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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

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