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Deriving tetration from selfroot? - Ivars - 03/12/2008

Just an idea:

Is it possible to derive tetration and above starting from self root concept and defining tetration as an inverse to it?
With all branches? And then see the conection to [n]-tation? Self root seems not so abstract and infinite, yet it yields inverse tetration nicely. What if me make it an starting axiom?

Of course, to include branches of Lambert function and h, will have to accept that each self root has infinity of values.Some of which might be real.

One example:

h(i^(1/i))= i
h((1/i)^(i) = -i

So we get nicely and logically first to branches of Lambert Wo, W-1. They both correspond to one real input,

i^(1/i)=(1/i)^i=e^pi/2 , but we get 2 different self roots as we rearange x^(1/x) into (1/x)^x.

The next branches should be attainable by taking self roots of i = e^i 5pi/2 instead of e^(ipi/2), so MAYBE:

h(e^(i5pi/2)^(e-i5pi/2))= +- next 2 branches of W in combination with some branches of logarithm.

Superroot might be attainable by using negative values of -i=e^(I*3pi/2), 1/-i=e^(-(I*3pi/2) I have to check.

May be this is wrong again or done 200 times with no use, please let me know.

Ivars

Moderator's note: Splitted from "Generalized recursive operations"


RE: Generalized recursive operators - Ivars - 03/13/2008

GFR Wrote:@Ivars -
The role of the selfroot is essential. Nevertheless, it is important as a solution of the y = b[s]y functional equation. But, unfortunately (so to say) it is not its unique "functional root". Another solution can be found by considering y = b[s](b[s]y), or [/b]s-log(y) = y = b[s]y.

E.g. for rank 3 (exponentiation), we should consider (supposing base b) an expression such: [/b]log x = x = b^x. I. e. also other intersection points between the exp and the log, which will give other "branches" of the functional roots, are relevant. And, the same should be valid for other hyperops ranks.

GFR

HI Gianfranco,

But if You only look at infinite tetration for time being. Do we need those intermediary heights that add solutions?

x= y^(1/y) does mean y=x^y but does NOT mean y=x^x^y. because it does not include all solutions.

However, in infinite tetration, h( x^(1/x) ) does mean x, and h(1/x^x) does mean 1/x. The intermediates disappear. Are You sure they reappear in branches of W, ln as solutions for infinite tetration?

Or am I wrong againSad

Ivars


RE: Deriving tetration from selfroot? - Ivars - 03/20/2008

Claryfying few things for myself:

If we have complex number z , we can find second superroot of 1/z ssroot(1/z) = ln(1/z)/W(ln(1/z))= -ln(z)/W(-ln(z)) = 1/h(z)

So h(z)= 1/ssroot(1/z)

and we know that h(z^(1/z)) = z .

So infinite tetration is just an operation that is inverse (in a sense 1/) to finding of second superroot, which is a function of type

ssroot(1/z)^ssroot(1/z) =1/z= 1/ h(z^(1/z)) or

h(z^(1/z))= ssroot(1/z)^(-ssroot(1/z)))

where argument of h is selfroot of z, while on the right side we have selfroot of second superroot of 1/z.

as an example, let z= 1/e, than h ( (1/e)^(e))=1/e, ssroot(e) = 1/Omega, 1/e = (1/Omega)^(-1/Omega)

If we have complex number z/i , we can find second superroot of i/z ssroot(1/(z/i)) = ln(1/(z/i))/W(ln(1/(z/i)))= -ln(z/i)/W(-ln(z/i)) = 1/h(z/i)

So h(z/i)= 1/ssroot(i/z)

and we know that h((z/i)^(i/z)) = z/i .

ssroot(i/z)^ssroot(i/z) =i/z= 1/ h((z/i)^(i/z)) or

h((z/i)^(i/z))= ssroot(i/z)^(-ssroot(i/z))

And finally, if we have complex number z/w-division of 2 complex numbers (this and next I have not checked):

h((z/w)^(w/z))=ssroot(w/z)^(-ssroot(w/z))

h((az+b)/(cz+d))^((cz+d)/(az+b))=ssroot((cz+d)/(az+b))^(-ssroot((cz+d)/(az+b)))

and may be even h(P/Q^(Q/P) where P,Q are complex polynomials of any power, perhaps max power of Q>maz power of P.

and h((S1/S2)^(S2/S1)) where S1, S2 are infinite powerseries, sometimes equivalent to analytic functions, sometimes divergent.

Now selfroots and second super roots are hopefully correctly linked via infinite tetration, and I will try to iterate function h backwards and forward as long as its argument can be expressed as z^(1/z). Just to see what happens.


Ivars


RE: Deriving tetration from selfroot? - Ivars - 03/20/2008

I tried to iterate infinite tetration h in both negative and positive directions and it seems that :

if h( z^1/z)) = z = then first negative iteration of h is z^(1/z), second

(z^(1/z))^(1/(z^(1/z)) and so on.

negative infinite iteration of (i^(1/i), but also then (i) also then 0,438283+0.3605924*i etc leads to 1
negative infinite iteration of h(e^(1/e)) then also e leads to 1,

Positive infinite iteration of h (i) -> h(h(h(h(......h(i))= 1 ( I only checked 30 iterations).

Does it mean that all infinite positive negative iterations of h(z) lead to 1?

Ivars


RE: Deriving tetration from selfroot? - Ivars - 03/21/2008

So it seems to me, that finding real and complex extensions to tetration (finite) is kind of symmetric to finding real and complex extensions to selfroots and superroots which are related to infinite tetration.

By extension I mean self root is usually made of 2 discrete parts: number x, its inverse 1/x and than exponentiation x^(1/x).
it is possible to extend it to 4, 8, 16...2^n discrete parts easily by repeated application. How to make it consist of non-integer number of parts? Real number? Complex?

Second superroot can be extended to 3,4,5 etc easily but not by simple iteration of f(x) =x^x as it will miss 3,5,7 and all odd superroots. First superroot is also not obvious. By looking at x^x^x = y where x is 3rd superroot, x^x^x^x =y, where now x is 4th superroot etc it is clear that odd superroots exist. How to extend this to fractional or real or complex number of superroots?

By analogy not by logic odd self roots would be: 3rd = x^(1/x) ^x =y, 5th x^(1/x)^x^(1/x)^x=y etc.

But what is then: x^((1/x)^x)? The change in grouping gives another value for 3rd self root- there are 2 of them.
For 5th self root, there are probably more different values of y by changing grouping.

Can it be done by looking into iterations?

if f(x)=x^(1/x), than f(f(x)) = (x^1/x))^(1/(x^(1/x));
if f(x) =x^x, than f(f(x) = (x^x)^(x^x):

Application of brackets or grouping of operations seems not to be trivial at all in these cases. Obviously, ((x^x)^x)^x is not equal to (x^x)^(x^x)=x^(x^x)^x= x^x^(x^x) which is not equal to x^(x^(x^x)), the last one being the biggest if x>1. So there are at least 3 4th superroots , 2 3rd superroots , 1 2nd superroot for any y >1. Interestingly, if we try with x=3, smallest of 4th superroots, 3^3^3^3 equals biggest of 3rd supperroots:

3^3^3^3 = 3^(3^3)= 7,6256E+12

The same is not true neither for 2, nor for 4 , nor probably any other real number which is not 3.

2^2^2^2=256 > 2^(2^2)=16
4^4^4^4=3,40282E+38 < 4^(4^4) = 1,3408E+154

Most of the effects of bracketing can be clearly seen by plotting self and superroots in polar coordinates, but what about analyzing them?

Ivars


RE: Deriving tetration from selfroot? - Ivars - 03/21/2008

Since 3 was interesting number, we can consider 3rd self roots:

3^(1/3)^3= 3 which is kind of not interesting even if it involves 2 complex 3rd roots of unity in the process, and:
3^((1/3)^3)=3^(1/27) = 1,041528498.. Which is more interesting (and 26 complex values).

Second iterate of self root of 3 would be than:

3^((1/3)^(1/(3^(1/3)))= 1,289058025.. but also 4 complex values.

With 2:

2^((1/2)^2)= 2^(1/4) = +-1,189207115... and 2 complex values
(2^(1/2))^(1/((2)^(1/2)))= +-1,277703768...

With 4

4^((1/4)^4) = 4^(1/256) = +-1,005429901... and 254 complex values but:

(4^(1/4))^(1/((4)^(1/4)))= 1,277703768..=(2^(1/2))^(1/((2)^(1/2))) and .. complex values

Anyway, function "Second iterate of self root of x" where f(x)=x^(1/x)

f(f(x)) = x^(1/x)^(1/(x^(1/x))) has maximum at e and it is = 1,290005369...

And values for x=2 and x=4 are the same, but not only-there are infinitely many such paired points at this stage. e seems to remain the value giving max of any positive iterate of self root.

Ivars


RE: Deriving tetration from selfroot? - Ivars - 03/22/2008

But if we add one more supperroot level (5th) it changes a bit:

3^3^3^3^3 = 3^3^(3^3) = 4,43426E+38 = 27^27
2^2^2^2^2= 2^(2^(2^2) = 65536

So 5 times exponentiated 2 =2[4]4; But 5 times exponenetiated 3 equals something that has no name -at least I do not know it - it is smaller(slower) then n=4 tetration of 3 but bigger(faster) than 4 times applied exponentation of 3.

Moving to 6th superroot with 2:

2^2^2^2^2^2 = 2^2^(2^(2^2)) = 4294967296= 4^4^4
3^3^3^3^3^3= 3^3^3^(3^3) = 8,719E+115

6 times exponentiation of 2 is smaller then 5 times tetration of 2 but equals something applied to 5 2-es that I have no name of.
6 times expnentiation of 3 equals 5 times something.

Ivars


RE: Deriving tetration from selfroot? - bo198214 - 03/22/2008

Ivars Wrote: 3^3^3^3^3 = 3^3^(3^3) = 4,43426E+38 = 27^27
2^2^2^2^2= 2^(2^(2^2) = 65536

So 5 times exponentiated 2 =2[4]4; But 5 times exponenetiated 3 equals something that has no name -at least I do not know it -

we call (x^x)^(x^x) balanced tetration, though I think there lacks still a symbol, so I use here [4B]:
x[4B] 2 = x^x
x[4B] 4 = (x^x)^(x^x)
x[4B] 8 = ((x^x)^(x^x))^((x^x)^(x^x))
Generally
x[4B] 1 = x
x [4B] 2^(n+1) = (x [4B] 2^n)^(x [4B] 2^n)

The topic was not really discussed yet on this board however, already mentioned in this posts: post1, post2.

Besides I think you forget several parens in the above quote, the convention is that if there are no brackets for exponentiation then the brackets are to the right side. 3^3^3 := 3^(3^3). What you meant was probably
(((3^3)^3)^3)^3=3^(3*3*3*3)=3^(3*3^3)=(3^3)^(3^3)=27^27


RE: Deriving tetration from selfroot? - Ivars - 03/22/2008

bo198214 Wrote:Besides I think you forget several parens in the above quote, the convention is that if there are no brackets for exponentiation then the brackets are to the right side. 3^3^3 := 3^(3^3). What you meant was probably
(((3^3)^3)^3)^3=3^(3*3*3*3)=3^(3*3^3)=(3^3)^(3^3)=27^27

Thank You;So balanced tetration is something in between repeated exponentiation from left and tetration.
Yes, I left brackets out thinking convention is from the left. But I meant what You write, exactly.

Ivars


RE: Deriving tetration from selfroot? - Ivars - 03/22/2008

I managed to plot in Excel first 6 integer iterations of f(x)=x^(1/x) for 1<=x<=10; I took only positive real values for f and its iterates.

What is obvious, for every x>e there is 1<x<e which has the same selfroot; most notably, 2^(1/2)=4^(1/4); but for e.g. x=3:

3^(1/3)= 1,442249...= 2,47805..^(1/2,47805..) meaning from h(x^(1/x))= x

h(1,442249..) = 2,47805.. if x<e
h(1,442249..)= 3 if x>e

So for any 1=<z=x^(1/x)<e^(1/e) there are 2 real values of infinite tetration.

Only for x=e there is exactly 1 value, which is also strange and requires further attention.

Since it seems that 1^(1/1) = lim n^(1/n) = infinity^(1/infinity) = 1 ,even for 1 there are 2 values

h (1) = 1 if x<e ,
h(1)= infinity if x>e;

So I understand that for x<1 there may be more interesting things. Also, going backwards with negative integer iterations might be instructive.


Ivars

[attachment=262]

Full range x>0 leads to a graph whihc looks like step function at x=1, continuing iterations to infinity will lead to step function? for x>0, I do not understand what happens with point x=e and 2 values of at x>1 and n-> infinity-do they all converge into one value =1? Andrew had somewhere in a thread a nice picture where these iterations where continued to negative.

[attachment=266]