Example: Obtain the Maclaurin’s expansion for

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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) . y(0) = log 1 = 0. , y 2 (0) = -1. Similarly. y 3 (0) = + 1. Example : Using the Maclaurin’s theorem find the

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## Example: Obtain the Maclaurin’s expansion for

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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)

y(0) = log 1 = 0

, y2(0) = -1

Similarly

y3(0) = + 1

Example: Using the Maclaurin’s theorem find the

expansion of y = sin-1 x upto the terms containing x5.

Solution:

 y12(1 – x2) = 1

Differentiating again and simplifying

y2(1 – x2) – xy1 = 0

Differentiating n times using Leibnitz’s theorem

- {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, …

Example: Apply Maclaurin’s theorem to find the

expansion upto x3 term for

Solution:

y2(0) = 0.

Similarly, y3(0)

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

Thereforey = 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 ,

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.

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 – 2y22 – 2y1y3,

y4(0)

y =

Therefore

by using Maclaurin’s theorem upto x7 term.

Solution:

y(0) = 0

y1(0) = 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

 (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) +

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)