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 Incoming links: Debye asymptotic for Legendre polynomials

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 30D / b / iii / Solution Created 2026-09-24 Updated 2026-09-25

At the two saddles,

\frac{z _{\pm}^{2} - 1}{z _{\pm} - t} = 2 z_{\pm} = 2 e^{\pm i θ}, f (z_{\pm}, t) = \frac{1}{z _{\pm} - t} = \mp \frac{i}{s i n θ},

and

∣ ϕ^{''} (z_{\pm}) ∣ = \frac{1}{s i n θ} .

Applying the method of steepest descent with the oriented angles from part (ii), and including the prefactor

1/ (2^{n + 1} πi)

, gives two conjugate contributions. If

Φ = (n + \frac{1}{2}) θ - \frac{π}{4},

their sum is

\frac{1}{2 π} \frac{2 π}{n s i n θ} (e^{i Φ} + e^{- i Φ}) .

Therefore the Debye asymptotic for Legendre polynomials is

P_{n} (cos θ) \sim \frac{2}{πn sin θ} cos (n θ + \frac{θ}{2} - \frac{π}{4}) .

Thus

A (θ) = \frac{2}{πn sin θ} .

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Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 336 / 1 / 1 / Solution Created 2026-09-24 Updated 2026-09-25

Write the Schläfli contour integral for Legendre polynomials as

P_{n} (μ) = \frac{1}{2 πi} \oint \frac{e ^{n ϕ (z)}}{2 ^{n} ( z - μ )} d z, ϕ (z) = lo g \frac{z ^{2} - 1}{z - μ} .

The saddle points satisfy

ϕ^{'} (z) = \frac{2 z}{z ^{2} - 1} - \frac{1}{z - μ} = 0, z^{2} - 2 μ z + 1 = 0.

For

μ = cos θ

, they are

z_{\pm} = e^{\pm i θ}

. Deforming the contour through both conjugate saddles and using the local Gaussian contributions from the method of steepest descent gives conjugate exponentials. Their sum is the Debye asymptotic for Legendre polynomials

P_{n} (cos θ) \sim \frac{2}{πn sin θ} cos [(n + \frac{1}{2}) θ - \frac{π}{4}] .

Both saddles are required for the real oscillatory answer.

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