Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 6 doc
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 6 pps
... values. For instance, (1
2
)
1/ 2
=
1
1/2
= 1 and
1
1/2
2
= ( 1)
2
= 1.
Example 6. 6.2 Consider 2
1/ 5
, (1 + ı)
1/ 3
and (2 + ı)
5 /6
.
2
1/ 5
=
5
√
2
e
ı2πk/5
, for k = 0, 1, 2, 3, 4
19 9
Example 6. 5 .1 ... ellipse.
18 7
6. 8 Hints
Complex Numbers
Hint 6 .1
Hint 6. 2
Hint 6. 3
Hint 6. 4
Hint 6. 5
Hint 6. 6
Hint 6. 7
The Complex Plane
Hint 6. 8...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 10 doc
... −2)
1/ 2
(z −3)
1/ 2
There are branch points at z = 1, 2, 3. Now we examine the point at infinity.
f
1
ζ
=
1
ζ
− 1
1
ζ
− 2
1
ζ
− 3
1/ 2
= ζ
−3/2
1 −
1
ζ
1 −
2
ζ
1 −
3
ζ
1/ 2
Since ... square root.
z
2
− 1
1/ 2
= (z + 1)
1/ 2
(z 1)
1/ 2
We see that there are branch points at z = 1 and z = 1. In particular we want the Arccos to be...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 1 pps
... y
2
((x − 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 ... 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 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
= ... x
direction.
f
(z) = u
x
+ ıv
x
f
(z) = 2
e
x
2
−y
2
(x cos(2xy) − y sin(2xy)) + 2
e
x
2
−y
2
(y cos(2xy) + x sin(2xy))
f
(z) = 2
e
x
2
−y
2
((x +...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 3 ppt
... are
u
x
= 3x
2
− 3y
2
− 2y,
u
y
= −6xy − 2x + 1.
The derivative of f(z) is
f
(z) = u
x
− ıu
y
= 3x
2
− 2y
2
− 2y + ı(6xy − 2x + 1).
On the real axis we have
f
(z = x) = 3x
2
− ı2x + ı.
Using ... max
a≤x≤b
|f(x)|.
466
with a, b and c complex-valued constants and d a real constant. Substituting z = x + ıy and expanding products
yields,
a
x
3
+ ı3x
2
y − 3xy
2
− ıy...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx
... z
z
3
z=−ı
+
2
2!
d
2
dz
2
(z
3
+ z + ı) sin z
z + ı
z=0
= 2 (−ı sinh(1)) + ıπ
2
3z
2
+ 1
z + ı
−
z
3
+ z + ı
(z + ı)
2
cos z
+
6z
z + ı
−
2( 3z
2
+ 1)
(z + ı)
2
+
2( z
3
+ z + ı)
(z ... formula.
C
z
z
2
+ 1
dz =
C
1 /2
z −ı
dz +
C
1 /2
z + ı
dz
=
1
2
2 +
1
2
2
= 2
3.
C
z
2
+ 1
z
dz =
C
z +
1
z
dz
=
C
z dz +
C
1
z
dz
= 0 + 2
= 2...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 5 pps
... 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)
Re(z)
R
R
2
1
C
r
1
r
2
z
C
C
C
1
2
z
Figure 12. 5: Contours for a Laurent ... closed form. (See Exercise 12. 9.)
N−1
n=1
sin(nx) =
0 for x = 2 k
cos(x /2) −cos((N−1 /2) x)
2 sin(x /2)
for x = 2 k
The partial sums have infinit...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 6 doc
... function.
Hint 12. 22
cos z = −cos(z − π)
sin z = −sin(z − 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
581
Solution 12. 22
cos z ... polynomial.
2 6 12 20
4 6 8
2 2
We s ee that the polynomial is second order. p(n) = an
2
+ bn + c. We solve for the coefficients.
a + b + c = 2
4a + 2b + c = 6
9...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 7 pdf
... 0.
|z|=3
z
1
z − ı /2
−
1
z − 2
+
c
(z − 2)
2
+ d
dz = 0
|z|=3
(z − ı /2) + ı /2
z − ı /2
−
(z − 2) + 2
z − 2
+
c(z − 2) + 2c
(z − 2)
2
dz = 0
2
ı
2
− 2 + c
= 0
c = 2 −
ı
2
Thus we see that ... − 2/ z
= −
1
z
∞
n=0
2
z
n
, for |2/ z| < 1
= −
∞
n=0
2
n
z
−n−1
, for |z| > 2
= −
−1
n=−∞
2
−n−1
z
n
, for |z| > 2
620...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 8 pot
... integrals.
1.
∞
0
ln
2
x
1 + x
2
dx =
π
3
8
2.
∞
0
ln x
1 + x
2
dx = 0
6 72
Hint 13.9
For the third part, does the integrand have a term that behaves like 1/x
2
?
Cauchy Principal Value for Contour ... −
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...
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
+...
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 3 Part 6 ppsx
... x
=
dx
dy
.
dy
dx
=
1
y
3
− xy
2
dx
dy
= y
3
− xy
2
x
+ y
2
x = y
3
Now we have a first order equation for x.
d
dy
e
y
3
/3
x
= y
3
e
y
3
/3
x =
e
−y
3
/3
y
3
e
y
3
/3
dy + c
e
−y
3
/3
Example 18 .3. 2 Consider ... equation of order
n − 1 for u
. Writing the derivatives of
e
u(x)
,
d
dx
e
u
= u
e
u
d
2
dx
2
e
u
= (u
+ (u
)
2
)
e
u
d
3
dx
3
e...
Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 9 doc
... the problem for u, (and
hence the problem for y).
As a check, then general solution for y is
y = −
1
3
cos 2x + c
1
cos x + c
2
sin x.
1115
We guess a particular solution of the form
y
p
= t
e
−t
(a ... y
2
=
e
3t
.
We compute the Wronskian of these solutions.
W (t) =
e
2t
e
3t
2
e
2t
3
e
3t
=
e
5t
We find a particular solution with variation of parameters.
y
p
= −...
Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 1 docx
... second equation.
c
1
r
2
1
+
1
r
2
(1 − c
1
r
1
)r
2
2
= 1
c
1
(r
2
1
− r
1
r
2
) = 1 − r
2
11 81
0.2 0 .4 0.6 0.8
1
1.2
0.3
0 .4
0.5
0.6
0.7
0.8
0.9
Figure 23.2: Plot of the solution and approximations.
Recall ... that
a
n
=
n/2
j=0
−
4j
2
−2j +1
(2j+2)(2j +1)
for even n,
0 for odd n.
11 93
c
1
=
1 − r
2
r
2
1
− r
1
r
2
=
1 −
1
√
5
2...
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