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lines 6-66 of file: xrst/theory/exp_reverse.xrst
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{xrst_begin exp_reverse}
{xrst_spell
   expm
}

Exponential Function Reverse Mode Theory
########################################

We use the reverse theory
:ref:`standard math function<reverse_theory@Standard Math Functions>`
definition for the functions :math:`H` and :math:`G`.
The zero order forward mode formula for the
:ref:`exponential<exp_forward-name>` is

.. math::

   z^{(0)}  =  F ( x^{(0)} )

and for :math:`j > 0`,

.. math::

   z^{(j)}  = x^{(j)} d^{(0)}
      + \frac{1}{j} \sum_{k=1}^{j} k x^{(k)} z^{(j-k)}

where

.. math::

   d^{(0)} = \left\{ \begin{array}{ll}
      0 & \R{if} \; F(x) = \R{exp}(x)
      \\
      1 & \R{if} \; F(x) = \R{expm1}(x)
   \end{array} \right.

For order :math:`j = 0, 1, \ldots` we note that

.. math::
   :nowrap:

   \begin{eqnarray}
   \D{H}{ x^{(j)} }
   & = & \D{G}{ x^{(j)} }  + \D{G}{ z^{(j)} } \D{ z^{(j)} }{ x^{(j)} }
   \\
   & = & \D{G}{ x^{(j)} }  + \D{G}{ z^{(j)} } ( d^{(0)} + z^{(0)} )
   \end{eqnarray}

If :math:`j > 0`, then for :math:`k = 1 , \ldots , j`

.. math::
   :nowrap:

   \begin{eqnarray}
   \D{H}{ x^{(k)} } & = &
   \D{G}{ x^{(k)} }  + \D{G}{ z^{(j)} } \frac{1}{j}  k z^{(j-k)}
   \\
   \D{H}{ z^{(j-k)} } & = &
   \D{G}{ z^{(j-k)} }  + \D{G}{ z^{(j)} } \frac{1}{j}  k x^{(k)}
   \end{eqnarray}

{xrst_end exp_reverse}