... x
2
, x
3
, x
4
} is independent, but not orthogonal in the interval [−1, 1]. Using Gramm-Schmidt orthogo-
12 84
1
2
3
4
5
6
-60
-40
-20
Figure 24. 3: log(error in approximation)
In Figure 24. 4 we ... =
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
3
+
1
√...
... solution and discuss in as much detail as possible what goes wrong.
13 64
Note that this formula is valid for m = 0, 1, 2, . .
Similarly, we can multiply by sin(mx) and integrate to solve for b
m
. ... +
n
2
x
2
− 2
n
3
sin(nx)
137 0
-3
-2
-1 1
2
3
-3
-2
-1
1
2
3
Figure 28.2: Graph of
ˆ
f(x).
would give a better approximation?
Least squared error fit. The most common crit...
... +
∆x
3
6
f
(x) +
∆x
4
24
f
(x
1
),
f(x −∆x) = f(x) −∆xf
(x) +
∆x
2
2
f
(x) −
∆x
3
6
f
(x) +
∆x
4
24
f
(x
2
),
where x ≤ x
1
≤ x + ∆x and x − ∆x ≤ x
2
≤ x.
Hint 3. 20
Hint 3. 21
a. ... 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.
Exam...
... =
ız
4
4
−
1
2z
2
ı
1+ı
=
1
2
+ ı
In this example, the anti-derivative is single-valued.
2.
C
sin
2
z cos z dz =
sin
3
z
3
ıπ
π
=
1
3
sin
3
(ıπ) −sin
3
(π)
= −ı
sinh
3
(π)
3
Again ... max
a≤x≤b
|f(x)|.
46 6
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...
... cosh z
z
3
sin z sinh z
=
1 −
z
2
2
+
z
4
24
− ···
1 +
z
2
2
+
z
4
24
+ ···
z
3
z −
z
3
6
+
z
5
120
− ···
z +
z
3
6
+
z
5
120
+ ···
=
1 −
z
4
6
+ ···
z
3
z
2
+ z
6
−1
36
+
1
60
+ ... ı /4) (n + 1))
z
n
+ d, for 2 < |z|
Solution 12.29
The radius of convergence of the series for f(z) is
R = lim
n→∞
k
3
/3
k
(k + 1)
3
/3
k+1
...
... polynomial are
α
1
=
1 +
1 − 3 /4
2
=
3
4
, α
2
=
1 −
1 − 3 /4
2
=
1
4
.
Thus our two series solutions will be of the form
w
1
= z
3 /4
∞
n=0
a
n
z
n
, w
2
= z
1 /4
∞
n=0
b
n
z
n
.
Substituting ... + 3)
For our first solution we have the difference equation
a
0
= 1, a
1
= 0, a
2
= 0, a
3
= 0, a
n +4
= −
a
n
(n + 4) (n + 3)
.
For our second solution,
b
0
= 0, b...
... the
formula to obtain information about the eigenvalues before we solve a problem.
Example 27 .4. 2 Consider the self-adjoint eigenvalue problem
−y
= λy, y(0) = y(π) = 0.
132 2
Example 27 .4. 1 ... eigenfunction
φ. Green’s formula states
φ|L[φ] − L[φ]|φ = 0
φ|λφ − λφ|φ = 0
(λ − λ)φ|φ = 0
Since φ ≡ 0, φ|φ > 0. Thus λ = λ and λ is real.
131 9
27.7 Hints
Hint 27.1
132 7
2...