@(@\newcommand{\W}[1]{ \; #1 \; }
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\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}} }
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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
