Question

FCS `y = tan_(fcs)( x; -1 ) = tan ( x -tan(x- tan ( x - ...`. How do you prove that, piecewise, at y = 1, the tangent is inclined at `33^o 41' 24''` to the x-axis, nearly?

Answer 1

See explanation and the wholesome Socratic graph.

Explanation:

The FCS generator is `y =tan ( x - y )`. Inversely,

`x = y + kpi + tan^(-1)y, k = 0, +-1, +-2, +-3, ..`.

At `y = 1, x = kpi + pi/4 +1, k = 0, +-1, +-2, +-3, ..= 1.76539,....`.

#(dx)/(dy) = 1 + 1/( 1 + y^2 )

#y' = 1/(dx)/(dy). And so,

`y' = ( 1 + y^2 )/( 2 + y^2 ) = 1 - 1/( 2 + y^2 ) in [ 1/2, 1).`

`= 2/3`, at y = 1, and so,

#the inclination of he tangent to the x-axis is

`tan^(-1)( 2/3 ) = 33.69^o = 33^o 41' 24''`, nearly. See graph for this

tangent `(y-1) = 2/3 (x-1.7654 )`
graph{(y-tan(x-y))((x-1.7654)^2+(y-1)^2-.01)((y-1) -2/3 (x-1.7654 ))=0[-10 6 -3 3]}