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Reference Text for Mathematical Physics
Deepanshu Bisht

Contents
1 Some Common Standard Math Formulas
1
...

1
...

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
7
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7
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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
...

11
...

11
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1

28
28
29
32

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

(6
...
13)

(6
...
15)

Fundamental theorems

The line integral of any gradient field doesn’t depend on path but just on the endpoints
...
dl = f (b) − f (a)

(6
...

ˆ

ˆ
(∇
...
da

(6
...
Since for a given boundary there can be infinite number of surfaces, it shows
that surface integral of curls is only dependent on the boundary of the surface
...
da =

A
...
18)

S

These operators are usually applied on scalar (gradient) and vector (divergence, curl) fields
...
f and g are scalar fields and k is just a scalar number

6
...
6

First derivatives

• Sum rules
∇(f + g) = ∇f + ∇g

(6
...
(A + B) = (∇
...
B)

(6
...
21)

• Multiplying by a constant rules
∇(kf ) = k∇f

(6
...
(kA) = k(∇
...
23)

∇ × (kA) = k(∇ × A)

(6
...
2
...
25)

∇(A
...
∇)B + (B
...
26)

∇
...
A) + A
...
27)

∇
...
(∇ × A) − A
...
28)

∇ × (f A) = f (∇ × A) − A × (∇f )

(6
...
∇)A − (A
...
B) − B(∇
...
30)

Second Derivatives

• Laplacian - Divergence of a Gradient
∇
...
31)

This has different expressions for different coordinate systems which are equivalent
...

• The curl of a gradient is always zero
∇ × (∇f ) = 0

(6
...
(∇ × v) = 0

(6
...
v) − ∇2 v

(6
...
1

Cartesian coordinates

Straightforward and easy to understand formulas for all things
...

The position vector of any point P is given by OP = xˆ
x + yy
ˆ + z zˆ where the unit vectors are x
ˆ,
y
ˆ and zˆ that are fixed in magnitude( = 1) and direction also
...
1)

• Equation of plane with unit normal vector n
ˆ and a given point on plane with position vector r0 :
(r − r0 )
...
2)

px + qy + sz = d n
ˆ = (p, q, s) r
...
3)

• In component form:
where d = perpendicular distance of plane from origin
• In intercept form:

x y z
+ + =1
a
b
c

(7
...
5)

• Relation b/w unit normal and intercepts:
n
ˆ∝

• Angle between two planes: l1 x + m1 y + n1 z = d1 and l2 x + m2 y + n2 z = d2 is:
l1 l2 + m1 m2 + n1 n2
p
cos θ = p 2
l1 + m21 + n21 l22 + m22 + n22

(7
...
(∇f ) =

∂2f
∂2f
∂2f
+ 2 + 2
2
∂x
∂y
∂z

(7
...
8)

7
...
2
...
1: Polar coordinate plane
The two coordinates used are: r, the distance from origin and θ the anticlock angle from the
reference axis: taken as the x-axis in cartesian coordinates
...
9)
 
y
θ = arctan
(7
...
11)

y = r sin θ

(7
...

rˆ = cos θx
ˆ + sin θy
ˆ

(7
...
14)

How base vectors change with time
dˆ
r
dθ ˆ
=
θ
dt
dt
dθˆ
dθ
= − rˆ
dt
dt

(7
...
16)

Motion in polar coordinates
r(t) = rˆ
r (θ)
v(t) =

7
...
17)
= rˆ

(7
...
19)

dA = rdrdθ

(7
...
21)

dV = r2 sin θdφdθdr

(7
...
23)

dV = ρdφdρdz

(7
...
4

Coordinate transformations and unit vectors

21

Chapter 8

Binomial theorem, Permutation and
Combinations
A way to visualize permutations and combinations
...
In the first pick we have 10 options, then for each
of those 10 paths, we have 9 options
...
Now each of those 9×10 different paths will branch into 8
different paths and so on
...
All those paths represent a different permutation
...
In the end if you have
encountered the same balls in your path, then all such paths will be considered equivalent
...
e
...
So there will only be 1 combination in this case
...
Form all possible 2 digit numbers from 5 different
digits = 5 P2
n!
n
(8
...
Like form a groups of 3 from a team of 10 =
C3
n!
n
Ck =
(8
...
3)
k

10

k=0

22

Chapter 9

Matrices and Determinants
We denote conjugate transpose by † Types of important matrices:
• Symmetric matrix: aij = aji or A0 = A
• Skew symmetric matrix: aij = −aji or A0 = −A
...
First one symmetric second on skew symmetric
• Unitary matrix: if for complex square matrix U , U U † = I, it is unitary
• Hermitian matrix: here, H = H †
...


9
...
e
...
And so det(cA) = cn A if
A is n × n matrix
• det(AB) = det(A)det(B)
• No such formula for determinant of sum of two matrices

23

9
...

Any function of a matrix A is evaluated by:


f (λ1 )
0

...

0 

 −1
(9
...


...
 S

...


...

0

0


...
The only difficulty occurs when either
of the expressions 0/0 or ∞/∞ results
...

Continue differentiating until the top and bottom limits are no longer both zero or both infinity
...
For a series with N
terms: a + (a + d) + (a + 2d) +
...
1)

In Geometric series ratio of any two consecutive terms is a constant r for the whole series
...
2)

when |r| < 1 and the series becomes infinite:
S=

10
...
3)

Infinite series

• An infinite series is a limit of a sequence of finite series and hence, if it exists will be a single value
...

25

• For convergent series pairwise addition,subtraction and scalar multiplication holds
...

• If two power series P (x) and Q(x) have regions of convergence that overlap to some extent then
the series produced by taking the sum, the difference or the product of P (x) and Q(x) converges
in the common region
...

• Multiplication of two infinite series goes like this:
! ∞ !
∞
∞
X
X
X
an
bn =
cn
n=0

where, cn =

n
P

n=0

(10
...
1
...
1
...
5)

(10
...
1
...
7)

Well known infinite series
∞
X
1
=
xn
1 − x n=0

(10
...
9)

26

∞
X
1
(n − 1)n n−2
=
x
(1 − x)3
2
n=2

(10
...

3!
5!

(10
...

2!
4!

(10
...

3
15
315

(10
...
(−1 < x < 1)
3
5
7

ex = 1 + x +
ln(1 + x) = x −

(1 + x)α

=
=

x2
+
...

2
3

1 + αx +
∞
X

α(α − 1)x2
+
...
α Ck xk

(10
...
15)

(10
...
17)

α∈R

(10
...

2
24

(10
...
then

1 = cosh x × (a0 + a1 x + a2 x2 +
...
20)

And now by multiplication and comparison we can find all the coefficients of 1/ cosh x series

10
...

2
3




1+z
z3
z5
Ln
=2 z+
+
+
...
21)
(10
...
1

Gamma and Beta Functions

11
...
1

Gamma Function

ˆ

∞

e−x xn−1 dx

Γ(n) =

(11
...
2)

• For n a positive integer or zero
Γ(n + 1) = n!

(11
...
And using that in this form of the gamma function: Γ(n) = 2 0 e−x x2n−1 dx
 
√
1
= π
(11
...
1
...
5)

0

β(l, m) = β(m, l)
ˆ ∞
xl−1
β(l, m) =
dx
(1 + x)m+l
0


1
1−2m
β(m, m) = 2
β m,
2

11
...
3

(11
...
7)
(11
...
9)

(11
...
2

Special Functions

11
...
1

Legendre

Legendre Differential Equation
(1 − x2 )y 00 − 2xy 0 + n(n + 1)y = 0

(11
...
12)

General Solution
It is the linear combination of two independent solutions found from the power series method around
x=0
y(x) = a0 y1 (x) + a1 y2 (x)
(11
...


(11
...


(11
...
y2 (x)
remains an infinite series
• Similarly when its a positive odd integer y2 (x) terminates after xn and y2 (x) becomes a polynomial
• These polynomials with suitable choice of a0 and a1 such that Pn (1) = 1 for all n are called
Legendre Polynomials
Some Legendre polynomials
P0 (x) = 1

(11
...
17)

1
(3x2 − 1)
2
1
P3 (x) = (5x3 − 3x)
2
1
P4 (x) = (35x4 − 30x2 + 3)
8

(11
...
21)

P2 (x) =

(11
...
20)

Rodrigues’ Formula
Pn (x) =
General Formula
This is derived using Binomial Theorem and Rodrigue’s Formula
Pn (x) =

n
1 X (−1)r (2n − 2r)! n−2r
x
2n r=0 r!(n − r)!(n − 2r)!

29

(11
...
23)

n=0

Recurrence relations for the Polynomials
(n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x)

(11
...
25)

(n)Pn (x) =

−

0
Pn−1
(x)

0
0
(2n + 1)Pn (x) = Pn+1
(x) − Pn−1
(x)

Pn0 (x)

=

0
xPn−1
(x)

+ nPn−1 (x)

(1 − x2 )Pn0 (x) = n(Pn−1 (x) − xPn (x))
(2n + 1)(1 − x

2

)Pn0 (x)

(11
...
27)
(11
...
29)

2
δmn
2n + 1

(11
...
2
...
31)

where n is real
Equation is solved by Frobenius method about point x = 0 - a regular singular point
• Roots of indicial equation, k = ±n
Recurrence Relation
ar =

−ar−2
(k + r)2 − n2

(11
...
33)

By taking
a0 =
we get
Jn (x) =

∞
X
r=0

J−n (x) =

∞
X
r=0

1
2n Γ(n + 1)

 n+2r
(−1)r
x
r!Γ(n + r + 1) 2

 −n+2r
(−1)r
x
r!Γ(−n + r + 1) 2

called Bessel function of first kind of order n and −n respectively
Case 2: n = 0
Here we get two solutions:


x2
x4
J0 (x) = a0 1 − 2 + 2 2 −
...
4

30

(11
...
35)

(11
...
37)

called Bessel function of first kind of order 0
  2





x4
x
1
1 4
x2
−
1
+
x
y1 (x) = a0 log(x) 1 − 2 + 2 2 −
...

2
2
...
42
2
called Bessel function of second kind of order 0
Case 3: n is an integer
The first solution is Jn (x)
...
38)

(11
...
40)

which is called Bessel function of second kind of order n
Generating function
Jn (x) is the coefficient of z n in the expansion of:
x

e 2 (z−1/z)

(11
...
42)

sin(x sin θ) = 2J1 sin(θ) + 2J3 sin(3θ) + 2J5 sin(5θ)

(11
...


(11
...


(11
...
46)

xJn0 = −nJn + xJn−1

(11
...
2
...
48)
(11
...
50)
(11
...
52)

H0 (x) = 1

(11
...
54)

H2 (x) = 4x2 − 2

(11
...
56)

H4 (x) = 16x4 − 48x2 + 12

(11
...
2
...
58)

L0 (x) = 1

(11
...
60)
2

L2 (x) = 1 − 2x +

x
2

3
x3
L3 (x) = 1 − 3x + x2 −
2
6

11
...
61)
(11
...
1

Fourier Series

• For function f (x) with period P = L and defined in range −L/2 to L/2

∞ 
a0 X
2nπx
2nπx
f (x) =
+
+ bn sin
an cos
2
L
L
n=1
where
an =

2
L

ˆ

L/2

(12
...
2)

f (x) sin

2nπx
dx
L

(12
...
And
bn =

2
L

ˆ

L/2

−L/2

Odd functions
• Contain only sine terms
...
So all bn = 0

12
...
1

Symmetry about quarter period

• If f (x) is even about L/4 then all aodd = 0 and all beven = 0
• If f (x) is odd about L/4 then all aeven = 0 and all bodd = 0

12
...
2

Parseval’s Identity
1
L

12
...
3

ˆ

L

{f (x)}2 dx = a20 +
−L

∞
X

(a2n + b2n )

(12
...
e
...


33

(12
...
2

Fourier integral and transforms

Fourier cosine integral which applies to an even function f (x)
ˆ
ˆ ∞
2 ∞
f (x) cos wx dx f (x) =
A(w) cos wx dw
A(w) =
π 0
0
Fourier sine integral which applies to an odd function f (x)
ˆ
ˆ ∞
2 ∞
B(w) =
f (x) sin wx dx f (x) =
B(w) sin wx dw
π 0
0

(12
...
7)

For a general function f (x), we have representation in terms of its fourier integral
...

11
...
8)
0

A(w) =

1
π

ˆ

∞

f (v) cos wv dv, B(w) =
−∞

1
π

ˆ

∞

f (v) sin wv dv

(12
...
2
...
10)
(12
...
12)

0

ˆ
2 ∞ ˆ
fs (k) sin kx dk
π 0
r ˆ ∞
2
fˆc (k) =
f (x) cos kx dx
π 0
r ˆ ∞
2
f (x) =
fˆc (k) cos kx dk
π 0
r

f (x) =

(12
...
14)
(12
...
16)
(12
...
18)
(12
...
20)
(12
...
Relating to central limit theorem and gaussian fitting etc
X denotes true value of a quantity and x
¯ is the mean of that quantity from observed data
...
1)
hxi =
−∞

With ei = xi − X and E = x
¯ − X we have by definition:
2
σm
= hE 2 i

(13
...
3)

Also,
E2 =

1 XX
1 X 2
ei ej
e
+
i
n2
n2 i j

(13
...
5)

(13
...
7)
n
2
2
σ =
hs i
(13
...
9)
n−1
Since we don’t know hs2 i we work with s2 : from one sample and have the approx but very useful formulae:

1/2
n
σ≈
s
(13
...
11)
n−1
which we use in error estimate of a particular quantity
...
12)
i=1

i=1

35

• An experiment involves different outcomes
...
Like getting an odd number is an event for a dice throwing
expt
...
The occurence
of A can change the prob of occurence of B and vice versa
...
13)

Whereas if they are dependent then
P (A ∩ B) = P (A)P (B|A)

P (A ∩ B) = P (B)P (A|B)

(13
...

And similary for the second equation from here bayes’ theorem is easily derived
...
15)

and so,
P (A|B) =

P (A)P (B|A)
¯ (B|A)
¯
P (A)P (B|A) + P (A)P

(13
...
Like total number of heads in 10 trials of an experiment is well defined
...
And each value of the random variable will have its
probability
...
e
...

• A probability is associated with each value of a random variable or a prob density can be defined
for continous random variables
...
If these two are
independent then
P (x < X ≤ x + dx, y < Y ≤ y + dy) = f (x, y)dxdy = f (x)g(y)dxdy

(13
...
i
...
there is 50% chance that X will be lower or higher than its median value
• From the given probability distribution of X we can find the probability distribution of Y = F (X)
i
...
another random variable that is a function of X
...
18)
dy
where these are the prob density functions
...
1

Popular discrete and continous distributions

13
...
1

Binomial distribution

Here there are two possible outcomes of the expt labelled success and failure with probabilities p and
q =1−p
X = no
...
The probability of X = x is given by:
P (X = x) =n Cx px q n−x


p n−x
P (X = x + 1) =
P (X = x)
q x+1

(13
...
20)

Mean = np
...
e
...
of heads obtained is
np = 10 × 0
...

variance of X = npq

13
...
2

Hypergeometric distribution

This is like a specific case of binomial distribution where the experiment does not have success or failure
but involves selection over multiple trials without replacement
...
21)

this is the probability of selecting x red balls out of M red balls in n trials
...
of balls are N so
N − M are white

37

13
...
3

Poisson distribution

This distribution is actually a discrete distribution cause the random variable is discrete
...
Like the number of calls that we will receive over some
particular time interval
...
22)
x!
This is the probability of receiving x number of calls in a time period of t given that λ is the average
number of calls per unit time
...
1
...

The probability density function is:


2 
1 x−µ
1
(13
...
Then with standard variable Z =
(X − µ)/σ


1
2
(13
...



x−µ
F (x) = Φ
(13
...
26)
P r(a < X ≤ b) = Φ
σ
σ

13
...
If head comes move forward, tail
comes move backwards
...
l = length of each step
...

• Pattern on analysis probabilities of different possible ns when N changes
...

38

• The number of ways to reach that final position:
With N = 5 and 3 heads and 2 tails the distinct permutations of H, H, H, T, T are: 5!/(3!×2!) = 20
...
Total routes = 25 (possibility of getting
head/tail in each of the 5 tosses) so
PN (n) = P5 (1) =

20
25

(13
...
28)

• Plotting n on x-axis and PN (n) on logarithmic y-axis, as n << N i
...
close to the origin the curve
is a parabola
• That is,
ln(PN (n)) = ln(C) −

n2
2N

⇒

PN (n) = Ce−n

2

/2N

(13
...

• Define random variables x1 , x2 ,
...

• Path endpoint n = x1 + x2 +
...
+ xN )2 i
Expand
hx1 x2 i and similar terms are zero, < x21 > etc terms are all 1
...
30)

Miscellaneous
• An angle of a regular polygon of n sides is:
2(n − 2)

π
2n

(13
...
+ an xn = 0

(13
...
33)

an−1
an

(13
...
αn = (−1)n
Sum of the roots:
α0 + α1 +
...


1
x

from x = 1 to

• A trick in integral Consider that we know:
ˆ

∞

0



π/2
sin w cos wx
dw = π/4

w

0
ˆ

So if we want to calculate:

0

∞

if 0 ≤ x < 1
if x = 1
if x > 1

sin 2w cos wx
dw
w

(13
...
36)

If we try to expand sin 2w, a difficult integration by parts comes up Simply substitute 2w = t then
we will get
ˆ ∞
sin t cos((t/2)x)
f (x) =
dt/2
(13
...
38)
f (x/2) =
t
0
And hence
ˆ
0

∞



π/2
sin 2w cos wx
dw = π/4

w

0

if 0 ≤ x/2 < 1
if x/2 = 1
if x/2 > 1

(13
...

• Proof of the infinite GP sum formula(found on stackexchange)
We have
∞
X
a0 ri−1 ≡ S
i=1

40

(13
...
)
= r(a0 r

⇒S

−1

= a0 + rS
a0
=
1−r

41

+ S)

(13
...
42)
(13
...
44)



---