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Reference Text for Mathematical Physics
Deepanshu Bisht

Contents
1 Some Common Standard Math Formulas
1
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1
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1
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1
1
2
3

2 Some less used formulae
2
...

2
...


5
5
5

3 Partial Derivatives
3
...

3
...


6
6
6

4 Differential Equations
7
4
...
7
4
...
10
4
...
10
5 Complex Analysis
11
5
...
11
5
...
12
5
...
14
6 Vector Calculus
16
6
...
16
6
...
16
7 Coordinate Systems
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8 Binomial theorem, Permutation and Combinations

19
19
20
20
21
22

9 Matrices and Determinants
23
9
...
23
9
...
24
10 Limits and Series
25
10
...
25
10
...
27
11 Standard Functions
11
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1

28
28
29
32

12 Everything Fourier
33
12
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33
12
...
34
13 Stats and Probability
35
13
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37
13
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38

2

About the document
This document is a go-to reference in need of mathematical formulae/equations and associated notes
which are directly or indirectly useful in doing physics
...


Chapter 1

Some Common Standard Math
Formulas
1
...
1
...
1)
(1
...
3)

sin(a + b) = sin(a) cos(b) + cos(a)sin(b)

(1
...
5)

cos(a + b) = cos(a) cos(b) − sin(a) sin(b)

(1
...
7)

tan(a + b) =

tan(a) + tan(b)
1 − tan(a) tan(b)

(1
...
9)

sin(2a) = 2 sin(a) cos(a)

(1
...
11)
(1
...
13)
(1
...
15)
(1
...
17)
(1
...
1
...
2

(1
...
20)
(1
...
22)
(1
...
24)

Derivatives
d
(sin x) = cos x
dx
d
(cos x) = − sin x
dx
d
(tan x) = sec2 x
dx
d
(cot x) = −cosec2 x
dx
d
(sec x) = sec x tan x
dx

(1
...
26)
(1
...
28)
(1
...
30)

d
1
(sin−1 x) = √
dx
1 − x2

(1
...
32)

d
1
(tan−1 x) =
dx
1 + x2
d
−1
(cot−1 x) =
dx
1 + x2
1
d
(sec−1 x) = √
dx
x x2 − 1
d
−1
(cosec−1 x) = √
dx
x x2 − 1
d x
(e ) = ex
dx
d
1
(log|x|) =
dx
x
 x 
d
a
= ax
dx loga

2

(1
...
34)
(1
...
36)
(1
...
38)
(1
...
3

Integrals

Note: Evaluating the integral by changing the variables through substitution is all well and good
until
...
That is, x = f −1 (t) is a multi-valued function
...
40)
0

If for this integral we make the substitution sin x = t, both upper and lower limits become zero and
integral turns out to be 0
...
This happens cause x = sin−1 (t)
is multivalued
...

For the substitution cos x = t this doesnt happen since cos x doesnt repeat in 0 ≤ x ≤ π

1
...
1

Algebraic

ˆ
xn dx =

xn+1
+c
n+1

(1
...
3
...
42)

px + q
A
B
=
+
(x − a)2
x − a (x − a)2

(1
...
44)

px2 + qx + r
A
Bx + C
=
+ 2
2
(x − a)(x + bx + c)
x − a x + bx + c

(1
...
46)

ln|x|dx = xln|x| − x + c

(1
...
48)
(1
...
50)
(1
...
52)
(1
...
3
...
54)

cot x dx = ln| sin x| + c

(1
...
56)

cosec x dx = ln|cosec x − cot x| + c

(1
...
58)

cot2 xdx = −cot x − x + c

(1
...
60)
(1
...
62)
(1
...
64)
(1
...
66)
(1
...
3
...
3
...
68)

−αx2

dx =

(1
...
70)
(1
...
72)

Chapter 2

Some less used formulae
2
...


2
...
1

Inverse trigonometry
cos−1 x + cos−1 x = π [−1 ≤ x ≤ 1]
sec

−1
−1

x + sec

−1

x = π |x| ≥ 1

−1

x=π x∈R
π
sin−1 x + cos−1 x = [−1 ≤ x ≤ 1]
2
π
tan−1 x + cot−1 x =
x∈R
2
π
cosec−1 x + sec−1 x = |x| ≥ 1
2
cot

2
...
1)
(2
...
3)
(2
...
5)
(2
...
7)

(2
...
1

Basics

Defined as:

∂f
f (x + ∆x, y) − f (x, y)
=
∂x
∆x
∂f
f (x, y + ∆y) − f (x, y)
=
∂y
∆y

(3
...
2)

for function f (x, y)
The total differential is given as: measures the total change in f , due to changes in x and y
df =

∂f
∂f
dx +
dy
∂x
∂y

(3
...

(3
...
Further for such an f ,
fxy = fyx so we get a condition for exactness
∂B
∂A
=
∂y
∂x

df = A(x, y)dx + B(x, y)dy

(3
...
2

Useful theorems




3
...
1

∂x
∂y

∂x
∂y
 
z




=
z

∂y
∂z

 
x

∂y
∂x

−1

∂z
∂x



(3
...
7)

y

Stationary points

The points of two variable functions f (x, y) at which the first partial derivatives become 0
...
1

Ordinary Differential Equations

4
...
1

Equations reducible to separable
dy
= f (ax + by + c)
dx
 
dy
y
=f
dx
x

(4
...
2)

Homogenous
f1 (x, y)dx + f2 (x, y)dy = 0

(4
...
4)

Change the variables: x = X + h, y = Y + k so equation becomes homogenous

4
...
2

Solution of exact equations

The equation is:
M dx + N dy = 0

(4
...
6)

(4
...
8)

(4
...
10)

• If ODE is such that:
f1 ydx + f2 xdy = 0
then
IF =

1
Mx − Ny

(4
...
12)

Linear first order equations
dy
+ P (x)y = Q(x)
dx
´

IF = e

(4
...
1
...
13)

Q(x)IF dx = C

(4
...
16)

where α is real
Put u(x) = y 1−α and the equation becomes linear in u
u0
+ p(x)u = q(x)
1−α

4
...
4

(4
...
18)

Find λ1 and λ2 : the roots of the auxiliary equation:
λ2 + aλ + b = 0

(4
...
20)

y(x) = (c1 + c2 x)eλx

(4
...
22)

Real and distinct

Real and equal

Complex

Then we have the following particular solutions
y1
y2

=

˙ 1 )x
e(a1 +ιb

(4
...
24)

=

ea1 x [cos(b1 x) − ι˙ sin(b1 x)]

(4
...
Subtract both and divide by 2˙ι
...
Their
general linear combination is general solution
y(x) = ea1 x [A cos(b1 x) + B sin(b1 x)]

4
...
5

(4
...
27)

m2 + (a − 1)m + b = 0

(4
...
29)

y(x) = (c1 + c2 ln x)xm

(4
...
31)

Real and equal

Complex

4
...
6

Linear second order Non homogeneous
y 00 + p(x)y 0 + q(x)y = r(x)

(4
...
33)

y 00 + ay 0 + by = r(x)

(4
...
If r(x) is sum of forms given in table, then assume yp also as sum of the corresponding yp
forms
Method of variation of parameters
If y1 and y2 are particular LI solutions of homgoneous solution yh then yp is:
ˆ
ˆ
y1 r(x)
y2 r(x)
dx + y2
dx
yp = −y1
W
W
Where W = y1 y20 − y2 y10 is wronskian

9

(4
...
2

Linear Systems of ODEs

The general form of such a system of ODEs is:
y10

=

a11 y1 + a12 y2 + · · · + a1n yn + g1 (t)

(4
...


...


...
37)

yn0

an1 y1 + an2 y2 + · · · + ann yn + gn (t)

(4
...
38)

There are n unknown variables and n linear ODEs
...
So in matrix form:
y0 = Ay + g
(4
...
2
...
41)

(4
...
43)

(4
...
3
4
...
1

(4
...
46)

(4
...
48)

Parabolic
Here we take ξ(x, y) = Constant and η = y which gives:
C ∗ uηη + D∗ uξ + E ∗ uη + F ∗ u = G∗

(4
...
50)

Elliptic

where, α =

ξ+η
2

and β =

ξ−η
2i

10

Chapter 5

Complex Analysis
Somehow, pretending that square roots of negative numbers made sense, even though they obviously did
not, could lead to sensible answers
...
But Bombellis calculation implied that there was more to
imaginaries than that
...

˙
eιπ
= −1

(5
...
But this equation should
not be interpreted as have we raised e to this weird power that doesn’t make sense
...
1

Basics

• Complex numbers are the most general set of numbers and include every other set of numbers
...

Both are equivalent and reinforces each other in beautiful ways
In Cartesian notation a complex variable is:
z = (x, y)
= x + ι˙y

(5
...
3)

z z¯ = |z|2 = x2 + y 2

(5
...
5)
(5
...
1
...
7)

sin(θ) =

˙
˙
eιθ
− e−ιθ
2˙ι

(5
...
9)

˙
˙
eιz
− e−ιz
2
z
e + e−z
cosh z =
2
z
e − e−z
sinh z =
2
cosh ι˙z = cos z

sin(z) =

5
...
2

(5
...
11)
(5
...
13)

sinh ι˙z = ι˙ sin z

(5
...
15)

De Moivre’s Theorem

˙
From this it can be shown with z = eιθ

zn +

5
...
3

1
= 2 cos nθ
zn

zn −

1
= 2˙ι sin nθ
zn

(5
...
17)

where k is an integer and will take values 0, 1, 2,
...
1
...
18)

Restricting to principal value by constraining the argument of z to lie between −πtoπ we get
singlevalued Ln(z)
The definition of complex number raised to a complex power is in terms of already defined complex
functions:
tz = ezln(t)
(5
...
2

Complex Functions
lim f (z) = ∞ if and only if lim

z→z0

z→z0

1
=0
f (z)

1
= w0
z
1
lim f (z) = ∞ if and only if lim
=0
z→∞
z→0 f (1/z)
lim f (z) = w0 if and only if lim f

z→∞

z→0

12

(5
...
21)
(5
...
2
...
23)

f 0 (z) exists at a point z0 = (x0 , y0 )
...
24)

f 0 (z0 ) = ux + ι˙vx

(5
...
26)
2
2
∂x
∂y
∂x
∂y
• The family of 2D curves (in xy plane) u(x, y) = constant and v(x, y) = constant intersect at right
angles to each other
...
we can make them single valued
and then use the analytic function analysis on it by concept of branch points and cuts
• Branch point is a point in Argand plane such that if z is varied in a closed loop enclosing the
branch point, f (z) doesn’t return to its original value, although z does since θ → θ + 2π doesn’t
affect z
...

• So, if we dont cross the branch cut, f (z) remains single-valued
...
2
...

• Pole: most imp isolated singularity
• If
f (z) =

g(z)
(z − z0 )n

such that g(z0 ) 6= 0 and is analytic in neighbourhood of z0 then its a pole of order n

13

(5
...
28)

z→z0

where a is non-zero then its a order n pole
• Removable singularity: if f (z) is 0/0 at z0 but the limit at z0 exists
...
3

Complex integrals

• Definition in terms of 4 real integrals
...
It is defined in parametric form
• In general depends on the end points AND the path taken and f (z)
• Cauchy’s thm: f (z) analytic and f 0 (z) is continuous at each point within and on the closed
contour C then
fi
f (z)dz = 0
(5
...

• Cauchy’s integral formula
1
f (z0 ) =
2π ι˙
f n (z0 ) =

n!
2π ι˙

fi

fi
C

f (z)
dz
z − z0

(5
...
31)

C

Complex taylor’s theorem: If f (z) is analytic inside and on a circle C of radius R centred at
z0 and for any point z inside C:
f (z) =

∞
X

an (z − z0 )n

n=0

an =

f (n) (z0 )
n!

(5
...
expand that in taylor series abt z0 and derive f (z)
f (z) =
Here:

a−p
a−1
+
...

p
(z − z0 )
z − z0
1
an =
2π ι˙

fi

f (z)
dz
(z − z0 )n+1

(5
...
34)

• Laurent series of any f (z) can be used at z = z0 to find the nature of z0
– If f (z) analytic at z0 , all an = 0 for n < 0
– if first non-zero term: am (z − z0 )m , m > 0 then z0 is a root of order m
– If not analytic then for some p, a−p 6= 0 and all coefficients below it are 0
...

– Value of a−1 is called residue of f (z) at z0
– If principle part has infinite terms then z0 is essential singularity

14

• Residues: For f (z) with pole of order m at z0 , its integral around a closed contour is:
fi
I=
f (z)dz = 2π ι˙a−1 = 2π ι˙ × (Residue of f(z) at z0)

(5
...
36)

This is found by expanding f (z) as a laurent series and seeing that only non-zero contribution to
the integral is by the term with coefficient a−1
• Residue for pole of order m is:

R(z0 ) = lim

z→z0


1
dm−1
m
[(z − z0 ) f (z)]
(m − 1)! dz m−1

(5
...
38)

• Or we can find residue by expanding f (z) in laurent series about z0 singularity
...
3
...
39)

j

Examples of standard integrals

1
...
40)

0

can be solved by expressing cosine, sine in terms of z by Demoivre’s thm and on unit circle around
origin, dθ = −˙ιz −1 dz
Convert to the equivalent complex contour integral and integrate
...
Next is infinite integrals

ˆ

∞

f (x)dx = 2π ι˙ × (sum of residues at poles with Im z above 0)
−∞

15

(5
...
1

Differentiation

Refere to section 1
...
2
6
...
1

d
dc
dA
(cA) = A + c
dt
dt
dt
d
dA
dB
(A
...
B + A
...
1)
(6
...
3)

Grad, Div, Curl
Gradient

In cartesian coordinates: the infinitesimal change in a scalar function f (x, y, z) is given by:
df

=
=

∂f
∂f
∂f
dx +
dy +
dz
∂x
∂y
∂z
∇f
...
4)
(6
...
2
...
2
...
6)

∇
...
7)

Divergence

Curl

x
ˆ
∂
∇ × A =  ∂x
 Ax

6
...
4

y
ˆ
∂
∂y

Ay


ˆ
z 
∂ 
∂z 
Az 

(6
...
9)

1 ∂
1 ∂A2
∂A3
(rA1 ) +
+
r ∂r
r ∂θ
∂z

(6
...
A =

16


ˆ
r
1  ∂
∇ × A =  ∂r
r
A1

rθˆ
∂
∂θ
rA2


ˆ
z 
∂ 
∂z 
A3 

(6
...
A =

6
...
5

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