# generalization of 2nd derivative test to multi dimensional when the hessian is inonclusive

We all know the 2nd derivative test in its original form, if $$f'(x)=0$$, then if $$f''(x)<0$$ the point is max, and if $$f''(x)>0$$ the point is min. We also know the generalization for the case is inconclusive with one variable: (I) https://en.wikipedia.org/wiki/Derivative_test#Higher-order_derivative_test

We also know the generalization for multi variable:(II) https://en.wikipedia.org/wiki/Second_partial_derivative_test#Functions_of_many_variables

The question is if there is a generalization of the two, say for multivariable if none of the conditions are met, then we can take the next derivative until we find a derivative which is not zero and use the conditions in (II) to decide whether it's a max or min.

So prove or disprove, one can just take nth derivative and check using the II conditions, that if an extremum is a max or min, or disprove via a counterexample .

• The following example, which I learned from Walter Rudin, might be relevant. The fourth-degree polynomial $(y-x^2)^2+y^4$ has a strict minimum at the origin. But if you modify it by subtracting a suitable higher-degree term like $2x^8$, the result no longer has a local minimum at the origin. Oct 21 '21 at 23:42
• @AndreasBlass I'm not quite sure what do you mean by that. Oct 22 '21 at 15:49

There cannot be such algorithm in general (As long as NP!=P) Suppose there was, let's call it A. the number of condition is constant. calculating matrix multiplication is polynomial. So A is polynomial. calculating n-th derivative is polynomial. Now use the algorithm A to find the if it is the minimum or not. so our algo is in P. But according to https://arxiv.org/abs/1012.0729 and other papers, finding local minimum is NP hard. so one of the following:

1. you prove P=NP or
2. A doesn't exist.

Sure! The general form of the Taylor's Theorem can be written as

\begin{aligned} f(x)= &f(x_0) +𝐃f(x_0)⋅(x-x_0) +\tfrac{1}{2}𝐃^2 f(x_0)⋅(x-x_0)^{⊗2}+⋯ \\&⋯+\tfrac{1}{k!}𝐃^k f(x_0)⋅(x-x_0)^{⊗k} +\tfrac{1}{(k+1)!}R_{k+1}(x)⋅(x-x_0)^{⊗ (k+1)} \end{aligned}

In the case when $$f\colon H→ℝ$$ is scalar, we may even express it as

\begin{aligned} f(x)= &⟨f(x_0)∣(x-x_0)^{⊗0}⟩_{H^{⊗0}} \\&+⟨𝐃f(x_0)∣(x-x_0)⟩_{H^{⊗1}} \\&+\tfrac{1}{2}⟨𝐃^2 f(x_0)∣(x-x_0)^{⊗2}⟩_{H^{⊗2}} \\&+⋯ \\&+\tfrac{1}{k!}⟨𝐃^k f(x_0)∣(x-x_0)^{⊗k}⟩_{H^{⊗k}} \\&+\tfrac{1}{(k+1)!}⟨R_{k+1}(x)∣(x-x_0)^{⊗ (k+1)}⟩_{H^{⊗(k+1)}} \end{aligned}

Using the induced inner product of the tensor product of Hilbert spaces. For example: the Frobenius inner product for $$m×n$$ matrices is the induces inner product on $$ℝ^m ⊗ ℝ^n$$. Convince yourself that the standard Hessian term $$xᵀ𝐇f(x_0)x$$ is identical to $$⟨𝐃^2f(x_0), x^{⊗2}⟩_{ℝ^n⊗ℝ^n}$$

We say a tensor $$𝐓∈H^{⊗2n}$$ is positive definite if and only if $$⟨𝐓∣x^{⊗2n}⟩_{H^{⊗2n}}≥0$$ for all $$x$$ with equality if and only if $$x=0$$.

Theorem Assume $$f\colon ℝ^n →ℝ$$ is $$2n$$ times continuously differentiable and $$𝐃^kf(x_0)=0$$ for $$k=1…(2n-1)$$ and $$𝐃^{2n}f(x_0)$$ is positive definite. Then $$f$$ has a strict local minimum at $$x_0$$.

Proof: It is more or less a replication of the regular proof. By Taylors formula:

\begin{aligned} f(x) - f(x_0)= \tfrac{1}{(2n)!}𝐃^{2n} f(x_0)⋅(x-x_0)^{⊗2n} +\tfrac{1}{(2n+1)!}R_{2n+1}(x)⋅(x-x_0)^{⊗ (2n+1)} \end{aligned}

Since $$𝐃^{2n} f(x_0)$$ is positive definite, the first term is always positive for $$x≠x_0$$. But it also dominates $$R$$ asymptotically as $$x→x_0$$. Hence, there must be an ε-ball $$U_ε(x_0)$$ around $$x_0$$ for which the RHS is positive for all $$x≠x_0$$. Hence $$f(x)>f(x_0)$$ locally.

• This is certainly true, but a proper generalization of the derivative test would have to address what happens in the following situation: suppose $x_0$ is a point where $Df(x_0)=0$ (so we have a critical point), but $D^2f(x_0)$ is non-zero, but say $D^2f(x_0)$ is not positive definite (and not negative-definite). In one-dimension of course this is impossible, but in higher dimensions it is. Oct 22 '21 at 0:15