lines 6-90 of file: xrst/theory/tan_forward.xrst

{xrst_begin tan_forward}

Tangent and Hyperbolic Tangent Forward Taylor Polynomial Theory
###############################################################

Derivatives
***********

.. math::
   :nowrap:

   \begin{eqnarray}
   \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{eqnarray}

If :math:`F(u)` is :math:`\tan (u)` or :math:`\tanh (u)`
the corresponding derivative is given by

.. math::

   F^{(1)} (u) = 1 \pm F(u)^2

Given :math:`X(t)`, we define the function :math:`Z(t) = F[ X(t) ]`.
It follows that

.. math::

   Z^{(1)} (t) = F^{(1)} [ X(t) ] X^{(1)} (t) = [ 1 \pm Y(t) ] X^{(1)} (t)

where we define the function :math:`Y(t) = Z(t)^2`.

Taylor Coefficients Recursion
*****************************
Suppose that we are given the Taylor coefficients
up to order :math:`j` for the function :math:`X(t)` and
up to order :math:`j-1` for the functions :math:`Y(t)` and :math:`Z(t)`.
We need a formula that computes the coefficient of order :math:`j`
for :math:`Y(t)` and :math:`Z(t)`.
Using the equation above for :math:`Z^{(1)} (t)`, we have

.. math::
   :nowrap:

   \begin{eqnarray}
   \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{eqnarray}

Setting the coefficients of :math:`t^{j-1}` equal, we have

.. math::
   :nowrap:

   \begin{eqnarray}
   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{eqnarray}

Once we have computed :math:`z^{(j)}`,
we can compute :math:`y^{(j)}` as follows:

.. math::

   y^{(j)} = \sum_{k=0}^j z^{(k)} z^{(j-k)}

{xrst_end tan_forward}