# Tetration Forum

Full Version: Tetration extension for bases between 1 and eta
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I believe I have found an analytic extension of tetration for bases 1 < b <= e^(1/e).

This is based on the assumption
(1) The function y=b^^x is a smooth, monotonic concave down function

Conjecture:

If assumption (1) is true then

${}^x b = \lim_{k\to \infty} (log_{b}^{ok}(x({}^k b- {}^{(k-1)} b)+{}^k b) )$ for $-1 \le x\le 0$

Some properties:

This formula converges rapidly for values of b that are closer one.

For base eta it converges to b^^x for all x but this is not true for the other bases.

Interestingly for b= sqrt(2) and x=1 it seems to be converging to the super square root of 2

I will try to post a proof in the next couple of days I just need some time to type it up.

Thanks
Hey Dan, thank you for this contribution.

The formula you mention is the inversion of Lévy's formula (which is generally applicable to functions with f'(p)=1 for a fixed point p of f).
See e.g. the tetration methods draft, formula 2.26 (where formula 2.27 is called Lévy's formula).

The formula is known to give the regular tetration for $b=e^{1/e}$. However it is new to apply it to $1. So I am really curious about your proof.

PS: for writing formulas in this forum please have a look at this post.
I'm not sure how rigorous this is but here it is

Proof:

Assumption
(1)The function $f(x)={}^x b$ is a smooth, monotonic concave down function

Based on (1) we can establish
(2)Any line can only pass through f(x) a maximum of 2 times
(3)The intermediate value theorem holds for the entire domain of f(x)

Take the region R bounded on the x axis by x=-1 and x=0 and bounded on the y axis by y=f(-1) and y=f(0) ( y=0 and y=1).
Because of (3) every value, x, has a corresponding value, f(x), on the interval.
If we take a point (x,y) in R and assume that it is on the curve f(x) we can then use the relation $f(x+1)=b^{f(x)}$ and obtain the new point $(x',y')$ which equals $(x+1,b^y)$. Applying this repeatedly we obtain the point $(x+k,{Exp}_b ^k (y))$.
We can now establish
(4)The point (x,y) in the region R is on the curve f(x) if $(x+k,{Exp}_b ^k (y))$ is not on the secant line that touches the curve at 2 other known points of f(x) for any value of k.

Now we will find the equation of the secant line that touches f(x) at 2 consecutive known points. Using the point slope formula we find the equation to be $g(x)= ({}^k b-{}^{(k-1)} b)(x-k)+{}^k b$. We must also note that if a point (x,y) is above the curve in the region R then $(x+k,{Exp}_b ^k (y))$ is above the curve for any value of k.
we shall now extend (4) to say
(5)The point (x,y) in the region R is on or above the curve f(x) if $g(x+k) < {Exp}_b ^k (y)$for any value of k

Finally if we take the limit as k approaches infinity we will find that the slope of the secant line approaches zero and therefore follows f(x) exactly because f(x) has an asymptote as x goes towards infinity. Therefore $g(x+k) = {Exp}_b ^k (y)$ for infinity large values of k

Now solving for y and taking the limit as k approaches infinity we obtain the desired result:

${}^x b = \lim_{k\to \infty} (log_{b}^{ok}(x({}^k b- {}^{(k-1)} b)+{}^k b) )$ for $-1 \le x\le 0$

q.e.d
Let me rephrase in my words:

We consider the linear functions $g_k$ on (k-1,k) determined by $g_k(k-1)=f(k-1)={^{ k-1}b}$ and $g_k(k)=f(k)={^k b}$, they are given by:
$g_k(x) = ({^k b} - {^{ k-1}}b)(x-k) + {^k b}$.

As f is concave $f(x+k) \ge g_k(x+k)$ for $x\in (-1,0)$, and as $\log_b$ is strictly increasing we have also $f(x) \ge \log_b^{\circ k} g_k(x+k)$.
On the other hand we know that $f(x+k)=\exp_b^{\circ k}(f(x))\uparrow a$ for $k\to\infty$, hence $a > f(x+k) > g_k(x+k)$ and $g_k(x+k)\uparrow a$ for $k\to\infty$.

We have now $f(x+k)-g_k(x+k)\downarrow 0$ for $k\to\infty$. But I think that does not directly show the convergence $f(x) - \log_b^{\circ k} g_k(x+k)\downarrow 0$.

Any ideas?
(11/07/2009, 09:31 AM)bo198214 Wrote: [ -> ]We have now $f(x+k)-g_k(x+k)\downarrow 0$ for $k\to\infty$. But I think that does not directly show the convergence $f(x) - \log_b^{\circ k} g_k(x+k)\downarrow 0$.

Any ideas?

Actually I think one can show that $\log_b^{\circ k} g_k(x+k)$ is strictly increasing with $k$ because $\log_b g_k(x+k)$ is concave and hence
$g_{k-1}(x+k-1) < \log_b g_k(x+k)$ and thatswhy
$\log_b^{\circ k-1} g_{k-1}(x+k-1)<\log_b^{\circ k} g_{k}(x+k)$
also it is bounded and hence must have a limit.

But now there is still the question why the limit $f(x)$ indeed satisfies
$f(x+1)=b^{f(x)}$?
(11/07/2009, 05:11 PM)bo198214 Wrote: [ -> ]But now there is still the question why the limit $f(x)$ indeed satisfies
$f(x+1)=b^{f(x)}$?

Actually it is not the case but we can obtain something very similar.
I just read in [1], p. 31, th. 10, that we have the following limit for a function
$f(x)=f_1 x+f_2x^2+f_3 x^3 + ...$ with $0:
$\lim_{n\to\infty} \frac{f^{\circ n}(t)-f^{\circ n}(\theta)}{f^{\circ n}(t)-f^{\circ n+1}(t)}=w\frac{1-f_1^w}{1-f_1}$ for $\theta = f^{\circ w}(t)$.

If we invert the formula we get
$f^{\circ w}(t) = \lim_{n\to\infty} f^{\circ -n}\left(w\frac{1-f_1^w}{1-f_1}\left(f^{\circ n+1}(t)-f^{\circ n}(t)\right)+f^{\circ n}(t)\right)$

In our case though we dont have f(0)=0 but there is some fixed point $z_f$ of $f$, $f(z_f)=z_f$.
In this case however the formula is quite similar, the only change is that $f_1=f'(z_f)$.

[1] Ecalle: Theorie des invariants holomorphes
If we set $f(x+1)=b^{f(x)}$ which is to say
$\lim_{k\to \infty} (log_{b}^{ok}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (log_{b}^{o(k-1)}(x({}^k b- {}^{(k-1)} b)+{}^k b) )$
then reduce it to
$\lim_{k\to \infty} (log_{b}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (x({}^k b- {}^{(k-1)} b)+{}^k b)$
Then just strait up plug in infinity for k
we get $log_{b} {}^\infty b = {}^\infty b$ which is the same as ${}^\infty b = {}^\infty b$

This is really weird because if i do $\lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) )$ for $b= sqrt2$
i get about 1.558 which is substantially larger then $sqrt2$

Can anyone else confirm that $\lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) ) \approx 1.558$ for $b= sqrt2$ ?
(11/07/2009, 11:30 PM)dantheman163 Wrote: [ -> ]If we set $f(x+1)=b^{f(x)}$ which is to say
$\lim_{k\to \infty} (log_{b}^{ok}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (log_{b}^{o(k-1)}(x({}^k b- {}^{(k-1)} b)+{}^k b) )$
then reduce it to
$\lim_{k\to \infty} (log_{b}((x+1)({}^k b- {}^{(k-1)} b)+{}^k b) ) = \lim_{k\to \infty} (x({}^k b- {}^{(k-1)} b)+{}^k b)$
Then just strait up plug in infinity for k
we get $log_{b} {}^\infty b = {}^\infty b$ which is the same as ${}^\infty b = {}^\infty b$

Slowly, slowly. The first line is what you want to show. You show from the first line something true, but you can also show something true starting from something wrong; so thats not sufficient. Also it seems as if you confuse limit equality with sequence equality.

Lets have a look at the inverse function $g=f^{-1}$,
$g(x)=\lim_{k\to\infty} \frac{\exp_b^{\circ k}(x)-{^k b}}{{^k b}-({^{k-1} b})}$ it should satisfy
$g(b^x)=g(x)+1$.

Then lets compute
$g(b^x)-g(x)=
\lim_{k\to\infty}\frac{\exp_b^{\circ k+1}(x)-{^k b}}{{^k b}-({^{k-1} b})} - \frac{\exp_b^{\circ k}(x)-{^k b}}{{^k b}-({^{k-1} b})}
=\lim_{k\to\infty} \frac{\exp_b^{\circ k+1}(x) - \exp_b^{\circ k}(x)}{{^k b}-({^{k-1} b})}
$

Take for example $x=1$ then the right side converges to the derivative of $b^x$ at the fixed point; and not to 1 as it should be.

This is the reason why the formula is only valid for functions that have derivative 1 at the fixed point, e.g. $e^{x/e}$, i.e. $b=e^{1/e}$.

Quote:This is really weird because if i do $\lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) )$ for $b= sqrt2$
i get about 1.558 which is substantially larger then $sqrt2$

Can anyone else confirm that $\lim_{k\to \infty} (log_{b}^{ok}(1({}^k b- {}^{(k-1)} b)+{}^k b) ) \approx 1.558$ for $b= sqrt2$ ?

Try the same with $b=e^{1/e}$ and it will work; but for no other base; except you use the modified formula I described before.
Oops... you may have noticed this just got a "1 star" rating. I just happened to accidentally click the mouse on the "rate" thing, sorry
Upon closer study i think i have found a formula that actually works.

${}^ x b = \lim_{k\to \infty} \log_b ^k({}^ k b (ln(b){}^ \infty b)^x-{}^ \infty b(ln(b){}^ \infty b)^x+{}^ \infty b)$

Also i have noticed that this can be more generalized to say,

if $f(x)=b^x$
then
$f^n(x)= \lim_{k\to \infty} \log_b ^k( Exp_b^k(x) (ln(b){}^ \infty b)^n-{}^ \infty b(ln(b){}^ \infty b)^n+{}^ \infty b)$

numerical evidence shows this to be true.

some plots

${}^x sqrt2$
[attachment=666]

half iterite of $sqrt2^x$

[attachment=667]

The graph falls apart at x=4 because $Exp_b^k(4.01)$ diverges as k goes to infinity.

thanks.

Edit: sorry for the huge pictures
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