$$f(x)=\frac{x}{2}-\frac{\sin(200πx)}{400π}$$
I'm not a mathematician. I'm making a heater which uses this formula to calculate the watt of given x. The problem is I need the inverse of this function between (0 and 0.01) to use in the PID algorithm. I couldn't get an inverse of it, so I asked Wolfram Alpha, which told me it doesn't have a closed mathematical form. I know that functions can be represented as Taylor series, so is it possible to get the inverse of this in terms of Taylor Series?
$$f(x)=\frac{x}{2}-\frac{\sin(200\pi x)}{400\pi}$$
Let $t=200\pi x$ and $y=400\pi f(x)$ to face $$y=t-\sin(t) \quad \text{with}\quad 0 \leq t \leq 2\pi \quad \text{and}\quad 0 \leq y \leq 2\pi$$ Expanding as a Taylor series $$y=\pi+2(t-\pi)+\sum_{n=1}^\infty \frac {(-1)^n}{(2n+1)!} (t-\pi )^{2 n+1}$$
Using power series reversion, then $$t=\pi+\sum_{n=0}^\infty a_n\, (y-\pi)^{2n+1}$$ where the first coefficients are $$\left\{\frac{1}{2},\frac{1}{96},\frac{1}{1920}, \frac{43}{1290240},\frac{223}{92897280}, \frac{60623}{32699842560 0},\cdots\right\}$$
Their numerators and denominators form respectively sequences $A362407$ and $A362406$ in $OEIS$.
To give an idea of the accuracy, consider the truncated series
$$S_p=\pi+2(t-\pi)+\sum_{n=1}^p \frac {(-1)^n}{(2n+1)!} (t-\pi )^{2 n+1}$$ and the corresponding infinite norm $$\Phi_p=\int_0^{2pi} \Big(y-S_p\Big)^2\, dt$$
$$\left( \begin{array}{cc} p & \Phi_p \\ 4 & 1.3232\times 10^{-5} \\ 5 & 4.6409\times 10^{-8} \\ 6 & 9.0993\times 10^{-11} \\ 7 & 1.0753\times 10^{-13} \\ 8 & 4.0580\times 10^{-17} \\ 9 & 2.0408\times 10^{-20} \\ \end{array} \right)$$
