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