HP 50G, Руководство пользователя

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Page 14-7
Applications of function HESS are easier to visualize in the RPN mode.
Consider as an example the function
φ
(X,Y,Z) = X
2
+ XY + XZ, we’ll apply
function HESS to function
φ
in the following example. The screen shots show the
RPN stack before and after applying function HESS.
When applied to a function of two variables, the gradient in level 2, when
made equal to zero, represents the equations for critical points, i.e.,
∂φ
/
∂
x
i
= 0,
while the matrix in level 3 represent second derivatives. Thus, the results from
the HESS function can be used to analyze extrema in functions of two variables.
For example, for the function f(X,Y) = X
3
-3X-Y
2
+5, proceed as follows in RPN
mode:
‘X^3-3*X-Y^2+5’
`
[‘X’,’Y’]
`
Enter function and variables
HESS Apply function HESS
SOLVE Find critical points
μ
Decompose vector
‘s1’
K
‘s2’
K
Store critical points
The variables s1 and s2, at this point, contain the vectors [‘X=-1’,’Y=0] and
[‘X=1’,’Y=0], respect
‘H’
K
Store Hessian matrix
J @@@H@@@ @@s1@@
SUBST
‚ï
Substitute s1 into H
The resulting matrix A has a
11
elements a
11
=
∂
2
φ
/
∂
X
2
= -6., a
22
=
∂
2
φ
/
∂
X
2
=
-2., and a
12
= a
21
=
∂
2
φ
/
∂
X
∂
Y = 0. The discriminant, for this critical point
s1(-1,0) is
Δ
= (
∂
2
f/
∂
x
2
)
⋅
(
∂
2
f/
∂
y
2
)-[
∂
2
f/
∂
x
∂
y]
2
= (-6.)(-2.) = 12.0 > 0. Since
∂
2
φ
/
∂
X
2
<0, point s1 represents a relative maximum.
Next, we substitute the second point, s2, into H:
J @@@H@@@ @@s2@@
SUBST
‚ï
Substitute s2 into H