Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 3 / 19C / b

Solution

Inductively,

p_{n}

is monic of degree

n

, because the leading term of

p_{n + 1}

comes only from

x p_{n}

. Assume

p_{0}, \dots, p_{n}

are mutually orthogonal. The definitions give

⟨ p_{n + 1}, p_{n} ⟩ = ⟨ x p_{n}, p_{n} ⟩ - α_{n} ⟨ p_{n}, p_{n} ⟩ = 0.

Using the preceding recurrence in the form

x p_{n - 1} = p_{n} + α_{n - 1} p_{n - 1} + β_{n - 1} p_{n - 2}

gives

⟨ x p_{n}, p_{n - 1} ⟩ = ⟨ p_{n}, x p_{n - 1} ⟩ = ⟨ p_{n}, p_{n} ⟩,

⟨ p_{n + 1}, p_{n - 1} ⟩ = 0

by the definition of

β_{n}

. If

j \leq n - 2

, then

x p_{j}

lies in the span of

p_{j - 1}, p_{j}, p_{j + 1}

, all orthogonal to

p_{n}

, and the remaining recurrence terms are also orthogonal to

p_{j}

. Hence

⟨ p_{n + 1}, p_{j} ⟩ = 0

. Induction proves that the recurrence defines monic orthogonal polynomials.

There are no discussions about this article yet.

There are currently no matching articles.