I'm interested in finding the Taylor expansion of $e^{is\sqrt{x+1}}$ with $x, s \in \mathbb{R}$, say for a branch cut at $x = 0$. More specifically I'm interested in putting a bound on the sum of the 1-norm of the coefficients. That is, if I have:
$$e^{is\sqrt{x+1}}=\sum_{n=0}^\infty a_n x^n$$
How can I bound $\sum_{n=0}^\infty \vert a_n\vert$ or even $\sum_{n=0}^M \vert a_n\vert$? From what I've gathered, each $a_n$ is determined using Bell's polynomials, but I'm really not sure how to deal with the Bell Polynomials.
Any help would be greatly appreciated! I would also be interested in being able to do the above in general for $e^{isf(x)}$.
Not sure that this is what you want but one can use Hardy inequality which says that if $g(z)=\sum_{n \ge 0} b_nz^n \in H^1$ then $\sum_{ n \ge 0}\frac {|b_n|}{n+1} \le \pi ||g||_1$
(where as usual $||g||_1=\frac{1}{2\pi}\int_{-\pi}^{\pi}|g(e^{i\theta})|d\theta$, where by a slight abuse of notation we denote by $g(e^{i\theta})$ the radial limit of $g$ which exists ae and is integrable on the unit circle by the assumption $g \in H^1$)
Taking $f(z)=e^{is\sqrt{z+1}}, g(z)=f'(z)=\frac{is}{2\sqrt {z+1}}e^{is\sqrt{z+1}}$ (where $f(0)=e^{is}$ to fix the principal square root branch in the unit disc) we notice that $g(z)=\frac{c(z)}{\sqrt {z+1}}$ with $c(z)$ continuous in the closed unit disc (since $\sqrt {1+z}$ is so and actually has an absolutely convergent Taylor series in the closed unit disc by easy estimates) hence $\frac{c(z)}{\sqrt {z+1}} \in H^{2-\epsilon}$ so in particular in $H^1$
Applying Hardy inequality to $g$ we get that if $f(z)=e^{is\sqrt{z+1}}=\sum_{n=0}^\infty a_n z^n$, so for $g(z)=\sum_{n=0}^\infty b_n z^n, a_{n+1}=\frac{b_n}{n+1}, n \ge 0$ we have that
$$\sum_{n \ge 1} |a_n| \le \pi ||g||_1$$ so $$\sum_{n \ge 0} |a_n| \le e^{-\Im s}+\pi ||g||_1$$
Now on the unit circle $z=e^{i\theta},1+z=2\cos \frac{\theta}{2}e^{i\theta/2}, \theta \in [-\pi, \pi]$ hence $ \sqrt {1+z}=\sqrt {2\cos \frac{\theta}{2}} e^{i\theta/4}$ by our principal branch choice for which $\sqrt {z+1}=2, z=1$ so one can find the maximum $c(s)$ of $|e^{is\sqrt{z+1}}|$ in terms of $\arg s$ so one can majorize $||g||_1$ by using $||g||_1=\frac{1}{2\pi}\int_{-\pi}^{\pi}|g(e^{i\theta})|d\theta \le \frac{|s|c(s)}{2}\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{d\theta}{\sqrt {2\cos \frac{\theta}{2}}}< 2|s|c(s)/\pi$ since the integral in $\theta$ is $7.41..$ (has actually a closed expression using $\Gamma$ function )
For example, if $s$ is real one has $\Re (is \sqrt {z+1})=-\sqrt {2\cos \frac{\theta}{2}}s\sin \theta/4$ and by symmetry, we can take $s$ negative so majorize $\sqrt {2\cos \frac{\theta}{2}}\sin \theta/4, 0 \le \theta \le \pi$ and find $c(s)=e^{c|s|}$ where $c$ is the maximum above, while if $s=ia, a \in \mathbb R$ one easily sees that $c(s)=1$ if $a>0$ and $c(s)=e^{|a|\sqrt 2}$ if $a<0$ so the final result will definitely depend on the argument of $s$
If one needs, one can definitely analyze $||g||_1$ more carefully but I suspect that the general form of the bound $K|s|e^{h(|s|,\arg s)}$ is the best one can get from this approach
Edit later based on the condition that $s$ is real; as noted above in this case the crude bound one gets is $2|s|e^{c|s|}$ where $c$ is the maximum of $\sqrt {2\cos \frac{\theta}{2}}\sin \theta/4, 0 \le \theta \le \pi$; a quick computation gets us that $c=1/2, \theta=2\pi/3$ so the crude bound is $2|s|e^{|s|/2}$. Now the exponential part can not be changed since the integrand will be close to this on a large enough arc near $\pm 2\pi/3$, but for large $|s|$ one probably can reduce the power of $s$ by a more careful estimate of the original integral.