09/04/2017, 11:12 PM
The context is asymptotics for real-analytic strictly rising f(x) , as x grows to + oo.
Let f(x) grow (asymptoticly) much faster than any polynomial.
So for large x
f(x) >> exp(a ln(x)) for any fixed a > 0.
But Also for large x we have
f(x) << exp^[h](x) for any h > 0.
Now if f(x) does grow faster than any elementary function can describe Then f(x) grows faster than
Exp^[k]( a ln^[k](x) ) for any fixed a >1 , k > 0.
Or equivalently
f(x) >> Exp^[k]( ln^[k](x) + a). For a,k > 0.
Combining
Exp^[k]( ln^[k](x) + a) << f(x) << Exp^[h](x) for any a,k,h > 0.
Now you might have assumed that such an f(x) does not exist.
But it does.
This fascinates me.
I think this deserves attention.
And ofcourse a " fake ".
Here is a possible solution I came up with while in kindergarten
f(x) = sexp^[1/2]( slog^[1/2](x) + 1 ).
So I encourage all investigations into this.
Regards
tommy1729
Let f(x) grow (asymptoticly) much faster than any polynomial.
So for large x
f(x) >> exp(a ln(x)) for any fixed a > 0.
But Also for large x we have
f(x) << exp^[h](x) for any h > 0.
Now if f(x) does grow faster than any elementary function can describe Then f(x) grows faster than
Exp^[k]( a ln^[k](x) ) for any fixed a >1 , k > 0.
Or equivalently
f(x) >> Exp^[k]( ln^[k](x) + a). For a,k > 0.
Combining
Exp^[k]( ln^[k](x) + a) << f(x) << Exp^[h](x) for any a,k,h > 0.
Now you might have assumed that such an f(x) does not exist.
But it does.
This fascinates me.
I think this deserves attention.
And ofcourse a " fake ".

Here is a possible solution I came up with while in kindergarten
f(x) = sexp^[1/2]( slog^[1/2](x) + 1 ).
So I encourage all investigations into this.
Regards
tommy1729