HP 49g+ User Manual

Page 464

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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], 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

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