Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 535
6.68 1 /k - 1 /(k + z) = z/k2 - z2/k3 + , and everything converges when
121 < 1.
6.69 Note that nL=, (1 + z/k)ePLlk = (nnfL)nPLe(lnn Hlllr. If f(z) = &(z!)
we find f(z)/z! + y � = H,.
6.70 For tan z, we can use tan z = cot z - 2 cot 22 (which is equivalent to the
identity of exercise 23). Also z/sin z = zcot z + ztan :Z has the power series
&o(-l)“P’(4n - 2)Bznz2”/(2n)!; and
tan2
In- = In “” - -lncosz
z
4*(4" -2)BLnz2"
= IL-‘)” (2n)(2n)! ’
n3 1
because -& In sin z = cot z and -& In cos z = -tan z.
4n(4n-l)B2,~2n
(2n)(2n)!
6.71 Since tan2z - sec22 = (sin2 + cosz)/(cosz - sinz), setting x = 1 in
(6.94) gives T, (1) = 2nT,, when n is odd, T,, (1) = 2"E, when n is even, where
1 /cos 2 = tn30 Ezn,-‘“/(2n)’ (The E, are called Euler numbers, not to be
confused with the Eulerian numbers (L) .)
6.72 2n+1(2n+' - l)B,+i/(n + l), if n > 0. (See (7.56) and (6.92); the
desired numbers are essentially the coefficients of 1 - tanhz.)
6.73 cot(z + rr) = cot z and cot(z + irr) = -tan z; hence the identity is
equivalent to
cot z = -in 2gl cot 3&z )
k=O
which follows by induction from the case n = 1. The stated limit follows since
zcot z + 1 as z -+ 0. It can be shown that term-by-term passage to the limit
is justified, hence (6.88) is valid. (Incidentally, the general formula
cot z = -; rcot q?
k=O
is also true. It can be proved from (6.88), or from
1
~ =
en2 - 1
k=O
which is equivalent to the partial fraction expansion of l/(2” - l).)