Multivariable calculus

Critical Points of sin x + sin y + sin(x+y)

Q: What does a discriminant of zero actually tell you?

Nothing about the point itself. It tells you the second order terms are inconclusive, which happens when the Hessian is singular. You then have to examine f directly, usually by restricting it to a few lines through the point and looking at the sign of the first nonzero term in the expansion. Here the restriction to y = x behaves like -t^{3}, and a sign change rules out both a maximum and a minimum.

Q: Should I call (pi, pi) a saddle?

Loosely, many textbooks would, since it is neither a maximum nor a minimum. But that verdict comes from the cubic behaviour, not from the discriminant, and this page reports what the second derivative test itself concludes, which is degenerate. Keeping those separate matters on an exam: if you write saddle, you should also show the line restriction that justifies it, because D = -1 and D = 0 are very different situations.

For f(x,y) = sin x + sin y + sin(x+y) the point (pi/3, pi/3) is a local maximum with D = 2.25, and (-pi/3, -pi/3) is a local minimum with D = 2.25. At (pi, pi) all three second partials vanish, so D = 0 and the test is inconclusive. Looking along the line y = x shows that point is neither a max nor a min.

f(x,y) = \sin x + \sin y + \sin(x+y)

$\left(1.0471975511965976, 1.0471975511965976\right)$ local maximumdiscriminant D = 2.25
$\left(-1.0471975511965976, -1.0471975511965976\right)$ local minimumdiscriminant D = 2.25
$\left(3.141592653589793, 3.141592653589793\right)$ degenerate, the test is inconclusivediscriminant D = 0

These are not necessarily all of them; the page says which it covers.

Reducing two equations to a quadratic in cosine

Each partial picks up the same $\cos(x+y)$ from the coupled term, so subtracting one equation from the other clears it.

\frac{\partial f}{\partial x} = \cos x + \cos(x+y), \qquad \frac{\partial f}{\partial y} = \cos y + \cos(x+y)

The difference gives $\cos x = \cos y$ , so $y = x$ or $y = -x$ , each up to a multiple of $2\pi$ . Take the diagonal branch first and substitute $y = x$ into the first equation.

\cos x + \cos 2x = 0 \implies 2\cos^{2}x + \cos x - 1 = 0 \implies (2\cos x - 1)(\cos x + 1) = 0

So $\cos x = \frac{1}{2}$ , giving $x = \pm\frac{\pi}{3}$ , or $\cos x = -1$ , giving $x = \pi$ . The branch $y = -x$ makes $\cos(x+y) = \cos 0 = 1$ and so needs $\cos x = -1$ again, which lands on $(\pi, -\pi)$ , the same point as $(\pi, \pi)$ once you shift $y$ by $2\pi$ . Nothing new comes from it.

One period therefore holds three distinct critical points, and the whole plane holds their translates by $2\pi$ in either variable.

Two clean verdicts and one the test refuses to give

The second partials are sines, all carrying a minus sign.

f_{xx} = -\sin x - \sin(x+y), \qquad f_{yy} = -\sin y - \sin(x+y), \qquad f_{xy} = -\sin(x+y)

At $\left(\frac{\pi}{3}, \frac{\pi}{3}\right)$ both $\sin\frac{\pi}{3}$ and $\sin\frac{2\pi}{3}$ equal $\frac{\sqrt{3}}{2}$ , so $f_{xx} = f_{yy} = -\sqrt{3}$ and $f_{xy} = -\frac{\sqrt{3}}{2}$ . Then $D = 3 - \frac{3}{4} = \frac{9}{4} > 0$ with $f_{xx} < 0$ , a local maximum with value $\frac{3\sqrt{3}}{2} \approx 2.598$ . Every sign flips at $\left(-\frac{\pi}{3}, -\frac{\pi}{3}\right)$ , so $D$ is the same $\frac{9}{4}$ but $f_{xx} = \sqrt{3} > 0$ and the point is a local minimum.

At $(\pi, \pi)$ the picture collapses. Both $\sin\pi$ and $\sin 2\pi$ are zero, so all three second partials are zero and $D = 0$ . The test returns no information whatsoever, and the honest classification is degenerate.

To find out what is really happening there, restrict $f$ to a line through the point. Along $y = x$ , write $x = y = \pi + t$ .

f(\pi + t, \pi + t) = -2\sin t + \sin 2t = -t^{3} + O(t^{5})

That cubic changes sign at $t = 0$ , so $f$ takes values above and below $f(\pi,\pi) = 0$ arbitrarily close to the point. It is neither a local maximum nor a local minimum. The cubic term is what the second derivative test could not see.

The mistake students make

The big one is treating $D = 0$ as a classification. It is not. A zero discriminant means the quadratic part of the Taylor expansion is flat in some direction, so the second order terms cannot settle the question and the answer lives in the cubic and higher terms. Reporting a point as a saddle because $D = 0$ is guessing, and reporting it as a minimum because $f_{xx} \ge 0$ is worse.

$D > 0$ and $f_{xx} > 0$ : local minimum.
$D > 0$ and $f_{xx} < 0$ : local maximum.
$D < 0$ : saddle.
$D = 0$ : the test says nothing, so go and look at $f$ directly.

The second slip is losing the factorisation. Students substitute $y = x$ and then try to solve $\cos x + \cos 2x = 0$ numerically instead of writing $\cos 2x = 2\cos^{2}x - 1$ and factoring the quadratic. The double angle identity turns a transcendental equation into $(2\cos x - 1)(\cos x + 1) = 0$ , and the root $\cos x = -1$ is precisely the degenerate point that a numerical solver would be most likely to skip.

Frequently asked questions

What does a discriminant of zero actually tell you?

Nothing about the point itself. It tells you the second order terms are inconclusive, which happens when the Hessian is singular. You then have to examine $f$ directly, usually by restricting it to a few lines through the point and looking at the sign of the first nonzero term in the expansion. Here the restriction to $y = x$ behaves like $-t^{3}$ , and a sign change rules out both a maximum and a minimum.

Should I call (pi, pi) a saddle?

Loosely, many textbooks would, since it is neither a maximum nor a minimum. But that verdict comes from the cubic behaviour, not from the discriminant, and this page reports what the second derivative test itself concludes, which is degenerate. Keeping those separate matters on an exam: if you write saddle, you should also show the line restriction that justifies it, because $D = -1$ and $D = 0$ are very different situations.