(sin )^(-1)(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]

1 Answer
Jul 10, 2018

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