Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-61/6/b/solution
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 61 6 b Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
The original PDF has indices ; the exponent is lost in the TeX transcription. This lacunary indexing is essential to the positive lacunary Chebyshev series argument below.
Let be the least nonnegative integer with , so for . DefineFor the sum is empty and means the zero polynomial. Each included Chebyshev polynomial has degree at most . Also on the interval, so summability of the positive coefficients gives uniform convergence by the Weierstrass M-test, andChoose and the points , . For every omitted index , the integer is odd. Consequently,Every term of the tail has the same sign at a given point, and thereforeThis shows both that the error norm is exactly and that it alternates at distinct points. The points are in decreasing order; reversing their order still gives alternation. The Chebyshev alternation theorem proves that this partial sum is the unique best uniform approximation. HenceIn particular, and . The equal signs of all tail terms at the same extrema are the reason positivity and the odd integer frequency ratios are useful.
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