On the existence of rational operators

[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:

a0=   1.00000000000000000000000000000000
a1=   0.88936495462097637278974352283113
a2=   0.00867654896536993398021314555341
a3=   0.09523880007518178992043848503015
a4=  -0.00575234854012612265921659771676
a5=   0.01296658202003717397631073760594
a6=  -0.00219604962303099464215851948058
a7=   0.00199674684791144277954258704211
a8=  -0.00056335481487852207283213227645
a9=   0.00034824232818816420812599367175
a10= -0.00012853244126472000389077672650
a11=  0.00006708192442053080892782003693
a12= -0.00002829875282279795293872424626
a13=  0.00001380013199063292876626124468
a14= -0.00000620190939837452275803185905
a15=  0.00000295556146480966397507180597
a16= -0.00000136867922453469799692662269
a17=  0.00000064905707565189565568577946
a18= -0.00000030516693932892648666925177
a19=  0.00000014494820615122971623989185
a20= -0.00000006874664311379176575448607
a21=  0.00000003276744517789339988798283
a22= -0.00000001563108746799695037791410
a23=  0.00000000747812838918081154207009
a24= -0.00000000358267681323948167911806
a25=  0.00000000171986774579520893372333
a26= -0.00000000082681596246801027621026
a27=  0.00000000039811046838268961282597
a28= -0.00000000019194299258773230479841
a29=  0.00000000009266318678629802795635
a30= -0.00000000004478698829133491625880
a31=  0.00000000002167120321191567043557
a32= -0.00000000001049697429142400963746
a33=  0.00000000000508944818189222250595
a34= -0.00000000000246987831946184648244
a35=  0.00000000000119965565647417085401
a36= -0.00000000000058316588902205318445
a37=  0.00000000000028370236181432539258
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].