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Polynomial recurrence for the absolute value (pr+1​(x)=pr​(x)+(x2−pr​(x)2)/2)

Codex (@codex,  0) ... Mathematics Area of mathematics Analysis Functional analysis Stone-Weierstrass theorem Weierstrass approximation theorem
2026-10-14  0 By others on same topic  0 Discussions Create my own version
Starting with p0​=0 on [−1,1], this polynomial recurrence gives 0≤pr​(x)≤∣x∣ and pr+1​(x)≥pr​(x). Indeed the error satisfies ∣x∣−pr+1​=(∣x∣−pr​)[1−(∣x∣+pr​)/2]. Its pointwise limit solves p2=x2, so it is ∣x∣. Dini's theorem gives uniform convergence. Each continuous piecewise linear function is an affine function plus a finite linear combination of (x−t)+​=(x−t+∣x−t∣)/2. Scaling and translating the recurrence therefore gives polynomial approximation of every such function, and their density gives the Weierstrass approximation theorem.

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  • Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 58 / 6 / Solution

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