Series Expansions: Maclaurin’s and Taylor’s Series

Series Expansions

Taylor’s series and Maclaurin's series expansion of a function which are infinitely differentiable.

Theorem 7.5

(a) Taylor’s Series

Let f ( x) be a function infinitely differentiable at x = a . Then f ( x) can be expanded as a series, in an interval ( x − a, x + a) , of the form

(b) Maclaurin’s series

If a = 0 , the expansion takes the form

Proof

The series expansion of f ( x) , in powers of ( x − a) , be given by

Substituting x = a gives A₀ = f (a) . Differentiation of (7) gives

Substituting x = a gives A₁ = f ′(a) . Differentiation of (8) gives

Substituting x = a gives A₂ = f ′′(a) / 2!. Differentiation of (9) gives

Differentiation of (10) (k − 3) times gives

Substituting x = a gives 

A_k =  f ^(k)(a) / k! 

which completes the proof of the theorem.

In order to expand a function around a point say x = a , equivalently in powers of ( x − a) we need to differentiate the given function as many times as the required powers and evaluate at x = a . This will give the value for the coefficients of the required powers of ( x − a) .

Example 7.30

Expand log(1+ x) as a Maclaurin’s series upto 4 non-zero terms for –1 < x ≤ 1.

Solution

Let f (x) = log(1+ x) , then the Maclaurin series of f (x) is f (x) = , where,  various derivatives of the function f (x) evaluated at x = 0 are given below:

Substituting the values and on simplification we get the required expansion of the function given by,

Example 7.31

Expand tan x in ascending powers of x upto 5^th power for – π/2 < x < π/2.

Solution

Let f (x) = tan x . Then the Mclaurin series of f ( x) is

Various derivatives of the function f ( x) evaluated at x = 0 are given below:

Now,

 f ′(x) = d/dx (tan x ) = sec²(x)

f’′(x) = d/dx (sec² ( x )) = 2 sec x â‹… sec x â‹… tan x = 2 sec²x â‹… tan x

f ′′′(x) = d/dx (2sec² ( x ). tan x) = 2 sec²(x ) â‹… sec²x + tan x â‹… 4 sec x â‹… sec x â‹… tan x

 = 2 sec⁴ x + 4 sec² x â‹… tan² x

f ^(iv)(x) = 8sec³ x â‹… sec x â‹… tan x + 4 sec² x â‹… 2 tan x â‹… sec² x + 8sec x â‹… sec x â‹… tan x â‹… tan² x

= 16 sec⁴ x tan x + 8sec² x â‹… tan³ x

f^(v)( x) = 16 sec⁴ x â‹… sec² x + 64 sec³ x â‹… sec x â‹… tan x â‹… tan x + 8sec² x â‹… 3 tan² x â‹… sec² x +16 sec x â‹… sec x â‹… tan x â‹… tan³ x

 = 16 sec⁶ x + 88sec⁴ x â‹… tan² x + 16 sec² x â‹… tan⁴ x .

Substituting the values and on simplification we get the required expansion of the function as

 tan x = x + 1/3 x³ + 2/15 x⁵ + .. … ;       âˆ’ Ï€/2 < x < Ï€/2

Example 7.32

Write the Taylor series expansion of 1/ x about x = 2 by finding the first three non-zero terms.

Solution

Let f (x) = 1/x, then the Taylor series of f ( x) is

Various derivatives of the function  f ( x) evaluated at x = 2 are given below.

Substituting these values, we get the required expansion of the function as