I have a recursive function that is defined like that:
$ x(\phi_0) = l_0$
$ x(\phi + \epsilon) = \frac{x(\phi)sin(C - \epsilon)}{sin(C)} $,
where $C$ is some constant, $\epsilon \to 0$
How can I find the explicit function $ x(\phi) $, which evaluates $ x $ for an arbitray $ \phi > l_0 $ ?
I have tried calculating $x$ by iterating with small $ \Delta\phi $ and I've got a good approximation, but I need to do it independently of the step size
The idea is to produce a differential equation from the given "recurrence" (which is not a recurrence, by the way). By subtracting a $x(\phi)$ and then dividing by $\epsilon$ we get
$$\frac {x(\phi + \epsilon) - x(\phi)} {\epsilon} = \frac {x(\phi) \sin (C - \epsilon) - x(\phi) \sin C} {\epsilon \sin C}$$
whence, by taking $\epsilon \to 0$ and making the change of variable $t = C - \epsilon$, we get
$$x'(\phi) = \frac {x(\phi)} {\sin C} \lim _{\epsilon \to 0} \frac {\sin (C - \epsilon) - \sin C} {\epsilon} = - \frac {x(\phi)} {\sin C} \lim _{\epsilon \to 0} \frac {\sin (C - \epsilon) - \sin C} {(C - \epsilon) - C} = \\ - \frac {x(\phi)} {\sin C} \lim _{t \to C} \frac {\sin t - \sin C} {t - C} = - \frac {x(\phi)} {\sin C} (\sin ' C) = - \frac {x(\phi)} {\sin C} \cos C = - x(\phi) \cot C$$
which can be rearranged as
$$\frac {x'(\phi)} {x(\phi)} = -\cot C .$$
Integrating this gives $\log |x(\phi)| = -\phi \cot C + K$ (with $K \in \Bbb R$ an integration constant), whence $x(\phi) = \Bbb e ^{- \phi \cot C} L$ with $L >0$ a constant. Evaluating in $\phi_0$ we get $x(\phi_0) = \Bbb e ^{- \phi_0 \cot C} L = l_0$, so $L = \Bbb e ^{\phi_0 \cot C} l_0$, whence finally $x(\phi) = \Bbb e ^{(\phi_0 - \phi) \cot C} l_0$.