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], respectively. The Hessian matrix is at level 1 at this point.
‘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
∆
= (
∂
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
