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Solutions to Problems; Chapter 8 369

Since x2 5 x, it follows that 1 - x 5 dn. Hence

--t + (1/2)(Jz+ 1) - (l/2)(& - 1) I J1-;cz

5 --t + (l/2)(45 + 1) + (l/2)(& - l),

which yields the result.

5.10. By Lagrange’s Formula,

P(X) = &Jk

(x - x0)... (x - X&1)(X - xk+l) *. * (x - x,)

k=O (tk - x0) ..... +k - 2,)

Now, Tn+l(x) = cos(n + 1)u = cos(n + 1) arccos x vanishes if and only if
x = cosuk for some k. Hence Tn+l(x) = c,+i(x - xi).. .(x - 2,). Thus,
when x # Xk for any k, we can write

p(x) = Tn+l(x) g (2 - Xk;;;+l(xk) ’

In the special case that each yk 5 1, we find that

1 = %+1(X) 2 l/[(X - Xk)T,f,+l(~k)].

k=O

Differentiating Tn+l(x) = cos(n + 1)u with respect to u yields

TA+,(x)(- sin u) = -(n + 1) sin(n + 1)~.

In particular, when x = xk, we have that u = ‘uk and

TA+,(Xk) = (n -k l)(-l)k/(SinUk)

and the result follows.

Solutions to Problems

Chapter 8

1. The equation is equivalent to (3y - 5)(9x - 3y - 5) = 34. Try out all
   the divisors 3y - 5 of 34 to obtain the solutions (x,y) = (-1, -4), (-1, l),
   (532) (5913).


2. 0 = a3-b3-c3 -3abc= (a-b-c)(a2+b2+c2+ab+ac-bc). Since the
   second factor never vanishes for a, b, c not all zero, a = b + c. This yields
   (a,hc) = (&I, 1).