Tetration Forum

Full Version: On the existence of rational operators
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Pages: 1 2
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...
I would like to see a plot in dependence of q. E.g. f(q) = 2 {q} 3.
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.

(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
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 :/
(12/14/2010, 01:41 AM)JmsNxn Wrote: [ -> ]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
(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:
-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}.
(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.
....
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.
Quote: 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=   0.00867654896536993398021314555342 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= -0.00000000000000000002085564130072
- Sheldon
(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).
(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.
....
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.
Quote: 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].
(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:
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

Pages: 1 2