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x(t)
near
t = 0
by the second order expansion
\[
x(t) = x^{(0)} + x^{(1)} * t + x^{(2)} * t^2 / 2
\]
It follows that for
k = 0 , 1 , 2
,
\[
x^{(k)} = \dpow{k}{t} x (0)
\]
\[
f(x) = 1 + x + x^2 / 2
\]
The corresponding second derivative function is
\[
\Dpow{2}{x} f (x) = 1
\]
x = .5
.
The corresponding operation sequence,
zero order forward sweep values,
and first order forward sweep values
are inputs and are used by a second order forward sweep.
t
,
at
t = 0
, for each variable in the sequence.
\[
v_i (t) = v_i^{(0)} + v_i^{(1)} * t + v_i^{(2)} * t^2 / 2
\]
We use
x^{(0)} = 1
, and
x^{(2)} = 0
so that second order differentiation
with respect to
t
, at
t = 0
,
is the same as the second partial differentiation
with respect to
x
at
x = x^{(0)}
.
Index |
| Zero |
| Operation |
| First |
| Derivative |
| Second |
1 | 0.5 |
|
v_1^{(1)} = x^{(1)}
| 1 |
v_1^{(2)} = x^{(2)}
|
v_1^{(2)} = 0
| ||||
2 | 1.5 |
|
v_2^{(1)} = v_1^{(1)}
| 1 |
v_2^{(2)} = v_1^{(2)}
|
v_2^{(2)} = 0
| ||||
3 | 0.25 |
|
v_3^{(1)} = 2 * v_1^{(0)} * v_1^{(1)}
| 1 |
v_3^{(2)} = 2 * (v_1^{(1)} * v_1^{(1)} + v_1^{(0)} * v_1^{(2)} )
|
v_3^{(2)} = 2
| ||||
4 | 0.125 |
|
v_4^{(1)} = v_3^{(1)} / 2
| .5 |
v_4^{(2)} = v_3^{(2)} / 2
|
v_4^{(2)} = 1
| ||||
5 | 1.625 |
|
v_5^{(1)} = v_2^{(1)} + v_4^{(1)}
| 1.5 |
v_5^{(2)} = v_2^{(2)} + v_4^{(2)}
|
v_5^{(2)} = 1
|
\[
\begin{array}{rcl}
1
& = &
v_5^{(2)} =
\left[ \Dpow{2}{t} v_5 \right]_{t=0} =
\left[ \Dpow{2}{t} f( x^{(0)} + x^{(1)} * t ) \right]_{t=0}
\\
& = &
x^{(1)} * \Dpow{2}{x} f ( x^{(0)} ) * x^{(1)} =
\Dpow{2}{x} f ( x^{(0)} )
\end{array}
\]
(We have used the fact that
x^{(1)} = 1
and
x^{(2)} = 0
.)
x = .1
,
what are the results of a zero, first, and second order forward sweep for
the operation sequence above;
i.e., what are the corresponding values for
v_i^{(k)}
for
i = 1, \ldots , 5
and
k = 0, 1, 2
.