Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 10 ppt
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 10 ppt
... principal value exists.
766
π
2
lim
z→0
(z −1)
2
(z −3 + 2
√
2) (z −3 2
√
2)
+ lim
z→3 2
√
2
(z −1)
2
z(z − 3 − 2
√
2)
.
1
0
x
2
(1 + x
2
)
√
1 − x
2
dx =
(2 −
√
2) π
4
Infinite Sums
Solution ... z
(z + 1)
2
, −1
.
2
∞
0
x
1 /2
log x
(x + 1)
2
dx + 2
∞
0
x
1 /2
(x + 1)
2
dx = 2 lim
z→−1
d
dz
(z
1 /2
log z)
2
∞
0
x
1 /2
log x
(x +...
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 − z
0
)
n
dz =
2
0
2 ... axis and is defined continuously on the real axis.)
Hint, Solution
481
C
Log z dz
≤
C
|Log z||dz|
=
π...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 9 ppt
... = 2 Res
z
2
(z
2
+ 1)
2
, z = ı
Res
z
2
(z
2
+ 1)
2
, z = ı
= lim
z→ı
d
dz
(z − ı)
2
z
2
(z
2
+ 1)
2
= lim
z→ı
d
dz
z
2
(z + ı)
2
= lim
z→ı
(z + ı)
2
2z − z
2
2(z + ı)
(z + ... 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...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 10 ppt
... p
2
(u
1
y
1
+ u
2
y
2
+ u
3
y
3
) + p
1
(u
1
y
1
+ u
2
y
2
+ u
3
y
3
)
+ p
0
(u
1
y
1
+ u
2
y
2
+ u
3
y
3
) = f(x)
u
1
y
1
+ u
2
y
2
+ u
3
y
3
+ u
1
L[y
1
] + u
2
L[y
2
] ... u
2
y
2
+ u
3
y
3
y
p
= u
1
y
1
+ u
1
y
1
+ u
2
y
2
+ u
2
y
2
+ u
3
y
3
+ u
3
y
3
= u
1
y
1
+ u
2
y
2
+ u
3
y
3
y...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 3 pptx
... y
) − (2x − y)(1 + 2y
)
(x + 2y)
2
=
5(y −xy
)
(x + 2y)
2
Substitute in the expression for y
.
= −
10( x
2
− xy −y
2
)
(x + 2y)
2
Use the original implicit equation.
= −
10
(x + 2y)
2
3.5 ... =
x
20
20 !
cos ξ
≤
|x|
20
20 !
.
x
20
/20 ! is plotted in Figure 3.13.
2
4
6
8
10
0 .2
0.4
0.6
0.8
1
Figure 3.13: Plot of x
20
/20 !.
Note that th...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 4 pptx
... cos x
=
0
2
= 0
lim
x→0
csc x −
1
x
= 0
109
Solution 3.15
a.
f
(x) = ( 12 −2x)
2
+ 2x( 12 − 2x)( 2)
= 4(x −6)
2
+ 8x(x − 6)
= 12( x 2) (x − 6)
There are critical points at x = 2 and x = 6.
f
(x) ... derivative exists and is nonzero for x = 2. At x = 2, the derivative does not exist and thus x = 2 is a
critical point. For x < 2, f
(x) < 0 and...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 7 ppt
... mapping.
24 3
2.
z + 2
2 − ız
=
x + ıy + 2
2 − ı(x − ıy)
=
x + ı(y + 2)
2 − y −ıx
=
x + ı(y + 2)
2 − y −ıx
2 − y + ıx
2 − y + ıx
=
x (2 − y) − (y + 2) x
(2 − y)
2
+ x
2
+ ı
x
2
+ (y + 2) (2 − y)
(2 − ... of z
2
is
|z
2
| =
z
2
z
2
= zz = (x + ıy)(x −ıy) = x
2
+ y
2
.
25 0
-2
-1
0
1
2
x
-2
-1
0
1
2
y
-5
0
5
-2
-1
0
1
2
x
-2
-1
0
1
2
y...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 8 ppt
... cut.
29 2
-2
0
2
x
-2
-1
0
1
2
y
2
4
-2
0
2
x
-2
0
2
x
-2
-1
0
1
2
y
0
2
4
-2
0
2
x
Figure 7 .20 : Plots of |cos(z)| and |sin(z)|.
Result 7.6.1
e
z
=
e
x
(cos y + ı sin y)
cos z =
e
ız
+
e
−ız
2
sin ... form. Denote any multi-valuedness explicitly.
2
2/5
, 3
1+ı
,
√
3 − ı
1/4
, 1
ı/4
.
Hint, Solution
28 7
-2
-1
0
1
2
x
-2
-1
0
1
2
y
0
0.5
1
-2...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 9 ppt
... Hint 7.19
Hint 7 .20
Hint 7 .21
Hint 7 .22
Hint 7 .23
Hint 7 .24
Hint 7 .25
1. (z
2
+ 1)
1 /2
= (z − ı)
1 /2
(z + ı)
1 /2
2. (z
3
− z)
1 /2
= z
1 /2
(z − 1)
1 /2
(z + 1)
1 /2
3. log (z
2
− 1) = log(z − 1) ... ı47
(
e
ız
+
e
−ız
) /2
(
e
ız
−
e
−ız
) /( 2)
= ı47
e
ız
+
e
−ız
= 47
e
ız
−
e
−ız
46
e
ı2z
−48 = 0
ı2z = log
24
23
z = −
ı
2
log
24
23
z = −
ı
2
ln...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 10 doc
... A branch of ((z
2
− 2) (z + 2) )
1/3
.
Now we determine w (2) .
w (2) =
2 −
√
2
1/3
2 +
√
2
1/3
(2 + 2)
1/3
=
3
2 −
√
2
e
ı0
3
2 +
√
2
e
ı0
3
√
4
e
ı0
=
3
√
2
3
√
4
= 2.
Note that we ... =
(6)(3) (2)
e
ı (2 n +2 n+π) /2
= ı6
We see that our choice of angles gives us the desired branch.
334
Since
−ı log(ı) = −ı
ı
π
2
+ 2 n
=
π
2
+ 2 n
and...
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