 Solution

ID: past-exam-of-the-approximation-theory-and-finite-element-methods-course-of-the-university-of-oxford/2009/a/1/d/solution

T_{0} = 1

T_{1} = x

, and

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

. Induction shows that

T_{j}

has degree

j

with leading coefficient

2^{j - 1}

for

j \geq 1

. The trigonometric definition also gives

T_{n}^{'} (cos θ) = n \frac{s i n ( n θ )}{s i n θ} .

Therefore

f_{n} (x) = \frac{2 ^{1 - n}}{n} (x^{2} - 1) T_{n}^{'} (x) .

The leading coefficient is

2^{1 - n} n^{- 1} \cdot n 2^{n - 1} = 1

, and the degree is

n + 1

. The endpoints are roots through

x^{2} - 1

; the other

n - 1

points

cos (s π / n)

are roots through

T_{n}^{'}

. They are distinct, so this monic polynomial is exactly

π_{n}

Differentiating the product

π_{n} (x) = (x - x_{s}) π_{n, s} (x)

x_{s}

gives

π_{n}^{'} (x_{s}) = π_{n, s} (x_{s}) = w_{s}^{- 1}

. For an interior node, differentiate the trigonometric form using

d / d x = - (sin θ)^{- 1} d / d θ

π_{n}^{'} (cos θ) = 2^{1 - n} [cot θ sin (n θ) + n cos (n θ)] .

θ_{s} = s π / n

, this is

2^{1 - n} n (- 1)^{s}

. At either endpoint, taking the limit makes the two terms equal, doubling the value:

π_{n}^{'} (x_{s}) = 2^{2 - n} n (- 1)^{s}

for

s = 0, n

After removing the common factor

2^{n - 1} / n

w_{s} = (- 1)^{s} δ_{s}, δ_{s} = 1 (1 \leq s < n), δ_{0} = δ_{n} = \frac{1}{2} .

These Chebyshev-Lobatto node weights differ in magnitude by only a factor of two.

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