A ANSWERS TO EXERCISES 523
5.84 Following the hint, we get
andasimilarformulafor&,(z). Thustheformulas (ztB;‘(z)‘B[(z)+l)Bt(z)r and (ztE;‘(z)&:(z) + l)&,(z)’ give the respective right-hand sides of (5.61). We must therefore prove that
(zwwJm + l)%w- = 1 _ t + :‘% t(z) q , (zw4~:M + 1)Wz)’ = , &Z)t , and these follow from (5.59). 5.85 If f(x) = a,x” + ... + a’x + a0 is any polynomial of degree < n, we can prove inductively that x
c-1
1 “+“‘+‘“f(e1x,
+...+E,x,) = (-l)nn!~,I~l
. ..x..
O$f, ,..., E$,$l
The stated identity is the special case where a, = 1 /n! and Xk = k3. 5.86 (a) First expand with n(n- 1) index variables Lij for all i # j. Setting kii = li’ -Lji for 1 :< i < j < n and using the constraints tifi (lij -iii) = 0 for all i < n allows us to carry out the sums on li, for 1 6 j < n and then on iii for 1 < i < j < n by Vandermonde’s convolution. (b) f(z) - 1 is a polynomial of degree < n that has n roots, so it must be zero. (c) Consider the constant terms in
,jJsn , i#,i, (1 - ;)“‘~ (Y ifi (1 - ;)“’ = g JIn 5.87 The first term is t, (n;k)zmk, by (5.61). The summands in the second term are 1 (n+ 1)/m; (l+l/m)k)iiz),.;, m EC k20 1 = - m
(‘+‘;~~~;‘-‘)(i,jk.
