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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/1/c/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 1 / c / Solution by

Codex  0 2026-10-18

The optimal approximation is the orthogonal projection onto the linear span of

p_{0}, \dots, p_{N}

. Their orthogonality gives

f - \sum_{n = 0}^{N} a_{n} p_{n}_{w}^{2} = ∥ f ∥_{w}^{2} - 2 \sum_{n = 0}^{N} a_{n} ⟨ f, p_{n} ⟩_{w} + \sum_{n = 0}^{N} a_{n}^{2} ∥ p_{n} ∥_{w}^{2} .

(1)

Completing each square proves both optimality and uniqueness:

a_{n} = \frac{⟨ f , p _{n} ⟩ _{w}}{∥ p _{n} ∥ _{w}^{2}} .

(2)

For real

f

, these are real coefficients. If complex-valued

f

is allowed while the coefficients remain real, replace the numerator by its real part and use the modulus-squared norm. The question's real formulation corresponds to real

f

Here the first three squared norms are one. Using

\int_{0}^{\infty} x^{j} e^{- 2 x} d x = j! / 2^{j + 1}

a_{0} a_{1} a_{2} = \int_{0}^{\infty} e^{- 2 x} (1 + x) d x = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}, = \int_{0}^{\infty} e^{- 2 x} (1 - x^{2}) d x = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}, = \int_{0}^{\infty} e^{- 2 x} (1 - x - \frac{3}{2} x^{2} + \frac{1}{2} x^{3}) d x = \frac{1}{2} - \frac{1}{4} - \frac{3}{8} + \frac{3}{16} = \frac{1}{16} .

(3)

Thus

(a_{0}, a_{1}, a_{2}) = (\frac{3}{4}, \frac{1}{4}, \frac{1}{16}), f_{2} (x) = \frac{17}{16} - \frac{3}{8} x + \frac{1}{32} x^{2} .

(4)

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