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dy = f.ForOne(x, j)
f
.
The syntax above sets
dy
to the
partial of @(@
F
@)@ with respect to @(@
x_j
@)@; i.e.,
@[@
dy
= \D{F}{ x_j } (x)
= \left[
\D{ F_0 }{ x_j } (x) , \cdots , \D{ F_{m-1} }{ x_j } (x)
\right]
@]@
f
has prototype
ADFun<Base> f
Note that the ADFun
object
f
is not const
(see ForOne Uses Forward
below).
x
has prototype
const Vector &x
(see Vector
below)
and its size
must be equal to
n
, the dimension of the
domain
space for
f
.
It specifies
that point at which to evaluate the partial derivative.
j
has prototype
size_t j
an is less than
n
,
domain
space for
f
.
It specifies the component of
F
for which we are computing the partial derivative.
dy
has prototype
Vector dy
(see Vector
below)
and its size is @(@
m
@)@, the dimension of the
range
space for
f
.
The value of
dy
is the partial of @(@
F
@)@ with respect to
@(@
x_j
@)@ evaluated at
x
; i.e.,
for @(@
i = 0 , \ldots , m - 1
@)@
@[@
.
dy[i] = \D{ F_i }{ x_j } ( x )
@]@
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.
f
contains the corresponding
Taylor coefficients
.
After a call to ForOne
,
the zero order Taylor coefficients correspond to
f.Forward(0,x)
and the other coefficients are unspecified.
true
, if it succeeds and false
otherwise.