 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/8/v/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 / 2013 / Paper 1 / 8 / v / Solution by

Codex  0 2026-10-18

For normalized Legendre polynomials,

P_{n} (1) = 1

and

P_{n} (- 1) = (- 1)^{n}

. Near the endpoints the integral factor has the asymptotics

\int^{x} \frac{d t}{( 1 - t ^{2} ) P _{n} ( t ) ^{2}} = - \frac{1}{2} lo g ∣1 - x ∣ + O (1) (x \to 1),

and

\int^{x} \frac{d t}{( 1 - t ^{2} ) P _{n} ( t ) ^{2}} = \frac{1}{2} lo g ∣1 + x ∣ + O (1) (x \to - 1) .

Thus an independent solution, for which

C_{2} \neq = 0

, has the logarithmic endpoint singularities

y = - \frac{C _{2}}{2} lo g ∣1 - x ∣ + O (1) (x \to 1), y = \frac{( - 1 ) ^{n} C _{2}}{2} lo g ∣1 + x ∣ + O (1) (x \to - 1) .

The constants are understood for a solution continued consistently across the ordinary interior points. Adding any multiple of

P_{n}

cannot cancel these logarithms. This explains why boundedness at an endpoint selects the Legendre polynomial and excludes the independent Legendre function of the second kind.

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