Tetration extension for bases between 1 and eta

[dantheman163]

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

[bo198214]

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 \( 1<b<e^{1/e} \). So I am really curious about your proof.

PS: for writing formulas in this forum please have a look at this post.

[dantheman163]

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

[bo198214]

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?

[bo198214]

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
\( f(x+1)=b^{f(x)} \)?

[bo198214]

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 \( 0<f_1<1 \):
\( \lim_{n\to\infty} \frac{f^{\circ n}(t)-f^{\circ n}(\theta)}{f^{\circ n}(t)-f^{\circ n+1}(t)}=w\frac{1-f_1^w}{1-f_1} \) 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

[dantheman163]

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 \( b= sqrt2 \) ?

[bo198214]

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. \( b=e^{1/e} \).

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.

[mike3]

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

[dantheman163]

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

\( {}^x sqrt2 \)
   

half iterite of \( sqrt2^x \)

   

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