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$\newcommand{\W}[1]{ \; #1 \; } \newcommand{\R}[1]{ {\rm #1} } \newcommand{\B}[1]{ {\bf #1} } \newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} } \newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} } \newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} } \newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }$
Second Order Forward Mode: Derivative Values

Syntax
y2 = f.Forward(1, x2)

Purpose
We use $F : B^n \rightarrow B^m$ to denote the AD function corresponding to f . The result of the syntax above is that for i = 0 , ... , m-1 ,       y2[i] $= F_i^{(1)} (x0) * x2 + \frac{1}{2} x1^T * F_i^{(2)} (x0) * x1$  where $F^{(1)} (x0)$ is the Jacobian of $F$, and $F_i^{(2)} (x0)$ is the Hessian of th i-th component of $F$, evaluated at x0 .

f
The object f has prototype       ADFun<Base> f  Note that the ADFun object f is not const. Before this call to Forward, the value returned by       f.size_order()  must be greater than or equal two. After this call it will be will be three (see size_order ).

x0
The vector x0 in the formula for y2[i] corresponds to the previous call to forward_zero using this ADFun object f ; i.e.,       f.Forward(0, x0)  If there is no previous call with the first argument zero, the value of the independent variables during the recording of the AD sequence of operations is used for x0 .

x1
The vector x1 in the formula for y2[i] corresponds to the previous call to forward_one using this ADFun object f ; i.e.,       f.Forward(1, x1) 
x2
The argument x2 has prototype       const Vector& x2  (see Vector below) and its size must be equal to n , the dimension of the domain space for f .

y2
The result y2 has prototype       Vector y2  (see Vector below) The size of y1 is equal to m , the dimension of the range space for f . Its value is given element-wise by the formula in the purpose above.

Vector
The type Vector must be a SimpleVector class with elements of type Base . The routine CheckSimpleVector will generate an error message if this is not the case.

Example
The file forward.cpp contains an example and test of this operation. It returns true if it succeeds and false otherwise.

Special Case
This is special case of forward_order where $$\begin{array}{rcl} Y(t) & = F[ X(t) ] \\ X(t) & = & x^{(0)} t^0 + x^{(1)} * t^1 + \cdots, + x^{(q)} * t^q + o( t^q ) \\ Y(t) & = & y^{(0)} t^0 + y^{(1)} * t^1 + \cdots, + y^{(q)} * t^q + o( t^q ) \end{array}$$ and $o( t^q ) * t^{-q} \rightarrow 0$ as $t \rightarrow 0$. For this special case, $q = 2$, $x^{(0)}$ = x0 , $x^{(1)}$ = x1 , $X(t) = x^{(0)} + x^{(1)} t + x^{(2)} t^2$, and $$y^{(0)} + y^{(1)} t + y^{(2)} t^2 = F [ x^{(0)} + x^{(1)} t + x^{(2)} t^2 ] + o(t^2)$$ Restricting our attention to the i-th component, and taking the derivative with respect to $t$, we obtain $$y_i^{(1)} + 2 y_i^{(2)} t = F_i^{(1)} [ x^{(0)} + x^{(1)} t + x^{(2)} t^2 ] [ x^{(1)} + 2 x^{(2)} t ] + o(t)$$ Taking a second derivative with respect to $t$, and evaluating at $t = 0$, we obtain $$2 y_i^{(2)} = [ x^{(1)} ]^T F_i^{(2)} [ x^{(0)} ] x^{(1)} + F_i^{(1)} [ x^{(0)} ] 2 x^{(2)}$$ which agrees with the specification for y2[i] in the purpose above.
Input File: omh/forward/forward_two.omh