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