Three variable Jacobians Taylor approximation and extrema

A planar Hessian determinant test is not automatically a three-dimensional test. In three variables, vector-output Jacobians compose coordinate changes, the Hessian controls all spatial directions, and constrained extrema require every feasible boundary case. We work through these tools with explicit three-dimensional examples.

1 Jacobian and chain rule in space

For $F:U\subset\mathbb R^3\to\mathbb R^m$ defined on an open set, the Jacobian is the $m\times3$ matrix whose $(i,j)$ entry is $\partial F_i/\partial x_j$. For $C^1$ maps it is the derivative in

$$F(a+h)=F(a)+J_F(a)h+o(\|h\|).$$

Composing with a differentiable map $G$ gives $J_{F\circ G}=J_F(G)J_G$. The proof substitutes one linear approximation into the other, using that the intermediate increment is $O(\|h\|)$. Dimensions specify the correct matrix order.

For a scalar field $T(x,y,z)=x^2+2y^2+3z^2$ and trajectory $r(t)=(t,t^2,1-t)$,

$$\frac d{dt}T(r(t))=(2x,4y,6z)\cdot(1,2t,-1) =8t^3+8t-6.$$

Direct substitution gives $T(r(t))=2t^4+4t^2-6t+3$, confirming the derivative. The rate along the moving trajectory is not a single partial derivative.

2 Three-dimensional Taylor approximation

For $f\in C^2$ near $a$, its Hessian has entries $f_{x_ix_j}$ and is symmetric. Restricting to $\phi(t)=f(a+th)$ and integrating twice gives

$$f(a+h)=f(a)+\nabla f(a)\cdot h+\tfrac12h^TH_f(a)h+o(\|h\|^2).$$

If the Hessian operator norm is bounded by $M$ along the segment, the first-order error is at most $M\|h\|^2/2$. The proof uses the integral remainder $\int_0^1(1-t)h^TH_f(a+th)hdt$, so all directions are controlled together.

For the quadratic field above at $a=(1,1,1)$, the value is 6, gradient $(2,4,6)$ and Hessian $\operatorname{diag}(2,4,6)$. The increment is exactly

$$2h+4k+6\ell+h^2+2k^2+3\ell^2.$$

There is no higher-order remainder because the function is quadratic. For displacement $(0.01,-0.02,0.03)$, the linear prediction is 0.12 and the quadratic correction is 0.0036, giving exact change 0.1236.

3 Critical points and quadratic directions

At an interior differentiable extremum, all three partial derivatives vanish. At a stationary point $a$ of a $C^2$ function, with $\nabla f(a)=0$, if the symmetric Hessian is positive definite, its quadratic form has a positive minimum on the unit sphere; Taylor's small-o remainder is eventually smaller than that positive quadratic term. This proves a strict local minimum. Negative definite gives a strict local maximum; directions of both signs give a saddle. A semidefinite Hessian is inconclusive.

Positive definiteness can be checked by eigenvalues, completing squares, or Sylvester's criterion: all leading principal minors are positive. The criterion follows from successive square completion, equivalently a positive-diagonal triangular factorization. A positive determinant alone is insufficient in three dimensions: $\operatorname{diag}(1,-1,-1)$ has positive determinant but is indefinite.

For $f=(x-1)^2+2(y+1)^2+3(z-2)^2$, the unique global minimum is zero at $(1,-1,2)$. Its positive-definite Hessian is an independent local check, while the explicit sum of nonnegative squares proves the global result.

4 A complete three-variable constrained example

Maximize and minimize $V(x,y,z)=xyz$ on $x,y,z\ge0$, $x^2+y^2+z^2=3$. The feasible set is compact and $V$ continuous, so extrema exist. On its boundary, at least one coordinate is zero and $V=0$; thus the minimum is zero, attained on all boundary arcs.

At an interior point with positive coordinates, the regular sphere constraint has nonzero gradient. A Lagrange point satisfies

$$(yz,xz,xy)=\lambda(2x,2y,2z).$$

The multiplier condition can be justified directly for this sphere: every vector perpendicular to $a$ is tangent to the normalized curve $\sqrt3(a+tv)/\|a+tv\|$, whose derivative at zero is $v$. An extremum has zero objective derivative in every such direction, so its gradient is parallel to $a$, the constraint normal.

Multiplying the three equations by $x,y,z$ gives $xyz=2\lambda x^2=2\lambda y^2=2\lambda z^2$. Since all coordinates and $xyz$ are positive, $\lambda\ne0$, so the squared coordinates agree. Nonnegativity and the constraint force $x=y=z=1$, with volume 1. Comparing with the boundary values proves this is the global maximum. The multiplier equations alone would not have supplied the minimum on boundary arcs.

A complete three-dimensional Sylvester test

Let $A=(a_{ij})$ be a real symmetric $3\times3$ matrix and let $\Delta_j$ be its leading $j\times j$ determinant. Assume first $\Delta_1>0$ and set $B=(b_{ij})_{i,j=2,3}$ with $b_{ij}=a_{ij}-a_{1i}a_{1j}/a_{11}$. Completing the first square gives

$$x^TAx=\Delta_1\left(x_1+\frac{a_{12}}{a_{11}}x_2+\frac{a_{13}}{a_{11}}x_3\right)^2+b_{22}x_2^2+2b_{23}x_2x_3+b_{33}x_3^2.$$

Direct $2\times2$ expansion gives $b_{22}=\Delta_2/\Delta_1$; subtracting multiples of the first row in the determinant gives $\det A=\Delta_1\det B$. If also $\Delta_2>0$, complete one more square:

$$x^TAx=\Delta_1\ell_1(x)^2+\frac{\Delta_2}{\Delta_1}\left(x_2+\frac{b_{23}}{b_{22}}x_3\right)^2+\frac{\Delta_3}{\Delta_2}x_3^2,$$

where $\ell_1$ is the first linear expression above. These triangular linear coordinates are invertible. Therefore all three positive minors imply positive definiteness. Conversely, positive definiteness first gives $a_{11}=\Delta_1>0$. Restrict to $x_3=0$ and choose $x_1$ to make $\ell_1=0$; positivity for $x_2\ne0$ gives $\Delta_2/\Delta_1>0$. Finally take $x_3\ne0$ and choose $x_2,x_1$ successively to kill the first two squares. Positivity gives $\Delta_3/\Delta_2>0$. This proves necessity as well as sufficiency in the stated real symmetric three-dimensional setting.

The $C^2$ Taylor formula and its uniform little-o bound are exactly the general finite-dimensional estimate proved in the earlier optimization chapter, with $n=3$. Its first-order error bound follows there from the same integral identity. Neither statement needs a third derivative.

5 Graded practice

E1 Jacobian. For $F=(x+y,y+z,z+x)$, find its Jacobian and the gradient of $\|F\|^2$ at $(1,0,0)$.

E2 Sphere. Maximize and minimize $2x-y+2z$ on $x^2+y^2+z^2=1$.

E3 Saddle. Classify the origin for $f=xy+z^2$ using explicit directional changes rather than only a determinant.

E4 Simplex. Find maximum and minimum of $xyz$ on $x,y,z\ge0$, $x+y+z=6$, including all boundary possibilities.

6 Full solutions

S1. $J=\begin{pmatrix}1&1&0\\0&1&1\\1&0&1\end{pmatrix}$. Since the gradient is $2J^TF$, at the point $F=(1,0,1)$ it is $(4,2,2)$. Expanding the squared norm verifies the result.

S2. Cauchy–Schwarz gives absolute value at most $\|(2,-1,2)\|=3$. Maximum 3 occurs at $(2,-1,2)/3$ and minimum $-3$ at its negative. These vectors lie on the unit sphere, so the bound is attained.

S3. Along $(t,t,0)$ the value is $t^2>0$ for nonzero $t$; along $(t,-t,0)$ it is $-t^2<0$. Both approaches tend to the origin, so it is a saddle despite the nonnegative $z^2$ term.

S4. Boundary values are zero. In the positive interior, Lagrange gives $yz=xz=xy$, hence $x=y=z=2$ and value 8. Compactness ensures extrema exist, so comparison gives maximum 8 at that point and minimum 0 on the boundary. No unexamined negative-coordinate candidates are feasible.

Summary

Three-variable calculus requires matrix shapes and quadratic directions appropriate to three dimensions. Taylor controls simultaneous spatial changes, while Hessian definiteness—not determinant alone—classifies regular critical points. Compactness, regular constraints and explicit boundary comparison complete a global optimization argument.