advanced mathematical methods for scientists and engineers
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 1 pdf
Ngày tải lên :
06/08/2014, 01:21
... 1566
32.7.3 Cosine and Sine Transform in Terms of the Fourier Transform . . . . . . . . . . . . . . . . . . 1568
32.8 Solving Differential Equations with the Fourier Cosine and Sine Transforms . . . ... = p(x)/q(x) where p(x) and q(x) are rational quadratic
polynomials. Give possible formulas for p(x) and q(x).
Hint, Solution
12
33 The Gamma Function 1605
33.1 Euler’s Formul a . . . . . . . ... also contains the word Scientists or Engineers ” the advanced book may be quite suitable for
actually learning the material.
xxvii
43.3 The Method of Characteristics and the Wave Equation ....
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 2 ppt
Ngày tải lên :
06/08/2014, 01:21
... is the angle from a to b and n is a unit vector that is orthogonal to a and b and in the direction s uch that
the ordered triple of vectors a, b and n form a right-handed system.
29
a
b
b
θ
b
Figure ... x
z
yj
i
k
z
k
j
i
y
x
Figure 2.7: Right and left handed coordinate systems.
You can visualize the direction of a ì b by applying the right hand rule. Curl the fingers of your right hand in the
direction from ... arbitrary vectors a and b. We can write b = b
⊥
+ b
where b
⊥
is orthogonal to a and b
is
parallel to a. Show that
a ì b = a ì b
.
Finally prove the distributive law for arbitrary b and c.
Hint...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 3 pptx
Ngày tải lên :
06/08/2014, 01:21
... δ −
√
x is a decreasing function of x and an increasing function of δ for positive x and
δ. Bound this function for fixed δ.
Consider any positive δ and . For what values of x is
1
x
−
1
x + δ
> ... Consider y = x
3
and the point x = 0. The function is differentiable. The derivative, y
= 3x
2
is
positive for x < 0 and positive for 0 < x. Since y
is not identically zero and the sign ... a
n
> 0 for all n > 200, and lim
n→∞
a
n
= L, then L > 0.
4. If f : R → R is continuous and lim
x→∞
f(x) = L, then for n ∈ Z, lim
n→∞
f(n) = L.
5. If f : R → R is continuous and lim
n→∞
f(n)...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 4 pptx
Ngày tải lên :
06/08/2014, 01:21
... of x and an increasing
function of δ for positive x and δ. Thus for any fixed δ, the maximum value of
√
x + δ −
√
x is bounded by
√
δ.
Therefore on the interval (0, 1), a sufficient condition for ... −2)
−1/3
The first 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 for x > 2, f
(x) > 0. ... satisfying,
e(x, δ) ≤ (δ),
for all x in the closed interval. Since (δ) is continuous and increasing, it has an inverse δ(). Now note that
|f(x) − f(ξ)| < for all x and ξ in the closed interval...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 5 pdf
Ngày tải lên :
06/08/2014, 01:21
... −
1
4
cos
t
2
i −
1
4
sin
t
2
j.
See Figure 5.8 for plots of position, ve locity and acceleration.
Figure 5.8: A Graph of Position and Velocity and of Position and Acceleration
Solution 5.2
If r(t) has ... Figure 5.12.) We find the volume obtained by rotating the
172
Set x = 2 and x = −2 to solve for a and b.
Hint 4.16
Expanding the integral in partial fractions,
x + 1
x
3
+ x
2
− 6x
=
x + 1
x(x ... |r
(t)|.
Differentiation Formulas. Let f(t) and g(t) be vector functions and a(t) be a scalar function. By writing out
components you can verify the differentiation formulas:
d
dt
(f · g) = f
·...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 6 pps
Ngày tải lên :
06/08/2014, 01:21
... y
2
e
ı arctan(x,y)
.
Cartesian form is convenient for addition. Polar form is convenient for multiplication and
division.
Example 6.3.1 We write 5 + ı7 in polar form.
5 + ı7 =
√
74
e
ı arctan(5,7)
We ... u
0
, u
1
, u
2
and u
3
are real numbers and ı, and k are objects which satisfy
ı
2
=
2
= k
2
= −1, ı = k, ı = −k
and the usual associative and distributive laws. Show that for any quaternions ... 4.
6.3 Polar Form
Polar form. A complex number written in Cartesian form, z = x + ıy, can be converted polar form, z = r(cos θ +
ı sin θ), using trigonometry. Here r = |z| is the modulus and θ = arctan(x,...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 7 ppt
Ngày tải lên :
06/08/2014, 01:21
... real-variable
counterparts.
7.1 Curves and Regions
In this section we introduce curves and regions in the complex plane. This material is necessary for the study of
branch points in this chapter and later for contour integration.
Curves. ... function, f(z) = z. In Cartesian coordinates and Cartesian form, the function
is f(z) = x + ıy. The real and imaginary components are u(x, y) = x and v(x, y) = y. (See Figure 7.9.) In modulus
-2
-1
0
1
2
x
-2
-1
0
1
2
y
-2
-1
0
1
2
-2
-1
0
1
2
x
-2
-1
0
1
2
x
-2
-1
0
1
2
y
-2
-1
0
1
2
-2
-1
0
1
2
x
Figure ... arctangent that is between
0 and π. The domain and a plot of the selected values of the arctangent are shown in Figure 7.8.
CONTINUE.
7.4 Cartesian and Modulus-Argument Form
We can write a function...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 8 ppt
Ngày tải lên :
06/08/2014, 01:21
... in
Cartesian form and z = r
e
ıθ
in polar form.
e
u+ıv
= r
e
ıθ
We equate the modulus and argument of this expression.
e
u
= r v = θ + 2πn
u = ln r v = θ + 2πn
With log z = u + ıv, we have a formula for ... See
Figure 7.18 and Figure 7.19 for plots of the real and imaginary parts of the cosine and sine, respectively. Figure 7.20
shows the modulus of the cosine and the sine.
The hyperbolic sine and cosine. ... infinity and its only singularity is at
z = 1, the only possi bili ties for branch points are at z = 1 and z = ∞. Since
log
1
z −1
= −log(z −1)
and log w has branch points at zero and infinity,...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 9 ppt
Ngày tải lên :
06/08/2014, 01:21
... Hints
Cartesian and Modulus-Argument Form
Hint 7.1
Hint 7.2
Trigonometric Functions
Hint 7.3
Recall that sin(z) =
1
ı2
(
e
ız
−
e
−ız
). Use Result 6.3.1 to convert between Cartesian and modulus-argument form.
Hint ... on which f(0) = ı
√
6. Write out an explicit formula for the
value of the function on this branch.
Figure 7.33: Four candidate sets of branch cuts for ((z − 1)(z − 2)(z − 3))
1/2
.
Hint, Solution
294
... 4
.
Figure 7.48 first shows the branch cuts and their s tereographic projections and then shows the stereographic projections
alone.
Solution 7.21
1. For each value of z, f(z) = z
1/3
has three...
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Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 10 doc
Ngày tải lên :
06/08/2014, 01:21
... condition for the analyticity of f(z).
Let φ(x, y) = u(x, y) + ıv(x, y) where u and v are real-valued functions. We equate the real and imaginary parts
of Equation 8.1 to obtain another form for the ... ı)
−1/3
=
3
√
r
e
ıθ/3
1
3
√
s
e
−ıφ/3
1
3
√
t
e
−ıψ/3
=
3
r
st
e
ı(θ−φ−ψ)/3
we have an explicit formula for computing the value of the function for this branch. Now we compute f (1) to see if we
chose the correct ranges for the angles. (If not, we’ll just ... . . 2π), (2π . . . 4π), . . .}.
Now we choose ranges for θ and φ and see if we get the desired branch. If not, we choose a different range for one of
the angles. First we choose the ranges
θ ∈...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 1 pps
Ngày tải lên :
06/08/2014, 01:21
... equations for à and are satised if and only if the Cauchy-Riemann
equations for u and v are satisfied. The continuity of the first partial derivatives of u and v implies the same of
à and . Thus ... =
x
3
(1+ı)−y
3
(1−ı)
x
2
+y
2
for z = 0,
0 for z = 0.
Show that the partial derivatives of u and v with respect to x and y exist at z = 0 and that u
x
= v
y
and u
y
= −v
x
there: the Cauchy-Riemann ... function.
Solution 8.11
We write the real and imaginary parts of f(z) = u + ıv.
u =
x
4/3
y
5/3
x
2
+y
2
for z = 0,
0 for z = 0.
, v =
x
5/3
y
4/3
x
2
+y
2
for z = 0,
0 for z = 0.
The Cauchy-Riemann...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 2 pptx
Ngày tải lên :
06/08/2014, 01:21
... Consider analytic functions f
1
(z) and f
2
(z) defined on the domains D
1
and
D
2
, respectively. Suppose that D
1
∩ D
2
is a region or an arc and that f
1
(z) = f
2
(z) for all
z ∈ D
1
∩ D
2
. (See ... converges uniformly for D
1
= |z| ≤ r < 1. Since the derivative also converges in this domain, the function is
analytic there.
440
Figure 8.7: The velocity potential φ and stream function ψ for Φ(z) ... Substitute this expression for v into the equation for ∂v/∂x.
−y
e
−x
sin y − x
e
−x
cos y +
e
−x
cos y + F
(x) = −y
e
−x
sin y − x
e
−x
cos y +
e
−x
cos y
Thus F
(x) = 0 and F (x) = c.
v =
e
−x
(y...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 3 ppt
Ngày tải lên :
06/08/2014, 01:21
... contour and do the integration.
z − z
0
=
e
ıθ
, θ ∈ [0 . . . 2π)
C
(z − z
0
)
n
dz =
2π
0
e
ınθ
ı
e
ıθ
dθ
=
e
ı(n+1)θ
n+1
2π
0
for n = −1
[ıθ]
2π
0
for n = −1
=
0 for n = −1
ı2π for ... u
y
) dx dy + ı
D
(u
x
− v
y
) dx dy
= 0
Since the two integrands are continuous and vanish for all C in Ω, we conclude that the integrands are identically zero.
This implies that the Cauchy-Riemann ... ıφ dy)
=
D
(ıφ
x
− φ
y
) dx dy
= 0
Since the integrand, ıφ
x
− φ
y
is continuous and vanishes for all C in Ω, we conclude that the integrand is identically
zero. This implies that the Cauchy-Riemann...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx
Ngày tải lên :
06/08/2014, 01:21
... converges if and only i f for any > 0 there exists an
N such that |a
n
− a
m
| < for all n, m > N. The Cauchy convergence criterion is equivalent to the definition we had
before. For some ... 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
1
and C
2
.
C
(z
3
+ ... integral
C
e
zt
z
2
(z + 1)
dz.
There are singularities at z = 0 and z = −1.
Let C
1
and C
2
be contours around z = 0 and z = −1. See Figure 11.6. We deform C onto C
1
and C
2
.
C
=
C
1
+
C
2
520
11.4 Exercises
Exercise...
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Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 5 pps
Ngày tải lên :
06/08/2014, 01:21
... ,
converges for α > 1 and diverges for α ≤ 1.
Hint, Solution
564
Example 12.3.2 Convergence and Uniform Convergence. Consider the series
log(1 − z) = −
∞
n=1
z
n
n
.
This series converges for |z| ... necessary
condition for uniform convergence. The Weierstrass M-test can succeed only if the series is uniformly and absolutely
convergent.
Example 12.2.1 The series
f(x) =
∞
n=1
sin x
n(n + 1)
is uniformly and ... Thus this series is not uniformly convergent in the domain
|z| ≤ 1, z = 1. The series is uniformly convergent for |z| ≤ r < 1.
545
12.2.2 Uniform Convergence and Continuous Functions.
Consider...
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