 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/8/d/solution

N = ∣ n ∣

is an integer, the recurrence has zero numerator at

j = N

. Choosing the parity containing

a_{N}

therefore makes all later coefficients vanish. Conversely, for a nonzero polynomial of degree

N

x^{N}

(n^{2} - N^{2}) a_{N}

, so

n^{2} = N^{2}

is necessary.

Normalizing the terminating series by

y (1) = 1

T_{0} (x) = 1, T_{1} (x) = x, T_{2} (x) = 2 x^{2} - 1, T_{3} (x) = 4 x^{3} - 3 x .

Substitution verifies each differential equation and endpoint value. More generally,

T_{N} (x) = cos (N arccos x)

; its recurrence

T_{N + 1} = 2 x T_{N} - T_{N - 1}

also establishes polynomiality and the normalization.

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