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Solutions to Problems; Chapter 7 363

5.1. The Lagrange polynomial of degree not exceeding n - 1 which assumes

the values f(ui) at ei is

~f(Qi)(t-(fl).*.(tZi).*.(t-(1.)

i=l P’(ai)

and this must be the polynomial f(t). Hence the coefficient oft”-’ must

vanish and the result follows.

5.2. (a) The Lagrange polynomial of degree not exceeding 3 which assumes

the value t4 at t = a, b, c, d is

x04(2 - b)(t - c)(t - d)

(a - b)(a - c)(a - d).

This must be equal to t4 - (t - a)(t - b)(t - c)(t - d). Comparing the

coefficient of t3 yields the result.

(b) The quadratic polynomial which assumes the values u5, b5, c5 at a,

b, c respectively is

a5(t - b)(t - c) + b5(t - u)(t - c) + c5(t - u)(t - b)

(a - b)(u - c) (b - a)(b - c) (c - u)(c - 6) ’

We can determine this polynomial in another way as

t5 - (t3 - ut2 + vt - w)(t2 + rt + s)

where u = a + b + c, v = ab + bc + cu, w = abc and r and s are chosen to
make the coefficients of t4 and t3 vanish, i.e.

r=u

s=ru-v=u2-v.

Then the left side of the required identity is equal to the coefficient of t2,
namely

as required.

w - rv + su = w - 2uv + u3

5.3. The proof is by double induction. The result holds for k = 1 and any
n, as well as for n = 1 and any k. Suppose it has been established for
k= 1,2,...,r- 1 and any n as well as for k = r and 1 5 n < m - 1. Then

S,(m) = S,(m - 1) + S-l(m)

= m(m + l)...( m-l+r)[(m-1)(2m-2+r)+(r+2)(2m+r-

(r + 2)!

= m(m + 1) ~1. (m - 1 + r)(2m + r)(m + r)

(r + 2)!