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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

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Example: Obtain the Maclaurin’s expansion for

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  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) =

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