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+--- Thread: On the existence of rational operators (/showthread.php?tid=546)
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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} (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)
![[Image: Rationaloperators.png?t=1292529818]](http://i233.photobucket.com/albums/ee309/jamesuminator/Rationaloperators.png?t=1292529818)
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
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
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.
-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.
...
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.
-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.
...
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}.
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.
....
The essential axioms are as follows:
f(x)=A(m,n) = 2 {m-1} 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?
q:log(2 {1+q} n) = q:log(2) * n
q:log(2 {q} n) = q:log(2) + q:log(n)
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= 0.00867654896536993398021314555342
a3= 0.09523880007518178992043848503015
a4= -0.00575234854012612265921659771675
a5= 0.01296658202003717397631073760594
a6= -0.00219604962303099464215851948058
a7= 0.00199674684791144277954258704212
a8= -0.00056335481487852207283213227644
a9= 0.00034824232818816420812599367176
a10= -0.00012853244126472000389077672649
a11= 0.00006708192442053080892782003694
a12= -0.00002829875282279795293872424625
a13= 0.00001380013199063292876626124469
a14= -0.00000620190939837452275803185904
a15= 0.00000295556146480966397507180597
a16= -0.00000136867922453469799692662269
a17= 0.00000064905707565189565568577947
a18= -0.00000030516693932892648666925176
a19= 0.00000014494820615122971623989185
a20= -0.00000006874664311379176575448606
a21= 0.00000003276744517789339988798283
a22= -0.00000001563108746799695037791409
a23= 0.00000000747812838918081154207009
a24= -0.00000000358267681323948167911806
a25= 0.00000000171986774579520893372333
a26= -0.00000000082681596246801027621026
a27= 0.00000000039811046838268961282597
a28= -0.00000000019194299258773230479840
a29= 0.00000000009266318678629802795635
a30= -0.00000000004478698829133491625880
a31= 0.00000000002167120321191567043557
a32= -0.00000000001049697429142400963746
a33= 0.00000000000508944818189222250595
a34= -0.00000000000246987831946184648244
a35= 0.00000000000119965565647417085401
a36= -0.00000000000058316588902205318444
a37= 0.00000000000028370236181432539258
a38= -0.00000000000013811825249928203946
a39= 0.00000000000006728838247350718829
a40= -0.00000000000003280308641198027785
a41= 0.00000000000001600150582457032312
a42= -0.00000000000000781025880698437811
a43= 0.00000000000000381431246327820127
a44= -0.00000000000000186381177420545697
a45= 0.00000000000000091119686946232850
a46= -0.00000000000000044569412126376943
a47= 0.00000000000000021810563400669687
a48= -0.00000000000000010678088335635441
a49= 0.00000000000000005230084084687218
a50= -0.00000000000000002562741201964736
a51= 0.00000000000000001256245687431084
a52= -0.00000000000000000616043558311725
a53= 0.00000000000000000302210047493470
a54= -0.00000000000000000148306782571966
a55= 0.00000000000000000072805147810109
a56= -0.00000000000000000035752527943085
a57= 0.00000000000000000017562645305590
a58= -0.00000000000000000008629920538158
a59= 0.00000000000000000004241825349287
a60= -0.00000000000000000002085564130072RE: 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 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).
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.
....
The essential axioms are as follows:
f(x)=A(m,n) = 2 {m-1} 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?
q:log(2 {1+q} n) = q:log(2) * n
q:log(2 {q} n) = q:log(2) + q:log(n)
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.
- SheldonCode: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
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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. \( \text{Taylor}(z-1)=\text{slog}_2(z) \), 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:
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