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y0 = f.Forward(0, x0)
y0 = f.Forward(0, x0, s)
f
.
The result of the syntax above is
@[@
y0 = F(x0)
@]@
See the FunCheck discussion
for
possible differences between @(@
F(x)
@)@ and the algorithm that defined
the operation sequence.
f
has prototype
ADFun<Base> f
Note that the ADFun
object
f
is not const
.
After this call to Forward
, the value returned by
f.size_order()
will be equal to one (see size_order
).
x0
has prototype
const Vector& x0
(see Vector
below)
and its size must be equal to
n
, the dimension of the
domain
space for
f
.
s
is not present, std::cout
is used in its place.
Otherwise, this argument has prototype
std::ostream& s
It specifies where the output corresponding to PrintFor
,
and this zero order forward mode call, will be written.
y0
has prototype
Vector y0
(see Vector
below)
and its value is @(@
F(x)
@)@ at
x = x0
.
The size of
y0
is equal to
m
, the dimension of the
range
space for
f
.
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.
= x0
,
@(@
X(t) = x^{(0)}
@)@, and
@[@
y^{(0)} = Y(t) = F[ X(t) ] = F( x^{(0)} )
@]@
which agrees with the specifications for
y0
in the purpose
above.