Download presentation

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) =

Similar presentations

OK

Day 3 Notes. 1.4 Definition of the Trigonometric Functions OBJ: Evaluate trigonometric expressions involving quadrantal angles OBJ: Find the angle.

Day 3 Notes. 1.4 Definition of the Trigonometric Functions OBJ: Evaluate trigonometric expressions involving quadrantal angles OBJ: Find the angle.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Download ppt on 15 august Free ppt on conductors and insulators Ppt on career options in humanities Ppt on natural numbers list Ppt on opera web browser Ppt on success and failure of reconstruction Ppt on polynomials and coordinate geometry graph Ppt on network switching modes Cell surface display ppt on ipad Training ppt on customer service