Answers to Exercises; Chapter 7 355
is a polynomial of degree k, each of its terms can be written as a linear
combination of factorial powers and the result follows.
1.17. (a) Use Exercise 16. We can write
f(t) = b,t(“) + b,-&-‘) +... + bit + Lo.By repeated application of the A-operator, we find that Anf(t) = n!b, and
A”+lf(t) = 0.
The converse is not true. If f(t) = sin 27rt, then Af(t) = f(t + 1) -f(t) =
0 for each t and all differences vanish.
(b) Abi = (b - l)b, so that A”b’ = (b- l)“b is nonzero. Hence bt is not
a polynomial.
1.18. (a) The Lagrange Polynomial which assumes the value f(j) at t = j
(j=O,l,..., k) is given by
2 f(i)t(t - 1)... (t - i + l)(t - i - 1) 1.. (t - k)
i=l i(i - l)..e(2)(1)(-l)(-2) .e.(-k+i)
’Since this is the unique polynomial of degree not exceeding k with the
assigned values, the result follows.
(b) The expression is the Lagrange polynomial with the value f(aj) when
t = Uje
1.19. (a) (1/2)t(t + l), or, more generally,(l/n!)t(t+ l).+.(t +n - 1).(b) Express f(t) as in Exercise 18(a). If n > k^1 i^2 0, then(:)( n;i;‘)is an integer, while if n < 0 2 i 5 k,(y)(“;i;‘)=(-l)i( InI+;- ) .(-l)k-i ( lilt: )
is an integer. Hence, for all n, the coefficient of f(i) in the expansion for
f(n) is an integer and the result follows.
1.20.
h(t) = 2 ( : ) = kt(‘)/r!
r=O t-=0
h(n + 1) = 2”+l - 1.
2.2. (b) (i) l/2; (ii) l/2; (iii) 314.