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x(t)
and
\varepsilon(t) ]
near
t = 0
by the second order expansions
\[
\begin{array}{rcl}
x(t) & = & x^{(0)} + x^{(1)} * t + x^{(2)} * t^2 / 2
\\
\varepsilon(t) & = & \varepsilon^{(0)} + \varepsilon^{(1)} * t
+ \varepsilon^{(2)} * t^2 / 2
\end{array}
\]
It follows that for
k = 0 , 1 , 2
,
\[
\begin{array}{rcl}
x^{(k)} & = & \dpow{k}{t} x (0)
\\
\varepsilon^{(k)} & = & \dpow{k}{t} \varepsilon (0)
\end{array}
\]
exp_eps(
x,
epsilon)
with x is equal to .5
and epsilon is equal to .2.
For this case, the mathematical function for the operation sequence
corresponding to exp_eps
is
\[
f ( x , \varepsilon ) = 1 + x + x^2 / 2
\]
The corresponding second partial derivative with respect to
x
,
and the value of the derivative, are
\[
\Dpow{2}{x} f ( x , \varepsilon ) = 1.
\]
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^{(1)} = 1
,
x^{(2)} = 0
,
use
\varepsilon^{(1)} = 1
, and
\varepsilon^{(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_2^{(2)} = x^{(2)}
| 0 | |||||
2 | 0.5 |
v_2^{(1)} = 1 * v_1^{(1)}
| 1 |
v_2^{(2)} = 1 * v_1^{(2)}
| 0 | |||||
3 | 0.5 |
v_3^{(1)} = v_2^{(1)} / 1
| 1 |
v_3^{(2)} = v_2^{(2)} / 1
| 0 | |||||
4 | 1.5 |
v_4^{(1)} = v_3^{(1)}
| 1 |
v_4^{(2)} = v_3^{(2)}
| 0 | |||||
5 | 0.25 |
v_5^{(1)} = v_3^{(1)} * v_1^{(0)} + v_3^{(0)} * v_1^{(1)}
| 1 |
v_5^{(2)} = v_3^{(2)} * v_1^{(0)} + 2 * v_3^{(1)} * v_1^{(1)}
| 2 | |||||
6 | 0.125 |
v_6^{(1)} = v_5^{(1)} / 2
| 0.5 |
v_6^{(2)} = v_5^{(2)} / 2
| 1 | |||||
7 | 1.625 |
v_7^{(1)} = v_4^{(1)} + v_6^{(1)}
| 1.5 |
v_7^{(2)} = v_4^{(2)} + v_6^{(2)}
| 1 |
\[
\begin{array}{rcl}
1
& = &
v_7^{(2)} =
\left[ \Dpow{2}{t} v_7 \right]_{t=0} =
\left[ \Dpow{2}{t} f( x^{(0)} + x^{(1)} * t , \varepsilon^{(0)} ) \right]_{t=0}
\\
& = &
x^{(1)} * \Dpow{2}{x} f ( x^{(0)} , \varepsilon^{(0)} ) * x^{(1)} =
\Dpow{2}{x} f ( x^{(0)} , \varepsilon^{(0)} )
\end{array}
\]
(We have used the fact that
x^{(1)} = 1
,
x^{(2)} = 0
,
\varepsilon^{(1)} = 1
, and
\varepsilon^{(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 , 7
and
k = 0, 1, 2
.