# A Mathematical Allegory

We have defined function $\arcsin$ as a set:

$\{(x, y) | \sin(y) =x, \frac{-\pi}{2} \le y \le \frac{\pi}{2}\}.$

By definition,

$\arcsin(-1) = \frac{-\pi}{2}, \arcsin(0)=0, \arcsin(1) = \frac{\pi}{2}$

and

$\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}}.$

It means that $\arcsin(x)$ is the unique solution of

$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\quad\quad\quad(\star)$

where $y(-1)=-\frac{\pi}{2}, y(0)=0$ and $y(1)=\frac{\pi}{2}.$

To compute $\arcsin(x)$, we solve $(\star)$ for $y(x)$ as follows:

Integrate $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$ from $-1$ to $x$ gives

$\displaystyle\int\limits_{-1}^{x}\frac{dy}{dx} \,dx=\displaystyle\int\limits_{-1}^{x}\frac{1}{\sqrt{1-\xi^2}}\; d\xi\overset{\textbf{FTC}}{\implies}y(x) - y(-1) = \displaystyle\int\limits_{-1}^{x} \frac{1}{\sqrt{1-\xi^2}}\, d\xi.$

Therefore,

$y(x) = \displaystyle\int\limits_{-1}^{x}\frac{1}{\sqrt(1-\xi^2}\,d\xi + y(-1) \overset{y(-1)=\frac{-\pi}{2}}{=} \displaystyle\int\limits_{-1}^{x}\frac{1}{\sqrt{1-\xi^2}}\,d\xi - \frac{\pi}{2}.$

That is,

$\arcsin(x) = \displaystyle\int\limits_{-1}^{x} \frac{1}{\sqrt{1-\xi^2}}\;d\xi - \frac{\pi}{2}.$

To obtain $\arcsin(x), -1 < x < 1$, we numerically evaluate $\int\limits_{-1}^{x}\frac{1}{\sqrt{1-\xi^2}}\,d\xi$, using function ‘quad_qags’.

Fig. 1

The result is visually validated in Fig. 2.

Fig. 2

Note: ‘romberg’, another function that computes the numerical integration by Romberg’s method will not work since it evaluates $\frac{1}{\sqrt{1-x^2}}$ at $x=-1.$

Fig. 3

An alternate approach is to solve $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}, y(0)=0$ as an initial-value problem of ODE using ‘rk’ , the function that implements the classic Runge-Kutta algorithm.

Fig. 4 for $0 < x < 1$

Fig. 5 $-1

Putting the results together, we have

Fig. 6 $-1

However, we cannot solve $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}, y(-1)=\frac{-\pi}{2}$ using ‘rk’:

Fig. 7

Exercise-1 Compute $\arccos(x)$ for $x \in (-1, 1)$.

Exercise-2 Explain why ‘rk’ cannot solve $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}, y(-1)=\frac{-\pi}{2}$.