to see a plot in dependence of q. E.g. f(q) = 2 {q} 3.
Still thinking about it.

[tommy1729]

for starters i would like to comment that i find this whole thread suspiciously like my own recent threads

http://math.eretrandre.org/tetrationforu...hp?tid=543

http://math.eretrandre.org/tetrationforu...hp?tid=520

and i dont see what the new benefit or new intention of this similar idea is.

also , i dont see why we need a new ackermann clone.

furthermore im not totally sure everything is correct.

also , i dont think we should use totally new terminology just like that.

0 <= q <= 1
q:log(m {1} n) = q:log(m) {1-q} q:log(n)
q:log(m {2} n) = q:log(m) {2-q} n

These operators would have the following property; if S(q) returns the identity of any operator:
m {q} S(q) = m
q:log(m) + q:log(S(q)) = q:log(m)
therefore
q:log(S(q)) = 0
which is true regardless of logarithm base.
m {1+q} S(1+q) = m
q:log(m) * S(1+q) = q:log(m)
therefore:
S(1+q) = 1
and which in general becomes all operators greater than or equal to one have identity one.



Rational operators are not distributive over addition, however, as multiplication is to exponentiation and as exponentiation is to multiplication {q} is to {1+q} as {1+q} is to {q}
therefore:
(m {q} n) {1 + q} l = (m {1+q} l) {q} (n {1+q} l)

(m {1+q} n) {q} m = m {1+q} (n+1)


m {1+q} 0 = S(q)
since
q:log(m) * 0 = q:log(S(q)) = 0

and therefore:
r:log(m) = b {3} (slog(m) - r)

Where slog(x) is the inverse function of tetration.

now, since:
-r:log(r:log(x)) = x
m {q} n = -q:log(q:log(m) + q:log(n))
m {q} n = b {3} (slog( (b {3} (slog(m) - q)) + (b {3} (slog(n) - q))) + q)

m {1+q} n = -q:log(q:log(m) * n)
m {1+q} n = b {3} (slog((b {3} (slog(m)-q)) * n) + q)


Now, further observing the identity function:
since:
q:log(S(q)) = 0

b {3} (slog(S(q)) -q) = 0

slog(S(q)) - q = -1

slog(S(q)) = q - 1

S(q) = b {3} (q-1)

And now if the critical strip of tetration is defined as:
-1 <= f <= 0

b {3} f = f + 1

S(q) = q

and therefore:
m {q} q = m
Further notes:
Consider the function
A(x) = m {x} n

Which is a generalization of the Ackerman function, extending it to domain real.

Interesting results are:
A(x) = 2 {x} 2 peaks at A(1 - 1/ln(2)) = 4.248828844

The fact that it's squiggly bewilders me and leaves me in awe.
I believe it has something to do with the extension of tetration I use...

im not sure that S(q) = q and S(1+q) = 1 .. in fact , isnt that a contradiction already combining those two ?

( we want S(z) to be differentiable not ? )

you started to define your identities for 0 =< q =< 1 , what makes me doubt if the proclaimed things for q > 1 must hold.

where does 1 - 1/ln(2) come from btw ??

yes , it has to do with the extension of tetration you used , your slog is defined piecewise and linear, hence no nice properties are satisfied and no new slog is discussed.

regards

tommy1729

[JmsNxn]

Sorry, I have no clue what you linked me to? I don't think you understand what I am getting at. I see the similarity but I see nothing outright stating that these are operators inbetween addition multiplication and exponentiation.

And as I said, it's simply a hypotheses that S(q) = q, actually, it's less so than a hypothesis.

Consider: 0 <= q <= 1
m {q} S(q) = m

take the rational iterated log q times, and we get

q:log(m) + q:log(S(q)) = q:log(m)

therefore q:log(S(q)) = 0
And everywhere I've checked,
q:log(q) = 0


Absolutely S(g) = 1, if g >= 1

The proof for operators less than {2} is so incredibly simple:
m {1 + q} S(1+q) = m

Take the rational iterated log q times, and we get:

q:log(m) * S(1+q) = q:log(m)
Therefore S(1+q) = 1

It is a direct result of the axiom which states that logarithms are distributive over any operator less than or equal to {1}, and exclusive for operators greater than {1}. Exclusive in the sense that only the base is placed within the logarithm.

The function I stated was not an Ackerman clone, it's the Ackerman functioned defined for Real operator values over period [0, 2]. It's a generalization.

1 - 1/ln2 comes from a long and clunky proof, not really important enough to repeat it here.

To be honest, I never considered differentiating S(z). There is only a critical section [0,1] that is worth underlining, and its value is marked by
S(q) = b {3} (q -1)
And so therefore if tetration is not linear over this domain S(q) is dependent upon a logarithm base.

And we should never consider the quantity q > 1, because operators follow recursion and and adding 1 to q is like stepping one step up on the recursion ladder. q should always be the rational part of a number.

I thank you for reading this over and I didn't mean to step on your toes if you already had this idea. I've been trying to get at the heart of what rational operators are for two years now :/

[sheldonison]

And now if the critical strip of tetration is defined as:
-1 <= f <= 0

b {3} f = f + 1

S(q) = q

and therefore:
m {q} q = m
Further notes:
Consider the function
A(x) = m {x} n

Which is a generalization of the Ackerman function, extending it to domain real.
...

I'm a little slow at catching on to the jist of your post, but the Ackermann function is A(m=4,n)=2^^(n+3) - 3, or roughly base(2) tetration for m=4. So I assume you're trying to define an extension to the Ackermann function for real numbers, where A(x) = m {x} n, where "m" is the base, and x is a rational operator.
So,
A(x=2)= m (2) n = m*n.
A(x=3)= m (3) n = m^n.
A(x=4)= m (4) n = m^^n

Is this the basic idea, where we are extending it to allow for for real values of "x" as well? Then Henryk's request is to see a graph of
f(q) = 2 {q} 3.
so f(2)=2*3=6, f(3)=2^3=8, f(4)=2^^3=16 ..... Sounds interesting!

I don't think the linear approximation for the critical strip for [-1..0] for tetration is a good idea. There are many approaches to extending tetration to real numbers, that are analytic on the complex plane, and they all seem to agree with each other.
- Sheldon

[JmsNxn]

And now if the critical strip of tetration is defined as:
-1 <= f <= 0

b {3} f = f + 1

S(q) = q

and therefore:
m {q} q = m
Further notes:
Consider the function
A(x) = m {x} n

Which is a generalization of the Ackerman function, extending it to domain real.
...

I'm a little slow at catching on to the jist of your post, but the Ackermann function is A(m=4,n)=2^^(n+3) - 3, or roughly base(2) tetration for m=4. So I assume you're trying to define an extension to the Ackermann function for real numbers, where A(x) = m {x} n, where "m" is the base, and x is a rational operator.
So,
A(x=2)= m (2) n = m*n.
A(x=3)= m (3) n = m^n.
A(x=4)= m (4) n = m^^n

Is this the basic idea, where we are extending it to allow for for real values of "x" as well? Then Henryk's request is to see a graph of
f(q) = 2 {q} 3.
so f(2)=2*3=6, f(3)=2^3=8, f(4)=2^^3=16 ..... Sounds interesting!

I don't think the linear approximation for the critical strip for [-1..0] for tetration is a good idea. There are many approaches to extending tetration to real numbers, that are analytic on the complex plane, and they all seem to agree with each other.
- Sheldon

Yes, everything you posted was completely right, except I like to begin our operation sequence with {0} as addition. My argument is because there are no well defined operators below {0} (besides successorship which is invalid with rational numbers). However, I do have a few "suggestions" on how to extend to negative and complex operators.

To be honest, I had only chosen the linear approximation model because it was on wikipedia, and I had the desperate urge to evaluate operators. It also makes the algebra simple. But now that I see the wavy lines I am a bit against it. But, if tetration is not linear over domain [0, 1] rational operators depend on a logarithm base for their identity--and it no longer has universality.

The essential axioms are as follows:
0<= q <= 1
q:log(m {1+q} n) = q:log(m) {1} n = q:log(m) * n
q:log(m {q} n) = q:log(m) {0} q:log(n) = q:log(m) + q:log(n)

They follow recursion, and therefore {q} can be thought of as multiplication, and {1+q} can be thought of as exponentiation. And therefore multiplication is to addition as exponentiation is to multiplication.
Therefore, rational tetration, which occurs over domain (2, 3] can be defined for natural numbers as recursive {1+q}.

[sheldonison]

Yes, everything you posted was completely right, except I like to begin our operation sequence with {0} as addition.
....
The essential axioms are as follows:

Consider the case for base 2, m=2. This would be the Ackermann function (as modfied by Buck, so that A(4,n)=tetration, A(3,n)=exponentation....). The goal is to extend this Ackermann function to real numbers. Here, by your notation, m=q+1.
f(x)=A(m,n) = 2 {m-1} n = 2 {q} n

Continuing, I am modifying your post base for m=2.

0<= q <= 1
q:log(2 {1+q} n) = q:log(2) * n
q:log(2 {q} n) = q:log(2) + q:log(n)

What does q:log mean? How do we use the equations to determine what the value of q:log(2) is? How do we determine what q:log(n) is?

Presumably, this leads to the value for 2 {q} n. How does the value for 2 {q} n depend on the values for \( \text{sexp}_2(n) \)= 2 {3} n? Here is a 60 term Taylor series for \( \text{sexp}_2(n) \), if that helps generate the graph of f(q) = 2 {q} 3.

Code:

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- Sheldon

[bo198214]

But, if tetration is not linear over domain [0, 1] rational operators depend on a logarithm base for their identity--and it no longer has universality.

Ya that was my first feel of your construction: "piecewise".
I mean its better than nothing, but usually one wants a smooth or better analytic function. I.e. that the function x {y} z is analytic in all 3 arguments.
Usually piecewise constructions aren't analytic at the gluing points (i.e. here at the integers).
Nonetheless its the first construction I heard of.
The requirements to such operators were discussed already on the board somewhere (perhaps later I can find these threads).

[JmsNxn]

Yes, everything you posted was completely right, except I like to begin our operation sequence with {0} as addition.
....
The essential axioms are as follows:

Consider the case for base 2, m=2. This would be the Ackermann function (as modfied by Buck, so that A(4,n)=tetration, A(3,n)=exponentation....). The goal is to extend this Ackermann function to real numbers. Here, by your notation, m=q+1.
f(x)=A(m,n) = 2 {m-1} n = 2 {q} n

Continuing, I am modifying your post base for m=2.

0<= q <= 1
q:log(2 {1+q} n) = q:log(2) * n
q:log(2 {q} n) = q:log(2) + q:log(n)

What does q:log mean? How do we use the equations to determine what the value of q:log(2) is? How do we determine what q:log(n) is?

Presumably, this leads to the value for 2 {q} n. How does the value for 2 {q} n depend on the values for \( \text{sexp}_2(n) \)= 2 {3} n? Here is a 60 term Taylor series for \( \text{sexp}_2(n) \), if that helps generate the graph of f(q) = 2 {q} 3.

Code:

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a38= -0.00000000000013811825249928203947
a39=  0.00000000000006728838247350718828
a40= -0.00000000000003280308641198027785
a41=  0.00000000000001600150582457032312
a42= -0.00000000000000781025880698437811
a43=  0.00000000000000381431246327820126
a44= -0.00000000000000186381177420545698
a45=  0.00000000000000091119686946232850
a46= -0.00000000000000044569412126376944
a47=  0.00000000000000021810563400669686
a48= -0.00000000000000010678088335635442
a49=  0.00000000000000005230084084687218
a50= -0.00000000000000002562741201964737
a51=  0.00000000000000001256245687431084
a52= -0.00000000000000000616043558311726
a53=  0.00000000000000000302210047493469
a54= -0.00000000000000000148306782571967
a55=  0.00000000000000000072805147810109
a56= -0.00000000000000000035752527943085
a57=  0.00000000000000000017562645305590
a58= -0.00000000000000000008629920538158
a59=  0.00000000000000000004241825349287
a60= -0.00000000000000000002085564130073

- Sheldon

Algebraically, 0<=q<=1

2 {q} n = -q:log(q:log(2) + q:log(n))
2 {1+q} n = -q:log(q:log(2) * n)
where:
q:log(x) = b {3} (slog(x) - q)
And q:log(x) is taken to mean the q'th iterate of log(x)



A taylor series expansion could only work if one also has a slog taylor series expansion. If you give me that I'd be happy to make a graph over domain [0, 2].

[sheldonison]

A taylor series expansion could only work if one also has a slog taylor series expansion. If you give me that I'd be happy to make a graph over domain [0, 2].

Here is the Taylor series. \( \text{Taylor}(z-1)=\text{slog}_2(z) \), which will converge nicely for z in the range [0..2]. If z<0, take \( z=2^z \) before generating Taylor(z-1)-1. If z>2, iterate \( z=\log_2(z) \), before generating Taylor(z-1)+n, so that z is in the range [0..2].

Code:

a0=   0.00000000000000000000000000000000
a1=   1.12439780182947880296975296510341
a2=  -0.01233408638319092919966757867732
a3=  -0.15195716580089316798328536602130
a4=   0.01868009944521288546047080416998
a5=   0.03456100685993161409190280063892
a6=  -0.00907417008961111769380973532974
a7=  -0.00882611191544351225979374105298
a8=   0.00382451721437283174576066832193
a9=   0.00228031089800741907723932214202
a10= -0.00151922346582286239408053757523
a11= -0.00055532576725948556607647219741
a12=  0.00058068759568845128571208222568
a13=  0.00011333875202118827889934233768
a14= -0.00021492130432643551427679982642
a15= -0.00001121186618462451210489139936
a16=  0.00007707957627653141354330216317
a17= -0.00000624892419462078938069406186
a18= -0.00002671099173826526447600018191
a19=  0.00000581456717530582001598703419
a20=  0.00000888533730151933862945265998
a21= -0.00000330763773421352876145208923
a22= -0.00000280211888032738989276581338
a23=  0.00000159184385525029832990193555
a24=  0.00000081754010898099012004646318
a25= -0.00000069935182173423145339560199
a26= -0.00000020858066529691830195782405
a27=  0.00000028862436851123339303428296
a28=  0.00000003862884977802212289870391
a29= -0.00000011328157592267567824609526
a30=  0.00000000094217657182114853689258
a31=  0.00000004247233495866309956742421
a32= -0.00000000615478181186900908929094
a33= -0.00000001519770422468823619166244
a34=  0.00000000440788391865597168670175
a35=  0.00000000515674874286180172029316
a36= -0.00000000238020450731772920188547
a37= -0.00000000163450373305911517165825
a38=  0.00000000113080753816867247217337
a39=  0.00000000046806124793717393704457
a40= -0.00000000049617938942328314300887
a41= -0.00000000011074640735137445965583
a42=  0.00000000020520322530692159331762
a43=  0.00000000001419679684872395334262
a44= -0.00000000008069713894112959557995
a45=  0.00000000000566866105603014575315
a46=  0.00000000003024789945770100323766
a47= -0.00000000000635476323844608537350
a48= -0.00000000001077582550414236272297
a49=  0.00000000000393825972677957845029
a50=  0.00000000000361455077455751078451
a51= -0.00000000000202059080878299162503
a52= -0.00000000000111781235812744791900
a53=  0.00000000000093621058575542170278
a54=  0.00000000000030319231965398624547
a55= -0.00000000000040462836021383451852
a56= -0.00000000000006151258021196048097
a57=  0.00000000000016547055506890328371
a58=  0.00000000000000095303168930585396
a59= -0.00000000000006439584830686473714
a60=  0.00000000000000855544017414425242
a61=  0.00000000000002385199982356358626
a62= -0.00000000000000663620464544180588
a63= -0.00000000000000836346245035952446
a64=  0.00000000000000374483888620781880
a65=  0.00000000000000273889585903612029
a66= -0.00000000000000184001327491851237
a67= -0.00000000000000081266892359162900
a68=  0.00000000000000083079153254456575
a69=  0.00000000000000020181213382841263
a70= -0.00000000000000035247402372018293
a71= -0.00000000000000002985852756369760
a72=  0.00000000000000014190787623066308
a73= -0.00000000000000000805476244793420
a74= -0.00000000000000005438469969331962
a75=  0.00000000000000001071535590487842
a76=  0.00000000000000001979662010087425
a77= -0.00000000000000000694989588600789
a78= -0.00000000000000000678704531046374
a79=  0.00000000000000000366768107666540
a80=  0.00000000000000000214963783636005