Mathematical Methods for Physicists: Supplement with Q&A
Paul Wenk
∗
Institut f¨ur Theoretische Physik, Universit¨at Regensburg, D-93040 Regensburg, Germany
(Dated: July 1, 2019)
Collection of frequently asked questions which came up during the course mathematical methods
for physicists at the University of Regensburg.
I. ANALYSIS
A. Partial Fraction Decomposition
The partial fraction decomposition is a way of expressing a rational function by a sum of fractions with a simpler
denominator. Let us assume f is a proper rational function, i.e., a rational function in which the degree of the poly-
nomial in the nominator is no greater than the degree of the polynomial in the denominator. Then the decompsition
can be written as follows
1
,
f(x) =
P (x)
Q(x)
=
M
X
i=1
j
i
X
k=1
A
ik
(x − a
i
)
k
+
N
X
i=1
k
i
X
k=1
B
ik
x + C
ik
(x
2
+ b
i
x + c
i
)
k
, x, a
i
, b
i
, c
i
∈ R . (1)
Here, the number of different real roots a
i
is M, indexed with i. Each root a
i
has a degeneracy j
i
. In general, also
complex root can appear. However, due to the complex conjugate root theorem the complex roots appear only in
complex conjugate pairs. Thus, to avoid the calculation with complex numbers one leaves quadratic polynomials
unfactorised, x
2
+ b
i
x + c
i
. The complex conjugate pairs can have also a degeneracy, here named by k
i
.
B. Dirac-Delta Distribution
Show that the following statement
Z
∞
−∞
dx f(x)δ(x) := lim
n→∞
Z
∞
−∞
dx f(x)δ
n
(x) = f(0) (2)
hold for a sequence of Gaussian distributions
δ
n
(x) =
n
√
π
e
−n
2
(x−x
0
)
2
(3)
Proof. We need the following preface:
(a) We evaluate
Z
∞
0
dx xe
−n
2
x
2
=
1
2n
2
Z
∞
0
dy e
−y
=
1
2n
2
(4)
where we used the substitution
y = (nx)
2
⇒ dy = 2xn
2
dx, dx =
1
2xn
2
dy. (5)
(b)
∀
n∈N
∗
:
Z
∞
−∞
dx δ
n
(x) = 1.
For the proof, see exercise sheet.
(c) The mean value theorem states that if f is a continuous function on a closed interval [a, b] there exists a ξ ∈ (a, b)
such that (see Fig. 1)
f
0
(ξ) =
f(b) − f(a)
b − a
. (6)
2
FIG. 1: Geometrical interpretation of the mean value theorem, Eq. 6.
Now we are ready to proof the statement by showing that the following expression vanishes,
∆
n
=
Z
∞
−∞
dx f(x)δ
n
(x) − f(0)
(7)
(b)
=
Z
∞
−∞
dx δ
n
(x)(f(x) − f(0))
(8)
≤
Z
∞
−∞
dx δ
n
(x)kf(x) − f(0)k (9)
(c)
≤ max
˜x∈R
kf
0
(˜x)k
Z
∞
−∞
dx δ
n
(x)|x − 0| (10)
(a)
= max
˜x∈R
kf
0
(˜x)k
1
√
π
1
n
. (11)
Therefore
lim
n→∞
∆
n
= lim
n→∞
max
˜x∈R
kf
0
(˜x)k
1
√
π
1
n
= 0 (12)
C. Euler’s formula
Here, we show an alternative proof of Euler’s formula. It uses the uniqueness of the definition of the natural
exponential function:
Theorem 1. The natural exponential function f(x) = e
λx
, with x, λ ∈ R is uniquely defined by
f
0
(x) = λf(x) ∧ f (0) = 1 . (13)
Proof. We know that ∀x: exp(x) 6= 0. Assume there are two functions with the properties
• f
0
(x) = f(x) and g
0
(x) = g(x),
• f(0) = g(0) = 1.
It follows that
f(x)
g(x)
0
=
f
0
(x)g(x) − f(x)g
0
(x)
g
2
=
f(x)g(x) − f(x)g(x)
g
2
= 0. (14)
Thus, f (x)/g(x) has to be a constant. But since f(x) and g(x) agree in x = 0 they agree everywhere.
Let us use the two properties which uniquely define the natural exponential function on w(x) = cos(x) + i sin(x):
3
Proof.
w
0
(x) = − sin(x) + i cos(x) = i(cos(x) + i sin(x)) = iw(x) (15)
and w(0) = 1. (16)
Thus, w(x) has to be equal to w(x) = e
ix
.
D. Total Time Derivative
Q: How to calculate
d
2
f(x,y)
dt
2
?
A: First, let us examine
d
2
f(x)
dt
2
=
d
dt
∂f (x)
∂x
| {z }
=:g(x)
dx
dt
|{z}
˙x
(17)
=
∂(g(x) ˙x)
∂x
˙x (18)
=
∂
2
f(x)
∂x
2
( ˙x)
2
+
∂f (x)
∂x
∂ ˙x
∂x
˙x . (19)
Let us now extend the problem and calculate the most general total time derivative of f(x(t), y(t)),
d
2
f(x(t), y(t))
dt
2
=
d
dt
∂
x
(f) ˙x
| {z }
g(x,y)
+ ∂
y
(f) ˙y
| {z }
m(x,y)
(20)
= ∂
x
(g(x, y) + m(x, y)) ˙x + ∂
y
(g(x, y) + m(x, y)) ˙y (21)
= [∂
2
x
(f) ˙x + ∂
x
(f)∂
x
( ˙x) + ∂
x
∂
y
(f) ˙y + ∂
y
(f)∂
x
( ˙y)] ˙x
+ [∂
2
y
(f) ˙y + ∂
y
(f)∂
y
( ˙y) + ∂
y
∂
x
(f) ˙x + ∂
x
(f)∂
y
( ˙x)] ˙y (22)
Here, one has to be carefull: The terms like ∂
x
( ˙y) are not necessarily zero, since ˙y can be a function of x and y.
E. Curl in curvilinear orthogonal coordinates
The curl of a vectorfield A : R
3
→ R
3
using the curvilinear orthogonal coordinates (q
1
, q
2
, q
3
) is given by
(∇ × A)
1
=
1
g
2
g
3
∂(g
3
A
3
)
∂q
2
−
∂(g
2
A
2
)
∂q
3
, (23)
(∇ × A)
2
=
1
g
3
g
1
∂(g
1
A
1
)
∂q
3
−
∂(g
3
A
3
)
∂q
1
, (24)
(∇ × A)
3
=
1
g
1
g
2
∂(g
2
A
2
)
∂q
1
−
∂(g
1
A
1
)
∂q
2
, (25)
with g
i
= |∂r/∂q
i
|. This can be written in a short form using a determinant,
∇ × A =
1
g
1
g
2
g
3
g
1
e
q
1
g
2
e
q
2
g
3
e
q
3
∂
∂q
1
∂
∂q
2
∂
∂q
3
g
1
A
1
g
2
A
2
g
3
A
3
. (26)
Here, one has to be careful by keeping in mind that the partial derivatives ∂/∂q
i
are operators and are in general
not commuting with the other factors. As a consequence, one is not free to apply Laplace’s formula to Eq. 26. That
means terms like g
i
A
i
∂
q
j
should not appear. The correct form should be equal to the formulas (∇ ×A)
i
given above.
4
F. Integrating Exact Differentials
1. Consideration for ω(x, y)
Assume the differential (sometimes called total derivative) dw to depend only on two variables, x, y,
dw(x, y) = ∂
x
w(x, y) dx + ∂
y
w(x, y) dy = A(x, y) dx + B(x, y) dy (27)
where A and B are given. We would like to recover w. Since w(x, y), we can write it in the most general form
w(x, y) = a(x) + b(y) + m(x, y) + c (28)
where c is some constant. Now calculate
w
1
(x, y) =
Z
A(x, y) dx = a(x) + m(x, y) + c
1
(y), (29)
w
2
(x, y) =
Z
B(x, y) dy = b(y) + m(x, y) + c
2
(x). (30)
This gives us
w
1
(x, y) + w
2
(x, y) = a(x) + b(y) + 2m(x, y) + c
1
(y) + c
2
(x) = w(x, y) + m(x, y) + c
1
(y) + c
2
(x). (31)
Thus, we get
w(x, y) = w
1
(x, y) + w
2
(x, y) − [m(x, y) + c
1
(y) + c
2
(x)] . (32)
On the other hand we also know that
∂
y
ω
1
(x, y) = B(x, y), (33)
∂
x
ω
2
(x, y) = A(x, y) (34)
which can be used to calculate the functions c
i
,
c
1
(y) =
Z
(B(x, y) − ∂
y
m(x, y)) dy, (35)
c
2
(x) =
Z
(A(x, y) − ∂
x
m(x, y)) dx. (36)
2. Example
Given:
dw(x, y) = (1 + y
2
) dx + 2xy dy (37)
We have
w
1
(x, y) =
Z
(1 + y
2
) dx = x(1 + y
2
) + c
1
(y), (38)
w
2
(x, y) =
Z
2xy dy = xy
2
+ c
2
(x) (39)
Thus, the mixed term is given by m(x, y) = xy
2
and we have
w(x, y) = x(1 + y
2
) + c
1
(y) + xy
2
+ c
2
(x) − (xy
2
+ c
1
(y) + c
2
(x)) (40)
= xy
2
+ x. (41)
5
II. LINEAR ALGEBRA
A. Q: What is the fastest algorithm to compute a determinant?
Some useful links: StackOverflow[1], StackOverflow[2].
B. Q: Where can I find more information about tensors?
• Gerd Fischer: Lineare Algebra, Vieweg: Chapter ”Dualitt und Tensorprodukte”
• lecture by Prof. Dr. Dullemond: syllabus
• J¨anich: Vektoranalysis, Springer: Chapter ”Tensorfelder”
C. I need more pictures to understand the concepts in lin. algebra! Some sources?
Beautifully made interactive course on linear algebra can be found e.g. at the immersivemath.com web page based
on the book by Jakob O Str¨om et al.
∗
paul.wenk@ur.de
1
For a proof see e.g. Feldman, International Mathematical Forum, Vol. 7, 2012, no. 31, 1517 - 1535.
