Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 8 pot
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 ... z
−1
)/ (2 )
dz =
C
2/ a
z
2
+ ( 2/ a)z − 1
dz
We factor the denominator of the integrand.
f(a) =
C
2/ a
(z − z
1
)(z − z
2
)...
Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 8 potx
... denominator.
s
2
+ s − 1
(s − 2) (s − ı)(s + ı)
We expand the function in partial fractions and then invert each term.
s
2
+ s − 1
(s − 2) (s − ı)(s + ı)
=
1
s − 2
−
ı /2
s − ı
+
ı /2
s + ı
s
2
+ s − 1
(s − 2) (s ... transform for t > 0.
1 485
• t
e
2t
is of exponential order α for any α > 2.
•
e
t
2
is not of exponential order α for any α.
• t
n
is of exponent...
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 2 Part 1 pps
... ations are
u
x
= v
y
and u
y
= −v
x
−(x − 1)
2
+ 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 ... point.
Example 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...
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 + ı...
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|
=
...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx
... 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
Solution 11.3
Let C be the ... integrals along C
1
and C
2
. (We could also
see this by deforming C onto C
1
and C
2
.)
C
=
C
1
+
C
2
We use the Cauchy Integral Formula to evaluate the integrals along C...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 5 pps
... 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 infinite discontinuities at x = 2 k, k ∈ Z. The partial ... n))
5.
∞
n=1
ln (2
n
)
ln (3
n
) + 1
6.
∞
n=0
1
ln(n + 20 )
7.
∞
n=0
4
n
+ 1
3
n
− 2
8.
∞
n=0
(Log
π
2)
n
9.
∞
n =2
n
2
− 1
n
4
− 1
10.
∞
n =2
n
2
(ln n)
n...
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
581
Solution 12. 22
cos z = −cos(z − π)
= −
∞
n=0
(−1)
n
(z −π)
2n
(2n)!
=
∞
n=0
(−1)
n+1
(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 +...
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
1...
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