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***bo198214*** · (This post was last modified: 12/15/2009, 01:40 AM by bo198214.)

(12/15/2009, 01:01 AM)dantheman163 Wrote: 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).
$f^{[w]}(z)=\lim_{k\to\infty}<br /> f^{[-k]}\left((1-\lambda^{w})\cdot p+\lambda^{w}\cdot f^{[k]}(z)\right)$
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 · 12/15/2009, 01:48 AM

(12/15/2009, 01:40 AM)bo198214 Wrote: Actually you rediscovered the Kœnigs formula (2.24 in the (unfinished) overview paper).
$f^{[w]}(z)=\lim_{k\to\infty}<br /> f^{[-k]}\left((1-\lambda^{w})\cdot p+\lambda^{w}\cdot f^{[k]}(z)\right)$
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 · 12/17/2009, 02:40 AM

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]

Filename: sin[x].gif Size: 5.64 KB 12/17/2009, 02:37 AM

e^x-1, x<0

Filename: e^x-1.gif Size: 4.63 KB 12/17/2009, 02:38 AM

thanks.

***bo198214*** · 12/17/2009, 10:59 AM

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 · (This post was last modified: 12/17/2009, 09:51 PM by dantheman163.)

(12/17/2009, 10:59 AM)bo198214 Wrote: 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)

Filename: 1x+1.gif Size: 7.09 KB 12/17/2009, 09:46 PM

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 · 12/19/2009, 05:06 AM

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*** · 12/19/2009, 10:55 AM

(12/19/2009, 05:06 AM)dantheman163 Wrote: 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 $f^s\circ f^t=f^{s+t}$ ?

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