06/20/2023, 08:39 PM

I want to point out that fixpoint methods are usually based on formula's like koenigs or the higher order taylor analogues.

So for instance with hyperbolic attracting fixpoint at 0 :

f(x) = 0 + a x + b x^2 + ...

with 0 < a < 1

f^[z](x) = lim f^[-m]( a^z f^[m](x) )

This used the linear taylor polynomial a x for f(x).

A faster converging method is to use iterations of a x + b x^2 ( mod x^3 if you want to trade speed for kinda closed form ).

Although this works nice it gives little insight.

But we can use taylor's theorem to express f(x) in terms of its derivatives and this gives better insights sometimes.

As an example I take again a ' linear approximation / method '.

f(x) = x/( 1 + ln(1+x) )

Now we clearly understand the approximation below is good and intuitive :

f^[2](x) = x/( 1 + ln(1+x) )^2

In general

f(x) = x * g(x)

g(x) = f(x)/x

Notice if f(x) = 0 + a x + ...

Then g(x) = a + ...

f^[z](x) = lim f^[-m]( (f(x)/x)^z f^[m](x) )

Higher order variants are also possible.

I rarely see this investigated.

Ofcourse regions of convergeance , branches and functional equation far away from the fixpoint may differ.

But that is also interesting.

But these are also analytic so are in the same category.

In fact they are the same functions but with different regions of convergeance , convergeance speed etc.

They even share the semi-group iso in general.

As example say

f(x) = x/( 1 + ln(1+x) )

f^[z](x) = lim f^[-m]( (1 + ln(1+x))^z f^[m](x) )

Or for

https://tetrationforum.org/showthread.php?tid=1750

f(x) = exp(x) - exp(-3/5 x)

f^[z](x) = lim f^[m]( ( (exp(x) - exp(-3/5 x))/x )^z f^[-m](x) )

( notice the signs of m are switched because f(x) is repelling here. The inverse of f(x) has no simple closed form hence the reason )

All in all it is not all that different, but the viewpoint and insight makes all the difference.

Keep in mind to use some common sense just as with the other methods.

For instance if x is large and m is small and z is large , this probably does not give a good approximation yet and requires more work and/or iterations and/or analytic continuation.

As a consequence I write :

exp^[z](x) = lim ln^[m] ( f^[2m]( ( (exp(x) - exp(-3/5 x))/x )^z f^[-2m](exp^[m](x)) ) )

regards

tommy1729

So for instance with hyperbolic attracting fixpoint at 0 :

f(x) = 0 + a x + b x^2 + ...

with 0 < a < 1

f^[z](x) = lim f^[-m]( a^z f^[m](x) )

This used the linear taylor polynomial a x for f(x).

A faster converging method is to use iterations of a x + b x^2 ( mod x^3 if you want to trade speed for kinda closed form ).

Although this works nice it gives little insight.

But we can use taylor's theorem to express f(x) in terms of its derivatives and this gives better insights sometimes.

As an example I take again a ' linear approximation / method '.

f(x) = x/( 1 + ln(1+x) )

Now we clearly understand the approximation below is good and intuitive :

f^[2](x) = x/( 1 + ln(1+x) )^2

In general

f(x) = x * g(x)

g(x) = f(x)/x

Notice if f(x) = 0 + a x + ...

Then g(x) = a + ...

f^[z](x) = lim f^[-m]( (f(x)/x)^z f^[m](x) )

Higher order variants are also possible.

I rarely see this investigated.

Ofcourse regions of convergeance , branches and functional equation far away from the fixpoint may differ.

But that is also interesting.

But these are also analytic so are in the same category.

In fact they are the same functions but with different regions of convergeance , convergeance speed etc.

They even share the semi-group iso in general.

As example say

f(x) = x/( 1 + ln(1+x) )

f^[z](x) = lim f^[-m]( (1 + ln(1+x))^z f^[m](x) )

Or for

https://tetrationforum.org/showthread.php?tid=1750

f(x) = exp(x) - exp(-3/5 x)

f^[z](x) = lim f^[m]( ( (exp(x) - exp(-3/5 x))/x )^z f^[-m](x) )

( notice the signs of m are switched because f(x) is repelling here. The inverse of f(x) has no simple closed form hence the reason )

All in all it is not all that different, but the viewpoint and insight makes all the difference.

Keep in mind to use some common sense just as with the other methods.

For instance if x is large and m is small and z is large , this probably does not give a good approximation yet and requires more work and/or iterations and/or analytic continuation.

As a consequence I write :

exp^[z](x) = lim ln^[m] ( f^[2m]( ( (exp(x) - exp(-3/5 x))/x )^z f^[-2m](exp^[m](x)) ) )

regards

tommy1729