1 / 17

Example: Obtain the Maclaurin’s expansion for

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

fulton-watkins
Télécharger la présentation

Example: Obtain the Maclaurin’s expansion for

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

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 , 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) = 32.12 = 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 Thereforey = 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 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.

  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. y = 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)

  16. Therefore f(x) =

More Related