06/13/2022, 11:09 PM

(06/13/2022, 10:46 PM)JmsNxn Wrote:(06/13/2022, 10:42 PM)tommy1729 Wrote:(06/12/2022, 11:00 PM)Catullus Wrote:(12/17/2009, 02:40 AM)dantheman163 Wrote: Sorry for continuing to post more limit formulas but I found another that I do not think has been mentioned before.This formula is known to produce Schröder iteration. The proof of that is at https://math.eretrandre.org/tetrationfor...6#pid10036.

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

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.

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.

But you could also have used the quadratic approximation : a^t x + k a^(t-1) ( a^t - 1 )/(a-1) x^2.

Now i assume the qaudratic approximation is faster AND different from schroder ( in general and here ).

And I assume your iteration formula is identical to schroder BUT FASTER , yet still slower than the qaudratic approximation ( though that is different ).

I also assume the origin is from approximating a derivative with a difference operator.

On the other hand , i suspect your iteration is only slightly faster than schroder because your formula is a bit longer and it is actually more like

" schroder in the future " meaning close to schroder with n + 1 or n + 2 instead of n.

and that +1 or +2 simply because you approximate product by derivate with more stuff.

I think the taylor series cut offs are optimal and cannot be improved substantially.

That explains my assumptions above.

IT also makes sense since your formula does not use the second derivative and thus * in a way * in part holds less information.

Those are alot of assumptions but i think you will agree.

I have not doubt about your claims , It is just that a formal proof might be interesting.

Such a proof must exist but might require some work.

IS THERE ANY SPECIFIC REASON TO INVESTIGATE THESE ALTERNATIVE ITERATION FORMULAS APART FROM CURIOUSITY ?

regards

tommy1729