To give An example that should work.
F(x) = Sum a_n x^n
With a_n = exp(-n^2)
To compute the fake a_n we consider
F_n(x) = Sum^n a_i x^i
Now we solve
(Alpha x^2)^[b] = a_n x^n
So alpha is close to a_n ^ (log_2(n-1))^(-1).
Or alpha ~ exp( - n^2 / log_2(n-1) )
And b ~ ln(n-1)/ln(2).
--
Max_x Integral_0^x exp(- t^2) exp(- (x-t)^2 ) dt =
Exp(- x^2/2 ) C.
Where C is a constant ( upper bound constant with respect to x ).
Therefore the correcting factor for taking sqrt is C.
And thus the correcting factor for taking r th root is C^log_2{r}.
Or r^( ln{C}\ln(2) ).
From the computation of b we get max{r} ~ 2^b.
Therefore D_n = (n-1)^( ln{C}\ln(2) )
Where D_n is An upper bound on the (final) correcting factor for a_n.
--
So we should have
Exp(- n^2) =< D_n min( f(x) / x^n ).
--
Something like that.
It looks a bit like using convolution and fourrier analysis , but its different.
Regards
Tommy1729
F(x) = Sum a_n x^n
With a_n = exp(-n^2)
To compute the fake a_n we consider
F_n(x) = Sum^n a_i x^i
Now we solve
(Alpha x^2)^[b] = a_n x^n
So alpha is close to a_n ^ (log_2(n-1))^(-1).
Or alpha ~ exp( - n^2 / log_2(n-1) )
And b ~ ln(n-1)/ln(2).
--
Max_x Integral_0^x exp(- t^2) exp(- (x-t)^2 ) dt =
Exp(- x^2/2 ) C.
Where C is a constant ( upper bound constant with respect to x ).
Therefore the correcting factor for taking sqrt is C.
And thus the correcting factor for taking r th root is C^log_2{r}.
Or r^( ln{C}\ln(2) ).
From the computation of b we get max{r} ~ 2^b.
Therefore D_n = (n-1)^( ln{C}\ln(2) )
Where D_n is An upper bound on the (final) correcting factor for a_n.
--
So we should have
Exp(- n^2) =< D_n min( f(x) / x^n ).
--
Something like that.
It looks a bit like using convolution and fourrier analysis , but its different.
Regards
Tommy1729

