Hei!
Har oppstått noen uklarheter som jeg håper dere kan hjelpe meg med.
[tex]A=f\underset{xx}{}[/tex]
[tex]B=f\underset{xy}{}[/tex]
[tex]C=f\underset{yy}{}[/tex]
[tex]D=AC-B^2[/tex]
Med følgede gitt mener NTNU sin lærebok at D>0 gir saddelpunkt og D<0 gir lokalt min/makspunkt.
UIO sine nettsider oppgir det motsatte, altså at D<0 gir saddelpunkt og D>0 gir lokat min/makspunkt.
Begge oppgir at gitt at det ikke er saddelpunkt, vil A>0 gi minpunkt og A<0 gi makspunkt.
Hvem har rett?
Hessiske determinant/Andrederivert test
Moderators: Vektormannen, espen180, Aleks855, Solar Plexsus, Gustav, Nebuchadnezzar, Janhaa
Jeg ville heller sett på generalisasjonen av andrederiverttesten, lånt av Wikipedia ( http://en.wikipedia.org/wiki/Second_par ... ative_test) dvs:
Functions of many variables
For a function f of more than two variables, there is a generalization of the rule above. In this context, instead of examining the determinant of the Hessian matrix, one must look at the eigenvalues of the Hessian matrix at the critical point. The following test can be applied at any critical point (a, b, ...) for which the Hessian matrix is invertible:
If the Hessian is positive definite (equivalently, has all eigenvalues positive) at (a, b, ...), then f attains a local minimum at (a, b, ...).
If the Hessian is negative definite (equivalently, has all eigenvalues negative) at (a, b, ...), then f attains a local maximum at (a, b, ...).
If the Hessian has both positive and negative eigenvalues then (a, b, ...) is a saddle point for f (and in fact this is true even if (a, b, ...) is degenerate).
In those cases not listed above, the test is inconclusive.[2]
Functions of many variables
For a function f of more than two variables, there is a generalization of the rule above. In this context, instead of examining the determinant of the Hessian matrix, one must look at the eigenvalues of the Hessian matrix at the critical point. The following test can be applied at any critical point (a, b, ...) for which the Hessian matrix is invertible:
If the Hessian is positive definite (equivalently, has all eigenvalues positive) at (a, b, ...), then f attains a local minimum at (a, b, ...).
If the Hessian is negative definite (equivalently, has all eigenvalues negative) at (a, b, ...), then f attains a local maximum at (a, b, ...).
If the Hessian has both positive and negative eigenvalues then (a, b, ...) is a saddle point for f (and in fact this is true even if (a, b, ...) is degenerate).
In those cases not listed above, the test is inconclusive.[2]
[tex]i \cdot i \cdot i \cdot i = i \cdot i \cdot (-1) = (-1) \cdot (-1) = 1[/tex]