Tetration extension for bases between 1 and eta

[bo198214]

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

Actually you rediscovered the Kœnigs formula (2.24 in the (unfinished) overview paper).
\( f^{[w]}(z)=\lim_{k\to\infty}
f^{[-k]}\left((1-\lambda^{w})\cdot p+\lambda^{w}\cdot f^{[k]}(z)\right) \)
for \( f(x)=b^x \). \( \lambda \) is the derivative at the fixed point \( p={^\infty b} \), which is \( \lambda=\ln(p)=\ln(b^p)=p\ln(b)=\ln(b){^\infty b} \).

Good work!

[dantheman163]

Actually you rediscovered the Kœnigs formula (2.24 in the (unfinished) overview paper).
\( f^{[w]}(z)=\lim_{k\to\infty}
f^{[-k]}\left((1-\lambda^{w})\cdot p+\lambda^{w}\cdot f^{[k]}(z)\right) \)
for \( f(x)=b^x \). \( \lambda \) is the derivative at the fixed point \( p={^\infty b} \), which is \( \lambda=\ln(p)=\ln(b^p)=p\ln(b)=\ln(b){^\infty b} \).

Good work!

ahh I knew it was something like that I had just never seen it before

[dantheman163]

Sorry for continuing to post more limit formulas but I found another that I do not think has been mentioned before.

\( f^n(x) = \lim_{k\to \infty} f^{-k}(\frac {f'(f^k(x))^n(f^k(x)-f(f^k(x)))+f(f^k(x))-f^k(x)f'(f^k(x))} {1-f'(f^k(x))}) \)

which is the same as

\( f^n(x) = \lim_{k\to \infty}f^{-k}(\frac {f'(u)^n(u-f(u))+f(u)-uf'(u)} {1-f'(u)})\\where\\u=f^k(x) \)


This works whenever a function has a regular attracting or repelling fixed point that it increases through. For it to work near a repelling fixed point you simply let k approach negative infinity.

Also note that \( \frac {f'(x)^n(x-f(x))+f(x)-xf'(x)} {1-f'(x)} \) gives a fairly decent aproximation of \( f^n(x) \) near a fixed point

Some pictures.
Red is \( f(x) \) blue is \( \frac {f'(x)^{1/2}(x-f(x))+f(x)-xf'(x)} {1-f'(x)} \) and green is \( f^{1/2}(x) \)

sin[x]
   

e^x-1, x<0
   


thanks.

[bo198214]

Not bad! How did you come to this formula? And what plot program do you use btw? The formula seems to be restricted to f with derivative 1 at the fixed point, right?

[dantheman163]

Not bad! How did you come to this formula? And what plot program do you use btw? The formula seems to be restricted to f with derivative 1 at the fixed point, right?

Basically I drew tangent lines to the curve and found the kth iterate for each one at the point of tangentcy. i used the fact that the kth iterate of \( f(x) = ax+b \) is \( a^k{x}+b\frac{1-a^k} {1-a} \).
the line tangent to the function at any point n is \( y=f'(n)x-nf'(n)+f(n) \) then setting \( a=f'(n) \) , \( b=-nf'(n)+f(n) \), and x=n then plunging in and simplifying we come up with the approximation formula.

now since this formula gives nice aproximations near a fixed point you can iterate the function to bring your value near the fixed point, then plug it into the formula, then un-iterate it to bring it back to where it was.

This formula does work with any function with an attracting or repelling fixed point even if the slope through it is not 1.
it may even work with functions with a negative slope (see pic)



\( f^{1/3}(x) \) if \( f(x)=1/x+1 \) (same colors)
   



I checked the values and it does satisfy \( f(f(f(x)))=1/x+1 \)


I use wz grapher to make the pictures. Its a free function grapher that you can download

[dantheman163]

Another use for this formula is that it converges very rapidly even if f(x) does not converge quickly to its fixed point.

For example I computed \( {}^{1/2} (e^{1/e}) \) using 10 iterations and got 1.257153126 which should be good to 7 decimal places. (using 11 iterations I get 1.257153143)

Thanks

[bo198214]

Another use for this formula is that it converges very rapidly even if f(x) does not converge quickly to its fixed point.

Thats great, as the standard Lévy formula is unacceptable slow.
But can you prove the formula. I.e. that it indeed describes an iteration semigroup,
i.e. that \( f^s\circ f^t=f^{s+t} \)?

[Catullus]

Sorry for continuing to post more limit formulas but I found another that I do not think has been mentioned before.

\( f^n(x) = \lim_{k\to \infty} f^{-k}(\frac {f'(f^k(x))^n(f^k(x)-f(f^k(x)))+f(f^k(x))-f^k(x)f'(f^k(x))} {1-f'(f^k(x))}) \)

which is the same as

\( f^n(x) = \lim_{k\to \infty}f^{-k}(\frac {f'(u)^n(u-f(u))+f(u)-uf'(u)} {1-f'(u)})\\where\\u=f^k(x) \)


This works whenever a function has a regular attracting or repelling fixed point that it increases through. For it to work near a repelling fixed point you simply let k approach negative infinity.

This formula is known to produce Schröder iteration. The proof of that is at https://math.eretrandre.org/tetrationfor...6#pid10036.

[tommy1729]

Sorry for continuing to post more limit formulas but I found another that I do not think has been mentioned before.

\( f^n(x) = \lim_{k\to \infty} f^{-k}(\frac {f'(f^k(x))^n(f^k(x)-f(f^k(x)))+f(f^k(x))-f^k(x)f'(f^k(x))} {1-f'(f^k(x))}) \)

which is the same as

\( f^n(x) = \lim_{k\to \infty}f^{-k}(\frac {f'(u)^n(u-f(u))+f(u)-uf'(u)} {1-f'(u)})\\where\\u=f^k(x) \)


This works whenever a function has a regular attracting or repelling fixed point that it increases through. For it to work near a repelling fixed point you simply let k approach negative infinity.

This formula is known to produce Schröder iteration. The proof of that is at https://math.eretrandre.org/tetrationfor...6#pid10036.

where is the proof ? 

copy the relevant part here ?

regards

tommy1729

[JmsNxn]

Sorry for continuing to post more limit formulas but I found another that I do not think has been mentioned before.

\( f^n(x) = \lim_{k\to \infty} f^{-k}(\frac {f'(f^k(x))^n(f^k(x)-f(f^k(x)))+f(f^k(x))-f^k(x)f'(f^k(x))} {1-f'(f^k(x))}) \)

which is the same as

\( f^n(x) = \lim_{k\to \infty}f^{-k}(\frac {f'(u)^n(u-f(u))+f(u)-uf'(u)} {1-f'(u)})\\where\\u=f^k(x) \)


This works whenever a function has a regular attracting or repelling fixed point that it increases through. For it to work near a repelling fixed point you simply let k approach negative infinity.

This formula is known to produce Schröder iteration. The proof of that is at https://math.eretrandre.org/tetrationfor...6#pid10036.

where is the proof ? 

copy the relevant part here ?

regards

tommy1729

Actually, you can see that Bo already showed it (he pointed out it's Kneser's iteration formula; i.e: Schroder iteration). I just gave a rough outline of how to prove it. It's pretty obvious if you think about it. The period in \(n\) will be that of the Schroder iteration, which is enough to conclude that it is Schroder's iteration (only one iteration has that period). I didn't exactly prove this, I proved if it converges it converges to Schroder's iteration. And even then, I mostly just sketched it. I also used the Ramanujan formula to show they must be the same too, but again, it was just a sketch.