# Tetration Forum

Full Version: Growth of superexponential
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Hi, it's me again,

I made an observation : for very small values of z, it seems likely that as b tends towards infinity, b^^z grows to infinity too, but rather slowly. I mean $\lim_{b \rightarrow \infty} {}^{z} b \rightarrow \infty$ for all $z > 0$. It's quite obvious for z > 1 because tetration grows much faster than exponentiation there. So, it would sufficient to consider z on the interval [0, 1]. Questions : 1) Is it straightforward from the definition of superexponential? If not, 2) Is it true/false? I am interested in a proof for both cases of #2, however.

Also, can somebody give me a plot of sexp'(x, 0.1) where x is the base( and the plot variable of course) and 0.1 is the height. sexp' means the derivative of superexponential function. It would be much appreciated.

Balarka
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That is indeed trivial. In fact so trivial that I doubted if I would even respond.

Notice that for (real) b >> (z+1) exp(1/e) and (real) z > 0 :

b^^z > z.

Hence lim b^^z >= lim z.

If you are not convinced of b^^z > z Consider that

1) b^z > z for z and b sufficiently large.

2) b^^0 (=1) > 0

3) the derivative of b^^z > 1 whereas the derivative of z is 1.

By induction that is clear.

However I must add that I did assume a nice tetration here.

In other words I assumed for c>>b that c^^z >> b^^z and that c^^z and b^^z are continu.

It always matters what type of tetration you speak of.

However they share similar properties e.g. all infinitely differentiable real solutions to exp^[1/2](x) agree on their values infinitely often.

Notice it is nicer if a half iterate is strictly rising , otherwise when it is both rising and decending taking a half derivitive of that is troublesome.

On a piece of paper that might make more sense to you.

Perhaps usefull to note is that in your question your value of b does not depend on 1/z. That is important since b^^0 = 1 , so if z goes to 0 much faster than b goes to oo ...

regards

tommy1729
tommy1729 Wrote:Hence lim b^^z >= lim z.

This is not that trivial as you think it is. I agree that limit b-->oo (b^^z) >= lim b-->oo (z) = z; the limit is hence greater than z. But that doesn't tells anything about the convergence of lim b-->oo b^^z, does it? at least I don't see it.

Thank you very much for your time,

Balarka
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(02/27/2013, 02:19 PM)Balarka Sen Wrote: [ -> ]I made an observation : for very small values of z, it seems likely that as b tends towards infinity, b^^z grows to infinity too, but rather slowly. I mean $\lim_{b \rightarrow \infty} {}^{z} b \rightarrow \infty$ for all $z > 0$
Its not a trivial question. And I had a difficult time getting my kneser.gp algorithm to converge for large bases, though it currently works for b>100000. Here is a related question, that may lead to a fruitful investigation path, and possibly a proof. Can you prove that the $\text{slog}_b(e)$ for arbitrarily large base=b approaches arbitrarily close to zero? There is a fairly simple linear approximation one can use for tetration for arbitrary bases, that is continuous, and has a continuous first derivative, and works surprisingly well. The estimation uses a straight line linear estimate between $\log_b(\log_b(e))$ and $\log_b(e)$. For example, for base e, the linear approximation is sexp(-1)=0, sexp(1)=1, with a straight line in between, and $\text{slog}_e(e)=1$.

edit: updated approximation equations, and plot
If I did my algebra correctly, than using the linear approximation for sexp for arbitrary bases leads to the estimate for bases>=e $\text{slog}_b(e)\approx \frac{1}{1+\log(\log(b))}=k$ and the region from k-1..k has an exponential approximation of $\text{sexp}_b(z)\approx\exp(\frac{z}{k})$. For z>k the approximation switches over to a double exponent. This exponential approximation assumes the linear region includes sexp(0), which is true if base>=e.

This approximation gives sexp(0)=1, and sexp(1)=b, which are both exact. Then you you could conjecture that for large enough bases (empirically, b>9), the actual sexp(k)>e. Also, I would conjecture that for b>9 the approximation is less than actual sexp(z) until z=1, where by definition the approximation is exactly correct once again.

Anyway, such an slog(e) approximation for large bases goes to zero, but very slowly. For b=googleplex=$10^{10^{100}}$, the approximation is slog(e)=0.0043. For b=10, the approximation is 0.545 and the correct $\text{slog}_{10}(e) \approx0.544$. For b=100, the approximation is 0.396 and the correct value is $\text{slog}_{100}(e) \approx0.374$. For b=100000, the approximation is 0.290, and the correct value is $\text{slog}_{100000}(e) \approx0.250$. Here is a graph of sexp_100000(z). The function is surprisingly well behaved in the region of interest. Here, $k\approx 0.29$, and the linear approximation region would be from -1.71 to -0.71. The actual sexp is in red, and the linear approximation is in green.
[attachment=1004]
- Sheldon

(02/27/2013, 06:40 PM)sheldonison Wrote: [ -> ]If I did my algebra correctly, than using the linear approximation for sexp for arbitrary bases leads to the estimate for $\text{slog}_b(e)=\frac{1}{1+\log(\log(b))}$, where with some algebra you could also come up with an approximation for the slope sexp_b(z)=e, and show that the slope gets arbitrarily large for superexponentially large bases, and then conjecture that for large enough bases, the approximation for slog(e) is an overestimate for the analytic slog, and that the slope for sexp(z) is increasing in this region....

That's an interesting approach! I was searching for a sequence diverges to infinity but grows slower than b^^z for z in the interval [0, 1], however. Finding such a sequence would be enough to prove the divergence of the limit we are concerned about. I will give it a try with the you suggested.

Thank you very much for your reply, it's much appreciated.

Balarka
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(02/27/2013, 06:40 PM)sheldonison Wrote: [ -> ]
(02/27/2013, 02:19 PM)Balarka Sen Wrote: [ -> ]I made an observation : for very small values of z, it seems likely that as b tends towards infinity, b^^z grows to infinity too, but rather slowly. I mean $\lim_{b \rightarrow \infty} {}^{z} b \rightarrow \infty$ for all $z > 0$
Its not a trivial question.
- Sheldon

Ok then I will post a proof soon.

tommy1729
picture this.

consider a C^2 function f(x) with

1) f(0) = 1 and f ' (0) > 1.

2) for x > 0 : f ' (x) > 0 and f '' (x) > 0.

3) f(1) = b and b > 2.

As b gets larger , what can you say about f(1,001) ?

Do you think f(1,001) is bounded as b grows or not ?

Once you solved this , notice how sexp_b(x) can be such a function if we set the conditions C^2 and 1) , 2) and 3).

Also notice that if we modify the conditions such that f ' (0) = 1 AND f '' (0) > 0 we get the same result.

( We do not get weird limits because of the property C^2 )

regards

tommy1729
And of course a sexp(x) with sexp(0) = 1 and a negative derivative for x > 0 is absurd. Hence the condition is logical.

tommy1729