09/09/2022, 02:37 AM
James is probably correct but im still a bit confused or unsatisfied.
one of the ideas is that g(-x) is the same function as g(x) for x close to 0.
Taylor expansion implies that ... but the radius is 0.
On the other hand i guess analysing g(-x) or g(x) should be the same since there is only ONE taylor expansion at 0.
then again there are two petals.
hmm
I prefer to positive side and modified laplace.
maybe im being silly or tired lol.
modified borel : f(x) = a_n x^n becomes a_n x^n / ( v^n n! )
for some v > 0.
B is modified borel , L is modified laplace
lets see
LB(g(x)) = modified borel summation g(x) =
integral from 0 to +oo
exp(-v t) B(g(x t) ) dt
this integral converges IFF
B(g(x t)) is bounded by C exp(v t).
Now g(x t) =< exp(x t) - 1.
also for some v :
B( g(x t) ) =< g(x t)
thus
B ( g(x t) ) =< exp(x t) - 1.
so if
exp(x t) - 1 < C exp(v t)
then
B(g(x t)) is bounded by C exp(v t) as desired.
In general beyond a certain size , the larger v is the smaller B( g(x t) ) becomes and the larger C exp(v t).
at the boundary we have
B(g(x t)) is asymptotic to C exp(v t).
which holds as v and x are about equal , lets call that positive number S.
so if x < S then B(g(x t)) is bounded by C exp(v t) as desired.
and that x implies a positive radius for B( g(x) ).
Also B( g(x t) ) for x > S is still smaller than exp(x t) -1.
hence
B ( g(x t) ) =< exp(x t) - 1.
and
exp(x t) - 1 < C exp(v t) for some C EQUAL OR LARGER THAN 1.
if x < v or x = v.
now take
v/10 < x < v
this works.
QED
someting like that.
thing is that B ( g(x t) ) might still be a divergent series at 0.
however when considered as expanded at x t > 0 then g or B(g) are not divergent anymore !
This however requires to show
B (g(A)) = B(g(Z))
in other words the expansion point of B(g) should not matter and be the same taylor... by continuation.
this is true for all expansion points larger than 0 because analytic functions have analytic borels.
but at 0 it is problematic.
this is somewhat running in circles ...
need to think
regards
tommy1729
one of the ideas is that g(-x) is the same function as g(x) for x close to 0.
Taylor expansion implies that ... but the radius is 0.
On the other hand i guess analysing g(-x) or g(x) should be the same since there is only ONE taylor expansion at 0.
then again there are two petals.
hmm
I prefer to positive side and modified laplace.
maybe im being silly or tired lol.
modified borel : f(x) = a_n x^n becomes a_n x^n / ( v^n n! )
for some v > 0.
B is modified borel , L is modified laplace
lets see
LB(g(x)) = modified borel summation g(x) =
integral from 0 to +oo
exp(-v t) B(g(x t) ) dt
this integral converges IFF
B(g(x t)) is bounded by C exp(v t).
Now g(x t) =< exp(x t) - 1.
also for some v :
B( g(x t) ) =< g(x t)
thus
B ( g(x t) ) =< exp(x t) - 1.
so if
exp(x t) - 1 < C exp(v t)
then
B(g(x t)) is bounded by C exp(v t) as desired.
In general beyond a certain size , the larger v is the smaller B( g(x t) ) becomes and the larger C exp(v t).
at the boundary we have
B(g(x t)) is asymptotic to C exp(v t).
which holds as v and x are about equal , lets call that positive number S.
so if x < S then B(g(x t)) is bounded by C exp(v t) as desired.
and that x implies a positive radius for B( g(x) ).
Also B( g(x t) ) for x > S is still smaller than exp(x t) -1.
hence
B ( g(x t) ) =< exp(x t) - 1.
and
exp(x t) - 1 < C exp(v t) for some C EQUAL OR LARGER THAN 1.
if x < v or x = v.
now take
v/10 < x < v
this works.
QED
someting like that.
thing is that B ( g(x t) ) might still be a divergent series at 0.
however when considered as expanded at x t > 0 then g or B(g) are not divergent anymore !
This however requires to show
B (g(A)) = B(g(Z))
in other words the expansion point of B(g) should not matter and be the same taylor... by continuation.
this is true for all expansion points larger than 0 because analytic functions have analytic borels.
but at 0 it is problematic.
this is somewhat running in circles ...
need to think
regards
tommy1729