# Tetration Forum

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Hello.

Without checking the whole forum through, I just wanted to throw in something from my experiences concerning tetration.
I hear there are some different possible solutions to this function, and I'm too curious if the following example is one of them:
While dabbling in this field of maths, I found a formula for 2^^x:
For 0 < x < 1, with n = 0.345627*(2-x), 2^^x ~~ [x*(2^n-1)+1]^(1/n)
(it's not 100% accurate, but ... about 99%, maybe some slight adjustments can fix that)
I could try and waste hours finding other parameters for other bases, but before I do that, I considered obtaining professional info from professional mathematicians.

martin
Hello Martin

martin Wrote:I hear there are some different possible solutions to this function, and I'm too curious if the following example is one of them:
While dabbling in this field of maths, I found a formula for 2^^x:
For 0 < x < 1, with n = 0.345627*(2-x), 2^^x ~~ [x*(2^n-1)+1]^(1/n)
(it's not 100% accurate, but ... about 99%, maybe some slight adjustments can fix that)

Why do you consider this formula as tetration?

We have some base requirements for tetration which are
2^^0 = 1
2^^(x+1) = 2^(2^^x)

I dont think that your formula satisfies these requirements.
Your formula satisfies the first few equalities:
x=0: 2^^0 = [0+1]^(1/n)= 1
x=1: 2^^1 = [1*(2^n-1)+1]^(1/n)=(2^n)^(1/n) = 2

for x=2, n=0 we have to take the limit for your formula to be defined:
$2\^\^2 = \lim_{n\to 0} [2(2^n-1)+1]^{1/n} = \lim_{n\to 0} [2^{n+1}-1]^{1/n}$ which seems indeed to be 4.

for x=3, n=-0.345627
$2\^\^ 3 = [3(2^n-1)+1]^{1/n}\approx 1.000002949\neq 2^{2^2}=2^4 = 16$
I know, I said this formula works for 0 < x < 1, the other values are naturally obtained
- for x<0 by iterated log[2^^(x+1)]/log(2)
- for x>1 by iterated 2^(x-1)

I just wanted to know if this is a way to obtain the values for non-integer x.

Thanks,
Martin
martin Wrote:I know, I said this formula works for 0 < x < 1, the other values are naturally obtained
- for x<0 by iterated log[2^^(x+1)]/log(2)
- for x>1 by iterated 2^(x-1)

So whats your requirement then? You can use any function on 0..1.
Of course, perhaps it should be continuous. And perhaps also differentiable, and perhaps also infinitely often differentiable and perhaps also analytic.

Hm, I didnt try it yet but I guess the formula is not infinitely often differentiable, did you test it? You only need to consider one patch point, for example 0 or 1. If it is infinitely differentiable there then also on all the other points. By this method by the way Andrew (which is also here on the forum) constructed an analytic solution. This is one thread that deals with this construction.
Sorry, I thought you already considered this type of formula or made a quick check about derivatives and such ... anyway, my "function" here is continuous and differentiable (as far as I understand what that means), but not infinitely often differentiable. The 3rd derivative has a slight peak near the integer points that I didn't manage to wipe away with a higher accuracy of my "parameter" 0.345627(2-x), or maybe Excel is just too inaccurate there. I've tried similar formulae before, but never got this close (e.g. had this peak already in the 2nd derivative).

Generally, for a better understanding, I was originally searching a function between x and x+1, such that log(f(x))/log(b) below x and b^f(x) above x+1 runs smoothly together to a continuous, monotonously increasing line on a graph. (I could explain more details, if you care to hear the whole story.)

I had in mind that this might be a formula, even if it only gives approximated values, that is not too complicated to calculate with. (Okay, I still would need to find a way to get a general formula to substitute this 0.345627-thingy for other bases.) Or, if there is a way to get that peak out of the 3rd derivative and eventually make it infinitely differentiable, whether the results are congruent to already existing results or they aren't.
martin Wrote:Sorry, I thought you already considered this type of formula or made a quick check about derivatives and such
No, absolutely not. New formula, new luck

Quote: ... anyway, my "function" here is continuous and differentiable (as far as I understand what that means), but not infinitely often differentiable. The 3rd derivative has a slight peak near the integer points that I didn't manage to wipe away with a higher accuracy of my "parameter" 0.345627(2-x), or maybe Excel is just too inaccurate there. I've tried similar formulae before, but never got this close (e.g. had this peak already in the 2nd derivative).

I really wonder how you actually came to *this* formula. If you are after finite differentiability, one rather would consider polynomials. With a polynomial of degree $d$ you can always get a $d$ times continuously differentiable function. One just have to determine numerically the coefficients. For example for $d=2$ let $p(x)=a+bx+cx^2$ the polynomial. We have to satisfy
$p^{(k)}(1)=\left(u^p\right)^{(k)}(0)$, for $k=0,1,2$, where $u$ is the base of our tetration. So well for $k=0$:
(*) $a+b+c = u^a$
then $k=1$
$(b+2cx)|_{x=1} = u^p \ln(u) (b+2cx) |_{x=0}$
(**) $b+2c = u^a \ln(u) b$
and now $k=2$
(***) $2c = u^p (\ln(u)(b+2cx))^2 + u^p \ln(u) 2c|_{x=0} = u^a (\ln(u)b)^2 + u^a \ln(u) 2c$

So these are 3 equations with 3 variables $a$, $b$, $c$ you can solve them numerically and voila you obtain a polynomial which gives a 2-times differentiable tetration.

Quote:I had in mind that this might be a formula, even if it only gives approximated values, that is not too complicated to calculate with. (Okay, I still would need to find a way to get a general formula to substitute this 0.345627-thingy for other bases.) Or, if there is a way to get that peak out of the 3rd derivative and eventually make it infinitely differentiable, whether the results are congruent to already existing results or they aren't.

Polynomials are not complicated to calculate with either, are they?
bo198214 Wrote:So these are 3 equations with 3 variables $a$, $b$, $c$ you can solve them numerically and voila you obtain a polynomial which gives a 2-times differentiable tetration.

The only problem with that approach, I see it now, is that the solutions need not to be real. As it is indeed in this case (the given equation system has only complex solutions). So it seems you have to tell us more about your formula ...
Oy, I'm not so good at explaining things, but I'll try anyway.

When I first tried to interpolate b^^x for non-integer x (and for easy cases like 2^^x or 3^^x), I thought of the following:
For example, 3^^0=1, 3^^1=3, 3^^2=27 - at first, the values appear to grow linearly, then exponentially. So I came up with the idea of a "flexible arithmetic-geometric mean". I expected the function to grow like a linear function ax+b at one point x and like a geometric series a^x*b at x+1, but that didn't turn out to be the case. That was about three years ago.

Playing around with numbers and maths in general being my hobby, I hit upon the following last year: all the "common" mean values (arithmetic, geometric, quadratic and harmonic mean) can be expressed in the form [(a(1)^n+a(2)^n+...+a(m)^n)/m]^(1/n). For n=1, this is the arithmetic mean, for n=2 the quadratic, for n=-1 the harmonic, and, by analytic continuation or whatever you may call it, for n=0 the geometric mean. I assumed that, with such a flexible mean value calculation, it just had to work. All I had to do was figure out the appropriate parameter n for a given interval [x ... x+1]. But lately I started doubting this assumption as well.

Darn, now I begin to see what you meant ... if this formula works between x and x+1, why shouldn't it also work beyond this interval? Seems I'm being still too much of an amateur here.

bo198214 Wrote:Polynomials are not complicated to calculate with either, are they?

Polynomials are not bad, really. I'm just not familiar with making polynomials with degree >= 2 out of given values.
martin Wrote:Oy, I'm not so good at explaining things, but I'll try anyway.

and exercise makes you better and better

Quote:all the "common" mean values (arithmetic, geometric, quadratic and harmonic mean) can be expressed in the form [(a(1)^n+a(2)^n+...+a(m)^n)/m]^(1/n). For n=1, this is the arithmetic mean, for n=2 the quadratic, for n=-1 the harmonic, and, by analytic continuation or whatever you may call it, for n=0 the geometric mean.

Thats interesting! Can you give a proof for n=0, i.e that
$\lim_{n\to 0} \sqrt[n]{\frac{a_1^n+\dots+a_m^n}{m}} = \sqrt[m]{a_1 \dots a_m}$?

Quote: I assumed that, with such a flexible mean value calculation, it just had to work. All I had to do was figure out the appropriate parameter n for a given interval [x ... x+1]. But lately I started doubting this assumption as well.

So the formula $\sqrt[c(2-x)]{x(2^{c(2-x)}-1)+1}$ is the generalized mean of what? *headscratch*
bo198214 Wrote:Thats interesting! Can you give a proof for n=0, i.e that
$\lim_{n\to 0} \sqrt[n]{\frac{a_1^n+\dots+a_m^n}{m}} = \sqrt[m]{a_1 \dots a_m}$?

Erm... no. Hasn't this already been proven somewhere?

bo198214 Wrote:So the formula $\sqrt[c(2-x)]{x(2^{c(2-x)}-1)+1}$ is the generalized mean of what? *headscratch*

It should merely interpolate 2^^x between x=0 and x=1 - the graph seemed smooth enough (together with the other iterated values beyond) to assume it represented the actual values to at least four decimal digits, but in other tables I've seen, only three decimal digits match.
Perhaps it's just my light-hearted view on all these numbers...

By the way, I achieved slightly better results when I changed that n=0.345627(2-x) into n=0.691-0,3450832*x. Gosh, I think this is getting a bit out of hand here...
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