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22
00
0
0
0
)1(11cos
cos1sin
1sin)(
1
cos)(01
cos)(1
nn
n
n
nx
ndxnx
nn
nxx
dxnxxdxdxnxxfa
n
n
nnxdxxbn
1sin)(
10
1
2 sin1
cos)1(1
4)(
n
n
nxn
nxn
xf
La función f es continua en (−, ) excepto en x = 0. Así su
serie de Fourier converge en x = 0 a:
La serie es una extensión periodica de la
función f. Las discontinuidades en x = 0,
2, 4, … convergen a:22
)0()0( ff
Y las discontinuidades en
x = , 3, … convergen a:
220
2)0()0( ff
02
)0()( ff
xxxSxxSS 2sin2
1sincos
2
4,sincos
2
4 ,
4 321
Secuencia de sumas parciales y su representación gráfica
Evaluar donde el contorno C
es el círculo |z|= 2.
C
z
dzzz
e34 5
iz
ezzi
zz
ez
dz
di
zfidzzz
e
z
z
z
z
C
z
12517
)5(
)178(lim
)5(lim
!21
2
)0,)((Res25
3
2
0
33
2
2
0
34
A Question
• In part (b) of Example 2 in Sec.19.3, we showed that the Laurent series of f(z) = 1/z(z – 1) valid for 1 < |z| is
The point z = 0 is an isolated singularity of f and the Laurent series contains an infinite number of terms involving negative inter powers of z. Does it mean that z = 0 is an essential singularity?
...111
)(432
zzzzf
• The answer is “NO”. Since the interested Laurent series is the one with the domain 0 < |z| < 1. From part (a) of that example, we have
Thus z = 0 is a simple pole for 0 < |z| < 1.
...11
)( 2 zzz
zf
C is positively oriented circle | z – 2| = 1.
Integrand is analytic everywhere except z=2 and z=0. Find Laurent series of f(z) in the disk 0 < | z – 2 | < 2
Residue of f at the isolated singular point z0
C
dzzz 4)2(
1
16
1
)2(
1
2
1)( Res
41zz 0
C
dzzzi
bzf
)2|2|0()2(2
)1(
32
1
1
)2(2
1
)2(
1 4
0144
zz
zzzzn
nn
n
ALTERNATIVE METHOD
Residue of f at the isolated singular point 2 is the coefficient of 1/(z–2).
16
1)( Res
0zzzf
4
324
4
2344
)2(
)2()2()2()2(
)2(
1
)2|2|0()2()2()2()2()2(
1
zz
zEzzDzzCzBzzA
zz
zz
E
z
D
z
C
z
B
z
A
zz
Solve for A, B, C, D, E by setting coefficients of z, z2, z3, z4 equal to 0.A + E = 0(A(z – 2) – Az)(z – 2)3 = – 2A(z – 2)3
D – 2A = 0(– 2A (z – 2) + 2A z)(z – 2)2 = 4A (z – 2)2
4A + C = 0(– 4A z + 4A (z – 2)) (z – 2) = –8A(z – 2)B – 8A =0 8A z – 8A (z – 2) = 16 A = 1A = 1/16, E = – 1/16
Partial Fraction Expansion Review
BBz
zA
zz
zz
Az
BzA
zz
zz
z
B
z
A
zz
)2(
2
1
)2(
)2(,2When
)2(2
1
)2(,0When
)2()2(
1
75.1
5.025.01)1(1
1)2(
1,1Pick
)2(
)2()2(
2
1
)2(
)2(,2When
)2()2(4
1
)2(,0When
)2()2()2(
1
2
22
2
2
22
22
B
BCBA
zzz
CCz
zB
z
zA
zz
zz
Az
Cz
z
BzA
zz
zz
z
C
z
B
z
A
zz
C is positively oriented circle | z – 2| = 1.Integrand is analytic everywhere except z=2 and z=0. Find Laurent series of f(z) in the disk 0 < | z – 2 | < 2
Residue of f at the isolated singular point z0
C
dzzz 4)2(
1
16
1
)2(
1
2
1)( Res
41zz 0
C
dzzzi
bzf
n
n
n
z
zzz
zzz
zzzzzzz
zzzzzzzz
)2(2
1
32
1
...2
)2(
2
)2(
2
21
32
1
2/)2(1
1
32
1
12/)2(
1
32
11
)2(
16/1
)2(
8/1
)2(
4/1
)2(
2/1
2)2(
16/1
)2(
1
)2|2|0()2(
16/1
)2(
8/1
)2(
4/1
)2(
2/116/1
)2(
1
0
3
3
2
2
2344
2344
All terms have positive exponents
b1