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.In general, thereare + 1) ways to represent where the products range overprimes.-l/j) Thesecondsumis,=similarly,2.32 =n(x-n) = (x-l) (x-3) + If 2n 1 x c vanish, andc cancels with in the limit of the other terms.Therefore the secondpartial sum is =5.35 (a)(b)A ANSWERS TO EXERCISES 5155.36 The sum of the digits of m n is the sum of the digits of m plus thesum of the digits of n, minus p 1 times the number of carries, because eachcarry decreases the digit sum by p 1.5.37 Dividing the first identity by n! yields =dermonde s convolution.The second identity follows, for example, from theformula = if we negate both x and5.38 Choose c as large as possible such that n.Then 0 nand continue in the same fashion.replace n by n=Conversely, any such representation is obtained in this way.(We can do thesame thing withn = 0 a,for any fixed m.)5.3 9 f o rall mn 0, by induction on m + n.The boxedsentence=5.4 0on the)other sideof this pageis true.5.41 k)! (n + k + + which is+ 15.42 We treat n as an indeterminate real variable.Gosper s method withq(k) = k 1 and r(k) = k 1 -n has the solution s(k) + 2); hencethe desired indefinite sum is (-1 AndThis exercise, incidentally, implies the formula1n - ln+ka  dual to the basic recurrence (5.8).5.43 After the hinted first step we can apply (5.21) and sum on k.Then(5.21) applies again and Vandermonde s convolution finishes the job.(A com-binatorial proof of this identity has been given by Andrews There s aquick way to go this identity to a proof of explained inexercise516 ANSWERS TO EXERCISES5.44 Cancellation of factorials shows thatso the second sum isI times the first.We can show that the first sumiswhenever n b, even if m a: Let a and b be fixedand call the first sum S( m,  1).Identity (5.32) covers the case n = b, andwe have S(m,n) = 1) + S(m- since+The result follows by induction onm+ n, since = 0 when n b and the case m = 0 is trivial.By symmetry,the formula holds whenever m a, even if n b.5.45 According to = If this form isn t  closedenough, we can apply (5.35) and get (2n + 1)5.46 By this convolution is the negative of the coefficient ofin Now 1) = hence= + + By the binomialtheorem,(1 =so the answer is 1) + 1).5.47 It s the coefficient of in =where = 1 + by (5.61).5.48 + 2,1; n + 2; a special case of (5.111).5.49 Saalschiitz s identity yields5.50 The left-hand side isk + a + m- 1mand the coefficient of isA ANSWERS TO EXERCISES 517by Vandermonde s convolution5.51 (a) Reflection gives 2a; 2) = (-1 a, 2a; 2).(Inciden-tally, this formula implies the remarkable identity f(0) = 0, whenf(n)(b) The term-by-term limit is plus an addi-tional term for k = 2m 1: the additional term is(-m).(-1) (1).(m) 1).(-1) (2m-2=hence, by this limit is the negative of what we had.5.52 The terms of both series are zero for k N.This identity correspondsto replacing k by N k.Notice that5.53 When b the left side of (5.110) is 1 22 and the right side is(1 independent of a.The right side is the formal power series1which can be expanded and rearranged to give 1 + but therearrangement involves divergent series in its intermediate steps when =so it is not legitimate.5.54 If m + n is odd, say 2N we want to show thatlim F1Equation applies, since -m + > -m + and the denominatorfactor = is infinite since N m; the other factors are finite.Otherwise m + n is even; setting n = m 2N we haveFby The remaining job is to show that(N (m-N)!m!518 ANSWERS TO EXERCISESand this is the case x = N of exercise 22.5.55 Let Q(k) = R(k) =Then t(k+ = where P(k) = Q(k) -R(k)is a polynomial.5.56 The solution to = is s(k)hence 1) + C.Also42=5.57 We have = Thereforewe let p(k) = k+ q(k) = (k- r(k) = k.The secret function s(k)must be a constant and we have=hence = z) and = z).The sum is(The special case z = 1 was mentioned in (5.18); the general case is equivalent5.58 If m 0 we can replace by and derive the formulaThe summation factor is therefore appropriate:We can unfold this to getT= +.Finally = so(H, (It s also possible to derivethis result by using generating functions; see Example in Section 7.5.)5.5 9which is= = (m-A ANSWERS TO EXERCISES5.60 is the case m = n of+5.61 Let = q and n mod p = The polynomial identity (x 1+ 1 (mod p) implies that(mod p).The coefficient of on the left is On the right it s whichis just (( because 0 65.62 = [ Pobierz całość w formacie PDF ]