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Infinite tetration and superroot of infinitesimal
Hi, Henrik!

bo198214 Wrote:
GFR Wrote:Actually, the beautiful formula I propose is:
ssqrt(x) = ln(x) / W(ln(x)), which, for x = 1/2, gives:

ssqrt(1/2) = ln(1/2) / W(ln(1/2)) = 0.26289282802173525.. + 0.4996694356833174.. i

That reminds me to update the wikipedia tetration article as there is no formula for the square super root yet.

And indeed your formula is also a solution:
We first see that your formula is equal to which is a solution to :
as is the inverse function of .

Please don't forget the "lower branch" of y = ssqrt(x), which can be obtained via the second (-1 level) branch of Lambert Function.
(see the attachment)

By calling "plog(x)" [Product Logarithm] the logical union of the two real branches of the Lambert Function, i.e. W(0) and (W-1), i.e.:
plog(z) = W(0,z) OR W(-1,z), we may write:

y = ssqrt(x) = ln(x)/plog(ln(x))

I would be grateful if you could kindly mention this in the revised Wikipedia page. This formula can give the two upper and lower values of y = ssqrt(x). Please see also, in red, the fronteer of the x domain where y is real [x >= 1/e]. [© of GFR and KAR ....]. The apex point has coordinates:
x = e^(-1/e) = 0.692200628..
y = 1/e = 0.367879441..

Thank you in advance.


Attached Files
.pdf   Ssqrt_x_.pdf (Size: 6.43 KB / Downloads: 267)
This formula has been however present in Lambert function page as a solution of

x^x =z


Also Henryk's equivalent formula was there..

Well, the use of the Lambert Function, for finding back x in y = x^x, is an old affair. The point is that, the present Wikipedia formula seems to covers only the "upper branch" of ssqrt(x), if we only use the W(0) "classical" branch, as it seems to be indicated, afaik (as far as I know), in the the first wikiplot of the W article..

A more complete ssqrt formula should also cover its "lower branch". The definition of "plog(x)" as the logical union of the two W(0) and W(-1) product logarithm real branches will do the business. But, I am sure that Henryk will amend and upgrade the tetration article, accordingly.

Like it!

Ops ... sorry! Once again!

the same situation is found in the definition of h(b), inverse of b = h^(1/h). In fact, we know that the "classical" solution is:

if: b = h^(1/h)
then: h = - W(-lnb)/lnb

We also know that, if we only take the W(0) branch of the Lambert Function, we only get the lower branch of h(b), which is not sufficient. Therefore, I propose to use the plog(z) operator, always meaning: plog(z) = W(0,z) OR W(-1,z), and write:

h(b) = plog(-lnb)/(-lnb) = b#(+oo)

See the attachment.

We can see (as we also know) that:
h(1/a) . ssqrt(a) = h(a) . ssqrt(1/a) = 1


Attached Files
.pdf   Imftowplot.pdf (Size: 11.15 KB / Downloads: 209)
Infinite tetration (if it has a real limit, i.e. in the range ) however always converges to the lower branch of .
I agree that it has a real limit in the Eta-Beta range [Eta = e^(1/e), Beta = e^(-e) !!]. Nevertheless, we must admit that there is also another real upper branch. Think of the famous example of b = sqrt(2). We have two heights h(sqrt(2)) = {2, 4}.
In fact, by applying b = h^(1/h), we have: 2^(1/2) = 4^(1/4) = sqrt(2).

The lower branch is reachable starting from points (-1,0) and (0,1) by applying b#(x+1) = b^(b#x). The upper branch is unreachablke (big mystery), but still it satisfies the same conditions. Always in the case of b = sqrt(1/2), if the upper branch is 4, we have that:
if: sqrt(2) # x = 4,
then: sqrt(2) ^ (sqrt(2) # x) = sqrt(2) ^ 4 = 4 , for any "x" !!
I mean that the upper branch defines (according to my point of view) a second constant real numerical value in y = b^x, in the 1-to-Eta domain of b.

I might be wrong. But, I still believe it. We always have the same problems with two-valued "functions", one of which, in this particular case, should have a "constant" value. I think we already discussed about the example y = x^2. A lot of orthodox mathematicians say that it is not inversible (!?!), and that sqrt(4) = 2, like all the pocket calculators indicate. Neverthaless, we know that sqrt(4) = {-2, +2}.

This time, I go and drink a "koke", or a "cola", as you say in Germany.

GFR Wrote:A lot of orthodox mathematicians say that it is not inversible (!?!),

"coke" or "latte macchiato" ... Smile

If tetration is defined as iterated exponentiation, with
y = a_n^a_(n-1)^...^a_2^a_1^a_0
and a_1=a_2=a_3=...=a_n =b, a_0=a , thus y = b^...^b^b^a
then the partial-expression argument, from which the multivaluedness of tetration is denied, is removed by a clean definition.

We should initiate an international campaign on this subject Smile
(Are NGO's allowed as consultants at MAA?)

Gottfried Helms, Kassel
I think I have to intervene and to clarify the basics.
1. There is the concept of a function in mathematics: We assign to every argument of the domain exactly one value of the codomain of the function. An example is with the domain of the real numbers and also (!) with the domain (and codomain!) being the positive real numbers. In the same way is a function defined on the positive real numbers and especially is a function defined on .

Those above functions can be (uniquely) developed into power series (at points of their domain) and hence being continued somewhat outside the real line on the complex plane. This continuation is performed along a path and usually the continuation along two different paths gives the same value at the common end point, however in certain cases - if you start at point wind the path around a singularity and finish again at - the continuation value is different from the starting value, and these different values are to be said the branches of the function .

For example if you take the function (single valued!) , you can develop it at 1 by
Surely there is a (/the only) singularity at 0.
If you continue the function to -1 along a path above 0 you get and if you continue the function to -1 along a path below 0 you get .

To avoid this scenario you slice the complex plane at the negative real axis. This means no path for continuation is allowed to cross the negative real axis, more precisely the argument of such a path must be continuous and completely lie in .
With this sliced complex plane by continuation you get exactly one value of for each point on the complex plane, particularely you get .

From there you can get to the branches. You can continue the function around 0 and get the values where is the winding number of the path, i.e. how often it does anticlockwise cross the negative real axis (minus the clockwise crossings). This is the same as the number of discontinuities in going from to (minus the discontinuities going from to ).

For the logarithm you use also this sliced complex plane (which guaranties a unique value everywhere except 0) and for the branches you get where is again the winding number.

Now back to this is a function (single valued!) on . It can be developed there into a power series (did Ioannis derive a powerseries?) and hence continued to the complex plane.
It has (the only?) singularity at and hence it would make sense to define it uniquely on the at sliced complex plane, more precisely a path for continuation must satisfy .
Starting fromt this definition would be the continuation for a path with winding number in the above sliced plane.

However I am not completely sure if that defintion of sliced plane is compatible with the definition of the branches of the lambert function, such that we could write:

or if we even should take into account the branches of the logarithm.
Perhaps someone takes pity on that somewhat tedious task.
Thank you Henryk and Gottfried.

Nevertheless, how about the theory of multi-valued functions ?

and/or (and ... or/and ... or ...) the theory of catastropes (by René Thom) ? See:
http://en.wikipedia.org/wiki/Catastrophe_theory ?

Are the "things" covered by those theories invertible? (... thank you, Gottfried!).

"Latte makkiato" is very good! An international campaign will certainly coming soon into the picture. We are in the appropriate times. Which NGO, Gottfried? By the way, MAA is the name of a car insurance company, in France. Wink

Come on, my friends, keep it going!


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