Question

`(sin )^(-1)` is the piecewise-wholesome inverse sine operator. The FCS y = `(sin )^(-1)(x+(sin )^(-1)(x+(sin )^(-1)(x+...)))`. How do you find the amplitude and the period, of this FCS wave?

Definition of `(sin)^(-1)X`:
`Y = (sin)^(-1)X = kpi + (-1)^ksin^(-1)X, k = 0, +-1, +-2, +-3, ..., Y in [kpi-pi/2, kpi+pi/2]`

Answer 1

Axis of the wave: x + y = 0. Wave is bounded by the parallels `x + y = +- 1`. Easily the amplitude is half of the width, `1/sqrt 2` and the period is `2sqrt2pi`.

Explanation:

Definition of piecewise-wholesome `(sin)^(-1)X`:

`Y = (sin)^(-1 )X = kpi + (-1)^k sin^(-1) X,`

`k = 0, +-1, +-2, +-3, ..., Y in [ k pi-pi/2, k pi + pi/2 ]`.

https://socratic.org/questions/on-what-interval-is-the-identity-sin-1-sin-x-x-valid#639442

The FCS wave, with axis `x + y = 0` and demarcation, marked by

`x + y = +- 1`, is created by

`x = sin y - y`. See uniform-scale graph. Observe the points of

inflexion aligned in x + y = 0 and crests and troughs aligned in

`x + y = +- 1`.
graph{(x-sin y + y)(x+y)((x+y)^2-1)=0}

Graph of just one wave from `y = sin^(-1)( x + y )` instead:

graph{(y-arcsin (x+y))(x+y)((x+y)^2-1)=0} .

The points of the meet with the axis are

`+- ( k pi, - k pi ), k = 0, +-1, +-2, +-3, ...`

The amplitude = half-width between `x + y = +- 1` is `1/sqrt2`.

Period = 2 ( distance between two consecutive axial points)

`= 2sqrt2 pi`.

Plots revealing a crust, a trough, period length and width.

graph{(x-sin y + y)(x-y)((x+y)^2-1)((x-3.14)^2+(y+3.14)^2-.01)((x+3.14)^2+(y-3.14)^2-.01)((x+0.57)^2+(y-1.57)^2 -.01)((x-0.57)^2+(y+1.57)^2 -.01)((x-0.5)^2 +(y-0.5)^2-.01)((x+0.5)^2 +(y+0.5)^2-.01)=0[-7 7 -3.5 3.5]}

Also see

https://socratic.org/questions/defining-the-wholesome-inverse-operator-sin-1-by-y-sin-1-x-k-pi-1-k-sin-1-x-y-in#63919