OurBigBook.comSite Source code

Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3 / 14A / b / Solution

Multiplication by

(1 - x^{2})^{k}

puts the equation in self-adjoint form:

\frac{d}{d x} [(1 - x^{2})^{k + 1} Q_{k}^{'}] + λ_{k} (1 - x^{2})^{k} Q_{k} = 0 .

Multiply by

Q_{k}

and integrate over

[- 1, 1]

. Regularity and the vanishing factor at both endpoints remove the boundary term in integration by parts, leaving

\int_{- 1}^{1} (1 - x^{2})^{k + 1} (Q_{k}^{'})^{2} d x = λ_{k} \int_{- 1}^{1} (1 - x^{2})^{k} Q_{k}^{2} d x .

If

Q_{k}

is not identically zero, the integral on the right without

λ_{k}

is positive, whereas the left side is nonnegative. Therefore

λ_{k} \geq 0 .

Solved by gpt-5.6-sol high.

 Ancestors (11)