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Example: Obtain the Maclaurin’s expansion for
log (1 + sin x) upto first three terms.
Solution:
y = log (1 + sin x)
y1 
y2 =
1
1  sin x
cos x
y(0) = log 1 = 0
y 1 (0) 
1
1 0
(1  sin x ).(  sin x )  co s x (co s x )
Similarly
(1  sin x )
2
y3(0) = + 1
=1
, y2(0) = -1
x
T herefore y  0 

1

x

1
x
x
2
(  1) 
2!
2

2!
x
x
3
1
3!
3
 ...
3!
Example: Using the Maclaurin’s theorem find the
expansion of y = sin-1 x upto the terms containing x5.
Solution: y 1 
1
2
(1  x )
 y12(1 – x2) = 1
Differentiating again and simplifying
y2(1 – x2) – xy1 = 0
Differentiating n times using Leibnitz’s theorem
n (n  1)


2
y n (  2) 
 y n  2 (1  x )  n.y n  1 (  2 x ) 
2!


- {xyn+1 + nyn} = 0
 (1 – x2) yn+2 – (2n + 1)xyn+1 – n2yn = 0
For x = 0, we obtain
yn+2(0) – n2 yn(0) = 0  yn+2(0) = n2yn(0)
y(0) = 0; y1 (0) = 1; y2(0) = 0.
y4(0) = 22. y2(0) = 0 (taking n = 2).
y6(0) = 0, y8(0) = 0, …
Taking n = 1, 3, 5…we get
y3(0) = 12.y(0) = 12
y5(0) = 32 y3(0) = 32.12 = 32, …
S ubstituting y = 0 + x.1 +
x
3
3!
2
2
1 +
1 .3
5!
2
5
x + ...
Example: Apply Maclaurin’s theorem to find the
expansion upto x3 term for
y
e
x
1 e
x
Solution:
y (0) 
1
11
x
y1 

1
2
x
x
(1  e )e  e .e
x 2
(1  e )
x
,
y 1 (0 ) 
1
(1  1)
2

1
4
x 2
y2 
x
x
x
(1  e ) .e  e .2(1  e ).e
x
,
x 4
(1  e )
Similarly, y3(0)  
y2(0) = 0.
1
8
2
3
x  1
T herefore y   x 
0
    ...
2
4
2!
3!  8 
1
1
x
Example: Expand y = ex sin x upto x3 term
using Maclaurin’s theorem.
Solution:
y(0) = 0
y1 = ex cos x + ex sin x = ex cos x + y, y1(0) = 1
y2 = - ex sin x + ex cos x + y1 = 2ex cos x,y2(0) = 2
y3 = 2ex cos x – 2ex sin x, y3(0) = 2
Therefore y = y0 + xy1(0) + (x2/2!) y2(0)
+ (x3/3!)y3(0) + ….
Example: Expand y = log (1+tan x) up to x3 term using
Maclaurin’s theorem.
Solution: Given 1 + tan x = ey ,
ey. y1 = sec2 x,
y(0) = 0
y1(0) = 1
ey y2 + y1 ey = 2 sec x sec x tan x,
y2(0) = -1
ey(y3 + y2) + (y2 + y1)ey
= 2.2 sec x. sec x tan x + 2 sec2 x sec2 x
y3(0) = 2 + 2 = 4.
T herefore y = y 0 + xy 1 (0) +
x
2
2!
y 2 (0)+
x
3
3!
y 3 (0) + ...
Example: Obtain Maclaurin’s expression for
y = f(x) = log (1 + ex) up to x4 terms.
Solution:
y(0) = log (1 + 1) = log 2
ey = 1 + ex
Differentiating we get ey. y1 = ex
y1 = ex-y
y1(0) = e0-y(0) = e-log2 = ½
y2 = e(x-y).(1 – y1)
y2(0) = y1(0) (1 –y1(0)) = ½(1 – ½) = 1/4
y3 = y2(1 – y1) + y1(- y2) = y2 – 2y1y2
y3(0) = 0
y4 = y3 – 2y2 – 2y1y3, y4(0)  
2
1
8
Therefore y = log 2 
1
x
2
E xa m p le : E x p a n d lo g

x +
1 x
2

4 2!
1+ x
2
1 x
4
 ...
8 4!

by using Maclaurin’s theorem upto x7 term.
Solution:
y1 
y(0) = 0
1
x
y1(0) = 1
1 x
2

1 


 
2
1  x 
x
1
1 x
2
T herefore y 1 1  x
2
1
 y12 (1 + x2) = 1
Differentiating w.r.t ‘x’
(1 + x2) y2 + xy1 = 0; y2(0) = 0
Taking nth derivative on both sides:
(1 + x2) yn+2 + n.2x. yn+1

n (n  1)
2!
2.y n + xy n+ 1 + n.1.y n = 0
 (1 + x2) yn+2 + (2n + 1) xyn+1 + n2 yn = 0
For x = 0; yn+2 (0) = -n2yn(0)
y3(0) = -12.y1 (0) = -1
y4(0) = -22 y2(0) = 0
y5(0) = -32 y3(0) = 9
y6(0) = - 42y4 (0) = 0, y7(0) = - 52y5 (0) = -225
Therefore y = y(0) + x.y1(0) +
x
x
3
x
+ 9
3!
5
-
5!
225
x
2
2!
y 2 (0) + ...
7
x + ...
7
Example : Find the Taylor’s series expansion of the
function about the point /3 for f(x) = log (cos x)
Solution:
f(x) = f(a) + (x – a) f1(a)

(x  a )
2!
2
f
11
(a)+
(x  a )
3!
3
f
111
(a) + ...
H ere a 

.
3

1

f   = log cos = log
3
2
3


f    (  tan x )    tan   3
x
3
3
3
1
f

2
2 
= -4
   (  sec x )  = - sec
x
3
3
3
11 
f

   8 3
3
111 
Therefore f(x) =
2


x  
  1   
3  11   
 
f     x  f   
f  
3 3
2!
3 
3
(x 


3
3!
)
3
f

 
3
111 