524 ANSWERS TO EXERCISES
Since ~06j
= m(--l)‘[k=mL], these terms sum to
(l+l/mW -n- 1 (-z”)k XC
k>n/m
mk-n--l
)
(m+l)k-n- 1 ) (-zm)k = t (” -kmk)pk k =Q k>n/m
k>n/m
Incidentally, the functions ‘B,,,(zm) and L2i+‘z~B1+II,(L2~+‘~)‘~m are the m+l complex roots of the equation w”‘+’ - wm = z”l. 5.88 Use the facts that Jr(e it - e nt) dt/t = Inn and (1 - ee’)/t $ 1. (We have (“,) = O(kmx ‘) a,s k + 00, by (5.83); so this bound implies that Stirling’s series tk sk (i) converges when x > -1. Hermite [155] showed that the sum is In r( 1 + x).) 5.89 Adding this to (5.19) gives ~~‘(x+y)~+’ on both sides, by the binomial theorem. Differentiation gives I sentence on the
other side of this page is not selfreferentia/.
and we can replace k by k + m +
1
and apply (5.15) to get
& (m;:: k) (-‘I; ‘) (-X)m+l+ky--l-k-n
In hypergeometric form, this reduces to
which is the special case (a, b, c, z) = (n + 1, m + 1 + r, m + 2, -x/y) of the reflection law (5.101). (Thus (5.105) is related to reflection and to the formula in exercise 52.) 5.90 If r is a nonnegative integer, the sum is finite, and the derivation in the text is valid as long as :none of the terms of the sum for 0 < k < r has zero in the denominator. Otherwise the sum is infinite, and the kth term (k ml- ‘) / ( k -i-l) is approximately k” ’ (-s - l)!/(-r - l)! by (5.83). So we
