# SPDX-License-Identifier: EPL-2.0 OR GPL-2.0-or-later
# SPDX-FileCopyrightText: Bradley M. Bell <bradbell@seanet.com>
# SPDX-FileContributor: 2003-24 Bradley M. Bell
# ----------------------------------------------------------------------------

{xrst_begin reverse_identity}
{xrst_spell
   rclr
}

An Important Reverse Mode Identity
##################################
The theorem and the proof below is a restatement
of the results on page 236 of
:ref:`Bib@Evaluating Derivatives` .

Notation
********
Given a function :math:`f(u, v)` where :math:`u \in \B{R}^n`
we use the notation

.. math::

   \D{f}{u} (u, v) = \left[ \D{f}{u_1} (u, v) , \cdots , \D{f}{u_n} (u, v) \right]

Reverse Sweep
*************
When using :ref:`reverse mode<reverse_any-name>`
we are given a function :math:`F : \B{R}^n \rightarrow \B{R}^m`,
a matrix of Taylor coefficients :math:`x \in \B{R}^{n \times p}`,
and a weight vector :math:`w \in \B{R}^m`.
We define the functions :math:`X : \B{R} \times \B{R}^{n \times p} \rightarrow \B{R}^n`,
:math:`W : \B{R} \times \B{R}^{n \times p} \rightarrow \B{R}`, and
:math:`W_j : \B{R}^{n \times p} \rightarrow \B{R}` by

.. math::
   :nowrap:

   \begin{eqnarray}
      X(t , x) & = & x^{(0)} + x^{(1)} t + \cdots + x^{(p-1)} t^{p-1}
      \\
      W(t, x)   & = &  w_0 F_0 [X(t, x)] + \cdots + w_{m-1} F_{m-1} [X(t, x)]
      \\
      W_j (x)   & = & \frac{1}{j!} \Dpow{j}{t} W(0, x)
   \end{eqnarray}

where :math:`x^{(j)}` is the *j*-th column of :math:`x \in \B{R}^{n \times p}`.
The theorem below implies that

.. math::

   \D{ W_j }{ x^{(i)} } (x) = \D{ W_{j-i} }{ x^{(0)} } (x)

A :ref:`general reverse sweep<reverse_any-name>` calculates the values

.. math::

   \D{ W_{p-1} }{ x^{(i)} } (x)  \hspace{1cm} (i = 0 , \ldots , p-1)

But the return values for a reverse sweep are specified
in terms of the more useful values

.. math::

   \D{ W_j }{ x^{(0)} } (x)  \hspace{1cm} (j = 0 , \ldots , p-1)

Theorem
*******
Suppose that :math:`F : \B{R}^n \rightarrow \B{R}^m` is a :math:`p` times
continuously differentiable function.
Define the functions
:math:`Z : \B{R} \times \B{R}^{n \times p} \rightarrow \B{R}^n`,
:math:`Y : \B{R} \times \B{R}^{n \times p }\rightarrow \B{R}^m`,
and
:math:`y^{(j)} : \B{R}^{n \times p }\rightarrow \B{R}^m` by

.. math::
   :nowrap:

   \begin{eqnarray}
      Z(t, x)  & = & x^{(0)} + x^{(1)} t + \cdots + x^{(p-1)} t^{p-1}
      \\
      Y(t, x)  & = & F [ Z(t, x) ]
      \\
      y^{(j)} (x) & = & \frac{1}{j !} \Dpow{j}{t} Y(0, x)
   \end{eqnarray}

where :math:`x^{(j)}` denotes the *j*-th column of
:math:`x \in \B{R}^{n \times p}`.
It follows that
for all :math:`i, j` such that :math:`i \leq j < p`,

.. math::
   :nowrap:

   \begin{eqnarray}
   \D{ y^{(j)} }{ x^{(i)} } (x) & = & \D{ y^{(j-i)} }{ x^{(0)} } (x)
   \end{eqnarray}

Proof
*****
If follows from the definitions that

.. math::

   \begin{array}{rclr}
   \D{ y^{(j)} }{ x^{(i)} } (x)
   & = &
   \frac{1}{j ! } \D{ }{ x^{(i)} }
      \left[ \Dpow{j}{t} (F \circ  Z) (t, x)  \right]_{t=0}
   \\
   & = &
   \frac{1}{j ! } \left[ \Dpow{j}{t}
      \D{ }{ x^{(i)} } (F \circ  Z) (t, x)
   \right]_{t=0}
   \\
   & = &
   \frac{1}{j ! } \left\{
      \Dpow{j}{t} \left[ t^i ( F^{(1)} \circ Z ) (t, x) \right]
   \right\}_{t=0}
   \end{array}

For :math:`k > i`, the *k*-th
partial of :math:`t^i` with respect to :math:`t` is zero.
Thus, the partial with respect to :math:`t` is given by

.. math::
   :nowrap:

   \begin{eqnarray}
   \Dpow{j}{t} \left[ t^i ( F^{(1)} \circ Z ) (t, x) \right]
   & = &
   \sum_{k=0}^i
   \left( \begin{array}{c} j \\ k \end{array} \right)
   \frac{ i ! }{ (i - k) ! } t^{i-k} \;
   \Dpow{j-k}{t} ( F^{(1)} \circ Z ) (t, x)
   \\
   \left\{
      \Dpow{j}{t} \left[ t^i ( F^{(1)} \circ Z ) (t, x) \right]
   \right\}_{t=0}
   & = &
   \left( \begin{array}{c} j \\ i \end{array} \right)
   i ! \Dpow{j-i}{t} ( F^{(1)} \circ Z ) (t, x)
   \\
   & = &
   \frac{ j ! }{ (j - i) ! }
   \Dpow{j-i}{t} ( F^{(1)} \circ Z ) (t, x)
   \\
   \D{ y^{(j)} }{ x^{(i)} } (x)
   & = &
   \frac{ 1 }{ (j - i) ! }
   \Dpow{j-i}{t} ( F^{(1)} \circ Z ) (t, x)
   \end{eqnarray}

Applying this formula to the case where :math:`j`
is replaced by :math:`j - i` and :math:`i` is replaced by zero,
we obtain

.. math::

   \D{ y^{(j-i)} }{ x^{(0)} } (x)
   =
   \frac{ 1 }{ (j - i) ! }
   \Dpow{j-i}{t} ( F^{(1)} \circ Z ) (t, x)
   =
   \D{ y^{(j)} }{ x^{(i)} } (x)

which completes the proof

{xrst_end reverse_identity}
