$Id: tan_forward.omh 3272 2014-05-20 13:50:52Z bradbell $
/* --------------------------------------------------------------------------
CppAD: C++ Algorithmic Differentiation: Copyright (C) 2003-14 Bradley M. Bell

CppAD is distributed under multiple licenses. This distribution is under
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                    Eclipse Public License Version 1.0.

A copy of this license is included in the COPYING file of this distribution.
Please visit http://www.coin-or.org/CppAD/ for information on other licenses.
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$begin tan_forward$$
$spell
	Taylor
$$

$index tan, forward theory$$
$index theory, tan forward$$
$index forward, tan theory$$

$section Tangent and Hyperbolic Tangent Forward Taylor Polynomial Theory$$

$head Derivatives$$
$latex \[
\begin{array}{rcl}
\tan^{(1)} ( u ) & = & [ \cos (u)^2 + \sin (u)^2  ] / \cos (u)^2 
\\ 
& = & 1 + \tan (u)^2 
\\
\tanh^{(1)} ( u ) & = & [ \cosh (u)^2 - \sinh (u)^2 ] / \cosh (u)^2 
\\ 
& = & 1 - \tanh (u)^2 
\end{array}
\] $$
If $latex F(u)$$ is $latex \tan (u)$$ or $latex \tanh (u)$$
the corresponding derivative is given by 
$latex \[
	F^{(1)} (u) = 1 \pm F(u)^2
\]$$
Given $latex X(t)$$, we define the function $latex Z(t) = F[ X(t) ]$$.
It follows that
$latex \[
Z^{(1)} (t) = F^{(1)} [ X(t) ] X^{(1)} (t) = [ 1 \pm Y(t) ] X^{(1)} (t)
\] $$
where we define the function $latex Y(t) = Z(t)^2$$.

$head Taylor Coefficients Recursion$$
Suppose that we are given the Taylor coefficients 
up to order $latex j$$ for the function $latex X(t)$$ and 
up to order $latex j-1$$ for the functions $latex Y(t)$$ and $latex Z(t)$$.
We need a formula that computes the coefficient of order $latex j$$
for $latex Y(t)$$ and $latex Z(t)$$.
Using the equation above for $latex Z^{(1)} (t)$$, we have
$latex \[
\begin{array}{rcl}
\sum_{k=1}^j k z^{(k)} t^{k-1}
& = &
\sum_{k=1}^j k x^{(k)} t^{k-1}
\pm
\left[ \sum_{k=0}^{j-1} y^{(k)} t^k        \right]
\left[ \sum_{k=1}^j k x^{(k)} t^{k-1}  \right]
+
o( t^{j-1} )
\end{array}
\] $$
Setting the coefficients of $latex t^{j-1}$$ equal, we have
$latex \[
\begin{array}{rcl}
j z^{(j)} 
=
j x^{(j)} 
\pm
\sum_{k=1}^j k x^{(k)} y^{(j-k)}
\\
z^{(j)} 
=
x^{(j)} \pm \frac{1}{j}  \sum_{k=1}^j k x^{(k)} y^{(j-k)}
\end{array}
\] $$ 
Once we have computed $latex z^{(j)}$$,
we can compute $latex y^{(j)}$$ as follows:
$latex \[
y^{(j)} = \sum_{k=0}^j z^{(k)} z^{(j-k)}
\] $$

$end