Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 1 pps
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 1 pps
... 1)
2
+ y
2
)
2
=
−(x − 1)
2
+ y
2
((x − 1)
2
+ y
2
)
2
and
2( x − 1) y
((x − 1)
2
+ y
2
)
2
=
2( x − 1) y
((x − 1)
2
+ y
2
)
2
The Cauchy-Riemann equations are each identities. The first partial derivatives ... 8.4.3 1/ sin (z
2
) has a second order pole at z = 0 and first order poles at z = (nπ)
1 /2
, n ∈ Z
±
.
lim
z→0
z
2
sin (z
2
)
= lim
z→...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx
... equal.
516
upper bound for the series.
∞
1
1
x
2
dx ≤
∞
n =1
1
n
2
≤ 1 +
∞
1
1
x
2
dx
1 ≤
∞
n =1
1
n
2
≤ 2
1
2 3
4
1
1
2 3
4
1
Figure 12 . 1: Upper and Lower bounds to
∞
n =1
1/n
2
.
In general, ... 0
517
11 .5 Hints
Hint 11 .1
Use the argument theorem.
Hint 11 .2
Hint 11 .3
To evaluate the integral, consider the circle at infinity.
Hint...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 5 pps
... Annulus.
Example 12 . 6 .1 Find the Laurent series expansions of 1/ (1 + z).
For |z| < 1,
1
1 + z
= 1 +
1
1
z +
1
2
z
2
+
1
3
z
3
+ ···
= 1 + ( 1)
1
z + ( 1)
2
z
2
+ ( 1)
3
z
3
+ ···
= 1 − z ... n)
n
11 .
∞
n =2
( 1)
n
ln
1
n
12 .
∞
n =2
(n!)
2
(2n)!
13 .
∞
n =2
3
n
+ 4
n
+ 5
5
n
− 4
n
− 3
5 62
Im(z)
Re(z)
R
R
2
1
Im(z)...
Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 1 pps
... < 1
1
4
(1 + t + x)
2
1 < x + t < 0
1
4
(1 − t
2
+ 2t (1 − x) + x (2 − x)) x −t < 1, 0 < x + t
1
2
(2t − t
2
− x
2
) 1 < x − t, x + t < 1
1
4
(1 − t
2
+ 2t (1 + x) − x (2 + ... − 1) πx
2L
−
∞
m =1
n =1
g
mn
(2m − 1) π
2L
2
+
nπy
H
2
2
√
LH
sin
(2m − 1) πx
2L
cos
nπy
H
=
∞
m =1
2
LH
sin
(2m − 1) πξ...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 2 pptx
... =
1
2
Log
x
2
+ y
2
+ ı Arctan(x, y).
4 52
2. We calculate the first partial derivatives of u and v.
u
x
= 2
e
x
2
−y
2
(x cos(2xy) − y sin(2xy))
u
y
= 2
e
x
2
−y
2
(y cos(2xy) + x sin(2xy))
v
x
= ... an arc and that f
1
(z) = f
2
(z) for all
z ∈ D
1
∩ D
2
. (See Figure
9.4.) Then the function
f(z) =
f
1
(z) for z ∈ D
1
,
f
2
(z) for z ∈ D
2...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 3 ppt
... =
2
0
e
ınθ
ı
e
ıθ
dθ
=
e
ı(n +1) θ
n +1
2
0
for n = 1
[ıθ]
2
0
for n = 1
=
0 for n = 1
2 for n = 1
2. We parameterize the contour and do the integration.
z − z
0
= 2 +
e
ıθ
, θ ∈ [0 . . . 2 )
C
(z ... 1
|dz|
≤
θ
0
+π
θ
0
R
2
e
2
1
R
2
e
2
+1
|R dθ|
≤ R
θ
0
+π
θ
0
R
2
+ 1
R
2
− 1
dθ
= πr
R
2
+ 1
R...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 6 doc
... π)
Hint 12 . 23
CONTINUE
Hint 12 . 24
CONTINUE
Hint 12 . 25
Hint 12 . 26
Hint 12 . 27
Hint 12 . 28
Hint 12 . 29
Hint 12 . 30
CONTINUE
5 81
Solution 12 . 22
cos z = −cos(z − π)
= −
∞
n=0
( 1)
n
(z −π)
2n
(2n)!
=
∞
n=0
( 1)
n +1
(z ... convergent, so is the series.
Solution 12 . 4
∞
n =1
( 1)
n +1
n
=
∞
n =1
1
2n − 1
−
1
2n
=
∞
n =1
1
(2n − 1) (2n)
<
∞
n =1...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 7 pdf
... lim
→0
+
2 −
1
e
ıθ
− 1
ıe
ıθ
dθ
= lim
→0
+
log(e
ıθ
− 1)
2 −
639
−
1
z − 2
=
1 /2
1 − z /2
=
1
2
∞
n=0
z
2
n
, for |z /2| < 1
=
∞
n=0
z
n
2
n +1
, for |z| < 2
−
1
z − 2
= −
1/ z
1 − 2/ z
= −
1
z
∞
n=0
2
z
n
, ... negative
direction.
648
(b)
1
z (1 −z)
=
1
z
+
1
1 − z
=
1
z
−
1
z
1
1 − 1/ z
=
1
z...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 8 pot
... −
n 1
k=0
lim
z→
e
ıπ (1+ 2k)/n
log z + (z −
e
ıπ (1+ 2k)/n
)/z
nz
n 1
= −
n 1
k=0
ıπ (1 + 2k)/n
n
e
ıπ (1+ 2k)(n 1) /n
= −
ıπ
n
2
e
ıπ(n 1) /n
n 1
k=0
(1 + 2k)
e
2 k/n
=
2
e
ıπ/n
n
2
n 1
k =1
k
e
2 k/n
=
2
e
ıπ/n
n
2
n
e
2 /n
1
=
π
n ... lim
z→0
cos z
1
= 1
3.
1 + z
2
z(z 1)
2
There is a first order pole at z = 0 and a sec ond order pole at...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 9 ppt
... c
n
k =1
(z −z
k
). Let m be
696
2.
−
1
1
1
x
3
dx = lim
→0
+
−
1
1
x
3
dx +
1
1
x
3
dx
= lim
→0
+
−
1
2x
2
−
1
+
−
1
2x
2
1
= lim
→0
+
−
1
2( −)
2
+
1
2( 1)
2
−
1
2( 1)
2
+
1
2
2
= ... C.
C
log
2
z
1 + z
2
dz = −
π
3
4
∞
0
ln
2
r
1 + r
2
dr +
0
∞
(ln r + ıπ)
2
1 + r
2
dr = −
π
3
4
2
...
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