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final uniqueness condition ... probably

Wed, 28 Jul 2010 17:30:53 +0200

d f^2 / d x^2 sexp(slog(x) + k) > 0 for all real x and all real k with 0 < k < Q where Q is any nonzero positive real.

assuming sexp resp slog to be C^2 of course , i 'believe' this condition implies analytic as well.

is this equivalent to d f^2 / d x^2 sexp(x) > 0 for all positive real x ?

i assume because of the substitution x = sexp(y)

sexp(slog(x) + k) = sexp(y + k) d sexp(y) ...

i believe the analogue uniqueness condition for other function as exp(x) that dont have a real fixpoint , map R to R and safisfy f ' (real) > 0 and f '' (real) > 0 to hold.

(this is an improved version of an earlier thread.)

regards

tommy1729

[Regular tetration] norming fixpoint-dependencies

Wed, 28 Jul 2010 16:04:48 +0200

If we do regular tetration having a "nice" base 1 So, for base b=sqrt(2) we have fp0=2 and fp1=4 .

Iterations on the real axis constitute -at a first sight- three different segments

Code:

.            seg1         seg2          seg3

.      --------------|-------------|-------------

.        (0)  ..   (fp0)   ..    (fp1)   ..    (+oo)

.        (0)  ..    (2)    ..     (4)    ..    (+oo)     // for instance for base sqrt(2)

.      --------------|-------------|-------------

.         x1->y1         x2->y2       x3->y3

For some x1 of seg1 we may iterate with infinite height to some y1, but y1 will remain in seg1. However, negative heights may result in imaginary results or even in the log(0)-singularity, so I didn't include that range in the sketch above.

For some x2 in seg2 we can iterate with arbitrary heights to some y2, but again are confined to y2 in seg2.

For some x3 in seg3 we can iterate with arbitrary negative height to some y3 and -in principle- also to arbitrary positive heigt, but practically encounter numerical overflow very soon.

The powerseries for regular tetration can be developed around fp0 or fp1. Let's call that tet0 and tet1 for shortness.

Then if we look at the schroeder-function of tet0 for all x1 we get negative values and for all x2 we get positive values. Thus we can map the set of seg1 to that of seg2 by negating the schroeder-value. This means for instance, that x1 = 1 gets mapped to x2 = 2.467914... and because the height-limits of seg2 are -infinity and +infinity we could use that value x2 to define a norm for that segment, so in seg2 the value x2=2.467914 could be said has (real) height 0 by definition.

We have a similar problem with seg3: here too we have infinity at both height-limits. But we can repeat the norming-process, now using the schroeder-values of tet1. We compute the schroeder-value of x2 using tet1 and compute x3 by negating that value. However, without further measures we get infinity here.
If we reduce the height of x2 by 2, then we get x3 = 417.234406762

So we have the segments with the normed heights

Code:

.            seg1                   seg2                 seg3

.     --------------------|-----------------------|-------------

.             .. 1 ..   (fp0)   ..              (fp1)   ..            (+oo)

.             .. 1 ..    (2)    .. 2.467 ...     (4)    .. 417 ...    (+oo)     // b=sqrt(2)

.      -------------------|-----------------------|-------------

.              h(x1)=0  oo|oo     h(x2)=0      -oo|***************    // set norms for tet0

.        *****************|oo     h(x2)=0      -oo|-oo  h(x3)=?=-2    // set norms for tet1

.

Unfortunately this has two asymmetries: the tet0 and tet1 have somehow opposite sign; but more inconvenient is, that we cannot have the same height-norm.

What we can do is to shift left and use x1=0 as reference. We get then

Code:

.         x1 = 0     x2 = 2.606584    (x3=417.2344)

.

.            seg1                   seg2                 seg3

.     --------------------|-----------------------|-------------

.        (0)  .. 0 ..   (fp0)   ..              (fp1)   ..            (+oo)

.        (0)  .. 0 ..    (2)    .. 2.606 ...     (4)    .. 417 ...    (+oo)     // b=sqrt(2)

.      -------------------|-----------------------|-------------

.             h0(x1)=-1 oo|oo    h0(x2)=-1     -oo|***************    // set norms for tet0

.        *****************|oo    h1(x2)=-1     -oo|-oo h1(x3)=?=-2    // set norms for tet1

.

and still x3 computed by x2 seems to become infinite. We may reduce again x2 by height 1 to get the usable x3-value of 417.2344...
We cannot reduce x1 by one more height, but my proposal here is to use x1 = b^^-1.5 as reference value.
Then we have, for base b=sqrt(2) the reference-values for height -1.5

Code:

.       x1 = -1.33729937324   x2 =  2.68345013524     x3 =  3465302.30778           

.

.            seg1                   seg2                 seg3

.     -----------------------|-----------------------|-------------

.             ..      ..   (fp0)   ..              (fp1)   ..             (+oo)

.             ..-1.33 ..    (2)    .. 2.683 ...     (4)    .. 3465302. ...(+oo)     // b=sqrt(2)

.      ----------------------|-----------------------|-------------

.           h0(x1)=-1.5    oo|oo    h0(x2)=-1.5   -oo|***************    // set norms for tet0

.          ******************|oo    h1(x2)=-1.5   -oo|-oo h1(x3)=-1.5    // set norms for tet1

.

The inversion of sign of the schroeder-function-value is essentially the iteration with an imaginary height. For notation I introduce now u0 = ln(fp0) and u1 = ln(fp1)

If we have, for some x, the schroeder-value s, then the schroeder-value of the h'th iterate of x is s*u^h and the negation of sign can be achieved by supplying the according complex value in h.
Using the different fixpoints and different u0 and u1 we can state this norming more explicitely

Code:

.        x1 = tet0(1,   -1.5)     

.        x2 = tet0(x1, Pi*I/ln(u0))

.        x3 = tet1(x2, Pi*I/ln(u1))

which define the heights -1.5 for the two tetrations in the three segments.

What is now interesting is, whether the observed wobbling of the tetrates in seg2 using the different fixpoints changes in some interesting way. I remember that the shifting of the height by a half-unit made some significant change in the wobbling when I considered the infinite alternating iteration series (tetra-series) in one of my older msgs, I'll have a look at it soon.

Gottfried

inspired by an equation

Wed, 28 Jul 2010 00:18:05 +0200

i was inspired by

f(z) = A^(Lz) = L^z.

which looks neat to me.

i havent really thought about it deep but here is an attempt :

A^(Lz) = L^z

=> A^L = L ; L^p = 1

p = 2pi i / (ln(A) L) ( the period )

=> A^L = L ; L^p = L^ 2pi i / (ln(A) L) = 1

=> L ln(A) = ln(L) => ln(A) = ln(L)/L

=> L^ 2pi i / ( ln(L) * L/L ) = 1

=> L^ 2pi i / ln(L) = 1

and now ? ln on both sides ?

=> 2 pi i / ln(L) * ln(L) = 0

=> 2pi i = 0 ??

certainly not all A and L satisfy A^(Lz) = L^z.

so where did the variables go to ?

is there no solution ?

i assume taking ln on both sides is the error

=> solve for L : L^ 2pi i / ln(L) = 1

and now ... Lambert W function ?

or is that equation already wrong because of branches and we need to return to

=> A^L = L ; L^p = L^ 2pi i / (ln(A) L) = 1

but thats again an equation in 2 complex variables

of course f(z) = A^(Lz) = L^z has the trivial solution f(z) = 1 or f(z) = 0.

but im looking for others.

maybe => 2 pi i / ln(L) * ln(L) = 0

means that all solutions for A must be integer and hence only the trivial solutions f(z) = 1 or f(z) = 0 exist.

i cant find or imagine any other.

also the generalizations of this equation inspire me.

analogues with double periodic functions , finding solutions in terms of other numbers ( 3d complex or other ) , replacing exp with another function and multiplication with its inv superfunction of that other function etc etc

this may be trivial sorry ...

( should have paid attention in class as a teenager ... if that was in class ... )

tommy1729

note to myself : post this kind of stuff in the general section tommy ! bo always puts it there so thats where it belongs

no svg upload

Tue, 27 Jul 2010 23:19:48 +0200

svg pics do not upload apparantly ...

True or False Logarithm

Tue, 27 Jul 2010 05:53:31 +0200

Hey,

the following function iapproaches - if the limit exists - the intuitive logarithm to base , i.e. the intuitive Abel function of developed at 1:

The question is whether this is indeed the logarithm, i.e. if for , provided that the limit exists at all.

It has a certain similarity to Euler's false logarithm series (pointed out by Gottfried here) as it can indeed be proven that for natural numbers (even for in difference to Euler's series):

if we now utilize that for then we get
f_n(b^m) = -\sum_{k=1}^n \left( n \\ k \right) (-1)^{k}\sum_{i=0}^{m-1} b^{ki}
= \sum_{i=0}^{m-1}\left(1-\sum_{k=0}^n \left(n\\k\right)(-1)^{k} b^{ki}\right)" align="middle" />

Hence

But is this true also for non-integer ? Do we have some rules like , or even ?

Another proof of TPID 6

Mon, 26 Jul 2010 00:51:54 +0200

(10/07/2009 12:03 AM)andydude Wrote: Conjecture

where such that

Discussion

To evaluate f at real numbers, an extension of tetration is required, but to evaluate f at positive integers, only real-valued exponentiation is needed. Thus the sequence given by the solutions of the equations

and so on... is the sequence under discussion. The conjecture is that the limit of this sequence is , also known as eta (). Numerical evidence indicates that this is true, as the solution for x in is approximately 1.44.

lim n-> oo x^^n = n conj : any real x = eta

since (eta+q) ^^ n grows faster than n for any positive q , we can use the squeeze theorem

lim q -> 0 eta =< x <= eta + q

hence x = eta

see also http://en.wikipedia.org/wiki/Squeeze_theorem

QED

regards

tommy1729

Ansus ?

Fri, 23 Jul 2010 16:26:54 +0200

My browser says: Ansus only a single post.
Did something happen? All msgs of Ansus deleted? (Would be a mess!)

Gottfried

plz explain kouznetsov slowly

Wed, 21 Jul 2010 23:47:55 +0200

plz explain the kouznetsov method very slowly to me.

step by step , without C++ code.

is it complicated or am i not smart enough ?

i believe it relates to some contour integrals , but i dont understand how.

i have the experience that most misunderstanding lie at the beginning so plz explain how to start.

bo did a great job explaining andrews slog , maybe he can explain this too and become the " ultimate tetration forum webmaster of all time "

sum(e - eta^^k): convergence or divergence?

Tue, 20 Jul 2010 11:30:47 +0200

From some fiddlings with the slog-subject I came across the question, whether this is divergent or convergent:

using

the sum:

Clearly the sequence of terms tends to zero because e is the fixpoint of iteration and in a first guess I thought that also the series converges. But the convergence of the sequence is slow and one needs a lot of terms to see a promising trend.
What I did is to look at the sequence of partial sums, each from zero to 2^n,

and that sequence {s_n} seem to increase, even slightly more than linear. at least if I look at the partial sums up to n = 12.

Here are the partial sums and the differences of order 1 to 3:

Code:

.  n  s_n          d1_n=s_n - s_(n-1)   d2_n=d1_n-d1_(n-1)   d3_n

.  1  4.00875692871  4.00875692871     4.00875692871      4.00875692871

.  2  5.58578587004  1.57702894134    -2.43172798737     -6.44048491607

.  3  7.77673131247  2.19094544242    0.613916501085      3.04564448845

.  4  10.5161613469  2.73943003447    0.548484592045   -0.0654319090397

.  5  13.6651189223  3.14895757539    0.409527540923    -0.138957051122

.  6  17.0811305635  3.41601164122    0.267054065824    -0.142473475099

.  7  20.6561843799  3.57505381638    0.159042175165    -0.108011890659

.  8  24.3207908875  3.66460650761   0.0895526912324   -0.0694894839327

.  9  28.0341817806  3.71339089304   0.0487843854235   -0.0407683058090

. 10  31.7736430757  3.73946129511   0.0260704020760   -0.0227139833474

. 11  35.5268806871  3.75323761140   0.0137763162852   -0.0122940857909

. 12  39.2873487126  3.76046802550  0.00723041410300  -0.00654590218217

How could we prove the divergence/convergence of the series S?

Gottfried

Logarithm reciprocal

Tue, 20 Jul 2010 05:13:56 +0200

Hello all you tetration brainies out there,

looking somewhat deeper into the intuitive Abel function of f(x)=b*x (which is supposed to be log_b(x) however unproven until now), I found a somewhat direct expression of the coefficients, which boils down to the following challenging question:

Let the sequence be defined recursively in the following way for 0" align="middle" />:
and for

Is ?

Does it converge? The following graph of the sequence for , leaves the question open:

(The messed up numbers on the left side are due to a bug in sage *sigh*)

An equivalent slightly nicer formulation of the problem

Let the sequence be defined recursively in the following way for 0" align="middle" />:
and for

Is ?

edit: this can be found now as TPID 9 in the open problems thread.

tommy's uniqueness conditions

Sat, 10 Jul 2010 23:15:39 +0200

tommy's uniqueness conditions

i will present 2 uniqueness conditions here.

however i dont know if the first uniqueness condition is actually truely unique .. the first is more like a conjecture.

also i dont know if the 2 uniqueness conditions are equivalent ...

i hope both are equivalent.

for clarity im talking about uniqueness of sexp resp slog for bases larger than sqrt(e).

i will not distinguish between sexp(0) = a or sexp(0) = b if they are equivalent apart from a lineair transform ( x + c ).

also i assume for both the conditions that they are analytic almost everywhere and have their fixpoints ( x = base^x ) at +/- oo i.

i partially already mentioned the first uniqueness condition before :

d f^n / d x^n sexp(slog(x) + k) > 0 for all positive integer n and all positive real k.

and probably ( i.e. if im not mistaken because of the local heat wave )this is true if and only if the following is true :

( i.e. i assume " equivalent to " )

d f^n / d k^n sexp(slog(x) + k) > 0 for all positive integer n and all positive real k.

i havent investigated this a lot yet ... basicly because of the intresting second uniqueness condition.

the second uniqueness condition could be a simple proof for the uniqueness of kouznetsov and thus be equivalent to kouznetsov's solution.

and at the same time shed some new insight on tetration and kouznetsov's solution.

the second uniqueness condition is this :

we search a uniqueness condition for sexp.

indirectly by finding a uniqueness condition for its inverse : slog.

in some sense this is the analogue of kouznetsov's uniqueness for sexp but for slog.

ok , the basic equation for slog(z) :

slog(base^z) = slog(z) + 1

lets take a certain base to simply matters : base e

slog(e^z) = slog(z) + 1

now clearly slog has period 2pi i.

now we introduce "another slog"

" another slog " = slog(x + v(x))

v(x) is meromorphic on C and has period 1 and v(0) = 0.

( we will later see that v(x) must be meromorphic rather than just entire )

so

slog(exp(x + v(x))) = slog(x + v(x)) + 1

letting y = x + v(x) we see that this is also a solution.

to explain further , why period 1 ?

well : sexp( x + v2(x) + 1 ) = exp ( sexp(x + v2(x)) )

it follows that with q = x + v2(x) we have another solution ;
" another sexp ".

but only if v2(x) has period 1. ( necc condition , not sufficient )

x + v(x) is the functional inverse of x + v2(x) and hence v(x) must have the period 1 as well.

but now the cruxial part occurs :

slog(exp(x + v(x))) = slog(x + v(x)) + 1

we now know that v(x) needs period 1 , but it also requires period 2pi i !

( again this can be shown by setting y = x + v(x) )

so any valid v(x) satisfies :

v(x) is complex differentiable.
v(0) = 0
v'(x) > 0
v(real) = real
v(x+2pi i) = v(x)
v(x+1) = v(x)

so v(x*) = v(x)* with fixpoint at 0 and v(x) is a non-constant meromorphic doubly periodic function.

such a function must have at least a pole and thus be meromorphic and not just entire BECAUSE :

( terminology : a prototype parallelogram (PP) is the vector span of its 2 periods i.e. with sides z , z + a , z + b , z + a + b where a and b are the periods )

A non-constant meromorphic doubly periodic function cannot be bounded on the prototype parallelogram (PP). For if it were it would be bounded everywhere, and therefore constant by Liouville's theorem.

also by picard's little , we know that it takes almost any value in such a parallelogram ( all but 1 actually ).

note that some values may be on the edge/sides though.

so we have almost all values of v(x) within (PP).

and thus slog(v(x)) contains almost all values of slog(x) in PP , even slog(oo) because of the pole in PP.

by pythagoras the diagonal length of PP is + sqrt ( 1 + (2pi)^2 ).

this means that in a circle with radius RT = + sqrt ( 1 + 4 pi^2 ) ,

x + v(x) takes on every value - no matter where the center of the circle is in the finite complex plane.

so if the elliptic function v(x) is not = 0 then

slog(x + v(x)) takes on every value of slog(x) in any circle with radius RT.

conversely , if F(x) = slog(x + v(x)) THUS ANY (complex differentiable) SOLUTION - only F(x) is given !! - such that F(x) does not take on every value of F(x) in a circle with radius RT that does not cross a branch point and F(x*) = F(x)* , then v(x) = 0 and hence F(x) is UNIQUE up to a linear substitution of x.

thus we have a uniqueness condition for slog(x).

since sexp is the unique inverse of slog up to branches , this means we also have a uniqueness condition for sexp(x).

--------------------------------
(note) that the one value not allowed for sexp(x + v2(x)) is a fixed point of base^x = x which kouznetsov calls 'L' and 'L*'.

if it does happen , the story is over before the end of my condition ; it cannot longer be Coo in the neighbourhood of the finite value x that maps sexp(x + v2(x)) to L or L*.

this might translate to another condition to slog , but i will leave in the sexp condition form for simplicity , since it is not in contradition to my slog condition.

basicly this is just a condition of Coo.
--------------------------------

as a sidenote , i would like to say that some other uniqueness conditions will follow from this one , so that ' incompleteness ' or ' incompatibility ' is not neccessarily true.

i wont go into details , since most of *those* conditions are not so intresting and havent occcured in this forum yet.

i know the presentation of this post could be better , but its really hot today and i have little time.

i guess you guys - as tetration experts - can follow the main ideas in it anyways.

i assume my solution to tetration for real bases b > sqrt(e)

- or at least its analytic continuation - will satisfy at least one of these 2 uniqueness conditions.

it needs to be emphased that the domain and image of slog(x) needs to be investigated , since ' all values of slog(x) ' in some region is then better understood.

basicly because slog(x) is not entire , im not sure if we get almost all values ... hmmm ...

nevertheless , regardless of that outcome , the result above remains indepent of that.

thinking about it , i think we do get almost all values because slog(x) when defined for all complex x , it has singularities and/or pseudo fractal like structures + all exponents of them ; thus 'many values'.

regards

tommy1729

fibonacci like

Sat, 10 Jul 2010 13:27:45 +0200

i was thinking about a fibonacci like recursion.

f(x) = f(x - 1) + f(x + i)

together with some initial conditions ( you choose )

perhaps old hat ... but i dont recall a closed form solution ( though its really hot here )

regards

tommy1729

An incremental method to compute (Abel) matrix inverses

Fri, 09 Jul 2010 07:31:15 +0200

I fiddled a bit around with Gottfried's suggestion of LU decomposition of the Abel matrix (though in the end the formula is independent of the LU decomposition).

The annoying thing about calculating the intuitive Abel function (by solving the equation Ax=b where A is the Abel matrix, x the powerseries development of the Abel function and b=(1,0,...)) that if you want to increas the matrix size you have to solve the complete equation again without being able to use your previous solution.

Now I found a way how you can compute the inverse of the matrix by using the Abel matrix. I dissect the matrix as follows, for brevity I set :

A_n=\left(\begin{array}{ccc|c}
\phantom{1}& &\phantom{1} & \\
&A& &\acute{a}\\
& & & \\\hline
&\grave{a}& &a_n
\end{array}\right)
" align="middle" />

means column vector and means row vector.

The final incremental formula is then:

1)(-\grave{a}A^{-1}\oplus 1)}{a_n-\grave{a}A^{-1}\acute{a}}" align="middle" />

Where means adding the entry 1 to the vector and is extended to a nxn matrix by filling with 0's.

The deriviation is perhaps too uninteresting and cumbersome to put, but I can post it if inquired.

levy ecalle koenigs ?

Fri, 09 Jul 2010 00:31:23 +0200

for parabolic iteration i use the carleman matrix method.

what other methods exist ?

bo mentioned " levy and ecalle " but i dont know about their parabolic iteration formula's ...

one can get many series expansions from the carleman matrix method , but i would like to see " totally different from carleman methods ".

in fact i would like to see a limit like superfunction type of solution to parabolic iteration , something similar to koenigs non-parabolic iteration solution.

does there exist a method for all fixpoints ( parabolic or non-parabolic ) in terms of a limit , not using carleman ?

im sorry , i tried looking on the internet , but found nothing apart from stuff that is equivalent to carleman ...

if i recall correct , koenigs is not necc analytic. are there similar methods that only work when the solution is analytic ?

sorry , im a number theorist getting old

regards

tommy1729

A functional equation

Mon, 05 Jul 2010 20:28:20 +0200

Hi,

Does anyone have an idea how to solve the following functional equation?

Thanks

weird series expansion

Mon, 05 Jul 2010 17:32:28 +0200

consider F(x) a real-analytic function mapping R+ to R+ with only a single fixpoint, the one at 0.

let g(x,n) be g(g(x,n-1),1) and g(x,1) = arcsinh(x/2)

F(x) is strictly nondecreasing. F(x) grows slower than x^2 + 1.

let a_n be real.

the following weird series expansion has come to my mind :

a0 + a1 x + a2 g(x,1) + a3 g(x,2) + a4 g(x,3) + ...

and many questions pop up.

regards

tommy1729

Self tetraroot constructed via Newton series interpolation

Sat, 03 Jul 2010 10:42:03 +0200

Hi.

I thought of a novel approach to aid in the "divination" of what the tetration function for real heights is: the self-tetraroot function. This function is defined so that . It is analogous to the self-root function , only with tetration instead of exponentiation. It gives the base of the tetrational with a given fixed point.

It can be evaluated at the integers using numerical root-finding methods, and when the points are plotted, it reveals a scatter that slowly, but regularly (i.e. no noticeable "bumpiness") decays to a fixed value about 1.444782 as .

This slow regular decay made me wonder if it would be a good candidate for interpolation via the Newton series. Newton series, when applied to tetrationals with bases in , yields the regular iteration.

The Newton series for a function at a point is

where is the falling factorial and is the unit forward difference operator.

Doing this for gives the following graph. Convergence is dog slow -- I needed 300 or so terms and over 100 decimals of accuracy just to get error that I'm confident is less than . This is very bad. Is there some way to accelerate the convergence of Newton series?

As you can see, it sort of looks like the graph of the self-root. I superimposed both graphs:

The maximum is at about or so (maybe if you want to push it as this approximation may be good to 5 places -- may -- it's so dog-slow, where it reaches a value of (maybe ). I presume this is the base where "pentation" goes from convergent to explosive growth, and so is to pentation what is to tetration. Flipping the self-tetroot along and taking the lower branch would then yield a graph of .

For , I get a tetra-self-root of ( rounded from whose last digit may be right or near). Plugging this into the regular iteration to try and compute suggests whatever "fractional iteration" this is creating may agree with it, but this is a dodgy bet with just 4 decimals. Yet if it does agree, then perhaps, since this function shows no "bumps" when the base crosses , maybe the regular iteration really doesn't have a natural boundary after all. But again, 4 decimals, 5 at best, geez... far too little to even bet seriously... If we could compute more, we might be able to get a better idea if it agrees with regular, and if it also agrees with the Cauchy integral (for the bases from to 1.635...), hence helping to "divine" if these are really "good" methods to use and if they are actually analytic continuations of each other (meaning that the STB is not a natural boundary of the regular iteration.).

What do you think of this function? Especially the similarity between the shape of its graph and that of the selfroot. One could almost imagine a continuous spectrum of similar functions in between them -- "fractional-rank hyperoperations", anyone?

Tetrate fractals

Fri, 02 Jul 2010 16:18:01 +0200

I was wondering if anyone had done any reasearch on fractals of the form . As you can see, it's similar to the Mandelbrot set, but the operators have been raised a level. Using Chaos Pro, I've gathered the following images for n=2, and later I'll go back for other n.

It seems like something similar to the hyperpower fractal, but here you can see the "leaves" appear to stick out from beneath the mess...

The above mess seems to have no pattern whatsoever, but you look closely, you'll see little blobs like this:

Also, when generating higher values of n, keep in mind that you must change the perturbation point to 1, otherwise you'll get a whole lot of black.

some questions about sexp

Tue, 29 Jun 2010 00:18:26 +0200

im going to ask some questions about sexp , and i mean all proposed solutions of sexp.

(question 1)
does the following hold :

d f^n / d x^n sexp(slog(x) + k) > 0 for all positive integer n and all positive real k ?

(question 2)

since slog(z) (base e) has period 2pi i why doesnt sexp(z) look like a log spiral ?

or does it , like having a branch cut at real x < -2 ?

(question 3)

what happens to limit cycles and n-ary fixpoints ??

sure we can set the fixpoints exp(L) = L at oo i but how about the fixpoints of exp(exp(.. q)) = q and limit cycles of the exp iterations ...

e.g. let e^q1 = q2 , e^q2 = q1 , if we want half-iterates , 1/3 iterates and sqrt(2) iterates to have the same fixpoints , this is a problem , not ?

perhaps we can 'hide' L at +/- oo i ( like kouznetsov ) and 'hide' the other points at - oo ??

(question 4)

do all 'analytic in the neigbourhood of the positive reals' sexp have the fixpoints exp(L) = L at oo i ?

( i know that [L,sexp(slog(L)+o(1))] cannot be in the analytic zone , but maybe [L,sexp(slog(L)+o(1))] isnt part of the analytic zone )

(question 5)

slog(z) base x is not holomorphic in x in a domain containing the interval [a,b] if eta is between a and b.

why is that ? i know that the real fixpoint dissappears but still ...

sorry if those are FAQ or trivial Q.

regards

tommy1729

Integration?

Mon, 28 Jun 2010 21:16:05 +0200

Hello!

I tried integrating the function x^^n for any nonnegative integer n. I wanted to share my method and results, so that you can check fofr accuracy (if I've done it correctly, it'd be a rare occurance - I haven't taken calculus as a class yet, so I went off of what I've read in books).

First we write

Then we can expand on the indefinite integral in the parentheses.

Once we have calculated this out, we can use substitution.

If we repeat this process n-1 times, then we get

Notice that I defined i_0 = 1. This merely simplifies the notation.
I haven't yet completed that general integral, but what matters is that it can be completed on a case-by-case basis through integration by parts. This is demonstrated for n=2...

Integrating by parts,

If we do this indefinitely, we end up with

Plugging in,

Bernoulli's integral, the so-called "Sophomore's Dream", is calculated as

Note that 0^0 is 1, so the second summation only has a non-zero value when n=k.

If anyone has any corrections, or if this has been done before, I'd love to know. Also, it'd be cool to see graphs of these functions from anyone who has Maple or Mathematica (or other amazing program).

Edit: Ugh. Just noticed that in the last few equations, I wrote the summations wrong. Rest assured, it's from k=0 to infinity.