Second Order Forward Mode: Derivative Values
Keywords: second order forward mode: derivative values two syntax purpose f x0 x1 x2 y2 vector example special case

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Second Order Forward Mode: Derivative Values
Syntax
`y2 = f.Forward(1, x2)`
Purpose
We use F : B n → 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 ) ( x 0 ) * x 2 + 1 2 x 1 T * F i ( 2 ) ( x 0 ) * x 1 `
` where F ( 1 ) ( x 0 ) is the Jacobian of F , and F i ( 2 ) ( x 0 ) 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 Y ( t ) = F [ X ( t ) ] X ( t ) = x ( 0 ) t 0 + x ( 1 ) * t 1 + ⋯ , + x ( q ) * t q + o ( t q ) Y ( t ) = y ( 0 ) t 0 + y ( 1 ) * t 1 + ⋯ , + y ( q ) * t q + o ( t q ) and o ( t q ) * t - q → 0 as t → 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