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+--- Thread: using (5/2)x - (5/6)x^3 + ... (/showthread.php?tid=1744)
using (5/2)x - (5/6)x^3 + ... - tommy1729 - 04/25/2023
Hi
I decided to dedicate some time on variants of the 2sinh(x) method.
As you know 2sinh(x) can be use up to bases exp(1/2) because 2sinh(x) = 2x + ... and 2sinh(x/2) = x + ... ( parabolic fixpoint and a function still above id(x) ).
But having a function close to exp(x) and being e x + ... is not enough to extend to base eta ( e^(1/e) ) because underestimates of eta^x get multiple real fixpoints ...
So lowering the base with pure lower asymptotics is doomed to fail in a sense.
But certainly the base exp(1/2) can be lowered.
Another motivation is we know the 2sinh type methods satisfy the semi-group iso.
( although there is still talk about that )
Now I have done so before in the past.
But I wanted the function near 0 to not only have higher f ' (0) but also a missing x^2 term in its maclauren, without having a huge negative x^3 term.
This makes it much more linear near the origin.
Just to be clear many nice properties/conditions for 2sinh type methods exist, but combining them all is impossible, choices need to be made.
( what explains why I investigated multiple cases )
One of the main ideas is to use exponential sums.
And to get closer to exp(x) than 2sinh(x) does.
Here I also wanted the function to be 2 pi i periodic as well.
And I wanted to start relatively simple ; no 50 terms of exponential terms and no complicated numbers.
Rather simple fractions.
So I decided to consider this
f(x) = (5/2)x - (5/6)x^3 + ...
Then I had to find such an f(x) as good approximation of exp(x).
And not too complicated.
So I ended up with
f(x) = exp(x) - (5/4)*exp(-2x) + (1/4)*exp(-4x)
which has maclauren
f(x) = (5/2)x - (5/6)x^3 +(15/8 )x^4 - ...
f_4(x) = (5/2)x - (5/6)x^3 +(15/8 )x^4
f_5(x) = (5/2)x - (5/6)x^3 +(15/8 )x^4 - (43/24)x^5
(f_n (x) is truncated maclauren )
f_*(x) = ( f_4(x) + f_5(x) ) /2 = (5/2)x - (5/6)x^3 +(15/8 )x^4 -(43/48 )x^5
So f_*(x) is an estimate for small x
plot [(5/2)x - (5/6)x^3 +(15/8 )x^4 -(43/48 )x^5,exp(x),x]
I plotted f_*(x) and it looks nice, it seems to almost touch the exp(x).
Since f(x) is so linear lowering the base makes more sense by using f(x/a).
Keep in mind it is a better approximation for exp(x) for x > 0 , but worse than 2sinh(x) for x < 0.
f(x) is also 2 pi i periodic and has only one real fixpoint being 0.
f(0) = 0
and going to +infinity :
lim |[exp(x)^2 - f(x)^2]| = lim |(exp(x) - (5/4)*exp(-2x) + (1/4)*exp(-4x))^2 - exp(2x)| = 0
and
lim (exp(x) - (5/4)*exp(-2x) + (1/4)*exp(-4x))^3 - exp(3x) = -15/4
So f(x) is pretty close to exp(x).
Ofcourse f(x) is also entire.
***
What bothers me with f(x) is however its behaviour for negative x.
at first it is ok but then it starts going to positive , so we get a kind of 2cosh type function instead of 2sinh.
This makes things harder and inconvenient.
Maybe a 3term exponential sum is insufficient and we need a 4 or 5 term.
Another drawback is that the coefficients are not all non-negative.
Keep in mind that I said satisfying all desired properties might not be possible and choices need to be made.
Just sharing some ideas
regards
tommy1729
RE: using (5/2)x - (5/6)x^3 + ... - tommy1729 - 04/26/2023
(5/2) x - (9/8 ) x^2 + ...
has a solution
exp(x) - exp(-x) + exp(-2x) - exp(- 5/2 x)
To infinity
lim (exp(x) - exp(-x) + exp(-2x) - exp(- 5/2 x))^2 - (2 sinh(x))^2 = 0
lim (exp(x) - exp(-x) + exp(-2x) - exp(- 5/2 x))^3 - (2 sinh(x))^3 = 3
This behaves nicer for negative x too.
exp(x) - exp(-x) + exp(-2x) - exp(- 5/2 x) is the new favorite for now.
regards
tommy1729
RE: using (5/2)x - (5/6)x^3 + ... - tommy1729 - 04/26/2023
Ok we can get sharper
f(x) = exp(x) - exp(-x) + (7/10)(exp(- 5/4 x) - exp(- 9/4 x))
f(x) = (27 x)/10 - (49 x^2)/40 + (459 x^3)/320 - (2597 x^4)/3840 + (34329 x^5)/102400 + O(x^6)
(Taylor series)
This f(x) is periodic with 8 pi i, and bijective on R
and f'(0) = 2.7 what is close to e.
regards
tommy1729
RE: using (5/2)x - (5/6)x^3 + ... - tommy1729 - 04/29/2023
I will analyse f(x) = exp(x) - exp( - e x) in the next thread.
regards
tommy1729