 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ib/paper-4/5a/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-05

Evaluate

R_{n}

x = 1

using

(x^{2} - 1)^{n} = (x - 1)^{n} (x + 1)^{n}

. In the Leibniz rule for the

n

th derivative, every term vanishes at

1

except the term putting all

n

(x - 1)^{n}

. Therefore

R_{n} (1) = n! 2^{n}

. Combining this with

P_{n} (1) = 1

gives the normalisation in the Rodrigues formula:

α_{n} = \frac{1}{2 ^{n} n !}, P_{n} (x) = \frac{1}{2 ^{n} n !} \frac{d ^{n}}{d x ^{n}} (x^{2} - 1)^{n} .

For

n = 0

, the formula gives

α_{0} = 1

, as required. As an independent check, the leading term of

R_{n}

(2 n)! x^{n} / n!

, so the resulting leading coefficient of

P_{n}

(2 n)! / (2^{n} (n!)^{2})

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