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Tangent and Hyperbolic Tangent Forward Taylor Polynomial Theory
Derivatives tan ( 1 ) ( u ) = [ cos ( u ) 2 + sin ( u ) 2 ] / cos ( u ) 2 = 1 + tan ( u ) 2 tanh ( 1 ) ( u ) = [ cosh ( u ) 2 - sinh ( u ) 2 ] / cosh ( u ) 2 = 1 - tanh ( u ) 2 If F ( u ) is tan ( u ) or tanh ( u ) the corresponding derivative is given by F ( 1 ) ( u ) = 1 ± F ( u ) 2 Given X ( t ) , we define the function Z ( t ) = F [ X ( t ) ] . It follows that Z ( 1 ) ( t ) = F ( 1 ) [ X ( t ) ] X ( 1 ) ( t ) = [ 1 ± Y ( t ) ] X ( 1 ) ( t ) where we define the function Y ( t ) = Z ( t ) 2 .
Taylor Coefficients Recursion
Suppose that we are given the Taylor coefficients up to order j for the function X ( t ) and up to order j -1 for the functions Y ( t ) and Z ( t ) . We need a formula that computes the coefficient of order j for Y ( t ) and Z ( t ) . Using the equation above for Z ( 1 ) ( t ) , we have ∑ k = 1 j k z ( k ) t k -1 = ∑ k = 1 j k x ( k ) t k -1 ± [∑ k = 0 j -1 y ( k ) t k ] [∑ k = 1 j k x ( k ) t k -1 ] + o ( t j -1 ) Setting the coefficients of t j -1 equal, we have j z ( j ) = j x ( j ) ± ∑ k = 1 j k x ( k ) y ( j - k ) z ( j ) = x ( j ) ± 1 j ∑ k = 1 j k x ( k ) y ( j - k ) Once we have computed z ( j ) , we can compute y ( j ) as follows: y ( j ) = ∑ k = 0 j z ( k ) z ( j - k )

Input File: omh/appendix/theory/tan_forward.omh