As noted in the OP, $\sin (d/dx) = (\exp (id/dx) - \exp (-id/dx))/(2i)$, which casts the operator as a combination of two shift operators, $$ \sin (d/dx) = \frac{1}{2i} (f(x+i) - f(x-i)) $$ The convergence radius of the Taylor expansion of $f$ around $x$ will have to include $x+i$ and $x-i$.

As noted in the OP, $\sin (d/dx) = (\exp (id/dx) - \exp (-id/dx))/(2i)$, which casts the operator as a combination of two shift operators, $$ \sin (d/dx) f(x) = \frac{1}{2i} (f(x+i) - f(x-i)) $$ The convergence radius of the Taylor expansion of $f$ around $x$ will have to include $x+i$ and $x-i$.

As noted in the OP, $\sin (d/dx) = (\exp (id/dx) - \exp (-id/dx))/(2i)$, which casts the operator as a combination of two shift operators, $$ \sin (d/dx) = \frac{1}{2i} (f(x+i) + f(x-i)) $$

As noted in the OP, $\sin (d/dx) = (\exp (id/dx) - \exp (-id/dx))/(2i)$, which casts the operator as a combination of two shift operators, $$ \sin (d/dx) = \frac{1}{2i} (f(x+i) - f(x-i)) $$ The convergence radius of the Taylor expansion of $f$ around $x$ will have to include $x+i$ and $x-i$.