Mathematical Methods MT2017: Problems 3

John Magorrian, john.magorrian@physics.ox.ac.uk

Fourier series
1. Sketch graphs of the following functions in the range −2π < x < 2π given that all the functions are
periodic with period 2π:
(a) f (x) = |x|, −π < x < π;
 −π < x < π;
(b) f (x) = x,
0, −π < x < 0,
(c) f (x) =
x, 0 < x < π.

2. State whether the following functions are even, odd or neither even nor odd:
(a) sin x (b) cos x (c) sin2 x (d) cos2 x (e) sin 2x
(f) cos 2x (g) x cos x (h) sin x + cos x (i) x + sin x (j) ex .

3. Find the Fourier series for the function


0, −π < x < 0,
f (x) = (Q3.1)
sin x, 0 < x < π.

Fourier sine and cosine series

Non-periodic functions defined on an interval [a, b] can be represented as Fourier series by continuing them
outside [a, b] in order to make them periodic. For example, Figure 1 below shows a function f (x) defined on
the interval [0, a], while Figure 2 shows two different continuations of f (x). The first is even in x, while the
second is odd, but both coincide with f (x) in the interval [0, a].

0 a 2a
Figure 1: A function f (x) deinfed on the interval [0, a].

0 a 2a 4a
0 a 2a 4a
Figure 2: Two different periodic continuations of f (x) with periods 2a and 4a respectively.

4. Find (a) the Fourier sine series and (b) the Fourier cosine series for the function

f (x) = x sin x, 0 < x < π. (Q4.1)

In order to determine the Fourier sine series, you need to extend the function to the interval [−π, 0] by
requiring that f (−x) = −f (x) so that the function becomes odd. Similarly, to obtain the cosine series,
you need to extend the function so that f (−x) = f (x).

1
5. Find (a) the Fourier sine series and (b) the Fourier cosine series for the function f (x) = x2 on the
interval 0 < x < 2.

Generalised Fourier series

6. By applying the the Gram–Schmidt procedure to the list of monomials 1, x, x2 , ..., show that the first
2
three elements of an orthonormal basis for the space L2w (−∞, ∞) with weight function w(x) = √1π e−x
√
are e0 (x) = 1, e1 (x) = 2x, e2 (x) = √12 (2x2 − 1).
Explain why any well-behaved function f : R → C may be expressed as a series of the form
∞
1 X 1 2
f (x) = √ an en (x)e− 2 x , (Q6.1)
2π n=0

and obtain an expression for the coefficient ak in terms of f (x) and the ei (x).
The series on the RHS of (Q6.1) is a Gauss–Hermite series; the polynomials ei (x) turn out to be (nor-
malised) Hermite polynomials. Later in the course we define Hermite polynomials to be eigenfunctions
of a particular differential operator that satisfy a particular normalisation convention.

Fourier transforms and generalized functions

Recall that the Fourier transform of a function f (x) is defined to be
Z ∞
˜ 1
f (k) ≡ √ f (x)e−ikx dx.
2π −∞

7. Evaluate f˜(k) for

(a − x)/a2 , for 0 ≤ x ≤ a,
(
f (x) =(a + x)/a2 , for −a ≤ x ≤ 0, (Q7.1)
0 otherwise.
[You may find it helpful to sketch f (x).]
Sketch f˜(k) as a function of k, marking in particular (a) the value of f˜(k) at k = 0 (b) the position of
the first zeros on either side of k = 0.
Explain the reason for the value of f˜(0) in geometrical terms.

8. This problem is for students who have already taken the complex analysis short option (and they are
strongly advised to do it).
Evalue the Fourier transforms of the following functions for a > 0 (using the residue theorem where
appropriate):
2a 1
(a) f (x) = a2 +x2 , (b) f (x) = exp(−a|x|), (c) f (x) = cosh x .

9. Let f˜(k) be the Fourier transform of f (x). Show that

(a) the Fourier transform of f (ax) is a1 f˜(k/a)
(b) the Fourier transform of f (a + x) is eika f˜(k)
(c) the Fourier transform of eiqx f (x) is f˜(k − q)
(d) the Fourier transform of df ˜
dx is ik f (k)
˜
(e) the Fourier transform of xf (x) is i dfdk
(k)

2
10. The cross correlation of the functions f (x) and g(x) is another function
Z ∞
C(x) = f ? (x0 )g(x + x0 ) dx0 . (Q10.1)
−∞

Prove the Wiener–Kinchin theorem:

√
C̃(k) = 2π f˜? (k)g̃(k), (Q10.2)
where C̃(k) is the Fourier transform of C(x), and so on.

11. Let X be a continuous random variable having probability density function (PDF) fX (x). The char-
acteristic function of X is defined to be
Z ∞
φX (t) ≡ E eitX = eitx fX (x) dx.
 
(Q11.1)
−∞

Verify the following properties of characteristic functions:

(a) If Z = aX + b, where a and b are real numbers, then the characteristic function of Z is
φZ (t) = eitb φX (at). (Q11.2)
Using this result, write down the characteristic function of a normally distributed (i.e., Gaussian)
random variable that has mean µ and standard deviation σ.
(b) The characteristic function of the sum Z = X + Y of independent random variables X and Y is
φZ (t) = φX (t)φY (t). Show this the following two ways: (i) by substituting Z = X + Y into the
definition (Q11.1) of φZ (t); (ii) by using the fact the PDF of Z is given by (see Problem Sheet 2)
Z ∞
fZ (z) = fX (x)fY (z − x) dx (Q11.3)
−∞

and using standard results about Fourier transforms. If X and Y were not independent, would
it still be true that φX+Y (t) = φX (t)φY (t)?
(c) Assuming that the moments E[X k ] exist, the characteristic function can be expanded as
(it)2
φX (t) = 1 + itE[X] + E[X 2 ] + · · · (Q11.4)
2
We can use these results to obtain a very important result in probability theory, the central limit
theorem. Suppose we have n independent, identically distributed random variables, X1 ,...,Xn , and
are interested in the PDF of their sample average, Sn = n1 (X1 + · · · + Xn ). Use the results above to
p
show that, as n becomes large, the PDF of the rescaled variable Zn = (Sn − E[Sn ])/ var[Sn ] tends to
a Gaussian of zero mean and unit variance.

12. A direct application of Fourier transform techniques is Fraunhofer diffraction. See the discussion of
p.443 of Riley, Hobson & Bence. Consider a beam of monochromatic light of wavelength λ incident on a
two-dimensional screen of width 2Y . The direction of the incident beam is specified by the wavevector
~k. The magnitude of ~k is given by wavenumber k = |~k| = 2π .
λ

Y ~k 0

~k θ
0 x

−Y

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 The essential quantity in the Fraunhofer diffraction pattern is the dependence of the observed amplitude
(and hence intensity) on the angle θ between the viewing direction (which we specify through a vector
~k 0 and the direction of the incident beam ~k. We suppose that at the position (0, y) the amplitude of
the transmitted light is f (y) per unit length in the y direction (note that f (y) may be complex). The
function f (y) is called an aperture function. Both the screen and the beam are assumed infinite in the
z direction. The total light amplitude at a position

~r0 = x0~ex + y0~ey , (Q12.1)

with x0 > 0 will be the superposition of all Huygens’ wavelets originating from the various parts of the
screen. For very large |~r0 | these can be treated as plane waves to give
h i
Z Y exp i~k 0 · (~r0 − y~ey )
A(~r0 ) = dy f (y) . (Q12.2)
−Y |~r0 − y~ey |
h i
Here the factor exp i~k 0 · (~r0 − y~ey ) represents the phase change undergone by the light in travelling
from the point y~ey on the screen to the point ~r0 . The denominator represents the reduction in amplitude
with distance. Writing ~k 0 = k cos θ~ex + k sin θ~ey and assuming that |~r0 |  Y we can approximate
|~r0 − y~ey | ≈ |~r0 | and hence approximate (Q12.2) as

exp(i~k 0 · ~r0 ) Y
Z
A(~r0 ) = dy f (y) exp [−iky sin θ] . (Q12.3)
|~r0 | −Y

Now using that f (y) = 0 for |y| > Y (remember, the slit has half-width Y ) we can extend the integration
limits to ±∞. Then the intensity in direction θ is given by

2π  2 2π ˆ
˜
I(θ) = |A|2 = f (k sin θ) ≡ I(θ). (Q12.4)

|~r0 |2 |~r0 |2

ˆ
Use this to calculate I(θ) for
(a) the “single finite slit” diffraction pattern, given by the aperture function

1, for |x| < b/2,
f (x) = (Q12.5)
0, otherwise;

(b) The “two narrow slits” diffraction pattern, given by the aperture function

f (x) = δ(x + a/2) + δ(x − a/2); (Q12.6)

(c) The “two finite slits” diffraction pattern, given by


 1, for a−b a+b
2 ≤x≤ 2 ,
f (x) = 1, for − 2 ≤ x ≤ − a−b
a+b
2 ,
(Q12.7)

0, otherwise,

where b < a. How is the result of (c) related to those for (a) and (b)? Can you explain this in terms of
the convolution theorem?