Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 168 1 i Solution Created 2026-09-24 Updated 2026-09-24
Write , , , and . The functions form the p-biased product measure orthonormal basis, so the Fourier expansion is . The normalized discrete derivative of a Boolean function satisfiesApplying Parseval identity and then exchanging two finite sums gives
The noise operator on the Boolean hypercube acts diagonally on the same basis: . Hence the noise stability isIts derivative isTaking the right-hand value at leaves exactly the linear Fourier weight , while taking the left-hand value at gives .
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 168 2 ii Solution Created 2026-09-24 Updated 2026-09-24
To prove it, put . Part (i), applied to each discrete derivative of a Boolean function , givesOn the other hand, expanding the noise stability in Fourier coefficients givesChoose and defineThe preceding bounds make the low-degree Fourier mass omitted by at most , while the hypothesis makes the high-degree mass at most . Thus . Finally,which gives the asserted bound on .
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 168 2 iv Solution Created 2026-09-24 Updated 2026-09-24
For a Boolean-valued , each discrete derivative of a Boolean function takes values in and has degree at most . If depends on coordinate , then is nonzero, so part (iii) givesSince has degree at most , the Fourier formula for total influence and Parseval identity giveIf coordinates affect , then , so . Thus is a -junta, which is the Nisan-Szegedy junta theorem.