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Exercises 22.4 Homework

1.

(Theoretical) The Chebyshev polynomials of the second kind \(U_n\) can be defined as follows: \(U_0(x) = 1\text{,}\) \(U_1(x) = 2x\text{,}\) and after that,

\begin{equation} U_{n+1}(x) = 2xU_n(x) - U_{n-1}(x)\label{eq-chebyshev-recursion-U}\tag{22.4.1} \end{equation}

(This is the same recursive formula as for \(T_n\text{,}\) but it produces different polynomials because \(U_1\) is different from \(T_1\text{.}\)) Prove that

\begin{equation} U_{n}(\cos\theta)\sin \theta = \sin ((n+1)\theta)\label{eq-chebyshev-cosine-U}\tag{22.4.2} \end{equation}

for all \(n\) and all \(\theta\text{.}\)

Hint: follow the proof of (22.2.2).

2.

(Theoretical) Use (22.4.2) to prove that the Chebyshev extreme points \(x_k = \cos(\pi k/n)\) with \(k=1, \dots, n-1\) are the roots of polynomial \(U_{n-1}\text{.}\)

(This is where the name “Chebyshev points of the second kind” comes form.)

3.

Suppose we want to draw a smooth parametric curve through 10 given points on xy-plane. The points are represented by vectors of their x- and y-coordinates:

X = [1 3 5 6 4 1 0 1 3 5];
Y = [1 0 0 2 4 4 6 9 9 7];

The command plot(X, Y) can join these points by line segments, but that does not make a smooth curve. To obtain a smooth curve, adapt Example 22.3.2 as follows:

  • Your script should create two interpolating polynomials (px and py) on the interval \([-1, 1]\text{.}\) One is computed from the values X and the other from the values Y.
  • At the end of the script, the curve through the given points can be displayed with
    t = linspace(-1, 1, 1000);
    plot(X, Y, 'r*',  px(t), py(t), 'b')

The plot should show a smooth S-shaped curve passing through the points marked by asterisks.