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Tetration extension for bases between 1 and eta
#21
(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.



which is the same as




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.

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
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#22
(06/13/2022, 11:09 PM)tommy1729 Wrote: IS THERE ANY SPECIFIC REASON TO INVESTIGATE THESE ALTERNATIVE ITERATION FORMULAS APART FROM CURIOUSITY ?

regards

tommy1729

I see no fucking reason. That's why I explained this is just Schroder, and obviously Schroder....
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#23
I just wonder if and when the quadratic approximation gives a different result than the linear approximation.

Im aware of the one period theta relating all solutions but that does not help at first sight.

maybe im asking a trivial thing ...
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#24
(06/15/2022, 11:59 PM)tommy1729 Wrote: I just wonder if and when the quadratic approximation gives a different result than the linear approximation.

Im aware of the one period theta relating all solutions but that does not help at first sight.

maybe im asking a trivial thing ...

Well, if we're using Taylor expansion and dealing with infinitesimal very well we can have some kinda insight into it.
First we should know the basic rule that, suppose f is an analytic function, derivative (attractive) s at fixed point L \(f(z)=L+s(z-L)+O(z-L)^2\) and thus the Schroder function \(\sigma_f(z)=z-L+O(z-L)^2\) and \(\sigma_f^{-1}(z)=z-L+O(z-L)^2\) assuming \(\sigma'(L)=1\) and thus we get \(f^n(z)=L+s^n(z-L)+O(z-L)^2\).

And now let k be arbitrarily large number, we must have \(f^{-k}(f^t(f^k(z)))=f^t(z)\) even k extends to infinity in limit, or we say \(f^n(z)=\lim_{k\to\infty}{f^{-k}(f^n(f^k(z)))}\), and we assume the limit \(\lim_{k\to\infty}{f^k(z)}\) exist and equal to \(L\).
Recall \(f^n(f^k(z))=L+s^n(f^k(z)-L)+O(f^k(z)-L)^2=L+s^n(f^k(z)-L)+O(s^{2k}(z-L)^2)\) and thus \(f^{-k}(L+s^t(f^k(z)-L))=L+s^{-k}(s^t(f^k(z)-L))+O(s^{-2k}s^{2(k+t)}(z-L)^2)=L+s^t(z-L)+O(s^{2t}(z-L)^2)\), so as z approaches L, k goes to infinity, the limit \(\lim_{k\to\infty}{f^{-k}(L+s^t(f^k(z)-L))}\) exists.
Let's assume this limit equal to \(g(z,t)\). As t=0, we easily get by plugging t=0 in the limit: \(g(z,0)=z\).
And now we consider \(f^v(g(z,t))=\lim_{k\to\infty}{f^{-k+v}(L+s^t(f^k(z)-L))}=\)\(\lim_{k+v\to\infty}{f^{-k}(L+s^t(f^{k+v}(z)-L))}=\lim_{k\to\infty}{f^{-k}(L+s^t(f^k(f^v(z))-L))}=g(f^v(z),t)\)

Now we remind that we can rewrite the expansion as \(\lim_{z\to L}{\frac{f^t(z)-L}{s^t(z-L)}}=1\), and we can easily know the sequence \(\{z,f(z),f^2(z),f^3(z),\dots\}\) converge to some \(L\) (maybe infinity as well), so \(\lim_{k\to\infty}{\frac{f^t(f^k(z))-L}{s^t(f^k(z)-L)}}=1\)

Then we show directly by \(g(f^v(z),t)=\lim_{k\to\infty}{f^{-k}(L+s^t(f^{k+v}(z)-L))}=\)\(\lim_{k\to\infty}{f^{-k}(L+s^t(f^v(f^k(z))-L))}=\lim_{k\to\infty}{f^{-k}(L+s^t(s^v(f^k(z)-L)))}=g(z,t+s)\). So we arrive at $$f^t(z)=\lim_{k\to\infty}{f^{-k}(L+s^t(f^k(z)-L))}$$
This holds true for all such z that \(\{z,f(z),f^2(z),f^3(z),\dots\}\) converge to \(L\).

Now remind \(f^k(z)\) is very close to the fixed point \(L\), thus we rearrange the terms in the formula,
First we note and should denote that, \(\lim_L:=\lim_{k\to\infty}=\lim_{f^k(z)\to L}=\lim_{u=f^k(z),k\to\infty}\).
Then we have basically: \(\lim_L{f'(f^k(z))}=f'(L)=s\) and \(z\to L,f(z)-z\sim L+s(z-L)-z=(s-1)(z-L)\)
hence the formula reads \(\lim_L{f^{-k}(\frac{s^n(u-f(u))+f(u)-su}{1-s})}=\lim_L{f^{-k}(\frac{s^n(u-L)(1-s)+f(u)-su}{1-s})}\) \(=\lim_L{f^{-k}(\frac{s^n(u-L)(1-s)+L+s(u-L)-su}{1-s})}=\lim_L{f^{-k}(L+s^n(u-L))}=f^n(z)\)
Tah-dah!


btw there maybe many mistakes in this proof in limitation but the main idea is this
Regards, Leo Smile
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