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

11

y cos x 1 sin x

1

1y (0) =1

1 0

2 2

(1 sin x).( sin x) cos x(cos x)y =

(1 sin x)

, y2(0) = -1

y3(0) = + 1Similarly

2 3x x xTherefore y 0 ( 1) 1

1 2! 3!

2 3x x x...

1 2! 3!

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

1 2

1y

(1 x )

Solution:

y12(1 – x2) = 1

Differentiating again and simplifying

y2(1 – x2) – xy1 = 0

Differentiating n times using Leibnitz’s theorem

2n 2 n 1 n

n(n 1)y (1 x ) n.y ( 2x) y ( 2)

2!

- {xyn+1 + nyn} = 0

(1 – x2) yn+2 – (2n + 1)xyn+1 – n2yn = 0

For x = 0, we obtainyn+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, …

3 2 22 5x 1 .3

Substituting y = 0 + x.1+ 1 + x +...3! 5!

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

x

x

ey

1 e

Solution:

1 1y(0)

1 1 2

x x x x

1 x 2

(1 e )e e .ey ,

(1 e )

1 2

1y (0)

(1 1)

1

4

x 2 x x x x

2 x 4

(1 e ) .e e .2(1 e ).ey ,

(1 e )

y2(0) = 0.

Similarly, y3(0) 1

8

2 31 1 x x 1Therefore y x 0 ...

2 4 2! 3! 8

Example: Expand y = ex sin x upto x3 term using Maclaurin’s theorem.

y(0) = 0

Solution:

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.

Given 1 + tan x = ey ,Solution: 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) = 2 + 2 = 4.

2 3

0 1 2 3x x

Therefore y = y + xy (0) + y (0)+ y (0) + ... 2! 3!

Example: Obtain Maclaurin’s expression for y = f(x) = log (1 + ex) up to x4 terms.

y(0) = log (1 + 1) = log 2

Solution:

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 – 2y22 – 2y1y3, y4(0) 1

8

Therefore y =2 41 1 x 1 x

log 2 x ...2 4 2! 8 4!

2: Expand log x + 1+ xExample

by using Maclaurin’s theorem upto x7 term.

1 2

1y

x 1 x

2

x1

1 x

2

1

1 x

y(0) = 0Solution:

y1(0) = 1

21Therefore y 1 x 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 nn(n 1)

2.y + xy + n.1.y = 02!

(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) + 2

2x

y (0) + ...2!

3 57x x 225

x + 9 - x + ...3! 5! 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)2 3

11 111(x a) (x a)f (a)+ f (a) + ...

2! 3!

Here a .3

1f =log cos = log

3 3 2

1

x3

f ( tan x) tan 33 3

11 2 2

x3

f ( sec x) = - sec =-4 3 3

111f 8 33

2

1 11

3

111

x3

f x f f3 3 3 2! 3

(x )3 f

3! 3

Therefore f(x) =