550
- By using multiplication of series, prove the following.
(a) (1 + z + z^2 + · · ·)^2 = 1+2z + 3z^2 + · · ·, lzl < 1
(b) 2sinzcosz = sin2z, Jzl < oo
2z^3 22 z^4 22 z^5
(c) ezcosz = l+z-- - --- --+ .. · Jzl < oo
. 3! 4! 5!
Chapter 8
(d)(Arctanz)
2
=z
2
-(1+~); +(1+~+~) z;-.. ·(lzJ<l)
- By division of series prove the following and find the radius of
convergence:
sinz z^3 z^5
(a) cos2z = z+l131 +^361 5! + ...
cos 2z z^2 z^4
(b) cos 3z =^1 +^5 2! +^205 4! + .. ·
- Use the 00 Weierstrass double series theorem to show that:
q2m z^00 q2k zk
(a) L 1-q2mz = L 1-q2k for JqJ < 1, Jzl < JqJ-
2
m=l k=l
oo m oo 2k-1
(b) ~ L._, 1 + z z2m -- ~(-l)k-l L._, 1 _ z z2k-1 for Jzl < 1
m=l k=l
- Expand each of the following functions to terms in z^5 and find the
radius z of convergence.
(a) ee (b) ecosz
( c) sin( sin z) ( d) Log cos z
- Carry the reversion of the following series to terms in w^5 and find the
region of convergence.
z3 zs
(a) w = sinz = z -
31
+
51
(b) w = Arctan z = z -! z^3 +! z^5 - ...
3 5
( c) w = 2z + 3z^2 + 4z^3 + · · · + ( n + 1 )zn + · · ·
- Show that:
oo 22nB
( ) a I _,ogsmz. = L ogz + ~ ""'( -l)n ( 2n 2n^0 J J
2 n)( 2 n)! z , < z < 7l"
(b) I L ~( )n-1^2
2
n(2
2
n - 2)B2n 2n I I
,ogtanz = ogz+ L., -1 ( )( )' z , 0 < z <
n=l 2n 2n.
1
·-7!"
2
oo 22n(22n - l)B 1
(c) Logcoshz = ~ ( 2 n)( 2 n)!
2
n z
2
n, lzl < 27!"
- By noting tha-t
