Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 569
which is 2O(f(n) tkzO If(k)/), so this case is proved. (b) But in this case if
a - b, = aPn, the convolution (n + 1 )aPn is not 0( 01 “).
n -
9.25 s,/(3t) = ~;4Lq2n+l)F we may restrict the range of summation
to 0 < k 6 (logn)‘, say. In this range nk = nk(l - (i)/n + O(k4/n2)) and
(2n + l)k = (2n)k(l + (“;‘)/2n+ O(k4/n2)), so the summand is
Hence the sum over k is 2 -4/n + 0( 1 /n2). Stirling’s approximation can now
be applied to (y) = (3n)!/(2n)!n!, proving (9.2).
9.26 The minimum occurs at a term Blm/(2m) (2m- 1 )n2”-’ where 2m z
2rrn + 3, and this term is approximately equal to 1 /(rceZnnfi ). The absolute
error in Inn! is therefore too large to determine n! exactly by rounding to an
integer, when n is greater than about e2n+‘.
9.27 We may assume that a # - 1. Let f(x) = x”; the answer is
f
k=l
n”+l
km = C,+ -
a+1
na-2k+l + 0~~” -2m 1).
(The constant C, turns out to be -1.)
9.28 Take f(x) = xlnx in Euler’s summation formula to get
A. nnL:2+n/:+1/12e~n~i4(1 + qn-2)) ,
where A z 1.282427 is “Glaisher’s constant!’
9.29 Let f(x) = xP1 lnx. Then fiZmi (x) > 0 for all large x, and we can write
n Ink
ET = y+lnS+z+Bn+, 0