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MATH20602

Numerical Analysis 1

Problem Sheet 3

Part A

(3.1) (Chebyshev Polynomials) For n > 0 and y e [ · 1. 1] deﬁne the Chebyshev polynomial of degree n by

Tn (y) = cos(n arccos(y))．

1. Show that the Tn satisfy the recurrence relation

Tn+1(y) · 2yTn (y) + Tn]1(y) = 0

for n > 1. Conclude from this that Tn (y) is a polynomial of degree n with leading term 2n]1yn for n > 1.

2. Using MATLAB or Python, plot the polynomials Tn for n = 0.．．．. 10 and ﬁnd an explicit expression for the roots.

3. Let y0 .．．．. yn be the roots of the polynomial Tn+1(y). Given a function f e C(n+1)([ · 1. 1]) and the unique interpolation polynomial gn (y) of f at the points yi (0 < i < n), show that the interpolation error can be bounded as

Mn+1

If(y) · gn (y)I <

where Mn+1 = max If(n+1)(y)I.

x-[]1 , 1]

4. Using MATLAB or Python, plot on the interval [ · 1. 1], for n = 0.．．．. 10:

● The function f(y) = 1/(1 + 25y2 );

● The unique interpolation polynomial gn (y) of f at equispaced points y0 .．．．. yn on [ · 1. 1];

● The unique interpolation polynomial gn (y) of f at the roots y0 .．．．. yn of Tn+1(y).

Can the error bound derived in part 3 explain what is observed in this experi- ment?

Part B

(3.2) (Newton’s divided differences) 娄

1. For the data

(yi . -i ) = ( · 1. · 1). (0. 3). (2. 11). (3. 27).

compute the divided difference table and use it to construct g3 (y).

Evaluate g3 ( ·2).

2. Add an extra point y = 4. - = 19 to the list above. Form the divided difference table for these data, and derive the quartic polynomial r(y) that interpolates these ﬁve points.

(3.3) 娄 Given n + 2 points (yi . -i ), 0 < i < n + 1, with yi yj for i j, let: g(y) be the interpolation polynomial in Pn for the points (yi . -i ), 0 < i < n; r(y) be the interpolation polynomial in Pn for the points (yj . -j ), 1 < j < n+1.

Show that

(y · y0 )r(y) · (y · yn+1)g(y)

yn+1 · y0

is the interpolation polynomial in Pn+1 for the points (yk . -k ), 0 < k < n + 1.

(3.4) Show that the polynomial gn (y) interpolating f (y) at yi can be written as

zk f (yk )/(y · yk )

gn (y) = .

zk /(y · yk )

k=0

for y yk where

zk = . k = 0.．．．. n．

j k

This is called the barycentric representation of gn (y). Comment on how efﬁcient it is to evaluate the barycentric form, compared to the Lagrange form.

HINT: Show that

n n f (yk )zk