Assume a curve represented by a function $f(b)=0.5(\sqrt{N-b^2}-b+1)$ with $1\leq b \leq \sqrt{\frac{N}{2}}$. I want to count the lattice points below this curve, more specifically, I would like a closed form expression. I can assume this figure to be a parabola and try curve fitting and then use the method given in this paper to sort of get an approximation. But I am not convinced. Can someone please offer me a better solution?
Thanks
If I understand what you are asking correctly, then the number of lattice points below the curve can be treated as a sort of "discrete integral".
Given that the number of lattice points below a point $(b,f(b))$ is, at most $f(b)$, the number of lattice points under the curve from $1$ to $\left\lfloor\sqrt\frac{N}{2}\right\rfloor$ could be obtained from summing the number of lattice points below each $f(n)$ for $n=1,\ldots,\lfloor \sqrt{\frac{N}{2}}\rfloor$
$$F(b)=\sum_{n=1}^{\left\lfloor \sqrt{\frac{N}{2}}\right\rfloor} \lfloor f(n)\rfloor$$
Where $F(b)$ coincides with the number of lattice points and $\lfloor\cdot\rfloor$ is the flooring function.
Edit: Okay, it seems that I messed up while entering this into Wolfram. After working it out by hand, the closest approximation to the graph of $F(b)$ is a parabola. However the closed form of the summation (if it exists) cannot be a polynomial of order less than 3. This is because the flooring function in the summation makes the graph asymetrical about the vertical line through the maximum for most $N$.