\newcommand{\W}[1]{ \; #1 \; }
\newcommand{\R}[1]{ {\rm #1} }
\newcommand{\B}[1]{ {\bf #1} }
\newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} }
\newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} }
\newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} }
\newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }This is cppad-20221105 documentation. Here is a link to its
current documentation
.
Tangent and Hyperbolic Tangent Forward Taylor Polynomial Theory
Derivatives
\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
F(u)
is
\tan (u)
or
\tanh (u)
the corresponding derivative is given by
F^{(1)} (u) = 1 \pm 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 \pm 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
\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
t^{j-1}
equal, we have
\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
z^{(j)}
,
we can compute
y^{(j)}
as follows:
y^{(j)} = \sum_{k=0}^j z^{(k)} z^{(j-k)}
Input File: omh/theory/tan_forward.omh
