Best approximation and Chebyshev equioscillation
Choose a norm before optimizing, prove a minimax certificate from alternating extrema, and distinguish optimality from interpolation.
On this page
The question for today
What you’ll understand
- Define best uniform polynomial approximation and justify existence.
- Apply the precise polynomial equioscillation criterion.
- Certify a minimax example using alternating residual extrema.
An approximation is good only relative to a stated error measure. A polynomial that minimizes average squared error can leave a large local error, while a uniform approximation controls the worst point. We begin with the latter problem on a closed interval, where a remarkable certificate replaces comparison against infinitely many competing polynomials: the largest residual must alternate in sign often enough.
A finite-dimensional optimization inside a function space
Let f be a real continuous function on a nondegenerate compact interval , and let contain real polynomials of degree at most n, with . Its best uniform error is the infimum of the maximum absolute residual. The word uniform matters: every point has equal authority to set the objective, regardless of how narrow the region of large error is. Changing the interval or introducing a weight changes the optimization problem.
A best polynomial exists. A minimizing sequence can be restricted to a bounded uniform-norm ball because its residuals are bounded and f is fixed. On the finite-dimensional space , all norms are equivalent, so its coefficient vectors are bounded and have a convergent subsequence. The corresponding polynomials converge uniformly; continuity of the norm then attains the infimum. Compactness of the interval and finite dimension each play a role here; arbitrary infinite-dimensional approximation sets need not have this property.
Alternating worst errors certify optimality
For positive error, the polynomial equioscillation theorem says p is the unique best approximant exactly when there are at least n+2 ordered points where its residual reaches the same maximal magnitude with alternating signs. We state the full characterization; the next argument proves its optimality direction. If f already belongs to the polynomial space, its best approximation is f itself with error zero, and the positive-error formulation is unnecessary.
Suppose another polynomial q had strictly smaller uniform error than p. At each alternating extremum, q−p would have to have the sign of f−p, otherwise q would fail to reduce that extreme residual. The intermediate value theorem would force a root of q−p between every consecutive pair, giving at least n+1 distinct roots. A nonzero polynomial of degree at most n cannot have so many roots. This contradiction proves no strict improvement is possible; uniqueness is an additional part of the full equioscillation theorem.
A certificate is stronger than a plot
Interpolating n+1 data points enforces zero residual at those nodes, whereas minimax approximation balances extreme residuals across the entire interval. The two can agree in special cases but solve different constraints. Chebyshev polynomials provide useful exact examples: has magnitude at most one on the interval and alternates between plus and minus one at its successive extremal points. A scaled copy can therefore expose an exact best error.
In computation, a dense residual plot suggests extremal locations but does not certify that a sharper spike was missed. A rigorous certificate also bounds the residual between the sampled points, perhaps by derivative information or validated interval computations. Numerical Remez methods search for alternating extrema, but a successful-looking iteration is not a proof without a checked error bound. Separate the mathematical optimization problem, its exact optimality certificate, and the numerical method used to search for that certificate.
Work it through
The best straight line for a quadratic
Find the best uniform approximation to by a polynomial of degree at most one on . Prove the error, rather than only guessing a horizontal line.
Try . Its residual lies between minus one half and plus one half on the whole interval, so its uniform error is one half.
At x=−1,0,1, the residuals are respectively +1/2,−1/2,+1/2. These are three ordered alternating extremes, exactly n+2 when the permitted degree is one.
Any strictly better line would need its difference from p to be positive at both endpoints and negative at zero, forcing two distinct roots in a nonzero degree-one polynomial. That is impossible. The equioscillation theorem also supplies uniqueness.
The unique best line is the constant one half, with maximum error one half; allowing a slope does not improve the worst case.
The living notebook
Explore the model
Interactive lab · function
Inspect the three extreme residuals
Pause and Reset, then set coefficient a = 1 and position x = −1; predict the displayed quadratic value is 1. Inspect x = 0 and x = 1 without playing, subtracting the proposed constant 0.5 on paper. Predict residuals +0.5,−0.5,+0.5, then compare. Reset and set both a = 0.5 and x = 1 to compare the scaled target. The instrument plots the quadratic, not an adjustable minimax approximant or its residual.
- 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 forAlternation needs equal extreme magnitudes and a global residual bound; three chosen sample values alone are not a general minimax test.
Pause & reason
Check your understanding
Your turn
Try it yourself
Show that is the best uniform degree-at-most-two approximant to on . Use and list enough alternating extremal points.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Express the residual as one quarter of T₃. Its extrema occur at cos(jπ/3), but write them in increasing order before checking signs.
Compare with the solution
The residual is , so its magnitude never exceeds one quarter. At , its values are respectively . These four ordered alternating extrema meet n+2 for n=2. Equioscillation therefore proves the stated polynomial is the unique best approximant and . Its degree is only one, which is allowed because the space includes all degrees at most two.
Keep the idea
Take it with you
- Best approximation depends on both the approximation space and the chosen norm.
- Positive-error polynomial minimax approximation has an n+2-point alternating certificate.
- A numerical residual plot helps find candidates but needs a global bound to certify them.
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