You are interpolating f(x) = 1/(1 + 25x²) on [−1, 1] using a degree-20 polynomial with 21 equally spaced nodes. You add 10 more equally spaced nodes to get a degree-30 polynomial. What most likely happens to the interpolation error near the endpoints?
AThe error decreases uniformly because more nodes always improve accuracy
BThe error near the endpoints increases dramatically due to the large values of |ω(x)| there for equally spaced nodes
CThe error stays roughly the same because both polynomials have the same function f
DThe error decreases near the endpoints but increases near the center
This is Runge's phenomenon. For equally spaced nodes, |ω(x)| is very small near the interval center but grows enormously near the endpoints. As you add more equally spaced nodes, this endpoint amplification gets worse, and the interpolating polynomial oscillates wildly there. The interpolation actually deteriorates near x = ±1 even as you add nodes. This is why node placement — not just node count — determines interpolation quality.
Question 2 Multiple Choice
In the interpolation error bound |E(x)| ≤ M/(n+1)! · |ω(x)|, which factor can a numerical analyst directly control by design choices?
AM = max|f^{(n+1)}|, by choosing a smoother function f
B|ω(x)| = |∏(x − xᵢ)|, by choosing where to place the interpolation nodes
CBoth M and |ω(x)| equally, since both depend on the polynomial degree
DNeither factor — both are determined entirely by the function f
M = max|f^{(n+1)}| depends only on the function being interpolated — you cannot change it. But |ω(x)| depends entirely on where you place the interpolation nodes, which is a design choice. This is why error analysis is a design tool: Chebyshev nodes are chosen precisely to minimize max|ω(x)| over the interval, giving the smallest possible worst-case error from the controllable factor.
Question 3 True / False
Replacing equally spaced nodes with Chebyshev nodes on [−1, 1] reduces the maximum value of |ω(x)| exponentially in n.
TTrue
FFalse
Answer: True
For n+1 Chebyshev nodes, max|ω(x)| over [−1, 1] equals 1/2ⁿ — exponentially small in n. For equally spaced nodes, the maximum of |ω(x)| grows rapidly. Chebyshev nodes cluster near the endpoints, where |ω| would otherwise be large, effectively spreading the node polynomial's values more evenly across the interval.
Question 4 True / False
Adding more interpolation nodes typically reduces the interpolation error for any function.
TTrue
FFalse
Answer: False
Runge's phenomenon shows this is false. For equally spaced nodes and certain functions (like Runge's function f(x) = 1/(1+25x²)), the interpolation error near the endpoints grows without bound as you add more nodes. The error depends on both M/(n+1)! (which decreases) and max|ω(x)| (which increases rapidly for equally spaced nodes). The product can diverge, making interpolation worse.
Question 5 Short Answer
The interpolation error bound |E(x)| ≤ M/(n+1)! · |ω(x)| separates into two factors. What distinct insight does each factor provide, and why does that separation matter practically?
Think about your answer, then reveal below.
Model answer: The factor M/(n+1)! measures the function's intrinsic complexity — how much its (n+1)-th derivative fluctuates — which is a property of f that the analyst cannot change. The factor |ω(x)| = |∏(x − xᵢ)| measures how the chosen node placement spreads the interpolation across the interval, which the analyst can control. The separation matters because it identifies what is within the analyst's power: by choosing Chebyshev nodes instead of equally spaced nodes, you minimize max|ω(x)|, dramatically reducing the worst-case error regardless of f's behavior.
This two-factor structure turns error analysis from a passive measurement into an active design tool. You cannot improve the bound by making f smoother, but you can choose nodes that keep |ω(x)| small everywhere — and the error bound tells you exactly how much improvement to expect.