UNIVERSITY OF DUBLIN
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TRINITY COLLEGE
Faculty of Science
school of mathematics
SF Mathematics
SF Theoretical Physics
SF Two Subject Mod
Hilary Term 2009
Course 231
Day Date, 2009
Dr. S. Ryan
ATTEMPT SIX QUESTIONS
Log tables are available from the invigilators, if required.
Non-programmable calculators are permitted for this examination,—please indicate the make
and model of your calculator on each answer book used.
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1. Express the function f (x) = |cos(x)| as a (real) Fourier series.
This has period π, so
2
an =
π
Z
π/2
2
dx| cos(x)| cos(nx) =
π
−π/2
Z
π/2
dx cos(x) cos(nx).
−π/2
For n = 0,
2
a0 =
π
Z
π/2
dx cos(x) =
−π/2
4
.
π
and
an
12
=
2π
Z
π/2
dx ei(n+1)x + ei(−n+1)x ,
−π/2
π/2
ei(−n+1)x 1 1 e(n+1)ix
+
=
,
π i n+1
−n + 1 −π/2
1
π
1
π
2
sin (n + 1)
+
sin (1 − n)
.
=
π n+1
2
−n + 1
2
π
π
For n odd, sin (n ±
1)
=
0.
For
n
even,
sin
(n
+
1)
= cos(nπ/2) = (−1)n/2
2
2
and sin (−n + 1) π2 = cos(nπ/2) = (−1)n/2 . So,
2
1
1
4
an =
+
.
(−1)n/2 =
π n + 1 −n + 1
1 − n2
So, the fourier series is
∞
2
4 X (−1)n
| cos(x)| = +
cos(2nx).
π π n=1 1 − 4n2
where n → 2n for n even.
3
2. Evaluate the integral
Z
dV
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√
3x2 + 3z 2 ,
D
where D is the solid bounded by y = 2x2 + 2z 2 and the plane y = 8.
You should begin by sketching the solid of interest.
Use spherical polars in the xz plane. Write x = r cos θ and z = r sin θ so that
x2 + z 2 = r2 . Then 2x62 + 2z 2 ≤ y ≤ 8 with 0 ≤ θ ≤ 2π and 0 ≤ r ≤ 2 (from z = 0
in 8 = 2x62 + 2z 2 ). Then
Z 2π Z 2 Z 8
Z
√
√
2
2
dθ
dr
dy 3x2 + 3z 2 ,
dV 3x + 3z =
0
0
2x62+2z 2
D
Z 2π Z 2 √
8
=
dθ
dr 3x2 + 3z 2 y ,
2x62+2z 2
0
0
Z 2π Z 2 √
=
dθ
dr 3x2 + 3z 2 (8 − (2x2 + 2z 2 )).
0
0
Now x2 + z 2 = r2 so we can rewrite the integral above as
Z
Z 2π Z 2 √
√
2
2
dV 3x + 3z =
dθ
dr 3r62(8 − 2r2 )r,
D
0 Z
0
√
= 2π 62dr 3(8r2 − 2r4 ),
0
2
√
8 3 2 5
,
32π r − r
=
3
5 0
√ 64 64
−
= 2π 3
3
5
√
256 3π
=
.
15
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3. An eigenfunction of the Fourier transform (fg
(k)), is a function f for which
Z ∞
1
dxf (x)e−ikx = λf (k),
fg
(k) =
2π −∞
where λ is a scalar called the eigenvalue.
1 2
Show that the Gaussian function f (x) = e− 2 x is such an eigenfunction.
Note: The gaussian integral formula is given in the useful information page and
may be used without proof
Consider the function
2xt−t2
Φ(x, t) = e
=
∞
X
tn
n=0
n!
Ht (x),
the so-called generating function for the Hermite polynomials Ht . By computing
1 2
1 2
the fourier transform of e− 2 x Φ(x, t) show that the functions fn (x) = e− 2 x Ht (x)
are eigenfunctions of the Fourier transform.
Determine the eigenvalues λn .
Work out fg
(k) by
fg
(k) =
=
=
=
=
2
Z ∞
1 2
1
dxe− 2 x e−ikx
2π −∞
r
1 π −ik2 /4( 1 )
2
e
2π 12
1√
2
2πe−k /2
2π
1 2
1
√ e− 2 k
2π
1
√ f (k).
2π
1 2
P∞
tn
n=0 n! Ht (x).
Taking the FT of e− 2 x Φ(x, t),
Z ∞
1 2
1
2
− 12 x2
F(e
Φ(x, t)) =
dxe− 2 x +2xt−t −ikx
2π −∞
Now, Φ(x, t) = e2xt−t =
k2
2
= e− 2 −2kit+t
∞
X
k2
(−it)n
=
e− 2 Hn (k)
.
n!
0
P∞
1 2 P∞
n
tn
−x2 /2
And the FT of e− 2 x
Hn (x)) tn! . Equatn=0 n! Ht (x) can be written as
n=0 F(e
ing like powers of n on the left and right sides of the FT gives
F(e−x
2 /2
Hn (x)) = (−i)n e−k
2 /2
Hn (k).
ie eigenfunctions as required and using the eigenfunction result above the eigenvectors are (−i)n .
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4. Prove that a continuous vector field F in an open and connected domain, is conservative if and only if it is path independent.
R
Given the vector field, F = yi − xj, determine the line integral C F · dl from (1, 0)
to (0, −1) along
• the straight line segment joining these points,
• three-quarters of the unit circle, centred at the origin traversed counterclockwise.
What do the results imply about the field F?
Proof in the notes.
• the straight line segment:
Parameterise the curve as r = (1 − t)i − ty, with 0 ≤ t ≤ 1. So, ∂r/∂t = −i − y
and also F = −ti + (t − 1)y . Then the line integral is
Z 1
Z
dt ((−t)i + (t − 1)y) · (−i − y)
F · dl =
0
C
Z 1
dtt + 1 − t
=
0
= 1.
• This time parameterise the curve as r = cos(t)i + sin(t)y giving ∂r/∂t =
− sin(t)i + cos(t)y and F = sin(t)i − cos(t)y. The line integral is
Z
3π/2
Z
F · dl =
C
(sin(t)i − cos(t)y) · (− sin(t)i + cos(t)y)
0
3π/2
Z
−1
=
0
= −
3π
.
2
The line integrals along the 2 paths differ so F is path-dependent and not
conservative.
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Some useful formulae
1. A function with period l has a Fourier series expansion
∞
a0 X
f (x) =
+
an cos
2
n=1
2πnx
l
+
∞
X
bn sin
n=1
2πnx
l
where
a0
an
bn
l/2
Z
2
=
l
dxf (x),
−l/2
Z l/2
2πnx
dxf (x) cos
,
l
−l/2
Z
2 l/2
2πnx
=
dxf (x) sin
.
l −l/2
l
2
=
l
2. A function with period l has a Fourier series expansion
f (x) =
∞
X
cn exp
n=−∞
where
1
cn =
l
Z
l/2
2iπnx
l
dxf (x) exp
−l/2
,
−2iπnx
l
.
3. The Fourier integral representation (or Fourier transform) is
Z ∞
f (x) =
dk fg
(k)eikx ,
−∞
Z ∞
1
g
f (k) =
dxf (x)e−ikx .
2π −∞
4. The Gaussian integral is
Z
∞
dxe
−∞
−ax2 +bx
r
=
π b2 /4a
e
,
a
for a > 0 and b ∈ C.
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