1. Example: Obtain the Maclaurin’s expansion for
log (1 + sin x) upto first three terms.
Solution:
y = log (1 + sin x)
y(0) = log 1 = 0
, y2(0) = -1
Similarly
y3(0) = + 1

2. Example: Using the Maclaurin’s theorem find the
expansion of y = sin-1 x upto the terms containing x5.
Solution:

3.  y12(1 – x2) = 1
Differentiating again and simplifying
y2(1 – x2) – xy1 = 0
Differentiating n times using Leibnitz’s theorem
- {xyn+1 + nyn} = 0

4.  (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, …

5. Taking n = 1, 3, 5…we get
y3(0) = 12.y(0) = 12
y5(0) = 32 y3(0) = = 32, …

6. Example: Apply Maclaurin’s theorem to find the
expansion upto x3 term for
Solution:

7. y2(0) = 0.
Similarly, y3(0)

8. 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) + ….

9. Example: Expand y = log (1+tan x) up to x3 term using
Maclaurin’s theorem.
Solution:
Given tan x = ey ,
y(0) = 0
ey. y1 = sec2 x,
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) = = 4.

10. 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

11. 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 – 2y22 – 2y1y3,
y4(0)

12. by using Maclaurin’s theorem upto x7 term.
Therefore
by using Maclaurin’s theorem upto x7 term.
Solution:
y(0) = 0
y1(0) = 1

13.  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

14.  (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

15. Therefore y = y(0) + x.y1(0) +
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)

17. Therefore f(x) =