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dy =
f.Forward(1,
dx)
F : B^n \rightarrow B^m
to denote the
AD function
corresponding to f.
The result of the syntax above is
\[
dy = F^{(1)} (x) * dx
\]
where
F^{(1)} (x)
is the Jacobian of
F
evaluated at
x
.
ADFun<
Base>
f
Note that the ADFun
object f is not const
.
Before this call to Forward
, the value returned by
f.size_taylor()
must be greater than or equal one.
After this call it will be will be two (see size_taylor
).
f.Forward(0,
x)
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 x.
const
Vector &
x
(see Vector
below)
and its size
must be equal to n, the dimension of the
domain
space for f.
Vector
dy
(see Vector
below)
and its value is
F^{(1)} (x) * dx
.
The size of dy
is equal to m, the dimension of the
range
space for f.