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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

RTn
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  • Important question: are you counting lattice points with $f(b)<0$ as negative? – R. Burton Mar 13 '19 at 17:19
  • No. In the interval $1\leq b \leq \sqrt{N/2}$, $f(b)$ is always positive. The choice of $N$ is thus made to ensure that. – RTn Mar 13 '19 at 17:21

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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 $N$.

R. Burton
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  • Well, indeed. But I don't get how you arrived at the Expression for $F(b)$. Assume $N=100$, $1\leq b \leq 6$, then the number of lattice points is $37$. For $b=6$, $f(b)=3$ and so according to your formula, $F(b)=21-\lfloor f(1)\rfloor$. This seems not right.. Am I missing something here? – RTn Mar 13 '19 at 16:58
  • Sorry, I forgot to add that the summation should run from your lower to upper bound, otherwise you have to take the difference, just like taking a definite integral. – R. Burton Mar 13 '19 at 17:02
  • I was wondering if there is a closed form expression for this instead of a summation. Like that which exists for a hyperbola using the Dirichlet method. It would be nice to have a closed form expression. – RTn Mar 13 '19 at 17:04
  • If there is a closed form, it will be for the summation $\sum_{n=1}^{\lfloor\sqrt{\frac{N}{2}}\rfloor}\lfloor f(n)\rfloor$. I'll see if I can fix it. – R. Burton Mar 13 '19 at 17:06
  • Oh wonderful. I will modify the original equation a bit to include a 0.5 as a multiplier. I tried taking a finite continuous integral of this function in the said interval and then subtracted the coefficient of $b^2$ that I obtained by curve fitting $f(b)$ to a parabola and it seems to work. The only thing is that it seems a bit arbitrary to do that. – RTn Mar 13 '19 at 17:09