 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2011/p/2/a/solution

x = cos θ

with

0 \leq θ \leq π

. The cosine addition identity gives

cos ((n + 1) θ) + cos ((n - 1) θ) = 2 cos θ cos (n θ),

hence

T_{0} = 1

T_{1} = x

, and

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

T_{n}

is a polynomial of degree

n

: the initial cases hold, and the leading term of

2 x T_{n}

has degree

n + 1

while the subtraction has lower degree. Its leading coefficient is

2^{n - 1}

for

n \geq 1

The same change of variables gives the weighted integral

\int_{- 1}^{1} \frac{T _{n} ( x )}{1 - x ^{2}} d x = \int_{0}^{π} cos (n θ) d θ = {π, 0, n = 0, n \geq 1.

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