Evaluating a polynomial that includes a little-o constant exponent

Question

In a paper I am reading, the following assertion is made:
N is an odd, positive composite number.
Let $I_1 = [\sqrt N - F^{2+o(1)} , \sqrt N + F^{2+o(1)}]$ hold the values of $x$ passed into the polynomial $Q(x)=x^2 - N$. Then the maximum absolute value of the $Q(x)$, for $x\in I$ is: $\sqrt NF^{2+o(1)}$.

EDIT: F is a smoothness bound.

How is this final value obtained?
When I pass $x = (\sqrt N + F^{2+o(1)}$) (which I believe to be the maximum value in $I$) into $Q$, I come up with $2\sqrt NF^{2+o(1)} + (F^{2+o(1)})^2$.

Answer 1

I assume the two following hypotheses were made (we don't actually know what $F$ is from your post):

• $F$ satisfies $F>1$;
• $F = o(N^{1/4})$;
• the asymptotics are taken with regard to $F\to\infty$, and $F$ is allowed to be a function of $N$.

If so, you get $$ 2\sqrt{N} F^{2+o(1)} + F^{4+o(1)} = 2(1+o(1))\sqrt{N} F^{2+o(1)} $$ and $$ 2(1+o(1))\sqrt{N} F^{2+o(1)} = \sqrt{N} F^{2+\frac{\ln (2(1+o(1)))}{\ln F}+o(1)}= \sqrt{N} F^{2+o(1)}. $$

In short: the $o(1)$ hides a lot.

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Edit: after looking at the paper (Section 1.1) [1]:

Indeed the $o(1)$ is taken with regard to $N\to\infty$ (or equivalently $F\to\infty$, as $F=L[\frac{1}{2},\frac{1}{2}] = \exp((\frac{1}{2}+o(1))\sqrt{\log N \log\log N})$ which goes to $\infty$ with $N$). Note also that $F$ is subpolynomial in $N$, i.e. $F=N^{o(1)}$ (asymptotically negligible in front of $N^\epsilon$ for any fixed $\epsilon > 0$).

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[1] Quadratic sieving, Thorsten Kleinjung. Math. Comp. 85 (2016), 1861-1873, 2015. DOI http://dx.doi.org/10.1090/mcom/3058