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 Incoming links: Mixed finite difference of a polynomial

Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 8G / b / Solution Created 2026-09-24 Updated 2026-09-25

The eigenvalue equation

D_{r} f = λ f

is

f (x + r) = (1 + λ) f (x) .

If

λ = - 1

, this forces

f (x + r) = 0

for every

x

, hence

f = 0

, so

- 1

is not an eigenvalue. If

c = 1 + λ \neq = 0

, choose arbitrary values on representatives of the cosets of

r Z

and extend by

f (t + k r) = c^{k} f (t), k \in Z .

This produces nonzero eigenfunctions. Choosing functions supported on distinct cosets gives infinitely many linearly independent eigenfunctions. Therefore

spec_{point} (D_{r}) = R ∖ {- 1},

and every eigenspace is infinite-dimensional.

Direct expansion gives

D_{r} D_{s} f (x) = f (x + r + s) - f (x + r) - f (x + s) + f (x),

which is symmetric in

r, s

. Hence

D_{r} D_{s} = D_{s} D_{r} .

Suppose a degree-

n

p

were a sum of

n

periodic functions,

p = f_{1} + \dots + f_{n}, D_{r_{i}} f_{i} = 0, r_{i} \neq = 0.

Apply the commuting product

D_{r_{1}} \dots D_{r_{n}}

. Every term on the right is killed by its corresponding factor, whereas for leading coefficient

a_{n}

the mixed finite difference of a polynomial gives

D_{r_{1}} \dots D_{r_{n}} p = n! a_{n} \prod_{i = 1}^{n} r_{i} \neq = 0.

This contradiction proves

p cannot be a sum of n periodic functions .

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