Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 6 pps

We use the definition of the Fourier coefficients to evaluate the integrals in the last sum.

π
−π
(f(x))
2
dx − πa
2
0
− 2π
∞

n=1

a
2
n
+ b
2
n

+
πa
2
0
2
+ π
∞

n=1


a
2
n
+ b
2
n

= 0
a
2
0
2
+
∞

n=1

a
2
n
+ b
2
n

=
1
π

π

−π
f(x)
2
dx
2. We determine the Fourier coefficients for f(x) = x. Since f(x) is odd, all of the a
n
are zero.
b
0
=
1
π

π
−π
x sin(nx) dx
=
1
π

−
1
n
x cos(nx)

π
−π
+

π

−π
1
n
cos(nx) dx
=
2(−1)
n+1
n
The Fourier series is
x =
∞

n=1
2(−1)
n+1
n
sin(nx) for x ∈ (−π . . . π).
We apply Parseval’s theorem for this series to find the value of

∞
n=1
1
n
2
.
∞

n=1
4
n

2
=
1
π

π
−π
x
2
dx
∞

n=1
4
n
2
=
2π
2
3
∞

n=1
1
n
2
=
π
2
6

1374
3. Consider f(x) = x
2
. Since the function is even, there are no sine terms in the Fourier series. The coefficients in
the cosine series are
a
0
=
2
π

π
0
x
2
dx
=
2π
2
3
a
n
=
2
π

π
0
x
2

cos(nx) dx
=
4(−1)
n
n
2
.
Thus the Fourier series is
x
2
=
π
2
3
+ 4
∞

n=1
(−1)
n
n
2
cos(nx) for x ∈ (−π . . . π).
We apply Parseval’s theorem for this series to find the value of

∞
n=1
1
n
4

.
2π
4
9
+ 16
∞

n=1
1
n
4
=
1
π

π
−π
x
4
dx
2π
4
9
+ 16
∞

n=1
1
n
4

=
2π
4
5
∞

n=1
1
n
4
=
π
4
90
1375
Now we integrate the series for f(x) = x
2
.

x
0

ξ
2
−
π
2
3

dξ = 4

∞

n=1
(−1)
n
n
2

x
0
cos(nξ) dξ
x
3
3
−
π
2
3
x = 4
∞

n=1
(−1)
n
n
3
sin(nx)
We apply Parseval’s theorem for this series to find the value of

∞

n=1
1
n
6
.
16
∞

n=1
1
n
6
=
1
π

π
−π

x
3
3
−
π
2
3
x

2
dx

16
∞

n=1
1
n
6
=
16π
6
945
∞

n=1
1
n
6
=
π
6
945
1376
Solution 28.2
1. We differentiate the partial sum of the Fourier series and evaluate the sum.
S
N
=
N

n=1

2(−1)
n+1
n
sin(nx)
S

N
= 2
N

n=1
(−1)
n+1
cos(nx)
S

N
= 2

N

n=1
(−1)
n+1
e
ınx

S

N

= 2

1 − (−1)
N+2
e
ı(N+1)x
1 +
e
ıx

S

N
= 

1 +
e
−ıx
−(−1)
N
e
ı(N+1)x
−(−1)
N
e
ıNx
1 + cos(x)

S


N
= 1 − (−1)
N
cos((N + 1)x) + cos(Nx)
1 + cos(x)
S

N
= 1 − (−1)
N
cos

N +
1
2

x

cos

x
2

cos
2

x
2

dS

N
dx
= 1 − (−1)
N
cos

N +
1
2

x

cos

x
2

1377
2. We integrate S

N
.
S
N
(x) − S
N
(0) = x −

x
0

(−1)
N
cos

N +
1
2

ξ

cos

ξ
2

dξ
x − S
N
=

x
0
sin

N +
1
2

(ξ −π)


sin

ξ−π
2

dξ
3. We find the extrema of the overshoot E = x − S
N
with the first derivative test.
E

=
sin

N +
1
2

(x − π)

sin

x−π
2

= 0
We look for extrema in the range (−π . . . π).

N +
1

2

(x − π) = −nπ
x = π

1 −
n
N + 1/2

, n ∈ [1 . . . 2N]
The closest of these extrema to x = π is
x = π

1 −
1
N + 1/2

Let E
0
be the overshoot at this p oint. We approximate E
0
for large N.
E
0
=

π(1−1/(N+1/2))
0
sin


N +
1
2

(ξ −π)

sin

ξ−π
2

dξ
We shift the limits of integration.
E
0
=

π
π/(N+1/2)
sin

N +
1
2

ξ

sin

ξ

2

dξ
1378
We add and subtract an integral over [0 . . . π/(N + 1/2)].
E
0
=

π
0
sin

N +
1
2

ξ

sin

ξ
2

dξ −

π/(N+1/2)
0
sin


N +
1
2

ξ

sin

ξ
2

dξ
We can evaluate the first integral with contour integration on the unit circle C.

π
0
sin

N +
1
2

ξ

sin

ξ
2

dξ =


π
0
sin ((2N + 1) ξ)
sin (ξ)
dξ
=
1
2

π
−π
sin ((2N + 1) ξ)
sin (ξ)
dξ
=
1
2
−

C


z
2N+1

(z −1/z)/(ı2)
dz
ız
= 


−

C
z
2N+1
(z
2
− 1)
dz

= 

ıπ Res

z
2N+1
(z + 1)(z −1)
, 1

+ ıπ Res

z
2N+1
(z + 1)(z −1)
, −1

= π

1

2N+1
2
+
(−1)
2N+1
−2

= π
1379
We approximate the second integral.

π/(N+1/2)
0
sin

N +
1
2

ξ

sin

ξ
2

dξ =
2
2N + 1


π
0
sin(x)
sin

x
2N+1

dx
≈ 2

π
0
sin(x)
x
dx
= 2

π
0
1
x
∞

n=0
(−1)
n
x
2n+1
(2n + 1)!

dx
= 2
∞

n=0

π
0
(−1)
n
x
2n
(2n + 1)!
dx
= 2
∞

n=0
(−1)
n
π
2n+1
(2n + 1)(2n + 1)!
dx
≈ 3.70387
In the limit as N → ∞, the overshoot is
|π −3.70387| ≈ 0.56.
Solution 28.3
1. The eigenfunctions of the self-adjoint problem
−y


= λy, y(0) = y(1) = 0,
are
φ
n
= sin(nπx), n ∈ Z
+
1380
We find the series expansion of the inhomogeneity f(x) = 1.
1 =
∞

n=1
f
n
sin(nπx)
f
n
= 2

1
0
sin(nπx) dx
f
n
= 2

−
cos(nπx)
nπ


1
0
f
n
=
2
nπ
(1 − (−1)
n
)
f
n
=

4
nπ
for odd n
0 for even n
We expand the solution in a series of the eigenfunctions.
y =
∞

n=1
a
n
sin(nπx)
We substitute the series into the differential equation.
y


+ 2y = 1
−
∞

n=1
a
n
π
2
n
2
sin(nπx) + 2
∞

n=1
a
n
sin(nπx) =
∞

n=1
odd n
4
nπ
sin(nπx)
a
n
=

4

nπ(2−π
2
n
2
)
for odd n
0 for even n
y =
∞

n=1
odd n
4
nπ(2 − π
2
n
2
)
sin(nπx)
1381
2. Now we solve the boundary value problem directly.
y

+ 2y = 1 y(0) = y(1) = 0
The general solution of the differential equation is
y = c
1
cos

√

2x

+ c
2
sin

√
2x

+
1
2
.
We apply the boundary conditions to find the solution.
c
1
+
1
2
= 0, c
1
cos

√
2

+ c
2
sin


√
2

+
1
2
= 0
c
1
= −
1
2
, c
2
=
cos

√
2

− 1
2 sin

√
2

y =
1
2


1 − cos

√
2x

+
cos

√
2

− 1
sin

√
2

sin

√
2x


We find the Fourier sine series of the solution.
y =
∞

n=1
a
n

sin(nπx)
a
n
= 2

1
0
y(x) sin(nπx) dx
a
n
=

1
0

1 − cos

√
2x

+
cos

√
2

− 1
sin

√

2

sin

√
2x


sin(nπx) dx
a
n
=
2(1 − (−1)
2
nπ(2 − π
2
n
2
)
a
n
=

4
nπ(2−π
2
n
2
)
for odd n

0 for even n
1382
We obtain the same series as in the first part.
Solution 28.4
1. The eigenfunctions of the self-adjoint problem
−y

= λy, y

(0) = y

(π) = 0,
are
φ
0
=
1
2
, φ
n
= cos(nx), n ∈ Z
+
We find the series expansion of the inhomogeneity f(x) = sin(x).
f(x) =
f
0
2
+
∞


n=1
f
n
cos(nx)
f
0
=
2
π

π
0
sin(x) dx
f
0
=
4
π
f
n
=
2
π

π
0
sin(x) cos(nx) dx
f
n
=

2(1 + (−1)
n
)
π(1 − n
2
)
f
n
=

4
π(1−n
2
)
for even n
0 for odd n
We expand the solution in a series of the eigenfunctions.
y =
a
0
2
+
∞

n=1
a
n
cos(nx)
1383
We substitute the series into the differential equation.

y

+ 2y = sin(x)
−
∞

n=1
a
n
n
2
cos(nx) + a
0
+ 2
∞

n=1
a
n
cos(nx) =
2
π
+
∞

n=2
even n
4
π(1 − n
2

)
cos(nx)
y =
1
π
+
∞

n=2
even n
4
π(1 − n
2
)(2 − n
2
)
cos(nx)
2. We expand the solution in a series of the eigenfunctions.
y =
a
0
2
+
∞

n=1
a
n
cos(nx)
We substitute the series into the differential equation.

y

+ 4y = sin(x)
−
∞

n=1
a
n
n
2
cos(nx) + 2a
0
+ 4
∞

n=1
a
n
cos(nx) =
2
π
+
∞

n=2
even n
4
π(1 − n
2

)
cos(nx)
It is not possible to solve for the a
2
coefficient. That equation is
(0)a
2
= −
4
3π
.
This problem is to be expected, as this boundary value problem does not have a solution. The solution of the
differential equation is
y = c
1
cos(2x) + c
2
sin(2x) +
1
3
sin(x)
1384
The boundary conditions give us an inconsistent set of constraints.
y

(0) = 0, y

(π) = 0
c
2

+
1
3
= 0, c
2
−
1
3
= 0
Thus the problem has no solution.
Solution 28.5
Cosine Series. The coefficients in the cosine series are
a
0
=
2
π

π
0
x
2
dx
=
2π
2
3
a
n
=

2
π

π
0
x
2
cos(nx) dx
=
4(−1)
n
n
2
.
Thus the Fourier cosine series is
f(x) =
π
2
3
+
∞

n=1
4(−1)
n
n
2
cos(nx).
In Figure 28.10 the even periodic extension of f(x) is plotted in a dashed line and the sum of the first five terms in the
Fourier series is plotted in a solid line. Since the even periodic extension is continuous, the cosine series is differentiable.

1385
-3
-2
-1 1
2
3
2
4
6
8
10
-3
-2
-1 1
2
3
-10
-5
5
10
Figure 28.10: The Fourier Cosine and Sine Series of f(x) = x
2
.
Sine Series. The coefficients in the sine series are
b
n
=
2
π


π
0
x
2
sin(nx) dx
= −
2(−1)
n
π
n
−
4(1 − (−1)
n
)
πn
3
=

−
2(−1)
n
π
n
for even n
−
2(−1)
n
π
n
−

8
πn
3
for odd n.
1386
Thus the Fourier sine series is
f(x) ∼ −
∞

n=1

2(−1)
n
π
n
+
4(1 − (−1)
n
)
πn
3

sin(nx).
In Figure 28.10 the odd perio d ic extension of f(x) and the sum of the first five terms in the sine series are plotted.
Since the odd periodic extension of f(x) is not continuous, the series is not differentiable.
Solution 28.6
We could find the expansion by integrating to find the Fourier coefficients, but it is easier to expand cos
n
(x) directly.
cos

n
(x) =

1
2
(
e
ıx
+
e
−ıx
)

n
=
1
2
n

n
0

e
ınx
+

n
1

e

ı(n−2)x
+ ··· +

n
n − 1

e
−ı(n−2)x
+

n
n

e
−ınx

If n is odd,
cos
n
(x) =
1
2
n


n
0

(
e

ınx
+
e
−ınx
) +

n
1

(
e
ı(n−2)x
+
e
−ı(n−2)x
) + ···
+

n
(n − 1)/2

(
e
ıx
+
e
−ıx
)

=

1
2
n

n
0

2 cos(nx) +

n
1

2 cos((n − 2)x) + ··· +

n
(n − 1)/2

2 cos(x)

=
1
2
n−1
(n−1)/2

m=0

n
m


cos((n − 2m)x)
=
1
2
n−1
n

k=1
odd k

n
(n − k)/2

cos(kx).
1387
If n is even,
cos
n
(x) =
1
2
n


n
0

(
e
ınx

+
e
−ınx
) +

n
1

(
e
ı(n−2)x
+
e
−ı(n−2)x
) + ···
+

n
n/2 − 1

(
e
ı2x
+
e
−i2x
) +

n
n/2



=
1
2
n

n
0

2 cos(nx) +

n
1

2 cos((n − 2)x) + ··· +

n
n/2 − 1

2 cos(2x) +

n
n/2

=
1
2
n


n
n/2

+
1
2
n−1
(n−2)/2

m=0

n
m

cos((n − 2m)x)
=
1
2
n

n
n/2

+
1
2
n−1
n

k=2

even k

n
(n − k)/2

cos(kx).
We may denote,
cos
n
(x) =
a
0
2
n

k=1
a
k
cos(kx),
where
a
k
=
1 + (−1)
n−k
2
1
2
n−1


n
(n − k)/2

.
1388
Solution 28.7
We expand f(x) in a cosine series. The coefficients in the cosine series are
a
0
=
2
π

π
0
x
2
dx
=
2π
2
3
a
n
=
2
π

π
0

x
2
cos(nx) dx
=
4(−1)
n
n
2
.
Thus the Fourier cosine series is
f(x) =
π
2
3
+ 4
∞

n=1
(−1)
n
n
2
cos(nx).
The Fourier series converges to the even periodic extension of
f(x) = x
2
for 0 < x < π,
which is
ˆ
f(x) =


x − 2π

x + π
2π

2
.
(· denotes the floor or greatest integer function.) This periodic extension is a continuous function. Since x
2
is an
even function, we have
π
2
3
+ 4
∞

n=1
(−1)
n
n
2
cos nx = x
2
for − π ≤ x ≤ π.
1389
We substitute x = π into the Fourier series.
π
2

3
+ 4
∞

n=1
(−1)
n
n
2
cos(nπ) = π
2
∞

n=1
1
n
2
=
π
2
6
We substitute x = 0 into the Fourier series.
π
2
3
+ 4
∞

n=1
(−1)

n
n
2
= 0
∞

n=1
(−1)
n+1
n
2
=
π
2
12
Solution 28.8
1. We compute the Fourier sine coefficients.
a
n
=
2
π

π
0
f(x) sin(nx) dx
=
2
π


π
0

cos x − 1 +
2x
π

sin(nx) dx
=
2(1 + (−1)
n
)
π(n
3
− n)
a
n
=

4
π(n
3
−n)
for even n
0 for odd n
1390
2. From our work in the previous part, we see that the Fourier coefficients decay as 1/n
3
. The Fourier sine series
converges to the odd periodic extens ion of the function,

ˆ
f(x). We can determine the rate of decay of the Fourier
coefficients from the smoothness of
ˆ
f(x). For −π < x < π, the o dd periodic extension of f(x) is defined
ˆ
f(x) =

f(x) = cos(x) − 1 +
2x
π
0 ≤ x < π,
−f(−x) = −cos(x) + 1 +
2x
π
−π ≤ x < 0.
Since
ˆ
f(0
+
) =
ˆ
f(0
−
) = 0 and
ˆ
f(π) =
ˆ
f(−π) = 0
ˆ

f(x) is continuous, C
0
. Since
ˆ
f

(0
+
) =
ˆ
f

(0
−
) =
2
π
and
ˆ
f

(π) =
ˆ
f

(−π) =
2
π
ˆ
f(x) is continuously differentiable, C

1
. However, since
ˆ
f

(0
+
) = −1, and
ˆ
f

(0
−
) = 1
ˆ
f(x) is not C
2
. Since
ˆ
f(x) is C
1
we know that the Fourier coefficients decay as 1/n
3
.
Solution 28.9
Cosine Series. The even periodic extension of f(x) is a C
0
, continuous, function (See Figure 28.11. Thus the
coefficients in the cosine series will decay as 1/n
2

. The Fourier cosine coefficients are
a
0
=
2
π

π
0
x sin x dx
= 2
a
1
=
2
π

π
0
x sin x cos x dx
= −
1
2
1391
a
n
=
2
π


π
0
x sin x cos(nx) dx
=
2(−1)
n+1
n
2
− 1
, for n ≥ 2
The Fourier cosine series is
ˆ
f(x) = 1 −
1
2
cos x − 2
∞

n=2
2(−1)
n
n
2
− 1
cos(nx).
-5 5
1
Figure 28.11: The even periodic extension of x sin x.
Sine Series. The odd periodic extension of f(x) is a C
1

, continuously differentiable, function (See Figure 28.12.
Thus the coefficients in the cosine series will decay as 1/n
3
. The Fourier sine coefficients are
a
1
=
1
π

π
0
x sin x sin x dx
=
π
2
1392
a
n
=
2
π

π
0
x sin x sin(nx) dx
= −
4(1 + (−1)
n
)n

π(n
2
− 1)
2
, for n ≥ 2
The Fourier sine series is
ˆ
f(x) =
π
2
sin x −
4
π
∞

n=2
(1 + (−1)
n
)n
(n
2
− 1)
2
cos(nx).
-5 5
1
Figure 28.12: The odd p eriodic extension of x sin x.
Solution 28.10
If ν = n is an integer, then the Fourier cosine series is the single term cos(|n|x). We assume that ν = n.
We note that the even periodic extension of cos(νx) is C

0
so that the series converges to cos(νx) for −π ≤ x ≤ π
1393
and the coefficients decay as 1/n
2
. We compute the Fourier cosine coefficients.
a
0
=
2
π

π
0
cos(νx) dx
=
2 sin(πν)
πν
a
n
=
2
π

π
0
cos(νx) cos(nx) dx
= (−1)
n


1
ν − n
+
1
ν + n

sin(πν)
The Fourier cosine series is
cos(νx) =
sin(πν)
πν
+
∞

n=1
(−1)
n

1
ν − n
+
1
ν + n

sin(πν) cos(nx).
We substitute x = 0 into the Fourier cosine series.
1 =
sin(πν)
πν
+

∞

n=1
(−1)
n

1
ν − n
+
1
ν + n

sin(πν)
π
sin πν
=
1
ν
+
∞

n=1
(−1)
n

1
ν − n
+
1
ν + n


Next we substitute x = π into the Fourier cosine series.
cos(νπ) =
sin(πν)
πν
+
∞

n=1
(−1)
n

1
ν − n
+
1
ν + n

sin(πν)(−1)
n
π cot πν =
1
ν
+
∞

n=1

1
ν − n

+
1
ν + n

1394
Note that neither cot(πν) nor 1/ν is integrable at ν = 0. We write the last formula so each side is integrable.
π cot πν −
1
ν
=
∞

n=1

1
ν − n
+
1
ν + n

We integrate from ν = 0 to ν = θ < 1.

ln

sin(πν)
ν

θ
0
=

∞

n=1

[ln(n − ν)]
θ
0
+ [ln(n + ν)]
θ
0

ln

sin(πθ)
θ

− ln π =
∞

n=1

ln

n − θ
n

+ ln

n + θ
n


ln

sin(πθ)
πθ

=
∞

n=1
ln

1 −
θ
2
n
2

ln

sin(πθ)
πθ

= ln

∞

n=1

1 −

θ
2
n
2


sin(πθ)
πθ
=
∞

n=1

1 −
θ
2
n
2

Solution 28.11
1. We will consider the principal branch of the logarithm, −π < (Log z) ≤ π. For −π < x < π, cos(x/2) is
positive so that ln(cos(x/2)) is well-defined. At x = ±π, ln(cos(x/2)) is singular. However, the function is
integrable so it has a Fourier series which converges except at x = (2k + 1)π, k ∈ Z.
ln

cos
x
2

= ln


e
ıx/2
+
e
−ıx/2
2

= −ln 2 + ln

e
−ıx/2
(1 +
e
ıx
)

= −ln 2 − ı
x
2
+ Log (1 +
e
ıx
)
1395
Since |
e
ıx
| ≤ 1 and
e

ıx
= −1 for (x) ≥ 0, x = (2k + 1)π, we can expand the last term in a Taylor series in
that domain.
= −ln 2 − ı
x
2
−
∞

n=1
(−1)
n
n
(
e
ıx
)
n
= −ln 2 −
∞

n=1
(−1)
n
n
cos(nx) − ı

x
2
+

∞

n=1
(−1)
n
n
sin(nx)

For −π < x < π, ln(cos(x/2)) is real-valued. We equate the real parts of the equation on this domain to obtain
the desired Fourier series.
ln

cos

x
2

= −ln 2 −
∞

n=1
(−1)
n
n
cos(nx), −π < x < π.
The domain of convergence for this series is (x) = 0, x = (2k + 1)π. The Fourier series converges to the
periodic extension of the function.
ln




cos
x
2



= −ln 2 −
∞

n=1
(−1)
n
n
cos(nx), x = (2k + 1)π, k ∈ Z
2. Now we integrate the function from 0 to π.

π
0
ln

cos
x
2

dx =

π
0


−ln 2 −
∞

n=1
(−1)
n
n
cos(nx)

dx
= −π ln 2 −
∞

n=1
(−1)
n
n

π
0
cos(nx) dx
= −π ln 2 −
∞

n=1
(−1)
n
n

sin(nx)

n

π
0
1396

π
0
ln

cos

x
2

dx = −π ln 2
3. We expand the logorithm.
1
2
ln




sin((x + ξ)/2)
sin((x − ξ)/2)





=
1
2
ln |sin((x + ξ)/2)| −
1
2
ln |sin((x − ξ)/2)|
Consider the function ln |sin(y/2)|. Since sin(x) = cos(x − π/2), we can use the result of part (a) to obtain,
ln



sin

y
2




= ln




cos

y −π
2






= −ln 2 −
∞

n=1
(−1)
n
n
cos(n(y −π))
= −ln 2 −
∞

n=1
1
n
cos(ny), for y = 2πk, k ∈ Z.
We return to the original function:
1
2
ln




sin((x + ξ)/2)
sin((x − ξ)/2)





=
1
2

−ln 2 −
∞

n=1
1
n
cos(n(x + ξ)) + ln 2 +
∞

n=1
1
n
cos(n(x − ξ))

,
for x ± ξ = 2πk, k ∈ Z.
1
2
ln





sin((x + ξ)/2)
sin((x − ξ)/2)




=
∞

n=1
sin(nx) sin(nξ)
n
, x = ±ξ + 2kπ
Solution 28.12
The eigenfunction problem associated with this problem is
φ

+ λ
2
φ = 0, φ(a) = φ(b) = 0,
1397