A nice exercise is perhaps this
Let f(f(x)) = exp(x)
for all x :
f ' (x) > 0
f ' ' (x) > 0
what is the lowest possible value of A ;
A = inf f(x)
or equivalent
A = f( - oo )
Does A have a closed form ?
That is a nice idea.
***
Following the above
consider the family of functions f_r(x) such that :
f_r^[r](x) = exp(x)
with r between 0 and 1.
And consider
for all x :
f_r ' (x) > 0
f_r ' ' (x) > 0
what are the lowest possible values of A( r ) ;
A( r ) = inf f_r(x)
or equivalent
A( r ) = f_r( - oo )
It is assumed that the lowest possible value A® for a given r is independant of the lowest possible values for other values of a different r.
This implies that there is a unique lowest function A of the variable r without contradictions : A( r ).
A( r ) = f_r(- oo) = sexp( slog(-oo) + r)
f_r ' (x) = d/dx sexp( slog(x) + r)
f_r ' ' (x) = d^2/d^2x sexp( slog(x) + r)
We then have a uniqueness condition ; A® uniquely defines sexp resp. slog by the above ( and the trivial sexp(x+1) = exp(sexp(x)) )
Food for thought.
regards
tommy1729
Let f(f(x)) = exp(x)
for all x :
f ' (x) > 0
f ' ' (x) > 0
what is the lowest possible value of A ;
A = inf f(x)
or equivalent
A = f( - oo )
Does A have a closed form ?
That is a nice idea.
***
Following the above
consider the family of functions f_r(x) such that :
f_r^[r](x) = exp(x)
with r between 0 and 1.
And consider
for all x :
f_r ' (x) > 0
f_r ' ' (x) > 0
what are the lowest possible values of A( r ) ;
A( r ) = inf f_r(x)
or equivalent
A( r ) = f_r( - oo )
It is assumed that the lowest possible value A® for a given r is independant of the lowest possible values for other values of a different r.
This implies that there is a unique lowest function A of the variable r without contradictions : A( r ).
A( r ) = f_r(- oo) = sexp( slog(-oo) + r)
f_r ' (x) = d/dx sexp( slog(x) + r)
f_r ' ' (x) = d^2/d^2x sexp( slog(x) + r)
We then have a uniqueness condition ; A® uniquely defines sexp resp. slog by the above ( and the trivial sexp(x+1) = exp(sexp(x)) )
Food for thought.
regards
tommy1729

