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Chapter: 12th Maths : UNIT 7 : Applications of Differential Calculus

Series Expansions: Maclaurin’s and Taylor’s Series

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

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 

Ak 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 5th 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 sec x tan x = 2 sec2x tan x

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

 = 2 sec4 x + 4 sec2 x tan2 x

f (iv)(x) = 8sec3 x sec x tan x + 4 sec2 x 2 tan x sec2 x + 8sec x sec x tan x tan2 x

= 16 sec4 x tan x + 8sec2 x tan3 x

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

 = 16 sec6 x + 88sec4 x tan2 x + 16 sec2 x 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 + .. … ;       − π/2 < x < π/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



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12th Maths : UNIT 7 : Applications of Differential Calculus


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