To distinguish if a critical point is a saddle point or not,. .

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SUMMARY

The discussion focuses on using the discriminant, defined as fxxfyy - fxy^2, to determine whether a critical point is a saddle point in multivariable calculus. The second derivative test is applied, where a positive discriminant indicates a local minimum, a negative discriminant indicates a local maximum, and a zero or mixed sign discriminant indicates a saddle point. The properties of the discriminant matrix D, including its diagonalizability and eigenvalues, are crucial for this analysis. The proof of the discriminant's utility in identifying saddle points is also outlined.

PREREQUISITES
  • Understanding of multivariable calculus concepts
  • Familiarity with second derivative tests
  • Knowledge of eigenvalues and eigenvectors
  • Basic matrix theory, specifically regarding symmetric matrices
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  • Study the properties of symmetric matrices and their eigenvalues
  • Learn about the second derivative test in multivariable calculus
  • Explore the concept of positive definite and negative definite matrices
  • Investigate applications of the discriminant in optimization problems
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Students and professionals in mathematics, particularly those studying calculus and optimization, as well as educators looking to explain the concepts of critical points and saddle points in multivariable functions.

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To distinguish a critical point is a saddle point or not, It is useful way to use discreminent.

The discreminent is fxxfyy-fxy^2.

What I want to know is how to prove the discreminent.
 
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alexio said:
To distinguish a critical point is a saddle point or not, It is useful way to use discreminent.

The discreminent is fxxfyy-fxy^2.

What I want to know is how to prove the discreminent.

This is the multivariate version of the second derivative test from calculus. If the second derivative is positive you are at a minimum. If the second der is Negative you are at a maximum.

Let D be the discriminant matrix, and h a 2x1 column vector. At a minimum h^T*D*h>0 for all small h indicates a minimum. This is an approximation to the function. Thus, D, which is diagonalizable since it is symmetric, is positive definite. Thus, all eigenvalues are real and positive. So, the determinant of D, which is what you call the discreminent, is positive. At a maximum, D is negative definite, which means both eigenvalues are negative. Hence, the discreminent is also positive. If D is neither positive definite or negative definite, the D has one positive and one negative eigenvalue meaning the discreminent is negative.

That's the basic idea of the proof. Hope it was helpful.

Take care,
Reverie
 

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