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I have struck on the following:

We can define 0^0 =0 or 0^0 = 1 = depending on the scale of 0 we use in first 0 and exponent ( based on various orders of infinitesimals ). if we denote 0 as dx, dy, than:

1) dx^dy = 0 if dx is infinitely smaller than dy, but dx=0, dy=0 dx+dy=0, dx/dy = 0 or

2) dx^dy=1 if dx is infinitely greater than dy, but dx=0, dy=0, dx+dy=0, dx/dy=infinity.

3) The only problem arises when dx=dy=0 and dx+dy=0, dx/dy=finite.
Then the problem 0^0 is solved axiomatically to fit other math.

but let us leave this case open.

if we look at h(0)= 0^0^0^0^............. infinite times:

and take partial exponents:

if 0^0 = 1, 0^(0^0)=0, 0^(0^(0^0))=0^(0^1)=0^0=1,

so we get oscillating values 1,0,1,0,1,0,1,0........

if 0^0=0, 0^(0^0)=0^0 = 0, 0^(0^(0^0))=0^(0^0) =0^0 =0

We get value 0 all the time.

4) So it might be correct to suppose that value of 0^0 when 0/0=finite value lies somewhere between these 2 cases.

But how to evaluate h(0) when 0^0=1?

5) If we look at sum 1-1+1-1+1-1+1..............we see that partial sums develop like 0,1,0,1 .....similar to case h(0) when 0^0=1.

We know that Euler summation gives values to those series =1/2. We might expect that something similar can work in case h(0) when 0^0=1.

6) but there is a difference:

we can look at 1-1+1-1+1 ....... as sum of

1+1+1+1+1.........= + infinity
and -1-1-1-1-1- = - infinity

So value 1/2 is actually attributed to sum -infinity+infinity= 1/2.

7) In case with tetration, we can look at :

0^1^0^1^0^1.............. = 0, 0^1=0
0^-1^-1^-1^-1.........= infinity , 0^-1=infinity
0^(-1)^0^-1^0^-1^0.........= either infinity or 0 or 1 depending on 0/0 and 0^0.
0^-1 = infinity.

minus infinity is never as result in exponentation of 0, so we are looking at value that would describe multiplication

0^(-1) ^-1^-1........* 0^(1)^1^1....... = 0^(0) which is something like 0*infinity so could be finite.

Cool the conjecture is, that there could exist a value atributable to the h(0) when 0^0=1 as a value of 0,1,0,1,0,1.............and that this value also gives the correct answer to 0^0 when 0/0 = 1.

h(0) when 0^0=0 is 0.

9) This value is not 1/2. What could it be? pi/2? pi/4? 1/e? i/2? sqrt(i)? something else?

The definition of h(z) by - W(-ln(0)/ln(0) does not seem very helpful here, but perhaps to ln(0) can be formally attributed some value as well?
There is something more sophisiticated with summing of divergent series.
Well, you may have an idea, to what a series is summable - but then you have to make sure, it makes sense in all other relations.

One important concept is that of partial sums.

One other important concept was, to see the series in question either as a power-series, where powers of x are cofactored with the coefficients like

s(x) = 1 - 1x + 1x^2 + 1x^3 - ... +

or

t(x) = 1^x - 2^x + 3^x - 4^x + ... - ...

and then considering the appropriate limit of the occuring expression using the concept of partial sums.

Note, that the notation 1 - 1 + 1 - 1 + ... - ... can occur as limit of very different powerseries and thus can be assigned an arbitrary value.
The standard case is assuming lim x->1 s(x) or lim x->0 t(x).

The interesting aspect is, that the limits in those definitions are the same.

lim x->1 s(x) is formally equal to the powerseries lim x->1 1/(1+x); and the latter can be analytically continued beyond the radius of convergence. But if , for instance,

1 - 1 + 1 - ... + ... is seen as limit of

u(x) = (1-x^2)/(1-x^3) = 1 - 1 x^2 + 1 x^3 - 1 x^5 + 1 x^6 - ... + ...

then the limit for lim x->1 u(x) = 2/3, so it would be that also 1 - 1 + 1 - 1 + ... - ... = 2/3
(see example in K.Knopp "Theorie und Anwendung unendlicher Reihen" (there is an english version available), chap XIII)

It may be misleading, that series like this without reference of their context are given as examples in books or online-media at all... I think, that, if there is no explicite reference is given, we have the (still ambiguous) "default", to interpret the series either as coefficients of a powerseries with consecutive powers (without holes) or as a zeta-series (as far as possible).
The series with which we deal in tetration are not automatically power- or exponential-series; for instance the Lambert-W, or the integrals involve coefficients with k^k, where k is the series-index. So we must be extra careful with the assignement of values to iterative expressions like lim k->inf a^^k or the like.
The problems of the ambiguity of 0^0 when seen as lim x->0 0^x or seen as lim y->0 y^0 are well known, and it is well known, that they don't converge to the same value (which would be necessary for a unique definition).

Gottfried

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Gottfried Wrote:Note, that the notation 1 - 1 + 1 - 1 + ... - ... can occur as limit of very different powerseries and thus can be assigned an arbitrary value.
The standard case is assuming lim x->1 s(x) or lim x->0 t(x).

I do not think these values are arbitrary. They follow from what series we use to get 1-1+1-1.......... - but no one have shown how they arise.

Quote:The problems of the ambiguity of 0^0 when seen as lim x->0 0^x or seen as lim y->0 y^0 are well known, and it is well known, that they don't converge to the same value (which would be necessary for a unique definition).

Gottfried

I believe the problem lies in the idea of limit itself. If ratio between 2 zeros could be any finite value, then of course 0^0 is difficult, except for cases I mentioned, when one of them is infinitely bigger than other- an infinitesimal of another order.

I am not a fan of limits.

But still, what about h(0)? May be that is easier than 0^0 as it involves infinitely many 0 so if their ratio is finite ( they are of same order of infinitesimal) , in infinite exponetiation it either loses its importance or becomes obvious?

And what about the idea that 0^0^0^0 is in fact:

0^1^1^1^1..........* 0^-1^-1^-1^-1............

is that correct?

And than of course 1 can be obtained in many ways, so not all 1 are equal for such exponentiation as well.
To answer your specific question, you should probably read either Knoebel's Exponentials Reiterated or McDonnell's Some Critical Points of the Hyperpower Function since they both talk about this.

Near zero, can be defined in one of three ways: as the limit of odd towers as the height tends to infinity (giving ), as the limit of even towers as the height tends to infinity (giving ), or as the inverse function of (giving ). Each one of these definitions gives a different answer as x approaches 0. The strange thing is that all definitions are equivalent for .

Andrew Robbins
Hej Andy,

andydude Wrote:To answer your specific question, you should probably read either Knoebel's Exponentials Reiterated or McDonnell's
Some Critical Points of the Hyperpower Function since they both talk about this.

Thanks, the second I can access, the first is pay for article- i can not afford to pay for articles as I sometimes need too many, and then some of those are not the right ones... I will have a look.

Quote:Near zero, can be defined in one of three ways: as the limit of odd towers as the height tends to infinity (giving ), as the limit of even towers as the height tends to infinity (giving ), or as the inverse function of (giving ). Each one of these definitions gives a different answer as x approaches 0. The strange thing is that all definitions are equivalent for .

Andrew Robbins

That is interesting, that is one of the reasons I thought the value could be 1/e -or e-e or (1/e)^(1/e) as h(1/e) = Omega -> I hoped there is a way to attribute value (or many different values) even if h(0) diverges as tetration. So that value would not be tetration, but - a value.In a way similar to series +1-1+1-1 ........ which are not summable in ordinary sense and than use these non-rigorous values of h(0) , h (z>e^1/e) to find new relations.

For that, may be there is a need/possibility for a starter to find an analoque of geometric series sums or values of a type 1/(1-x) or 1/(1+x) in tetration,which would work in converge nce region but give values outside it. I think Gotfrieds formula goes in that direction when all branches will be found. Just an idea.

Ivars