In single-variable calculus, the second-derivative test states that if $x$ is a real number such that $f'(x)=0$, then:

  1. If $f''(x)>0$, then $f$ has a local minimum at $x$.
  2. If $f''(x)<0$, then $f$ has a local maximum at $x$.
  3. If $f''(x)=0$, then the text is inconclusive.

But there's no need to despair if the second-derivative test is inconclusive, because there is the higher-order derivative test. It states that if $x$ is a real number such that $f'(x)=0$, and $n$ is the smallest natural number such that $f^{(n)}(x)\neq 0$, then:

  1. If $n$ is even and $f^{(n)}>0$, then $f$ has a local minimum at $x$.
  2. If $n$ is even and $f^{(n)}<0$, then $f$ has a local manimum at $x$.
  3. If $n$ is odd, then $f$ has an inflection point at $x$.

Similarly, in multivariable calculus the second-derivative test states that if $(x,y)$ is an ordered pair such that $\nabla f(x,y) = 0$, then:

  1. If $D(x,y)>0$ and $f_{xx}(x,y)>0$, then $f$ has a local minimum at $(x,y)$.
  2. If $D(x,y)>0$ and $f_{xx}(x,y)<0$, then $f$ has a local maximum at $(x,y)$.
  3. If $D(x,y)<0$, then $f$ has a saddle point at $(x,y)$.
  4. If $D(x,y)=0$, then the test is inconclusive.

where $D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2$ is the determinant of the Hessian matrix of $f$ evaluated at $(x,y)$.

My question is, what do you do if this test is inconclusive? What is the analogue of the higher-order derivative test in multivariable calculus?


This webpage states and proves a version of the higher-order derivative test that applies not only to functions defined on $\mathbb{R}^2$ or $\mathbb{R}^N$, but functions defined on arbitrary Banach spaces. First there is this theorem:

Theorem 38 (Higher derivative test). Let $A\subseteq $ be an open set and let f$:A\to\mathbb{R}$. Assume that $f$ is $(p-1)$ times continuously differentiable and that $D^p f(x)$ exists for some $p\ge 2$ and $x\in A$. Also assume that $f'(x),\dots,f^{(p-1)}(x)=0$ and $f^{(p)}(x)\ne 0$. Write $h^{(p)}$ for the $p$-tuple $(h,\dots,h)$.

  1. If $f$ has an extreme value at $x$, then $p$ is even and the form $f^{(p)}(x)h^{(p)}$ is semidefinite.
  2. If there is a constant $c$ such that $f^{(p)}(x)h^{(p)}\ge c > 0$ for all $|h|=1$, then $f$ has a strict local minimum at $x$ and (1) applies.
  3. If there is a constant $c$ such that $f^{(p)}(x)h^{(p)}\le c < 0$ for all (|h|=1), then $f$ has a strict local maximum at $x$ and (1) applies.

Then there is this corollary for the finite dimensional case, which is what we’re interested in:

Corollary 39 (Higher derivative test, finite-dimensional case). In Theorem 38, further assume that $E$ is finite-dimensional. Then $h\mapsto f^{(p)}(x)h^{(p)}$ has both a minimum and maximum value on the set $\{h\in E:|h|=1\}$, and:

  1. If the form $f^{(p)}(x)h^{(p)}$ is indefinite, then $f$ does not have an extreme value at $x$.
  2. If the form $f^{(p)}(x)h^{(p)}$ is positive definite, then $f$ has a strict local minimum at $x$.
  3. If the form $f^{(p)}(x)h^{(p)}$ is negative definite, then $f$ has a strict local maximum at $x$.

Here $f^{(p)}(x)$ denotes a tensor containing all the pure and mixed partial derivatives of $f$ of order $p$, evaluated at $x$.

  • $\begingroup$ It seems a general theorem not useful as a general test when hessian test fails. $\endgroup$ – gimusi Aug 2 '18 at 16:45
  • $\begingroup$ @gimusi Why isn’t it useful? It gives a precise procedure you can feed into a computer to determine whether a point is a local maximum or minimum. $\endgroup$ – Keshav Srinivasan Aug 2 '18 at 18:53
  • $\begingroup$ Sorry I thought you were looking for a test similar to Hessian test to check for max, min or saddle also by hand calculation.The possibility for numerical algorithms was already noticed by Robert Israel. Then if you are looking for that your problem is now solved, Bye $\endgroup$ – gimusi Aug 2 '18 at 19:49
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    $\begingroup$ @gimusi I'm not interested in numerical algorithms at all . I was looking for something that works when the Hessian test fails, just like you can use the higher-order derivative test if the second derivative test fails in single variable calculus. In what sense do you think corollary 39 is not useful when the Hessian test fails? $\endgroup$ – Keshav Srinivasan Aug 2 '18 at 20:10
  • $\begingroup$ How is this not similar to the Hessian test? $\endgroup$ – Keshav Srinivasan Aug 3 '18 at 1:25

In that case if the test is inconclusive there is not a general rule or test that always work. We need case be case to show what kind of critical point we have using some manipulation and inequalities.

As a simple example for


at $(x,y)=(0,0)$ the test is inconclusive but

$$f(x,y)=x^4-2x^2y^2+y^4=(x^2-y^2)^2\ge 0$$

  • $\begingroup$ Are you saying that the higher-order derivative test has no analogue in multivariable calculus? $\endgroup$ – Keshav Srinivasan Aug 2 '18 at 6:16
  • $\begingroup$ @KeshavSrinivasan Yes exactly, there is not an analogue test in multivariable calculus. $\endgroup$ – gimusi Aug 2 '18 at 6:20
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    $\begingroup$ How do you know there isn't an analogue? The higher-order derivative test springs from Taylor series, and Taylor series exist for multivariable functions, so why wouldn't you be able to derive an analogue? $\endgroup$ – Keshav Srinivasan Aug 2 '18 at 6:24
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    $\begingroup$ @gimusi None of that answers why a higher-order derivative test would not exist for multivariable functions. $\endgroup$ – Keshav Srinivasan Aug 2 '18 at 6:34
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    $\begingroup$ "The examples I given are aimed to show the techniques we use when hessian test fails." OK, but I'm not interested in that. I'm specifically interested in the higher-order derivative test. $\endgroup$ – Keshav Srinivasan Aug 2 '18 at 6:44

Consider a homogeneous polynomial $f(x_1, \ldots, x_n)$ of total degree $d > 0$ in $n$ variables. In order to tell that $(0,\ldots,0)$ is a local minimum, we would need to know that $f(x_1,\ldots,x_n) \ge 0$ for all $x_1,\ldots,x_n$. Unfortunately this is a difficult problem in general, and I'm pretty sure there are no very simple tests, although there are algorithms related to Hilbert's 17th problem.

  • $\begingroup$ There is a simple test, see my answer. $\endgroup$ – Keshav Srinivasan Aug 3 '18 at 14:47

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