Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 3 ppsx
Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 3 ppsx
... 12
β
α
2
− 1
< 0 for |β/α| < 1
= 0 for |β/α| = 1
> 0 for |β/α| > 1
Thus we see that
β
α
x may be a minimum for |β/α| ≥ 1 and may be a maximum for |β/α| ≤ 1.
Jacobi Condition. ... closed curve, and P (t) is a given entire function of t.
Exercise 48.41
Solve
−
1
0
f(t)
t − x
dt + −
3
2
f(t)
t − x
dt = x
where this equation is to hold for x in...
Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 3 pptx
... 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 the error is very small for x < 6, fairly small but non-negligible for x ≈ 7 and large for x > 8. The ... positive for −π < x < 0 and negative for 0 < x < π. Since the sign of y
goes from positive to negative, x = 0 is a
relative maxima. See Figure 3. 7.
Example 3....
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 3 ppt
... dz =
sin
3
z
3
ıπ
π
=
1
3
sin
3
(ıπ) −sin
3
(π)
= −ı
sinh
3
(π)
3
Again the anti-derivative is single-valued.
491
3.
e
x
(sin x cos y cosh y −cos x sin y sinh y)
Exercise 9 .3
For an analytic ... 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 − 3...
Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx
... integrand to s ee that there are singularities at the cube roots of 9.
z
z
3
− 9
=
z
z −
3
√
9
z −
3
√
9
e
ı2π /3
z −
3
√
9
e
−ı2π /3
Let C
1
, C
2
and C
3
be contours around z =
3
√
9, ... lower and
530
3.
C
√
z dz
1. Since sin (cos (z
5
)) is an analytic function inside the unit circle,
C
sin
cos
z
5
dz = 0
2.
1
(z 3) (3z−1)
has singulari ties at z =...
Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 3 pdf
... −x)
=
2
π
π
0
x(π −x) sin(nx) dx
=
2
π
2
n
3
(1 − (−1)
n
)
=
2
π
4
n
3
for odd n
0 for even n
Thus the expansion is
x(π −x) =
∞
n=1
oddn
8
πn
3
sin(nx) for x ∈ [0, π].
In the first graph of Figure ... =
1
√
π
x
−1
e
−x
2
−
1
√
π
∞
x
t
−2
e
−t
2
dt
=
1
√
π
x
−1
e
−x
2
−
1
√
π
−
1
2
t
3
e
−t
2
∞
x
+
1
√
π
∞
x
3
2
t
−4
e
−t
2
dt
=
1
√
π
e
−x
2
x
−1
−
1
2
x...
Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 3 potx
... the one, two and three term
approximations give relative errors of 0.01, 0.00 06 and 0.000 06, respectively.
167 9
Part V
Partial Differential Equations
168 0
Solution 34 .8
For this part we will use ... put
Equation 36 . 1 in the form
u
ξξ
= G(ξ, ψ, u, u
ξ
, u
ψ
). ( 36 . 6)
We require that the u
ξψ
and u
ψψ
terms vanish. That is β = γ = 0 in Equation 36 . 2. This gives us two...
Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 1 pps
... = ξ and time t = τ. The disturbance
influences the solution for all ξ − ct < x < ξ + ct and t > τ.
-1
-0.5
0
0.5
1
x
0
0.2
0.4
0 .6
0.8
1
t
0
0.2
0.4
-1
-0.5
0
0.5
1
x
0
0.2
0.4
0 .6
0.8
1
t
Figure ... t
Figure 45.1 shows the solution at t = 1/2 and t = 2.
-2
-1 1
2
0.1
0.2
0 .3
0.4
0.5
-4
-2 2
4
0.1
0.2
0 .3
0.4
0.5
Figure 45.1: The solution at t = 1/2 and t = 2.
Figure...
Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 2 potx
... = 3/ 4, t = 7/2 is,
u (3/ 4, 7/2) =
1
30
+
7
2
− 3
∞
n=1
1
n
4
π
4
cos(3nπ/2) cos(7nπ).
Note that the summand is nonzero only for even terms.
u (3/ 4, 7/2) =
53
15
−
3
16
4
∞
n=1
1
n
4
cos(3nπ) ... =
53
15
−
3
16
4
∞
n=1
1
n
4
cos(3nπ) cos(14nπ)
=
53
15
−
3
16
4
∞
n=1
(−1)
n
n
4
=
53
15
−
3
16
4
−7π
4
720
u (3/ 4, 7/2) =
12727
38 40
2 033
determine the coe...
Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 4 ppt
... x
1
√
c
2
− 1
The values of the integral for these two values of a are
√
2(x
1
)
3/ 2
−1 + 3c
2
± 3c
√
c
2
− 1
3( c ±
√
c
2
− 1)
3/ 2
.
2095
-3 -2
-1 1
2 3
-3
-2
-1
1
2
3
Figure 48.4: sinh(cx)
Weierstrass ... yields,
x(t) =
kt
2
(3T −t)
2
√
3T
3/ 2
−
1
2
gt
2
.
Applying the condition x(T ) = h gives us,
k
√
3
T
3/ 2
−
1
2
gT
2
= h,
1
4
g
2
T
4
−
k
3
T
3
+ ghT
2
+ h
2
= 0...
Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 5 pdf
... few are
λ
1
≈ 0 .67 8298
λ
2
≈ 7.27 931
λ
3
≈ 24. 930 2
λ
4
≈ 54.25 93
λ
5
≈ 95 .30 57
The eigenfunctions are,
φ
n
(x) = v
(1; λ
n
)u(x; λ
n
) − u
(1; λ
n
)v(x; λ
n
).
Solution 48 .32
1. First note ... equation is
φ(x) =
a + bx for λ = 0
a cos
√
λx
+ b sin
√
λx
for λ > 0
a cosh
√
−λx
+ b sinh
√
−λx
for λ < 0
We see that for λ = 0 and λ &l...
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