Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 6B (Numerical Analysis)

Table of contents
- a 6B
  - Solution a
- b 6B
  - Solution b

a

Solution

The nonzero orthogonal polynomials

Q_{0}, \dots, Q_{n}

have distinct degrees and form an orthogonal basis of

P_{n}

p_{n}^{*} = \sum_{k = 0}^{n} \frac{⟨ f , Q _{k} ⟩}{⟨ Q _{k} , Q _{k} ⟩} Q_{k} .

For each

j \leq n

,

⟨ f - p_{n}^{*}, Q_{j} ⟩ = ⟨ f, Q_{j} ⟩ - \frac{⟨ f , Q _{j} ⟩}{⟨ Q _{j} , Q _{j} ⟩} ⟨ Q_{j}, Q_{j} ⟩ = 0.

Thus the residual

f - p_{n}^{*}

is orthogonal to all of

P_{n}

.

For any

p \in P_{n}

, write

f - p = (f - p_{n}^{*}) + (p_{n}^{*} - p) .

The two terms are orthogonal, so the Pythagorean theorem in an inner-product space gives

∥ f - p ∥^{2} = ∥ f - p_{n}^{*} ∥^{2} + ∥ p_{n}^{*} - p ∥^{2} \geq ∥ f - p_{n}^{*} ∥^{2} .

Equality holds only for

p = p_{n}^{*}

. This proves the formula and uniqueness of the least-squares polynomial in an orthogonal-polynomial basis.

b

Solution

Part (a) shows that

f - p_{n}^{*}

is orthogonal to

p_{n}^{*} \in P_{n}

. Since

f = (f - p_{n}^{*}) + p_{n}^{*},

the Pythagorean theorem in an inner-product space immediately gives

∥ f ∥^{2} = ∥ f - p_{n}^{*} ∥^{2} + ∥ p_{n}^{*} ∥^{2} .

This is the norm decomposition associated with orthogonal projection onto a finite-dimensional subspace.

 Ancestors (9)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version