Past exam of the mathematics course of the University of Cambridge 2009 ib Paper 1 14B Solution Created 2026-09-24 Updated 2026-10-14
Write with . Substitution into the ordinary differential equation givesUnless the series terminates, its coefficient ratio is , so it converges for every finite . A coefficient first vanishes exactly when is a nonnegative integer , and then , . Thus is a polynomial precisely for , in which caseis the normalized Laguerre polynomial.
Multiplying the differential equation by gives . Multiply by , subtract the equation with the indices exchanged, and integrate over . The boundary term vanishes both at zero and infinity. ThereforeThis is the weighted orthogonality of the Laguerre polynomials.
Their leading coefficients show . Since have distinct degrees, they form a basis of that polynomial space. Hence , and comparison of leading coefficients gives . Likewise the degrees of and for run through and , respectively. A nontrivial linear combination cannot cancel its highest-degree term, so these functions are linearly independent.
For , this gives the basis of the cubic polynomial space, establishing the specified expansion. Its weighted integral is : integrates to one, and are orthogonal to , and integrates to zero. At the two roots of , the expansion reduces to . Solving these two equations for givesFor explicit verification, the recurrence gives , . Thus , are distinct positive roots and , . This is the two-node Gauss-Laguerre quadrature rule, exact for every cubic.