(maybe start a new thread , after this 'mess' ? )
ok , another attempt and a related question.
forgive me for mistakes or ignorance , its still hot here ...
quick vague sketch of the idea :
sexp(x) = a0 + a1/1! x + a2/2! x^2 + a3/3! x^3 + ...
sexp2(x) = sexp(x + v(x)) with v(x) a periodic entire function satisfying v(0) = 0 and v*(x) = v(x*)
proposed uniqueness by :
a0 = 1 , a0 < a1 < a2 < a3 < a4 < ...
a1/a0 > the base , a2/a1 > the base , ...
(assuming existance ! )
the idea is now to show that for any v(x) =/= 0 ( but still with v(0) = 0 ) this is violates for the 'new' sexp2.
v(x) = 0 + v1 x + v2/2! x^2 + ...
now we use the so-called general leibniz rule to get to a system of inequalities :
a1*(1+v1) > 0 => v1 > -1
a2*(1+v1) + a1(v2) > 0
... ( by leibniz rule ) ...
now for large n , a_n < n!
now something important : note that a_n must grow fast.
and thus a_n probably becomes the dominant factor of the inequalities , more specific of the n'th inequality.
unless v_n becomes dominantly small !
since v(x) is periodic , there must be oo n such that v_n is negative.
lets call those v'_m. ( the m'th ) and let v'm = m_n corresponding to v_n.
now dominantly small means appromimately :
lim n -> oo m_n = around = 1/ a_n = little less than = 1/(n-1)!
but is there a v(x) , such that lim n-> oo m_n is around (n-1)! ??
i doubt it.
but such a v(x) is necc to break the uniqueness condition.
i doubt such a v(x) exists because if those negative v_n shrink so fast , then the series seems to converge to fast to be periodic.
for instance bessel functions resemble such v_n , but bessel is not periodic but is 'brutely' approximated by sin(x)/sqrt(x) ...
but of course in math we need formal proof.
this could be trivial even high skool , but right now i cant remember such a thing ...
i guess such a thing isnt hard to prove or disproof , but im having a bad day ...
but maybe a bad day with a good idea ...
notice that my uniqueness criterion reduces to a_n > 1 once we proved that no non-violating entire v(x) exists.
but i could not have started this post with a_n > 1 because i needed to explain the importance and growth rate of a_n and 1/v_n.
i couldnt help noticing that taylor series seems more adequate than fourier series for periodic functions in this context ( for v(x) ) ...
taylor fourier 1-0.
high regards
tommy1729
ok , another attempt and a related question.
forgive me for mistakes or ignorance , its still hot here ...
quick vague sketch of the idea :
sexp(x) = a0 + a1/1! x + a2/2! x^2 + a3/3! x^3 + ...
sexp2(x) = sexp(x + v(x)) with v(x) a periodic entire function satisfying v(0) = 0 and v*(x) = v(x*)
proposed uniqueness by :
a0 = 1 , a0 < a1 < a2 < a3 < a4 < ...
a1/a0 > the base , a2/a1 > the base , ...
(assuming existance ! )
the idea is now to show that for any v(x) =/= 0 ( but still with v(0) = 0 ) this is violates for the 'new' sexp2.
v(x) = 0 + v1 x + v2/2! x^2 + ...
now we use the so-called general leibniz rule to get to a system of inequalities :
a1*(1+v1) > 0 => v1 > -1
a2*(1+v1) + a1(v2) > 0
... ( by leibniz rule ) ...
now for large n , a_n < n!
now something important : note that a_n must grow fast.
and thus a_n probably becomes the dominant factor of the inequalities , more specific of the n'th inequality.
unless v_n becomes dominantly small !
since v(x) is periodic , there must be oo n such that v_n is negative.
lets call those v'_m. ( the m'th ) and let v'm = m_n corresponding to v_n.
now dominantly small means appromimately :
lim n -> oo m_n = around = 1/ a_n = little less than = 1/(n-1)!
but is there a v(x) , such that lim n-> oo m_n is around (n-1)! ??
i doubt it.
but such a v(x) is necc to break the uniqueness condition.
i doubt such a v(x) exists because if those negative v_n shrink so fast , then the series seems to converge to fast to be periodic.
for instance bessel functions resemble such v_n , but bessel is not periodic but is 'brutely' approximated by sin(x)/sqrt(x) ...
but of course in math we need formal proof.
this could be trivial even high skool , but right now i cant remember such a thing ...
i guess such a thing isnt hard to prove or disproof , but im having a bad day ...
but maybe a bad day with a good idea ...
notice that my uniqueness criterion reduces to a_n > 1 once we proved that no non-violating entire v(x) exists.
but i could not have started this post with a_n > 1 because i needed to explain the importance and growth rate of a_n and 1/v_n.
i couldnt help noticing that taylor series seems more adequate than fourier series for periodic functions in this context ( for v(x) ) ...
taylor fourier 1-0.
high regards
tommy1729