Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 10 doc
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 1 Part 1 pdf
... . . . . . 10 04
18 .10 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 06
19 Transformations and Canonical Forms 10 18
19 .1 The Constant ... . . . 19 10
43 .10 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 11
44 Transform Methods 19 18
44 ....
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 2 ppt
... a and b are orthogonal, (perpendicular), or one of a and b are zero.
26
1
2
1
2
2
4
6
8
10
1
2
Figure 1. 11: Plots of f(x) = p(x)/q(x).
1. 6 Hints
Hint 1. 1
area = constant ×diameter
2
.
Hint 1. 2
A ... from 1 to n and are called free indices.
Example 2 .1. 1 Consider the matrix equation: A · x = b. We can write out the matrix and vectors explicitly.
a
11
··· a
1...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 3 pptx
... ( 1)
n 1
(n − 1) !, for n ≥ 1.
By Taylor’s theorem of the mean we have,
ln x = (x − 1) −
(x − 1)
2
2
+
(x − 1)
3
3
−
(x − 1)
4
4
+ ··· + ( 1)
n 1
(x − 1)
n
n
+ ( 1)
n
(x − 1)
n +1
n + 1
1
ξ
n +1
.
72
3.8 ... are
f
1
(x) = 1
f
2
(x) = 1 + x
f
3
(x) = 1 + x +
x
2
2
f
4
(x) = 1 + x +
x
2
2
+
x
3
6
The four approximations are graphed in Figure 3 .11 .
-1
-0.5...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 4 pptx
... thumb:
12 1
c.
ln
lim
x→+∞
1 +
1
x
x
= lim
x→+∞
ln
1 +
1
x
x
= lim
x→+∞
x ln
1 +
1
x
= lim
x→+∞
ln
1 +
1
x
1/ x
= lim
x→+∞
1 +
1
x
1
−
1
x
2
1/ x
2
= lim
x→+∞
1 +
1
x
1
= ... + x cos x
Solution 3 .12
Let y(x) = sin x. Then y
(x) = cos x.
d
dy
arcsin y =
1
y
(x)
=
1
cos x
=
1
(1 −sin
2
x)
1/ 2...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 5 pdf
... =
1
(0 − 1) (0 − 2)(0 − 3)
= −
1
6
b =
1
(1) (1 − 2) (1 − 3)
=
1
2
c =
1
(2)(2 − 1) (2 − 3)
= −
1
2
d =
1
(3)(3 − 1) (3 − 2)
=
1
6
1
x(x − 1) (x − 2)(x − 3)
= −
1
6x
+
1
2(x − 1)
−
1
2(x − 2)
+
1
6(x ... lim
δ→0
+
1 δ
0
1
(x − 1)
2
dx + lim
→0
+
4
1+
1
(x − 1)
2
dx
Hint 4 .18
1
0
1
√
x
dx = lim
→0
+
1
1
√
x
dx
Hint 4 .19
1...
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 ... 4
x
2
+ y
2
= 16 − 8
(x − 2)
2
+ y
2
+ x
2
− 4x + 4 + y
2
x − 5 = −2
(x − 2)
2
+ y
2
x
2
− 10 x + 25 = 4x
2
− 16 x + 16 + 4y
2
1
4
(x − 1)
2
+
1
3
y
2
=...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 7 ppt
... + ı| = 1.
2 31
Now we do the multiplication in modulus-argument, (polar), form.
1 + ı
√
3
10
=
2
e
ıπ/3
10
= 2
10
e
− 10 π/3
=
1
102 4
cos
−
10 π
3
+ ı sin
−
10 π
3
=
1
102 4
cos
4π
3
− ... y
2
.
250
-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
Figure 7 .10 : A few branches of arg(z).
-2
-1
0
1
2
x
-2
-1...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 8 ppt
... arguments:
1. log( 1) = log
1
1
= log (1) − log( 1) = −log( 1) , therefore, log( 1) = 0.
2. 1 = 1
1/2
= (( 1) ( 1) )
1/ 2
= ( 1)
1/ 2
( 1)
1/ 2
= ıı = 1, therefore, 1 = 1.
Hint, Solution
Exercise 7 .11
Write ... Cartesian form. Denote any multi-valuedness explicitly.
2
2/5
, 3
1+ ı
,
√
3 − ı
1/ 4
, 1
ı/4
.
Hint, Solution
287
-2
-1
0
1
2
x
-2
-1
0
1
2
y
0
0.5...
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) + ... z
1
= z
2
= 1.
Arg(( 1) ( 1) ) = Arg (1) = 0, Arg( 1) + Arg( 1) = 2π
Log(( 1) ( 1) ) = Log (1) = 0, Log( 1) + Log( 1) = ı2π
306
sign. This will change the value of arccos...
Từ khóa: