04/30/2023, 01:07 AM
I decided to dedicate some time on variants of the 2sinh(x) method.
As you know 2sinh(x) method can be used 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 )
In this thread I will investigate ways to get a lower bound for the bases where methods similar to the 2sinh method work.
Notice this is an 'extension' of the method , since by the semi-group iso it will give the same result for the larger bases.
ok lets start.
Lets think about what we essentially want ...
f(x) is an entire function and close to exp(x) and
We want
f(x) such that
f(0) = 0
f ' (0) = e
lim f(x)^m - exp(m x) exists for a high value of m.
f(x/t) has a single strict positive fixpoint ;
v > 0 , f(v/t) = v
If f(x) is a sum of exponentials then the simplest is the one with only 2 terms.
Trying to satisfy the above and using " greedy logic " we get to this :
f(x) = exp(x) - exp( - (e-1) x)
Lets see
f(x) is an entire function and close to exp(x) and
f(0) = 0
f ' (0) = e
Also for real A > 0 ( here A = 1 for f(x) )
lim ( exp(x) - A exp( - (e-1) x) )^e - exp( e x ) = - A e.
Or here
lim ( exp(x) - exp( - (e-1) x) )^e - exp( e x ) = lim f(x)^e - exp( e x ) = - e.
This power e is the max.
Even for sums of exp with more terms this power e is maximum.
So indeed
f(x) = exp(x) - exp( - (e-1) x)
is the function we need and want.
It seems to agree with the original 2sinh method for base e as expected.
So how much can we lower the base ?
Remember
f(x/t) has a single strict positive fixpoint ;
v > 0 , f(v/t) = v
so we want the fixpoint to be the only positive real one and we want t to be maximum.
Now In fact if there is only one real fixpoint than t is the max, since only one value of t gives a unique real strict positive fixpoint.
So our job is to find the constants v,t such that
v and t > 0 ,
f(v/t) = v
OR equivalently
v and t > 0 ,
exp(v/t) - exp ( - (e-1) v/t ) = v
Such an equation seems hard on one hand and open to many theorems/methods on the other hand.
Solving it could be done by generalized contour integrals( cauchy theorems from complex analysis ),
generalized fourier-budan methods for taylor series and generalized descartes rule of signs.
The key here is the known closed form taylor series for f(x) and the extensions of the above theorems for polynomials to taylors.
This is an interesting subject , yet it seems a simpler way is better.
We consider
f((x+v)/t) =
exp((x+v)/t) - exp( - (x+v)(e-1)/t )
at x = 0 we get
exp(v/t) (1 - exp(-e v/t) ) + ...
the terms are simple closed form patterns.
(1 - (e-1)^n exp(-e v/t) ) is a factor of every n th coefficient.
This is very insightfull and also teaches us where the functions higher derivatives change sign !
( at the zero's of (1 - (e-1)^n exp(-e v/t) ) , and this helps us with the above ideas such as contour , fourier-budan , descartes etc )
Despite all those interesting things, still a much better method exists.
NOTICE THAT WE KNOW
AT THE UNIQUE FIXPOINT WE HAVE DERIVATIVE = 1 !
so we solve
d/dx [ exp((x+v)/t) - exp( - (x+v)(e-1)/t ) ] = 1
This gives after some algebra :
v = t ln( 1/2 ( t - sqrt(t^2 - 4) ) )
v = t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
Lets investigate these 2 branches.
t ln( 1/2 ( t - sqrt(t^2 - 4) ) ) = t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
We can drop the t on both sides and we notice
( t - sqrt(t^2 - 4) ) = 1/( t + sqrt(t^2 - 4) ) )
knowing ln(1/x) = - ln(x) the baviour of these branches is easy :
t ln( 1/2 ( t - sqrt(t^2 - 4) ) ) = (-1) * t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
since t = 2 is a zero this means
t ln( 1/2 ( t - sqrt(t^2 - 4) ) ) = t ln( 1/2 ( t + sqrt(t^2 - 4) ) ) = 0 for t = 2
and other wise the branches are negatives of each other.
Now clearly
t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
is the positive branch.
So we can continue :
v = t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
Now lets return to the original equation and plug in the above identity for v !!
v and t > 0 ,
exp(v/t) - exp ( - (e-1) v/t ) = v
exp( t ln( 1/2 ( t + sqrt(t^2 - 4) ) ) / t ) - exp( - (e-1) t ln( 1/2 ( t + sqrt(t^2 - 4) ) ) / t ) = t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
Now t =/= 0 so we can cancel the t/t parts !
Also we can remove the first ln since exp(ln(x)) = id(x)
1/2 ( t + sqrt(t^2 - 4) ) - exp( - (e-1) ln( 1/2 ( t + sqrt(t^2 - 4) ) ) ) = t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
solving for t > 0 we get
t = 2
or
t = 2.44369396324203812176211410..
if my computation is correct.
( probably , pretty certain about 2.443693963242038121762114 , less about the last 2 digits )
1/t = 0.40921654472367086504976586583879
This means v = 1.59907...
and our lowest base , The Low Base Constant (LBC)
LBC =
exp(1/t) = 1.5056377230919...
by comparison the original 2sinh had exp(1/2) = 1.6487... and eta = exp(1/e) = 1.4446..
and ( exp(1/2) + exp(1/e) )/2 = 1.546694..
so this LBC base constant is closer to eta than to the lowest base for 2sinh.
A big improvement.
(ok ok , i did not compute the exact value for the lowest base possible for the original 2sinh method but took the best case exp(1/2), ignoring fixpoints
nevertheless since the actual base would be even larger the conclusion certainly follows ! )
So in conclusion
exp(x) - exp( - (e-1) / x) works for all bases >> 1.5056377230919...
The Low Base Constant :
LBS = 1.5056377230919...
It took me some time doing things by hand.
This number 1.5056... looks familiar ...
Did I compute it before ?
Did it occur he before ?
Do we know a closed form ?
I remember estimating the lowest base around 3/2 about 12 years ago or so ... which is close.
Sorry for not getting into details.
I need to sleep.
Comments are welcome.
regards
tommy1729
Tom Marcel Raes
" Truth is what does not go away when you stop believing in it "
tommy1729
As you know 2sinh(x) method can be used 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 )
In this thread I will investigate ways to get a lower bound for the bases where methods similar to the 2sinh method work.
Notice this is an 'extension' of the method , since by the semi-group iso it will give the same result for the larger bases.
ok lets start.
Lets think about what we essentially want ...
f(x) is an entire function and close to exp(x) and
We want
f(x) such that
f(0) = 0
f ' (0) = e
lim f(x)^m - exp(m x) exists for a high value of m.
f(x/t) has a single strict positive fixpoint ;
v > 0 , f(v/t) = v
If f(x) is a sum of exponentials then the simplest is the one with only 2 terms.
Trying to satisfy the above and using " greedy logic " we get to this :
f(x) = exp(x) - exp( - (e-1) x)
Lets see
f(x) is an entire function and close to exp(x) and
f(0) = 0
f ' (0) = e
Also for real A > 0 ( here A = 1 for f(x) )
lim ( exp(x) - A exp( - (e-1) x) )^e - exp( e x ) = - A e.
Or here
lim ( exp(x) - exp( - (e-1) x) )^e - exp( e x ) = lim f(x)^e - exp( e x ) = - e.
This power e is the max.
Even for sums of exp with more terms this power e is maximum.
So indeed
f(x) = exp(x) - exp( - (e-1) x)
is the function we need and want.
It seems to agree with the original 2sinh method for base e as expected.
So how much can we lower the base ?
Remember
f(x/t) has a single strict positive fixpoint ;
v > 0 , f(v/t) = v
so we want the fixpoint to be the only positive real one and we want t to be maximum.
Now In fact if there is only one real fixpoint than t is the max, since only one value of t gives a unique real strict positive fixpoint.
So our job is to find the constants v,t such that
v and t > 0 ,
f(v/t) = v
OR equivalently
v and t > 0 ,
exp(v/t) - exp ( - (e-1) v/t ) = v
Such an equation seems hard on one hand and open to many theorems/methods on the other hand.
Solving it could be done by generalized contour integrals( cauchy theorems from complex analysis ),
generalized fourier-budan methods for taylor series and generalized descartes rule of signs.
The key here is the known closed form taylor series for f(x) and the extensions of the above theorems for polynomials to taylors.
This is an interesting subject , yet it seems a simpler way is better.
We consider
f((x+v)/t) =
exp((x+v)/t) - exp( - (x+v)(e-1)/t )
at x = 0 we get
exp(v/t) (1 - exp(-e v/t) ) + ...
the terms are simple closed form patterns.
(1 - (e-1)^n exp(-e v/t) ) is a factor of every n th coefficient.
This is very insightfull and also teaches us where the functions higher derivatives change sign !
( at the zero's of (1 - (e-1)^n exp(-e v/t) ) , and this helps us with the above ideas such as contour , fourier-budan , descartes etc )
Despite all those interesting things, still a much better method exists.
NOTICE THAT WE KNOW
AT THE UNIQUE FIXPOINT WE HAVE DERIVATIVE = 1 !
so we solve
d/dx [ exp((x+v)/t) - exp( - (x+v)(e-1)/t ) ] = 1
This gives after some algebra :
v = t ln( 1/2 ( t - sqrt(t^2 - 4) ) )
v = t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
Lets investigate these 2 branches.
t ln( 1/2 ( t - sqrt(t^2 - 4) ) ) = t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
We can drop the t on both sides and we notice
( t - sqrt(t^2 - 4) ) = 1/( t + sqrt(t^2 - 4) ) )
knowing ln(1/x) = - ln(x) the baviour of these branches is easy :
t ln( 1/2 ( t - sqrt(t^2 - 4) ) ) = (-1) * t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
since t = 2 is a zero this means
t ln( 1/2 ( t - sqrt(t^2 - 4) ) ) = t ln( 1/2 ( t + sqrt(t^2 - 4) ) ) = 0 for t = 2
and other wise the branches are negatives of each other.
Now clearly
t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
is the positive branch.
So we can continue :
v = t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
Now lets return to the original equation and plug in the above identity for v !!
v and t > 0 ,
exp(v/t) - exp ( - (e-1) v/t ) = v
exp( t ln( 1/2 ( t + sqrt(t^2 - 4) ) ) / t ) - exp( - (e-1) t ln( 1/2 ( t + sqrt(t^2 - 4) ) ) / t ) = t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
Now t =/= 0 so we can cancel the t/t parts !
Also we can remove the first ln since exp(ln(x)) = id(x)
1/2 ( t + sqrt(t^2 - 4) ) - exp( - (e-1) ln( 1/2 ( t + sqrt(t^2 - 4) ) ) ) = t ln( 1/2 ( t + sqrt(t^2 - 4) ) )
solving for t > 0 we get
t = 2
or
t = 2.44369396324203812176211410..
if my computation is correct.
( probably , pretty certain about 2.443693963242038121762114 , less about the last 2 digits )
1/t = 0.40921654472367086504976586583879
This means v = 1.59907...
and our lowest base , The Low Base Constant (LBC)
LBC =
exp(1/t) = 1.5056377230919...
by comparison the original 2sinh had exp(1/2) = 1.6487... and eta = exp(1/e) = 1.4446..
and ( exp(1/2) + exp(1/e) )/2 = 1.546694..
so this LBC base constant is closer to eta than to the lowest base for 2sinh.
A big improvement.
(ok ok , i did not compute the exact value for the lowest base possible for the original 2sinh method but took the best case exp(1/2), ignoring fixpoints
nevertheless since the actual base would be even larger the conclusion certainly follows ! )
So in conclusion
exp(x) - exp( - (e-1) / x) works for all bases >> 1.5056377230919...
The Low Base Constant :
LBS = 1.5056377230919...
It took me some time doing things by hand.
This number 1.5056... looks familiar ...
Did I compute it before ?
Did it occur he before ?
Do we know a closed form ?
I remember estimating the lowest base around 3/2 about 12 years ago or so ... which is close.
Sorry for not getting into details.
I need to sleep.
Comments are welcome.
regards
tommy1729
Tom Marcel Raes
" Truth is what does not go away when you stop believing in it "
tommy1729
