## Description

MAT128A: Numerical Analysis, Section 2

1. Suppose that n is a nonnegative integer. You know that the function y(t) = cos(nt) satisfies the

second order differential equation

y:(t) + n

2

y(t) = 0 for all − π < t < π.

Use this observation to show that the function

Tn(x) = cos(n arccos(x))

is a solution of the equation

(1 − x

2

)y

00(x) − xy0

(x) + n

2

y(x) = 0 for all − 1 < x < 1.

Here, I am using y

0

to denote the derivative of y with respect to x and y9 to denote the derivative

of y with respect to t.

Hint: use the chain rule to compute

dy

dt and d

2y

dt2

in terms of

dy

dx and d

2y

dx2

.

2. Show that

Z 1

−1

Tn(x)Tm(x)

dx

?

1 − x

2

=

0 m 6= n

π m = n = 0

π

2 m = n 6= 0.

3. (a) Using the trigonometric identity

cos(nt) = cos((n − 1)t) cos(t) − sin((n − 1)t) sin(t),

show that

Tn(x) = xTn−1(x) − Un−1(x)

a

1 − x

2, (1)

where Un is defined via

Un(x) = sin(n arccos(x)).

(b) Use the trigonometric identity

sin(nt) = sin((n − 1)t) cos(t) + cos((n − 1)t) sin(t),

to show that

Un(x) = Tn−1(x)

a

1 − x

2 − Un−1(x)x. (2)

1

(c) Combine (1) and (2) to show that

Un(x)

a

1 − x

2 = Tn−1(x) + xTn(x). (3)

(d) Use (3) and (1) — replace n with n + 1 in (1) — to obtain the recurrence relation

Tn+1(x) = 2xTn(x) − Tn−1(x).

4. Suppose that n is a nonnegative integer. Show that

(1 − x

2

)T

0

n

(x) = nTn−1(x) − nxTn(x)

for all −1 < x < 1.

2