The smooth function f(x)=1/(1+25x²) on [−1,1] is a standard warning. Equally spaced high-degree interpolants can oscillate severely near the endpoints even while matching every node exactly. Increasing the degree is therefore not a universal repair. The problem is not necessarily rounding: the exact interpolation polynomials themselves can behave badly. Compare a node residual of zero with an off-node error before declaring success.
Predict
Predict whether changing from expanded coefficients to Newton form removes this phenomenon. It may improve computation, but the unique exact polynomial is the same, so its approximation behavior cannot change. Changing nodes or using piecewise low-degree approximations changes the approximation itself. This separates representation, conditioning and approximation strategy, three decisions that are often collapsed into one vague instruction to use a better polynomial.
Experiment
Use a small weight comparison to inspect data amplification.
Evaluation location
Cardinal weights
Sum of absolute weights
At a node
1,0,0
1
Between/outside selected nodes
−1/8,3/4,3/8
5/4
For the second row, choose data-error signs matching the weight signs to attain the bound with equal error magnitudes. This table is a local sensitivity example, not a numerical demonstration of the entire Runge sequence.
Understand
Let Iₙ be interpolation at n+1 nodes and Λₙ=maxₓΣ|Lⱼ(x)| on the interval. Perturbing data gives |δp(x)|≤max|δyⱼ|Σ|Lⱼ(x)|. For any polynomial q of degree at most n, Iₙq=q, so f−Iₙf=(f−q)−Iₙ(f−q). Taking norms gives ‖f−Iₙf‖∞≤(1+Λₙ)‖f−q‖∞, and then the best-approximation bound. Thus good approximation availability and controlled amplification both matter. Polynomial density alone supplies no rate to offset an increasing Λₙ.
Apply
Chebyshev polynomials satisfy Tₘ(cos θ)=cos(mθ), so |Tₘ(x)|≤1 on [−1,1]. For m=n+1 their roots are cos((2j+1)π/(2n+2)). Since Tₙ₊₁ has leading coefficient 2ⁿ, the monic node product is 2^(−n)Tₙ₊₁, uniformly bounded by 2^(−n). This controls the product in the interpolation remainder and clusters nodes near endpoints where equispaced interpolation is vulnerable. It does not bound high derivatives of f automatically. The minimax property of this monic product follows from alternating extrema: a smaller monic product would differ by degree at most n yet change sign between n+2 alternating extremal points, impossible by root counting.
The endpoint clustering formula is easy to inspect without a large plot. For three Chebyshev roots, the nodes are √3/2,0,−√3/2. Their monic product is x³−3x/4=T₃(x)/4, and its maximum absolute magnitude on [−1,1] is 1/4. Compare this with x(x−1)(x+1), the product for equally spaced nodes −1,0,1. The small example illustrates a node-product calculation; it does not itself establish the asymptotic Runge phenomenon.
Continue
In a practical comparison, report node placement, degree, off-node evaluation grid, maximum observed error and whether a theorem supplies a true bound. A sampled maximum is only a lower bound on the continuum maximum unless extra analysis is supplied. Barycentric evaluation is an important stable implementation technique, while spline interpolation limits degree locally. Neither removes the need to check assumptions. Next, cubic splines trade one global high-degree polynomial for low-degree pieces joined with smoothness conditions.
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At an evaluation point, cardinal weights are −1/4,3/2,−1/4. Find their absolute sum.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
If the absolute weight sum is 5 and every data error is at most 1/100, bound output error.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Derive |δp(x)|≤max_j|δy_j| Σ_j|L_j(x)|.
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Prove ‖f−I_nf‖∞≤(1+Λ_n) inf_deg(q)≤n ‖f−q‖∞ for interpolation operator I_n and Lebesgue constant Λ_n.
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Why do Chebyshev roots cluster near endpoints, and what does that accomplish?
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Repair the claim that Chebyshev interpolation converges uniformly for every continuous function solely because polynomials are dense.
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Reflect and continue
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