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 On the existence of rational operators JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 12/20/2010, 09:38 PM (This post was last modified: 12/20/2010, 09:43 PM by JmsNxn.) (12/20/2010, 08:28 PM)sheldonison Wrote: (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 it is. Taylor series for $\text{slog}_2(z-1)$, which will converge nicely for z in the range [0..2]. If z<0, take $z=2^z$ before generating slog(z-1)-1. If z>2, iterate $z=\log_2(z)$, before generating slog(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 window screen [xmin = 0, xmax =2, ymin=0, ymax = 10] I'm not liking the way this looks either (don't worry, I used the same algorithm you defined). Are there any other ways to extend tetration that I can try? Hopefully one of them will make the graph without any angles. If not, I guess it just expresses a very odd connection. The function 2 {x} 3 is defined piecewise, so I guess it's just the way it is :/. JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 03/11/2011, 07:23 PM (This post was last modified: 03/11/2011, 08:42 PM by JmsNxn.) I've come up with a new idea attached to rational operators. It first involves defining a new set of operators that behave as addition, Tommy_r already noted them and talked about the distribution law, he even talked a bit about what I'm getting at except he didn't fully get at it: 0 <= q <= 1; q:ln(x) = exp^[-q](x) x {-q} y = q:ln(-q:ln(x) + -q:ln(y)) I should prove some lemmas so we have something to work with: q:ln(x + x) = q:ln(x) {-q} q:ln(x) q:ln(2 * x) = q:ln(x) {1-q} q:ln(2) q:ln(x) {-q} q:ln(x) = q:ln(x) {1-q} q:ln(2) which by induction becomes x {1-q} q:ln(n) = x {-q} x {-q} x .... n times This is why I chose to index these operations with negative values because they do not obey the fundamental law of recursion. They do however behave as addition: q:ln(x * (y + a)) = q:ln(xy + xa) q:ln(x) {1-q} q:ln(y+a) = q:ln(xy) {-q} q:ln(xa) q:ln(x) {1-q} (q:ln(y) {1-q} q:ln(a)) = (q:ln(x) {1-q} q:ln(y)) {-q} (q:ln(x) {1-q} q:ln(a)) which simplified is: x {1-q} (y {-q} a) = (x {1-q} y) {-q} (x {1-q} a) Now we have a full ring of operators, exactly as there's addition (+) and multiplication (*) and exponentiation (^), now there's {-q} which behaves as addition, {1-q} which behaves as multiplication, and {2-q} which behaves as exponentiation. Given this final lemma we can create a new calculus. if S(x) is the identity function: q:ln(x + 0) = q:ln(x) {-q} q:ln(0) = q:ln(x) therefore S(-q) = q:ln(0) and therefore there is a pole at S(-1) and there is no identity for operator {-1} Now we must prove as 0 is to multiplication, S(-q) is to {1-q} (which holds for q=1, since technically S(-1) is infinity and x + inf = inf which is parallel to multiplication by zero equaling zero). -q:ln(x {1-q} q:ln(0)) = -q:ln(x) * 0 = 0 therefore: x {1-q} q:ln(0) = q:ln(0) and q:ln(0) = S(-q) with this, we can now create the fundamental theorem of logarithmic semi operator calculus: q:d/dx f(x) = lim h->S(-q) [f(x {-q} h) }-q{ f(x)] }1-q{ h Note that 0:d/dx f(x) = d/dx f(x), and that 1:d/dx f(x) is a pole of negative inf. Here are some laws, they turn out exactly the same: q:d/dx f(x) {-q} g(x) = q:d/dx f(x) {-q} q:d/dx g(x) which is the law that states differentiation is distributable over addition. q:d/dx f(x) {1-q} g(x) = [q:d/dx f(x) {1-q} g(x)] {-q} [f(x) {1-q} q:d/dx g(x)] which is the product rule. The chain rule is applicable, it becomes: q:d/dx f(g(x)) = q:d/dg(x) f(g(x)) {1-q} q:d/dx g(x) With this you can see that every operation you do with derivatives, you can do a parallel operation with just "lowered" operator power. Here comes the sweet part, we can now invent lower operator polynomials and then transform that knowledge into lower operator Taylor series and invent a definition of semi-operator analytic functions q:d/dx x {2-q} n = q:ln(n) {1-q} (x {2-q} (n-1)) This derivative is a result of the normal power rule proof, but at the end when you collect n amount of x^(n-1), you must convert that to q:ln(n) because {-q} is not proper recursive. Now we can develop a taylor series, if X(n=0, y, q) An = A1 {-q} A2 ... {-q} Ay and q:n! = q:ln(1) {1-q} q:ln(2) {1-q} ... q:ln(n) = q:ln(n!) q:0! = S(1-q) q:1! = S(1-q) f(x) = X(n=0, inf., -q) [S(1-q) }1-q{ q:n!] {1-q} q:d^n/dx^n f(a) {1-q} [(x }-q{ a) {2-q} n] We can now create the lower operator sine function, the lower operator exponential function which will have the same relationship: exp_q(x) = X(n=0, inf, -q) (S(1-q) }1-q{ q:n!) {1-q} (x {2-q} n) sin_q(x) = X(n=0, inf., -q) ((-S(1-q) {2-q} n) }1-q{ q:(2n+1)!) {1-q} (x {2-q} 2n+1) cos_q(x) = X(n=0, inf, -q) ((-S(1-q) {2-q} n) }1-q{ q:2n!) {1-q} (x {2-q} 2n+1) Which may look messy at first, is just the normal sin cos exp taylor series with every operator lowered by q and the q:ln taken of the factorial to account for the false recursion in operator {-q}. if we create J(q) such that J(q) {q} J(q) = -S(q) and therefore J(1) = i, and J(0) = 0. In a perfect world J(0.5) = 0.5 * i, but this is false. J(q) = -S(q) {1+q} 1/2 = -q:ln(q:ln(-S(q))*1/2) we can now generalize euler's formula to exp_q(x {1-q} J(1-q)) = cos_q(x) {-q} (J(1-q) {1-q} sin_q(x)) q:d/dx exp_q(x) = exp_q(x), as well I'm trying to look into extending all of analytic calculus into logarithmic semi-operator calculus. So far it's very difficult to take the "q-derivative" of any ordinary function. And because of a certain law, the lowered exponential function, b{2-q} x, is in-differentiable. There, that's all of it. I hope some of you have opinions. PS: I also wanted to ask, does anyone think that choosing an erratic base for the operators might make the transition of 2 {x} 3 more smooth unlike how it is now? It's a shame really because every other transformation is very very nice, like f(x) = 1 }1-q{ x, and varying q bit by bit. JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 03/19/2011, 05:36 PM (This post was last modified: 03/19/2011, 05:36 PM by JmsNxn.) I have a discovery that starts to legitimize logarithmic semi-operator calculus. given: q:d/dx f(x) = lim h->S(-q) [f(x{-q}h) }-q{ f(x)] }1-q{ h I can prove that q:d/dx e^x = e^x or that if we lower the operators involved in the difference quotient by any value 0 <= q < 1; and change h's limit from zero (the additive identity) to S(-q) (the lowered additive identity), e^x is still it's own output. « Next Oldest | Next Newest »

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