Is there an analytic function $f:\mathbb{C}\to\mathbb{C}$ such that $f(x) = \sin(\sqrt{x})$ Is there an analytic function $f:\mathbb{C}\to\mathbb{C}$ such that $f(x) =
\sin(\sqrt{x})$ for all $x \in \mathbb{R}≥0$ ? What if we replace $\sin$ by $\cos$ ? Can anyone give a hint, thnx for your help
 A: (I).For the sine: For non-negative real $r$ let $\sqrt r$ denote the standard real function.For non-negative real $r$ and for real $t\in (-\pi,\pi)$ let $f_1(r \exp (i t))=\sqrt r \exp (i t/2)$ and $f_2(r\exp (i t))=-\sqrt r \exp (i t/2).$   For $j\in \{1,2\}$ and for non-negative $\Re (x)$ let $g_j(x)=\sin (f_j(x))$. If $g(x)\in \{f_1(x),f_2(x)\}$ whenever $\Re(x)$  (the real part of $x$) is non-negative, then $g(x)$ cannot be continuous everywhere in $C$ no matter what $g(x)$ is for any negative real $x$.PROOF: Consider $S=\{\pi/2 \exp (i t): t\in (-\pi,\pi)\}.$ Suppose $g(\pi/2)=f_k(\pi/2).$   Suppose $g$ is continuous on $S$. Then $\forall x\in S (g(x)=f_k(x)).$  But then, as $x$ converges to $-pi/2$ along $S$ thru values satisfying $\arg (x)\in (0,\pi)$, the value of $g(x)$ converges to a member of $T=\{\pm \sin (i\sqrt {\pi/2})\}$ while if $x$ converges to $-pi/2$ along $S$ thru  values satisfying $\arg (x)\in (-\pi,0)$, the value of $g(x)$ converges to the OTHER member of $T.$ Which makes $g(x)$ discontinuous at $x=-\pi/2.$....(II).For the cosine: Let $h(x)=\sum_0^{\infty}(-x)^n/(2n)!$  Then $h(x)$ and $h(-x)$ are analytic and we could call one of them $\cos \sqrt x$, although the notation $\sqrt x$ is usually avoided in $C$ because of ambiguity and discontinuity.
A: If $f(x) = \sin (\sqrt x), x \ge 0,$ then
$$\frac{f(x) - f(0)}{x-0} = \frac{\sin (\sqrt x)}{x} = \frac{1}{\sqrt x}\frac{\sin (\sqrt x)}{\sqrt x}.$$
As $x\to 0^+,$ the expression on the right $\to \infty,$ hence $f'(0)$ does not exist. So the answer to your first question is no.
For the second question, I would use the power series for the cosine function as D. Ullrich suggested.
