How to calculate the inverse of $f(t) = \frac{(k/2)(\sin(t) + t) - t}{2π (k/2 - 1)}$?

Question

This function arises from a specific problem in optical engineering. I am modeling a light source whose amplitude is being modulated sinusoidally. I have derived this function which describes the cumulative quantity of light emitted over time:

$$f(t) = \frac{\left(\frac{k}{2}\right)(\sin(t) + t) - t}{2π\left(\frac{k}{2} - 1\right)}$$

where $k$ is a constant modulation depth, and $0 ≤ k ≤ 1$.

EDIT: This simplifies further to

$$f(t) = \frac{1}{2π}\left[\frac{k}{k-2}t + \sin(t)\right]$$

but I think that this does not affect my questions.

This function is also normalized, so that one period of oscillation delivers one measure of light, independent of $k$.

Here's a graph of $f(t)$ at various modulation depths over the interval $[0, 4π]$ to make the normalization more apparent.

Now, to model photon flux as a time series, I will need $f^{-1}(t)$. I need a general solution for all $t$.

EDIT: I am OK with a series approximation. See immediately below, I discuss $\arcsin$, which does not have an analytic solution.

I have searched for possible ways to derive my answer from the known $\arcsin$ series expansion. I think this may not be possible. Does a shortcut exist?

Alternately, I think I might have to apply the Lagrange Inversion Theorem. This theorem is new to me, but I suppose it is how the series for the inverse trig functions were derived in the first place. The Wikipedia entry jumps straight into the deep end of the pool, and I am unsure how to start. Are there simpler explanations of the Lagrange Inversion Theorem available, and/or examples of other functions being inverted that I might study?

Thanks for your suggestions.

Answer 1

You function can be reduced to $$ f(t) = w \sin(t) + v t $$ this function as shown in wolfran alpha does no have explicit inverse (or at least their inverse is a hell to derivate). And by Langrange inversion thoerem we get (If I didn't make astupid mistakes)

$$ t = t_0 + \sum_{n=1}^{\infty} \frac{(f(t) - f(t_0))^n}{(n + 1)!} v^{-n} $$ where $t_0$ is the offset

Answer 2

Rewriting the equation as $$f(t)=\frac{(2-k) t-k \sin (t)}{2 \pi (2-k)}$$ using the trivial bounds of the sine function, we have $$\frac{(2-k) t-k }{2 \pi (2-k)} \leq f(t) \leq \frac{(2-k) t+k }{2 \pi (2-k)}$$ which gives as bounds $$2\pi\, f(t)-\frac k{2-k} \leq t \leq 2\pi\, f(t)+\frac k{2-k}$$ One iteration of Newton method would give $$t_1=2 \pi f(t)-\frac{k \sin (2 \pi f(t))}{k \cos (2 \pi f(t))-(2-k)}$$

Trying for $f=1.75$ and $k=0.75$ would give, as an estimate, $t=\frac{7 \pi }{2}-\frac{3}{5}=10.3956$ while the solution given by Newton method is $t=10.4750$.