Questions: Chebyshev Nodes and Optimal Interpolation

Question 1

You interpolate f(x) = 1/(1 + 25x²) on [-1,1] using a degree-10 polynomial. With equally-spaced nodes, the interpolant oscillates wildly near the endpoints. After switching to Chebyshev nodes, the oscillation disappears. What feature of Chebyshev nodes explains this?

• A. Chebyshev nodes avoid the interval endpoints entirely, where error is largest
• B. Chebyshev nodes cluster near the endpoints, reducing |ω(x)| precisely where it would otherwise blow up
• C. Chebyshev nodes are more evenly spaced than equally-spaced nodes, distributing error uniformly
• D. Chebyshev nodes use a higher-degree polynomial near the endpoints to compensate for curvature

Answer: (B)

The interpolation error bound involves |ω(x)| = |(x−x_0)(x−x_1)···(x−x_n)|. With equally-spaced nodes, this product becomes very large near x = ±1, causing Runge oscillation. Chebyshev nodes are deliberately *denser* near the endpoints — they cluster where the error would otherwise explode. This is counterintuitive: the fix is not more uniform spacing but strategic non-uniform spacing that places nodes close to the danger zones. The nodes are not 'more uniform' — they are non-uniform in exactly the right way.

Question 2

What is the mathematical reason Chebyshev nodes are optimal for polynomial interpolation on [-1, 1]?

• A. They minimize the degree of the interpolating polynomial needed for a given accuracy
• B. They are the roots of T_{n+1}(x), which is the monic degree-(n+1) polynomial with smallest possible maximum on [-1,1]
• C. They minimize the average error ∫|f(x) − p(x)|dx rather than the maximum error
• D. They are equidistant in the Chebyshev metric, corresponding to the L² norm on [-1,1]

Answer: (B)

The node error product ω(x) is monic of degree n+1 (leading coefficient 1, since each root contributes one factor). The Chebyshev polynomial T_{n+1}(x)/2^n is the unique monic polynomial of degree n+1 with the smallest possible maximum on [-1,1] — the minimax property. Choosing the roots of T_{n+1} as the interpolation nodes makes ω(x) equal to this minimax polynomial, achieving the tightest possible error bound. No other choice of n+1 nodes can do better in the minimax sense.

Question 3

Chebyshev nodes are optimal because they distribute interpolation nodes uniformly across the interval, ensuring equal spacing between adjacent nodes.

• T. True
• F. False

Answer: False

The opposite is true. Chebyshev nodes are *non-uniformly* spaced: denser near the endpoints and sparser near the center. This deliberate non-uniformity is what makes them optimal. Uniform spacing is the source of Runge's phenomenon, not the solution to it. Chebyshev nodes are geometrically the projections of equally-spaced points on the upper semicircle onto the x-axis — a construction that naturally produces endpoint clustering.

Question 4

For smooth functions, using Chebyshev nodes guarantees that the interpolating polynomial converges to the function as the number of nodes increases, whereas equally-spaced nodes cannot provide this guarantee.

• T. True
• F. False

Answer: True

Convergence with equally-spaced nodes is not guaranteed even for smooth functions — Runge's example (1/(1+25x²)) is smooth but the equally-spaced interpolant diverges near the endpoints. Chebyshev nodes make max|ω(x)| = 1/2^n (where n+1 is the node count), which decreases fast enough to overcome the growth of higher derivatives for smooth functions, guaranteeing convergence. This is a major practical advantage: you can add more Chebyshev nodes and be confident the approximation improves.

Question 5

Why is minimizing max|ω(x)| = max|(x−x_0)(x−x_1)···(x−x_n)| the key problem in choosing interpolation nodes, and what makes this quantity controllable?

Model answer: The interpolation error satisfies |f(x) − p(x)| ≤ (max|f^(n+1)|/(n+1)!) · |ω(x)|. The first factor depends only on the function being interpolated and cannot be controlled by node placement. The second factor — max|ω(x)| — depends entirely on where the nodes are placed. Minimizing max|ω(x)| is therefore the only part of the error bound we can optimize. Since ω(x) must be monic of degree n+1 (its roots are exactly the nodes), choosing the roots of the minimax monic polynomial T_{n+1}(x)/2^n achieves the smallest possible value of max|ω(x)|.

This is why Chebyshev node optimality is not a heuristic improvement but a mathematically provable best strategy. The Chebyshev polynomial T_{n+1}(x)/2^n achieves maximum value 1/2^n on [-1,1], the smallest possible for any monic degree-(n+1) polynomial. Since ω(x) must be monic, choosing its roots to be the Chebyshev nodes makes ω(x) equal to this minimax polynomial. The result is the tightest possible interpolation error bound across the entire interval.