07/25/2021, 11:58 PM
An interesting idea is this :
Are all these " beta methods " equivalent ? as James asks.
And
Is there a way to accelerate the convergeance of the iterations ?
Series acceleration is well known but iteration acceleration not so much.
Those are 2 nice questions , but what is the interesting idea you might ask ??
Well that those 2 questions are related !!
let
f_1(s) = exp( t(1 s) * f(s-1))
f_2(s) = exp( t(2 s) * f(s-1))
and the resp analytic tetrations from them : F1(s) and F2(s).
Remember that tetration(s + theta(s)) is also tetration where theta(s) is a suitable analytic real 1-periodic function.
so F2(s) = F1(s + theta1(s)) , F1(s) = F2(s + theta2(s)).
BUT THIS ALSO IMPLIES THAT
f*_2(s) = exp( t( 2*(s + theta2(s)) ) * f(s-1)) .. RESULTING IN F2_*(s) is actually equal to
F2_*(s) = F2(s + theta(s)) = F1(s)
HENCE USING t(1 s) = t(s) is the same as using t( 2 s + 2 theta(s)) !!
So this relates to the main questions posed :
when are 2 solutions equal ?
How to accelerate convergeance ?
As for the acceleration , ofcourse the complexity and difficulty of theta and computing theta are key.
But numerically it is expected using t( 2 s + 2 theta(s) ) converges faster. ( because using t(2s) does converge faster than using t(s) ) .
---
Tom(s,v) = exp( t(v * s) * exp(Tom(s-1,v)) )
resulting in
tet(s+1,v) = exp( tet(s,v) ).
I Like that notation.
---
regards
tommy1729
Are all these " beta methods " equivalent ? as James asks.
And
Is there a way to accelerate the convergeance of the iterations ?
Series acceleration is well known but iteration acceleration not so much.
Those are 2 nice questions , but what is the interesting idea you might ask ??
Well that those 2 questions are related !!
let
f_1(s) = exp( t(1 s) * f(s-1))
f_2(s) = exp( t(2 s) * f(s-1))
and the resp analytic tetrations from them : F1(s) and F2(s).
Remember that tetration(s + theta(s)) is also tetration where theta(s) is a suitable analytic real 1-periodic function.
so F2(s) = F1(s + theta1(s)) , F1(s) = F2(s + theta2(s)).
BUT THIS ALSO IMPLIES THAT
f*_2(s) = exp( t( 2*(s + theta2(s)) ) * f(s-1)) .. RESULTING IN F2_*(s) is actually equal to
F2_*(s) = F2(s + theta(s)) = F1(s)
HENCE USING t(1 s) = t(s) is the same as using t( 2 s + 2 theta(s)) !!
So this relates to the main questions posed :
when are 2 solutions equal ?
How to accelerate convergeance ?
As for the acceleration , ofcourse the complexity and difficulty of theta and computing theta are key.
But numerically it is expected using t( 2 s + 2 theta(s) ) converges faster. ( because using t(2s) does converge faster than using t(s) ) .
---
Tom(s,v) = exp( t(v * s) * exp(Tom(s-1,v)) )
resulting in
tet(s+1,v) = exp( tet(s,v) ).
I Like that notation.
---
regards
tommy1729

