Higher-Order Terms at Degenerate Two-Variable Points

When the Quadratic Test Runs Out

A stationary point is a point where both first partial derivatives vanish. The two-variable Hessian determinant test classifies many stationary points by examining the quadratic terms near them. But if those terms vanish or are only semidefinite, they may not determine what the function does in every nearby direction.

At such a degenerate point, the function may still have a strict local minimum, a strict local maximum, or a saddle. Higher-order terms can settle the question. Carefully chosen paths can also reveal that a point is a saddle. The key is to distinguish a test that proves a classification from one that merely fails to find a counterexample.

Definition: A function \(f\) has a local minimum at \((a,b)\) if there is a neighborhood of \((a,b)\) in which \(f(x,y)\geq f(a,b)\). The minimum is strict if the inequality is strict at every other point in that neighborhood. A local maximum is defined by reversing the inequality. A stationary point that is neither a local minimum nor a local maximum is a saddle point.

Reading the Taylor Terms

Write a nearby point as \((a+u,b+v)\), where \(u\) and \(v\) are small. If \(f\) has continuous partial derivatives through order \(m\) near \((a,b)\), its Taylor expansion through degree \(m\) can be grouped into homogeneous polynomials:

f(a+u,b+v) = f(a,b)+P_1(u,v)+P_2(u,v)+\cdots+P_m(u,v) +o\!\left(\|(u,v)\|^m\right).

Here \(P_j\) contains the terms of total degree \(j\), and is homogeneous: \(P_j(tu,tv)=t^jP_j(u,v)\). At a stationary point, \(P_1=0\). The quadratic term is \(P_2(u,v)=\frac12 (u,v)\mathbf{H}(a,b)(u,v)^{\top}\), where \(\mathbf{H}\) is the Hessian matrix. If the Hessian is zero, or its quadratic form has zero directions, this term alone may not classify the point.

When all terms of degrees below \(m\) vanish and \(P_m\) is the first nonzero term, it is called the leading homogeneous term. Its sign often gives a decisive answer. The remainder notation means that the remainder divided by \(\|(u,v)\|^m\) tends to zero as \((u,v)\) tends to \((0,0)\).

Theorem: Suppose \(P_m\) is the first nonzero homogeneous term in the Taylor expansion at a stationary point. If \(P_m(u,v)>0\) for every \((u,v)\ne(0,0)\), the point is a strict local minimum. If \(P_m(u,v)<0\) for every nonzero \((u,v)\), it is a strict local maximum. If \(P_m\) takes both positive and negative values, the point is a saddle.

Why a Definite Leading Term Decides

The theorem follows because a homogeneous polynomial has a fixed scale along each ray from the origin. To see why the remainder cannot overturn a definite leading term, consider the positive case. On the unit circle, \(P_m\) is continuous and positive, so it has a positive minimum \(c\). For every nonzero \(h=(u,v)\), homogeneity then gives \(P_m(h)\geq c\|h\|^m\).

Bound the leading term.
Since \(P_m\) is positive on the unit circle, compactness gives a number \(c>0\) such that \(P_m(h)\geq c\|h\|^m\).

Control the remainder.
Because the remainder is \(o(\|h\|^m)\), for sufficiently small \(h\) its absolute value is less than \(\frac{c}{2}\|h\|^m\).

Compare the full change.
All lower terms vanish, so \(f(a+u,b+v)-f(a,b)\geq c\|h\|^m-\frac{c}{2}\|h\|^m>0\). This proves a strict local minimum.

The negative case follows by applying the same reasoning to \(-f\). If \(P_m\) has opposite signs at two nonzero vectors, follow the two corresponding rays: the leading term dominates the smaller remainder sufficiently close to the point, so the function takes values both above and below \(f(a,b)\). That is exactly the saddle condition.

Worked Example: A Quartic Strict Minimum

Worked Example: Classifying \(f(x,y)=x^4+y^4\)

Classify the stationary point at the origin.

First calculate the gradient: \(f_x=4x^3\) and \(f_y=4y^3\). Both vanish at \((0,0)\), and \(f(0,0)=0\). The second partial derivatives are \(f_{xx}=12x^2\), \(f_{yy}=12y^2\), and \(f_{xy}=0\), so the Hessian at the origin is the zero matrix. The quadratic test therefore gives no classification.

The first nonzero term is \(P_4(x,y)=x^4+y^4\). For every \((x,y)\ne(0,0)\), at least one of \(x^4,y^4\) is positive and neither is negative. Hence \(P_4(x,y)>0\). In fact, \(f(x,y)-f(0,0)=x^4+y^4>0\) for every nonzero point, so the origin is a strict local minimum (indeed, a global minimum).

Check: the function value at the origin is \(0\), and the exact difference from that value is \(x^4+y^4\), which is positive away from the origin. This confirms the classification directly.

Worked Example: A Quartic Saddle

Worked Example: Classifying \(g(x,y)=x^4-y^4\)

Again, consider the origin. The partial derivatives are \(g_x=4x^3\) and \(g_y=-4y^3\), so both vanish there. The second partial derivatives are \(g_{xx}=12x^2\), \(g_{yy}=-12y^2\), and \(g_{xy}=0\); thus the Hessian at the origin is zero.

The first nonzero homogeneous term is \(P_4(x,y)=x^4-y^4\). On the \(x\)-axis, where \(y=0\), \(g(t,0)=t^4>0\) for every \(t\ne0\). On the \(y\)-axis, where \(x=0\), \(g(0,t)=-t^4<0\) for every \(t\ne0\). These values occur arbitrarily close to the origin, so every neighborhood contains function values both above and below \(g(0,0)=0\). Therefore, the origin is a saddle.

Check: substituting \((t,0)\) gives \(t^4-0=t^4\), while substituting \((0,t)\) gives \(0-t^4=-t^4\). The two opposite signs verify the saddle classification.

When the Quadratic Term Is Semidefinite

The first nonzero homogeneous term is not always decisive. In particular, a nonnegative term can vanish along some directions. Along those directions, higher-degree terms may determine the behavior. One must then examine enough of the function to control all nearby points, or use paths that establish a saddle by exhibiting both signs.

Worked Example: A Quartic Term Along a Flat Direction

Classify \(h(x,y)=x^2+y^4\) at the origin. Its gradient is \(h_x=2x\), \(h_y=4y^3\), so \(\nabla h(0,0)=\mathbf{0}\). Its Hessian is \(\begin{pmatrix}2&0\\0&0\end{pmatrix}\), which is positive semidefinite but not positive definite. The quadratic term \(P_2=x^2\) vanishes on the \(y\)-axis, so the quadratic term by itself does not establish a strict minimum.

The full change in function value is \(h(x,y)-h(0,0)=x^2+y^4\). Both summands are nonnegative. If \((x,y)\ne(0,0)\), then either \(x\ne0\), making \(x^2>0\), or \(x=0\) and \(y\ne0\), making \(y^4>0\). Thus the difference is strictly positive at every nearby point other than the origin, and the origin is a strict local minimum.

Along the flat direction \(x=0\), the function is \(h(0,t)=t^4>0\) for \(t\ne0\); along \(y=0\), it is \(h(t,0)=t^2>0\). These path checks illustrate the different orders of growth, but the full sum-of-nonnegative-terms argument is what proves positivity in every direction.

Check: \(h(0,0)=0\), and direct subtraction gives \(h(x,y)-h(0,0)=x^2+y^4\), confirming the stated inequality everywhere.

Using Paths Carefully

A path test is especially efficient for disproving a local extremum. If a path approaching the point gives values above the function value and another gives values below it, the point is a saddle. A single path with values above the point rules out a local maximum, but it does not rule out a local minimum. Likewise, a single path with values below rules out a local minimum, but does not rule out a local maximum.

Worked Example: Opposite Signs Along Two Paths

Consider \(q(x,y)=x^2-y^4\) at the origin. Its partial derivatives are \(q_x=2x\) and \(q_y=-4y^3\), which both vanish at \((0,0)\). Its Hessian there is \(\begin{pmatrix}2&0\\0&0\end{pmatrix}\), so the quadratic term is semidefinite.

Along the path \((x,y)=(t,0)\), \(q(t,0)=t^2>0\) for \(t\ne0\). Along \((x,y)=(0,t)\), \(q(0,t)=-t^4<0\) for \(t\ne0\). Since \(t\) can be arbitrarily small, both signs occur in every neighborhood of the origin. Therefore the origin is a saddle.

Check: the function value at the origin is \(0\); the two substitutions give \(t^2-0=t^2\) and \(0-t^4=-t^4\), respectively. The signs are opposite for every nonzero \(t\), as required.

Paths are not a general substitute for proving an inequality in a whole neighborhood. Checking ten paths, or even every straight line, may miss behavior in other directions or along curved paths. In particular, if all tested paths give values above \(f(a,b)\), that does not prove a local minimum. For a minimum or maximum, establish a bound that applies to all sufficiently small \((u,v)\), or use a theorem whose hypotheses ensure such a bound.

Common pitfall: Seeing \(P_m(u,v)\geq0\) and concluding that there is a strict minimum. For example, \(P_2(x,y)=x^2\) is nonnegative but vanishes whenever \(x=0\). Its sign alone does not determine the function along that axis. In \(x^2-y^4\), the higher-order term is negative on the axis \(x=0\), and the point is a saddle. Check directions where the leading term vanishes before drawing a conclusion.

A Practical Classification Strategy

At a degenerate stationary point, organize the reasoning rather than stopping at an inconclusive Hessian test:

Translate the point to the origin in local coordinates if needed, and subtract the constant function value when comparing nearby values.
Write the first relevant Taylor terms, grouped by total degree. Identify which terms vanish and where the surviving terms vanish.
If the first nonzero homogeneous term is positive or negative in every nonzero direction, use its sign and the Taylor remainder to classify the point as a strict extremum.
If that term takes both signs, or if two paths give opposite signs, classify the point as a saddle.
If the leading term is semidefinite or vanishes in some directions, inspect higher-order behavior there. To prove an extremum, finish with an argument valid throughout a neighborhood.

This analysis matters because the Hessian describes only the quadratic contribution to local change. When that contribution is degenerate, the function can still curve upward, curve downward, or rise in some directions and fall in others at higher orders. The appropriate conclusion comes from the first terms that actually control the function, together with a justification that they dominate the remainder wherever needed.

Check Your Understanding

Use Taylor terms and path behavior to justify each answer.

For \(f(x,y)=x^4+y^4\), what is the first nonzero homogeneous term at the origin, and what does its sign show?
Why does \(x^4-y^4\) have a saddle at the origin even though its gradient vanishes there?
For \(x^2+y^4\), why does the quadratic term alone fail to prove a strict minimum, and what completes the argument?
What must two path tests show in order to prove that a stationary point is a saddle?
Why do positive values along several selected paths not, by themselves, prove a local minimum?