Is the “Higher Order Derivative Test” more informative than the “Second Order Derivative Test”

calculusderivativesfunctions

I was reading about the "Second Order Derivative Test" which can show whether a given (stationary) point is a minimum, maximum or a saddle point.

However, I have heard that the "Second Order Derivative Test" is not always conclusive – there are certain conditions in which the "Second Order Derivative Test".

This lead me to the "Higher Order Derivative Test" which if I have understood correctly, can do everything that the "Second Order Derivative Test" can do, and aditionally clarify certain inconclusive instances that the "Second Order Derivative Test" struggles with. Although there are still some instances in which the "Higher Order Derivative Test" can be inconclusive.

My Question:

  • Can the "Higher Order Derivative Test" be considered to be better than the "Second Order Derivative Test", i.e. provide informative results in all cases where the "Second Order Derivative Test" works and in additional cases?

  • For high dimensional multivariate functions, is it true that the "Higher Order Derivative Test" will have very complex matrices of partial derivatives – thus making it ineffecient to use in these cases, and thus favors the "Second Order Derivative Test"?

  • Finally, is it true that the "Second Order Derivative Test" and the "Higher Order Derivative Test" can not tell if a stationary point is a local minimum or a global minimum? The only way to tell if a stationary point is a global minimum would be to compare the value of the function at all stationary points and see which one of these stationary points produces the lowest value of the function?

Thanks!

Best Answer

  1. The "Second Order Derivative Test" is a special case of the "Higher Order Derivative Test" (let $n=1$ in the definition of HODT and you get SODT). So in that sense, HODT is strictly "better": it copes with the case that the second derivative is zero, by moving up to higher derivatives.
  2. Yes - but in any given case, either the HODT is exactly the SODT, or the SODT does not apply. So there's no "extra" complexity to the HODT.
  3. Correct. But your proposed method to see whether a stationary point is a global minimum is not enough: consider $y = x^3$.
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