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?

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Answer 1 · 2018-07-03T02:29:55

See explanation and the wholesome Socratic graph.

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]}

FCS y = tan_(fcs)( x; -1 ) = tan ( x -tan(x- tan ( x - ... 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'' 33^o 41' 24'' to the x-axis, nearly?