548 Chapter^8
and letting z 0 ---+ 0 we have
. = 22nB
· smz "C )n 2n 2 n
Log -z-= ~ -l (2n)(2n)! z ' lzl < 7f (8.5-11)
Note Whenever the function F(z) on the left-hand side of any of the
formulas above is not defined for z = 0, the definition is supposed to be
completed by analytic extension.
- Let
E2 2 En n
sech z = 1 + E 1 z + - z + · · · + - z + · · ·
2! n!
(8.5-12)for lzl < ~7f. Since sechz is an even function we must have Ei =Ea =
· · · = E2k-l = · · · = 0. By using the symbolic method (8.5-12) can be
written as
which gives
Hence
2 = e(E+i)z + e<E-l)z = 2 + [(E + 1) + (E - 1)] z
2 2 z2
+ [(E + 1) + (E -1) ]
21
+ · · ·+ [(E + 1r + (E - ltl zn + · · ·
n!(E + 1) + (E -1) = 0, (E + 1)2 + (E -1)^2 = 0,
(E + 1r +CE -1r = o,Reducing and passing back to subindices, we find that
... ,
2E1 = 0, E 2 + 1 = 0, Ea + 3E1 = 0, E4 + 6E 2 + 1 = 0, ... ,
E2m+1+(
2
m 2 +1)E2m-l + · · · + (2~:
1)E1 = 0 if n = 2m + 1
or
E2m + (2:)E2m-2 + ... + (2~7: 2)E2+1=0 if n = 2m
The equations involving the E'^8 with odd indices are linear homogeneous
in the E's, and since Ei = 0 it follows that Ea = Es = · · · = 0, as
anticipated. By solving the equations involving the E's with even indices,
we obtain successively
E2 = -1, E4 = 5, E6 = -61, Es= 1385, E10 = -50,521,