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ddy = f.ForTwo(x, j, k)
f
.
The syntax above sets
@[@
ddy [ i * p + \ell ]
=
\DD{ F_i }{ x_{j[ \ell ]} }{ x_{k[ \ell ]} } (x)
@]@
for @(@
i = 0 , \ldots , m-1
@)@
and @(@
\ell = 0 , \ldots , p
@)@,
where @(@
p
@)@ is the size of the vectors
j
and
k
.
f
has prototype
ADFun<Base> f
Note that the ADFun
object
f
is not const
(see ForTwo Uses Forward
below).
x
has prototype
const BaseVector &x
(see BaseVector
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 derivatives listed above.
j
has prototype
const SizeVector_t &j
(see SizeVector_t
below)
We use
p
to denote the size of the vector
j
.
All of the indices in
j
must be less than
n
; i.e.,
for @(@
\ell = 0 , \ldots , p-1
@)@, @(@
j[ \ell ] < n
@)@.
k
has prototype
const SizeVector_t &k
(see SizeVector_t
below)
and its size must be equal to
p
,
the size of the vector
j
.
All of the indices in
k
must be less than
n
; i.e.,
for @(@
\ell = 0 , \ldots , p-1
@)@, @(@
k[ \ell ] < n
@)@.
ddy
has prototype
BaseVector ddy
(see BaseVector
below)
and its size is @(@
m * p
@)@.
It contains the requested partial derivatives; to be specific,
for @(@
i = 0 , \ldots , m - 1
@)@
and @(@
\ell = 0 , \ldots , p - 1
@)@
@[@
ddy [ i * p + \ell ]
=
\DD{ F_i }{ x_{j[ \ell ]} }{ x_{k[ \ell ]} } (x)
@]@
BaseVector
must be a SimpleVector
class with
elements of type Base
.
The routine CheckSimpleVector
will generate an error message
if this is not the case.
SizeVector_t
must be a SimpleVector
class with
elements of type size_t
.
The routine CheckSimpleVector
will generate an error message
if this is not the case.
f
contains the corresponding
Taylor coefficients
.
After a call to ForTwo
,
the zero order Taylor coefficients correspond to
f.Forward(0, x)
and the other coefficients are unspecified.
true
, if it succeeds and false
otherwise.