Questions: Runge's Phenomenon

Question 1

A student interpolates f(x) = 1/(1+25x²) on [−1, 1] with 15 equally spaced nodes and observes large oscillations near x = ±1. She adds 10 more equally spaced nodes hoping to fix the problem. What will most likely happen?

• A. The oscillations near the endpoints will decrease as the polynomial better fits the smooth function
• B. The oscillations near the endpoints will get worse — adding more equally spaced nodes increases the maximum of the node product near the endpoints
• C. The oscillations will disappear because 25 nodes are always sufficient for any smooth function
• D. The error will decrease uniformly across the whole interval

Answer: (B)

This is Runge's phenomenon: for f(x) = 1/(1+25x²) with equally spaced nodes, the interpolation error grows without bound near the endpoints as the degree increases. Adding more equally spaced nodes makes the node product |ω_{n+1}(x)| larger near x = ±1, not smaller, because equally spaced nodes leave large gaps at the endpoints while packing interior nodes tightly. The function's complex singularities at x = ±i/5 amplify this problem. The correct response is not more nodes but different node placement (Chebyshev nodes) or a different interpolation strategy (splines).

Question 2

What is the fundamental reason that Chebyshev nodes reduce interpolation error compared to equally spaced nodes?

• A. Chebyshev nodes avoid placing points near the endpoints where the function may be undefined or have large values
• B. Chebyshev nodes minimize the maximum value of the node product |ω_{n+1}(x)| over the interval by clustering nodes near the endpoints where the product would otherwise be largest
• C. Chebyshev nodes are computed using a numerically stable algorithm that equally spaced nodes lack
• D. Chebyshev nodes are optimal only for polynomials with no complex singularities

Answer: (B)

The interpolation error bound is proportional to |ω_{n+1}(x)| = |(x−x₀)(x−x₁)···(x−xₙ)|. Equally spaced nodes leave large gaps near the endpoints, making this product large there. Chebyshev nodes cluster near the endpoints precisely because the endpoints are where the node product is hardest to control — they balance the product across the entire interval, reducing its maximum value to 2^{−n}. This is the minimax property: Chebyshev nodes minimize the worst-case node product over the interval.

Question 3

Runge's phenomenon can occur even for functions that are infinitely differentiable and have no singularities on the real interval being interpolated.

• T. True
• F. False

Answer: True

f(x) = 1/(1+25x²) is infinitely differentiable on [−1,1] — it has no real singularities on the real line. The problem arises from complex singularities at x = ±i/5, which are close to the real axis. These cause the higher derivatives to grow rapidly even though the function looks smooth. Runge's phenomenon is a reminder that smoothness on the real line is not sufficient to guarantee convergence of high-degree polynomial interpolation — the behavior in the complex plane also matters. This is precisely what makes the phenomenon surprising and important.

Question 4

Increasing the degree of a polynomial interpolant by adding more equally spaced nodes is typically a reliable strategy for improving approximation accuracy over the entire interval.

• T. True
• F. False

Answer: False

This is exactly what Runge's phenomenon disproves. For functions with complex singularities near the real axis (like 1/(1+25x²)), polynomial interpolation on equally spaced nodes diverges near the endpoints as the degree increases — the approximation actively gets worse, not better. The key lesson is that accuracy depends on both the number of nodes AND their placement. More nodes with poor placement can increase error. Chebyshev nodes or piecewise polynomial methods (splines) are needed for reliable high-accuracy approximation.

Question 5

Why does the placement of interpolation nodes matter as much as their number, and how does the node product |ω_{n+1}(x)| explain why equally spaced nodes cause oscillations near the endpoints of the interval?

Model answer: The interpolation error at a point x is proportional to |ω_{n+1}(x)| = |(x−x₀)(x−x₁)···(x−xₙ)|, the product of distances from x to each node. With equally spaced nodes on [−1,1], the interior nodes are tightly clustered but the endpoints have no nearby nodes to make the factors (x−xᵢ) small. Near x = ±1, several factors are large simultaneously, making the node product large. With more equally spaced nodes, this worsens because the interior becomes more crowded but the endpoint region remains relatively sparse. Chebyshev nodes fix this by clustering at the endpoints, making the factors near x = ±1 small and distributing the node product's size more uniformly across the interval.

The insight is that the node product cannot be made uniformly small — it must be large somewhere. Chebyshev nodes distribute this unavoidable largeness as evenly as possible (minimax). Equally spaced nodes stack all the largeness at the endpoints, which is the worst possible choice for bounding the maximum error.