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356 Answers to Exercises and Solutions to Problems

2.4. (a) Sketch graphs. pi(t) = 0. ps(t) = l/2, since the difference t2 - l/2

achieves its maximum absolute values with alternating signs 3 times at

t = -l,O, 1. p3(t) is of the form mt for some value of m, for it can be

seen from a sketch that there is a value of m strictly between 0 and 1 for

which It3 - mtl s ]l - ml for -1 5 t _< 1, and that t3 - mt assumes the

value f(1 -m) with alternating signs at the four points -1, -b, b, 1 where

0 < b < 1. The function (t3 - mt) + (1 - m) = (t + l)(t2 - t + 1 - m) is

nonnegative for -1 5 t 2 1 and assumes its minimum value 0 when t = 6.

Hence t2 -t-l-m = (t-b)2, so that the discriminant 1-4(1-m) = 4m-3

vanishes. Hence m = 314 and p3(t) = 3t/4.

(b) C1 = t, C, = (2t2 - 1)/2, C, = (4t3 - 3t)/4.

(c) ck = T$!k-l. Th e maximum absolute value of Ck and Tk on [-1, l]

is assumed at the k + 1 points t = cos ix/k where 0 < i 5 k.

2.5. (b)

B(f,n; t) = 5 ( ; ) t”(1 - t)n-k = [(l -t) + t]” = 1

k=O

%, n,t) = 2 k=. (t> ( ; )WY

(c) B(t2, n; t) = [(n - l)t2 + t]/n. (See Remark after Answer 2.6.)
2.6. (b) Since B(f, n; 0) = f(0) and B(f, n; 1) = f(l), if k # 1, the eigen-
function f(t) must satisfy f(0) = f(1) = 0.

n=l: k = 1 and f(t) = at + b

n=2: k = 1 and f(t) = at + b

k: = l/2 and f(t) = a(t2 -t) = ut(t - 1)

n=3: k = 1 and f(t) = at + b

k = 213 and f(t) = at(t - 1)

k = 219 and f(t) = a(2t3 - 3t2 + t) = at(t - 1)(2t - 1)

n=4: k = 1 and f(t) = at + b

k = 314 and f(t) = at(t - 1)

k = 3/B and f(t) = ut(t - 1)(2t - 1)

k = 3132 and f(t) = a(14t4 - 28t3 + 17t2 - 3t)

= at(t - 1)(14t2 - 14t + 3).