- #1
rvadd
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When discussing the second partial derivative test in multivariate calculus, a reference is usually made to an elusive "higher order test" that one must defer to in the case that the second partial derivative test fails. Does anyone know the general form of these higher order test?
My first intuition would be to analyze the bivariate cubic equation (as the second derivative test is proved oftentimes via lemmas involving the bivariate quadratic equation):
ax^{3}+bx^{2}y+cxy^{2}+dy^{3}
This approach seems difficult (involving completing a cube). In addition, I doubt the same properties will arise, since they rely on the positive values of squared expressions. At the same time, you might need to include the quadratic terms when analyzing the positive/negative/zero conditions of the concavity.
The other explanation frequently used to define the second partial derivative test is the determinant of the Hessian. This seems even worse than the previous approach because there are eight third partial derivatives and they wouldn't fit in a square matrix (which is necessary for a determinant). To generalize a Hessian, you would need to extend it into a tensor (contravariant rank 3 just as the Hessian is rank 2 and the Jacobian is rank 1 for f(x,y) ), which seems difficult once again because how then do you take the determinant (if you do need a determinant in the first place)?
Are these reasonable methods for accomplishing this generalization or am I missing some obscure/obvious simpler theorem?
My first intuition would be to analyze the bivariate cubic equation (as the second derivative test is proved oftentimes via lemmas involving the bivariate quadratic equation):
ax^{3}+bx^{2}y+cxy^{2}+dy^{3}
This approach seems difficult (involving completing a cube). In addition, I doubt the same properties will arise, since they rely on the positive values of squared expressions. At the same time, you might need to include the quadratic terms when analyzing the positive/negative/zero conditions of the concavity.
The other explanation frequently used to define the second partial derivative test is the determinant of the Hessian. This seems even worse than the previous approach because there are eight third partial derivatives and they wouldn't fit in a square matrix (which is necessary for a determinant). To generalize a Hessian, you would need to extend it into a tensor (contravariant rank 3 just as the Hessian is rank 2 and the Jacobian is rank 1 for f(x,y) ), which seems difficult once again because how then do you take the determinant (if you do need a determinant in the first place)?
Are these reasonable methods for accomplishing this generalization or am I missing some obscure/obvious simpler theorem?