Taylor series:
A Taylor series expansion is a representation of a function by an infinite
series of polynomials around a point.
Mathematically, the Taylor series of a function, f(x) around a point x =
a, is defined as
𝑓 (𝑎) (𝑥 − 𝑎)
𝑓(𝑥) =
𝑛!
Where 𝑓 is the 𝑛 derivative of 𝑓 and 𝑓 is the function 𝑓.
( )
So we can write 𝑓(𝑥) = 𝑓(𝑎) + 𝑓 (𝑎) (𝑥 − 𝑎) + !
(𝑥 − 𝑎) +
( ) ( )
(𝑥 − 𝑎) + (𝑥 − 𝑎) + … … ..
! !
Note: A Taylor series centered at a = 0 (around a point 𝑎 = 0) is specially
named a Maclaurin series
Application in Physics:
In case of simple harmonic motion for finding angular frequency of a
particle executing SHM:
For expression of a given mathematical function 𝑦 = 𝑓 (𝑥) around a
point 𝑥 = 0 Taylor’s theorem can be written as
𝑓′′(0) 𝑓′′′(0)
𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 (0) + 𝑥 +𝑥 + ⋯ ….
2! 3!
If x is taken as the displacement of a particle from its mean position and
the restoring force on particle depends on this 𝑥 by the function FR =
𝑓 (𝑥) then it can be given as
Restoring force on particle at a distance x from mean position is
𝑓′′(0) 𝑓′′′(0)
𝐹 = 𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 (0) + 𝑥 +𝑥 + ⋯ ….
2! 3!
Here, 𝑓(0)= 0 as at 𝑥 = 0 or mean position restoring force is zero and
for small displacement of particle higher power of 𝑥 can neglected so
restoring force can be given as
𝐹 = −𝑥 𝑓′(0)
[Negative sign shows the restoring nature]
Acceleration of particle during oscillation is
( )
𝑎= = −( )𝑥
Comparing this equation with general differential equation of SHM (𝑎 =
( ) ( )
−𝜔 𝑥) we get 𝜔 = 𝑜𝑟 𝜔 =
Hence time period of this SHM is 𝑇 = = 2𝜋 ( )
Problem: Two point charges with charge + Q are fixed at pts (0, r) and
(0, – r) on -𝑌 axis of a coordinate system as shown in figure. Another
small particle of mass m and charge – q is placed at origin of system,
where it stays in equilibrium. If this mass m is slightly displaced along +
x direction by a small distance x and released, show that it executes
SHM and find its time period of oscillations using Taylor series.
After simple calculation first find the restoring force on the particle and
the value will be 𝐹 = − ( ) /
( )
Now using Taylor series, we get 𝜔 =
𝑑𝐹
𝑓 (0) =
𝑑𝑥
3
(𝑟 + 𝑥 ) . 1 − 𝑥 . . (𝑟 + 𝑥 )
=2𝑘𝑞𝑄 2
(𝑟 + 𝑥 )
=2𝑘𝑞𝑄 =
Therefore 𝜔 =
Problem 2: In a given force field, the potential energy of a particle is
given as a function of its x-coordinates as
𝑝 𝑞
𝑈 (𝑥) = −
𝑥 𝑥
where a and b are positive constants. Find the period of small oscillations
of the particle about its equilibrium position in the field.
Sol.: Restoring force on particle 𝐹 = −(− + )
The equilibrium position of particle can calculated by taking F=0
− + = 0 or 𝑥 =
𝑆𝑜 𝑥 = position particle is in equilibrium.
Angular frequency of SHM of particle is given as
2𝑝
𝑓′( 𝑞 )
𝜔=
𝑚
Here 𝑓 = = − =
2𝑝
𝑓′( 𝑞 ) 𝑞
𝜔= =
𝑚 8𝑚𝑝