Past exam of the mathematics course of the University of Cambridge 2004 ib Paper 4 6C Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2005 ib Paper 2 18F b Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2006 ia Paper 1 8A i Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2007 ii Paper 1 2F ii Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2007 ii Paper 1 2F i Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2008 ib Paper 4 8D Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2009 ii Paper 2 2F b Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2010 ii Paper 4 2F Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 61 4 b Solution Created 2026-10-03 Updated 2026-10-07
The first Jackson theorem for periodic approximation asserts that a universal constant satisfiesfor every continuous -periodic function , where the infimum is over degree-at-most- trigonometric polynomials.
Apply it to the even function . If is a trigonometric polynomial approximating , its even parthas no larger error, because and the triangle inequality bounds each averaged error by . Every even trigonometric polynomial has the formHere the Chebyshev polynomials have algebraic degree , so has degree at most . Surjectivity of cosine givesConversely, every algebraic polynomial of degree at most yields such an even trigonometric polynomial. Taking infima therefore proves the exact identity , not merely an inequality. Combine this with the periodic theorem and the preceding modulus of continuity estimate:Symmetrization, the degree correspondence and the equality of norms justify every step in transferring the Jackson-type estimate.