Multivariable Taylor expansions and Hessians
Use second-order structure to classify stationary points and certify local approximation errors.
On this page
The question for today
What you’ll understand
- Derive the quadratic Taylor term along a displacement.
- Classify stationary points using Hessian definiteness and recognize inconclusive cases.
- Bound a first-order remainder using a neighborhood Hessian bound.
A tangent plane cannot distinguish a bowl from a saddle when both have zero slope at the point of contact. Second derivatives record how the first-order sensitivity changes and supply the missing curvature information. In several variables, that information is a matrix, but the meaningful quantity is its quadratic action on a displacement. Taylor's formula connects this action to local prediction, while explicit remainder assumptions decide when that prediction can be trusted.
Reduce a multidimensional displacement to a one-variable path
Let f be twice continuously differentiable on an open neighborhood of a. To study a nearby point a+h, consider the one-variable function φ(t)=f(a+th) along the segment. The chain rule gives φ′(0)=∇f(a) dotted with h and φ″(t)=h transposed times Hf(a+th) times h. Applying one-variable Taylor expansion yields the quadratic model. The factor one half comes from integrating the linear change of the first derivative; it is not a special adjustment invented for multivariable notation.
For a twice continuously differentiable real function, mixed partial derivatives agree, so the Hessian is symmetric. The off-diagonal contribution occurs twice inside the quadratic form before multiplication by one half. For two coordinates, the quadratic term is one half fxx h² plus fxy hk plus one half fyy k². Forgetting that doubling produces an incorrect coefficient for cross terms. With only weaker differentiability assumptions, mixed derivatives need not agree; the regularity hypothesis should be stated rather than inferred from notation.
Curvature tests require both stationarity and definiteness
At a stationary point, the linear term vanishes. A positive definite Hessian makes the quadratic term positive in every nonzero direction, with a uniform positive lower bound on the unit sphere. Continuity of the Hessian ensures that the smaller remainder cannot overturn that sign sufficiently near the point, giving a strict local minimum. A negative definite Hessian gives a strict local maximum. An indefinite Hessian supplies directions of both signs and therefore a saddle. These are local conclusions and do not establish global optimality.
A positive semidefinite Hessian alone does not settle the question. The functions x⁴+y⁴ and x⁴−y⁴ have the same zero Hessian at the origin, but one has a strict minimum and the other a saddle. Higher-order terms decide the difference. In two dimensions, the determinant test is a convenient summary: positive determinant together with a positive first diagonal entry gives positive definiteness; negative determinant gives indefiniteness. A zero determinant is inconclusive. Applying this test away from a stationary point does not classify a local extremum there.
An error bound must cover the whole segment
The Hessian at a single point describes leading curvature but does not bound a finite step unless surrounding curvature is controlled. The integral remainder averages Hessian values along the segment with weight 1−t. Taking an operator-norm bound M over that segment gives the displayed estimate. If third derivatives are uniformly bounded in an appropriate multilinear norm, a quadratic-model error can instead be bounded by a constant times the cube of displacement length. Each extra accuracy claim therefore comes with an extra regularity and boundedness requirement.
For a quadratic polynomial, the Hessian is constant and the second-order formula is exact. Such examples are excellent for checking signs and factors, but their zero higher-order remainder should not be generalized to nonlinear functions. In optimization, an eigenvector direction of a symmetric Hessian identifies a principal curvature, while a small eigenvalue can signal a nearly flat direction. It does not automatically indicate numerical error or a symmetry; either interpretation needs additional analysis of the function and the scale of its coordinates.
Work it through
Read a cross term correctly
For f(x,y)=x²+2xy+3y², classify the origin and compute f(0.1,−0.2) using its Hessian.
The gradient is (2x+2y,2x+6y), which vanishes at the origin. The Hessian is [[2,2],[2,6]], with positive first diagonal entry and determinant 12−4=8.
Those conditions make the symmetric Hessian positive definite, so the origin is a strict local minimum. Completing the square gives f=(x+y)²+2y², showing this example also has a unique global minimum.
For h=(0.1,−0.2), Hh=(−0.2,−1). Then h·Hh=−0.02+0.2=0.18, and half of it is 0.09. Direct evaluation gives 0.01−0.04+0.12=0.09.
The quadratic prediction is exactly 0.09 because the function is quadratic. The mixed term contributes −0.04, not −0.02.
The living notebook
Explore the model
Interactive lab · function
Inspect the exact curvature remainder on a slice
Reset, set coefficient a=1 and position x=0.5. Predict height 0.25 and tangent slope 1 for y=ax². Compare with a second fresh run using Reset, a=1, x=0.25: height quarters while slope halves. This is a one-dimensional quadratic slice with constant second derivative 2a; it does not display a general Hessian or certify a multidimensional minimum.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forSecond-order changes scale quadratically along a fixed displacement direction.
Pause & reason
Check your understanding
Your turn
Try it yourself
Classify the origin for q(x,y)=x²+4xy+y². Then use the bound ‖Hq‖op=6 to certify an upper bound on |q(h)| for ‖h‖≤0.1.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use the directions (1,1) and (1,−1) to expose the Hessian eigenvalues before applying the remainder bound.
Compare with the solution
The Hessian [[2,4],[4,2]] has eigenvalues 6 and −2, so the origin is a saddle. The constant and linear terms vanish there. The Hessian bound gives |q(h)|≤(6/2)‖h‖²≤0.03. Equality is attainable on a displacement of norm 0.1 in the normalized direction (1,1), where the positive curvature is 6. Indefiniteness does not prevent an absolute-value error bound.
Keep the idea
Take it with you
- A symmetric Hessian acts through a quadratic form with a one-half Taylor factor.
- Definite curvature classifies a stationary point locally; semidefinite curvature may not.
- A finite-step error bound requires curvature control along the entire relevant segment.
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