I'm interested in putting an asymptotic bound on $M$ (preferebly as tight as possible) where $\forall x \in \{0,1,..., N-1\}$:
$$\vert e^{i\sqrt{x}} - \sum_{m=-M}^M c_m e^{i\frac{2\pi m}{N}x}\vert \leq \epsilon$$
Calculating the coefficients $c_i$ through the definition gives me:
$$c_j = \frac{1}{N} \int_0^N e^{i(\sqrt{x}-\frac{2\pi j}{N}x)} dx = $$
$${\DeclareMathOperator{\erf}{erf}} \left[\frac{(-1)^{3/4} \sqrt{2} \sqrt{N} e^{\frac{i N}{8 \pi j}} \erf\left(\frac{(1+i)(N - 4 \pi j \sqrt{x})}{4\sqrt{\pi} \sqrt{j} \sqrt{N}}\right) + 4 i \sqrt{j} e^{i (\sqrt{x} - \frac{2\pi j x}{N})}}{8 \pi j^{3/2}}\right]_{x=0}^{x=N}$$
Which reduces to:
$$ \frac{(-1)^{3/4} \sqrt{2} \sqrt{N} e^{\frac{i N}{8 \pi j}} }{8 \pi j^{3/2}} \left(\erf\left( \frac{(1+i)\sqrt{N}}{4\sqrt{\pi} \sqrt{j}}-\sqrt{\pi} \sqrt{j}\right) - \erf\left( \frac{(1+i)\sqrt{N}}{4\sqrt{\pi} \sqrt{j} }\right)\right) + \frac{i}{2 \pi j} (e^{i\sqrt{N}}-1)$$
Where erf is the error function, I'm not sure how to bound the above, but when I look at norms of $\vert c_i \vert$ numerically using the inverse DFT, it's clear that they decay very fast:
How do I show this analytically? Any help on this would be greatly appreciated!