 Solution

ID: past-exam-of-the-analysis-course-of-the-university-of-oxford/2007/b/6/a/solution

Differentiating the Rodrigues identity gives

H_{0} = 1

and

H_{n + 1} = 2 t H_{n} - H_{n}^{'}

. Induct with the proposed finite sum. The coefficient at

t^{n + 1 - 2 j}

in this recurrence is the contribution from

2 t H_{n}

minus the derivative contribution with previous index

j - 1

\frac{n ! ( - 1 ) ^{j} 2 ^{n + 1 - 2 j}}{j ! ( n - 2 j )!} - \frac{n ! ( - 1 ) ^{j - 1} 2 ^{n - 2 j + 2}}{( j - 1 )! ( n - 2 j + 1 )!} .

Using a common denominator, this equals

(n + 1)! (- 1)^{j} 2^{n + 1 - 2 j} / [j! (n + 1 - 2 j)!]

. At an endpoint omit terms whose factorial indices would be negative; the same formula then covers the sole remaining term. Thus

H_{n} (t) = n! j = 0 \sum ⌊ n /2 ⌋ \frac{( - 1 ) ^{j} 2 ^{n - 2 j}}{j ! ( n - 2 j )!} t^{n - 2 j} .

The initial value and recurrence identify these polynomials with the given Rodrigues definition.

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