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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 y 3 (0)

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Presentation on theme: "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 y 3 (0)"— Presentation transcript:

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, y 2 (0) = -1 y 3 (0) = + 1 Similarly

2 Example: Using the Maclaurin’s theorem find the expansion of y = sin -1 x upto the terms containing x 5. Solution:

3  y 1 2 (1 – x 2 ) = 1 Differentiating again and simplifying y 2 (1 – x 2 ) – xy 1 = 0 Differentiating n times using Leibnitz’s theorem - {xy n+1 + ny n } = 0

4  (1 – x 2 ) y n+2 – (2n + 1)xy n+1 – n 2 y n = 0 For x = 0, we obtain y n+2 (0) – n 2 y n (0) = 0  y n+2 (0) = n 2 y n (0) y(0) = 0; y 1 (0) = 1; y 2 (0) = 0. y 4 (0) = 2 2. y 2 (0) = 0 (taking n = 2). y 6 (0) = 0, y 8 (0) = 0, …

5 Taking n = 1, 3, 5…we get y 3 (0) = 1 2.y(0) = 1 2 y 5 (0) = 3 2 y 3 (0) = = 3 2, …

6 Example: Apply Maclaurin’s theorem to find the expansion upto x 3 term for Solution:

7 y 2 (0) = 0. Similarly, y 3 (0)

8 Example: Expand y = e x sin x upto x 3 term using Maclaurin’s theorem. y(0) = 0 Solution: y 1 = e x cos x + e x sin x = e x cos x + y,y 1 (0) = 1 y 2 = - e x sin x + e x cos x + y 1 = 2e x cos x,y 2 (0) = 2 y 3 = 2e x cos x – 2e x sin x,y 3 (0) = 2 Therefore y = y 0 + xy 1 (0) + (x 2 /2!) y 2 (0) + (x 3 /3!)y 3 (0) + ….

9 Example: Expand y = log (1+tan x) up to x 3 term using Maclaurin’s theorem. Given 1 + tan x = e y, Solution: y(0) = 0 e y. y 1 = sec 2 x,y 1 (0) = 1 e y y 2 + y 1 e y = 2 sec x sec x tan x,y 2 (0) = -1 e y (y 3 + y 2 ) + (y 2 + y 1 )e y = 2.2 sec x. sec x tan x + 2 sec 2 x sec 2 x y 3 (0) = = 4.

10 Example: Obtain Maclaurin’s expression for y = f(x) = log (1 + e x ) up to x 4 terms. y(0) = log (1 + 1) = log 2 Solution: e y = 1 + e x Differentiating we get e y. y 1 = e x

11 y 1 = e x-y y 1 (0) = e 0-y(0) = e -log2 = ½ y 2 = e (x-y).(1 – y 1 ) y 2 (0) = y 1 (0) (1 –y 1 (0)) = ½(1 – ½) = 1/4 y 3 = y 2 (1 – y 1 ) + y 1 (- y 2 ) = y 2 – 2y 1 y 2 y 3 (0) = 0 y 4 = y 3 – 2y 2 2 – 2y 1 y 3, y 4 (0)

12 Therefore y = by using Maclaurin’s theorem upto x 7 term. y(0) = 0 Solution: y 1 (0) = 1

13  y 1 2 (1 + x 2 ) = 1 Differentiating w.r.t ‘x’ (1 + x 2 ) y 2 + xy 1 = 0; y 2 (0) = 0 Taking n th derivative on both sides: (1 + x 2 ) y n+2 + n.2x. y n+1

14  (1 + x 2 ) y n+2 + (2n + 1) xy n+1 + n 2 y n = 0 For x = 0; y n+2 (0) = -n 2 y n (0) y 3 (0) = -1 2.y 1 (0) = -1 y 4 (0) = -2 2 y 2 (0) = 0 y 5 (0) = -3 2 y 3 (0) = 9 y 6 (0) = y 4 (0) = 0, y 7 (0) = y 5 (0) = -225

15 Therefore y = y(0) + x.y 1 (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) f 1 (a)

16

17 Therefore f(x) =


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