On the existence of rational operators  Printable Version + Tetration Forum (https://math.eretrandre.org/tetrationforum) + Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) + Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) + Thread: On the existence of rational operators (/showthread.php?tid=546) Pages:
1
2

On the existence of rational operators  JmsNxn  12/14/2010 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} (x1) and therefore, connected to superfunctions: r:log(b {3} x) = b {3} (xr) 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) {1q} q:log(n) q:log(m {2} n) = q:log(m) {2q} 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} (x1) or more generally: r: (1+q);log(b {2+q} x) = b {2+q} (xr) r:q;log(b {1+q} x) = b {1+q} (xr) 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} (xr) 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} (q1) 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... RE: On the existence of rational operators  bo198214  12/19/2010 I would like to see a plot in dependence of q. E.g. f(q) = 2 {q} 3. Still thinking about it. RE: On the existence of rational operators  tommy1729  12/19/2010 for starters i would like to comment that i find this whole thread suspiciously like my own recent threads http://math.eretrandre.org/tetrationforum/showthread.php?tid=543 http://math.eretrandre.org/tetrationforum/showthread.php?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. (12/14/2010, 01:41 AM)JmsNxn Wrote: 0 <= q <= 1 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 RE: On the existence of rational operators  JmsNxn  12/19/2010 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 :/ RE: On the existence of rational operators  sheldonison  12/19/2010 (12/14/2010, 01:41 AM)JmsNxn Wrote: And now if the critical strip of tetration is defined as: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 RE: On the existence of rational operators  JmsNxn  12/20/2010 (12/19/2010, 08:23 PM)sheldonison Wrote:(12/14/2010, 01:41 AM)JmsNxn Wrote: And now if the critical strip of tetration is defined as: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. 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 identityand 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}. RE: On the existence of rational operators  sheldonison  12/20/2010 (12/20/2010, 02:16 AM)JmsNxn Wrote: Yes, everything you posted was completely right, except I like to begin our operation sequence with {0} as addition.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 {m1} n = 2 {q} n Continuing, I am modifying your post base for m=2. Quote: 0<= q <= 1What 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 = 2 {3} n? Here is a 60 term Taylor series for , if that helps generate the graph of f(q) = 2 {q} 3. Code: a0= 1.00000000000000000000000000000000 RE: On the existence of rational operators  bo198214  12/20/2010 (12/20/2010, 02:16 AM)JmsNxn Wrote: But, if tetration is not linear over domain [0, 1] rational operators depend on a logarithm base for their identityand 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). RE: On the existence of rational operators  JmsNxn  12/20/2010 (12/20/2010, 04:53 AM)sheldonison Wrote:(12/20/2010, 02:16 AM)JmsNxn Wrote: Yes, everything you posted was completely right, except I like to begin our operation sequence with {0} as addition.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. 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]. RE: On the existence of rational operators  sheldonison  12/20/2010 (12/20/2010, 07:01 PM)JmsNxn Wrote: 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. , which will converge nicely for z in the range [0..2]. If z<0, take before generating Taylor(z1)1. If z>2, iterate , before generating Taylor(z1)+n, so that z is in the range [0..2]. Code: a0= 0.00000000000000000000000000000000 