Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

376 views

Published on

differentiation and integration of power series

Published in:
Education

No Downloads

Total views

376

On SlideShare

0

From Embeds

0

Number of Embeds

4

Shares

0

Downloads

9

Comments

0

Likes

3

No embeds

No notes for slide

- 1. Differentiation and Integration of Power Series
- 2. Differentiation and Integration of Power Series Theorem (Derivative and integral of Taylor series) :
- 3. Σk=0 ∞ Differentiation and Integration of Power Series Let f(x) = P(x) = the Taylor series = ck(x – a)k for all x in the interval (a – R, a + R) where R is the radius of convergence, then Theorem (Derivative and integral of Taylor series) :
- 4. Σk=0 ∞ Differentiation and Integration of Power Series Let f(x) = P(x) = the Taylor series = ck(x – a)k for all x in the interval (a – R, a + R) where R is the radius of convergence, then Theorem (Derivative and integral of Taylor series) : I. the Taylor series of f '(x) is P'(x) = [ck(x – a)k]' with radius of convergence R and f '(x) = P'(x) for all x in the interval (a – R, a + R). Σk=0 ∞
- 5. Σk=0 ∞ Differentiation and Integration of Power Series Let f(x) = P(x) = the Taylor series = ck(x – a)k for all x in the interval (a – R, a + R) where R is the radius of convergence, then Theorem (Derivative and integral of Taylor series) : I. the Taylor series of f '(x) is P'(x) = [ck(x – a)k]' with radius of convergence R and f '(x) = P'(x) for all x in the interval (a – R, a + R). Σk=0 ∞ II. the Taylor series of ∫f(x)dx is ∫P(x)dx= ∫ck(x – a)kdx with radius of convergence R and ∫f(x)dx = ∫P(x)dx for all x in the interval (a – R, a + R). Σk=0 ∞
- 6. Σk=0 ∞ Differentiation and Integration of Power Series Let f(x) = P(x) = the Taylor series = ck(x – a)k for all x in the interval (a – R, a + R) where R is the radius of convergence, then Theorem (Derivative and integral of Taylor series) : I. the Taylor series of f '(x) is P'(x) = [ck(x – a)k]' with radius of convergence R and f '(x) = P'(x) for all x in the interval (a – R, a + R). Σk=0 ∞ II. the Taylor series of ∫f(x)dx is ∫P(x)dx= ∫ck(x – a)kdx with radius of convergence R and ∫f(x)dx = ∫P(x)dx for all x in the interval (a – R, a + R). Σk=0 ∞ III. ∫ f(x)dx = ∫ ck(x – a)kdx with the series converges absolutely and (c, d) is any interval in (a – R, a + R). Σk=0 ∞ c d c d
- 7. Example: Given the Mac series of Differentiation and Integration of Power Series Σ k=0 (2k+1)! (-1)kx2k+1∞ x – 3! x3 + 5! x5 + .. =7! x7 – Σ k=0 (2k)! (-1)kx2k∞ + 4! x4 6! x6 8! x8 +1 – – – .. =2! x2 sin(x) = cos(x) =
- 8. Example: Given the Mac series of [sin(x)]' = Differentiation and Integration of Power Series Σ k=0 (2k+1)! (-1)kx2k+1∞ x – 3! x3 + 5! x5 + .. =7! x7 – Σ k=0 (2k)! (-1)kx2k∞ + 4! x4 6! x6 8! x8 +1 – – – .. =2! x2 sin(x) = cos(x) = The derivative of the power series of sin(x) is
- 9. Example: Given the Mac series of [sin(x)]' = [ Differentiation and Integration of Power Series Σ k=0 (2k+1)! (-1)kx2k+1∞ x – 3! x3 + 5! x5 + .. =7! x7 – Σ k=0 (2k)! (-1)kx2k∞ + 4! x4 6! x6 8! x8 +1 – – – .. =2! x2 sin(x) = cos(x) = x – 3! x3 + 5! x5 + .. ]'7! x7 – The derivative of the power series of sin(x) is
- 10. Example: Given the Mac series of [sin(x)]' = [ Differentiation and Integration of Power Series Σ k=0 (2k+1)! (-1)kx2k+1∞ x – 3! x3 + 5! x5 + .. =7! x7 – Σ k=0 (2k)! (-1)kx2k∞ + 4! x4 6! x6 8! x8 +1 – – – .. =2! x2 sin(x) = cos(x) = x – 3! x3 + 5! x5 + .. ]'7! x7 – + 4! x4 6! x6 + …= 1 – –2! x2 The derivative of the power series of sin(x) is
- 11. Example: Given the Mac series of [sin(x)]' = [ Differentiation and Integration of Power Series Σ k=0 (2k+1)! (-1)kx2k+1∞ x – 3! x3 + 5! x5 + .. =7! x7 – Σ k=0 (2k)! (-1)kx2k∞ + 4! x4 6! x6 8! x8 +1 – – – .. =2! x2 sin(x) = cos(x) = x – 3! x3 + 5! x5 + .. ]'7! x7 – + 4! x4 6! x6 + …= 1 – –2! x2 = cos(x) The derivative of the power series of sin(x) is
- 12. Example: Given the Mac series of [sin(x)]' = [ Differentiation and Integration of Power Series Σ k=0 (2k+1)! (-1)kx2k+1∞ x – 3! x3 + 5! x5 + .. =7! x7 – Σ k=0 (2k)! (-1)kx2k∞ + 4! x4 6! x6 8! x8 +1 – – – .. =2! x2 sin(x) = cos(x) = x – 3! x3 + 5! x5 + .. ]'7! x7 – + 4! x4 6! x6 + …= 1 – –2! x2 = cos(x) The derivative of the power series of sin(x) is ∫sin(x)dx = The antiderivative of the power series of sin(x) is
- 13. Example: Given the Mac series of [sin(x)]' = [ Differentiation and Integration of Power Series Σ k=0 (2k+1)! (-1)kx2k+1∞ x – 3! x3 + 5! x5 + .. =7! x7 – Σ k=0 (2k)! (-1)kx2k∞ + 4! x4 6! x6 8! x8 +1 – – – .. =2! x2 sin(x) = cos(x) = x – 3! x3 + 5! x5 + .. ]'7! x7 – + 4! x4 6! x6 + …= 1 – –2! x2 = cos(x) The derivative of the power series of sin(x) is ∫sin(x)dx = ∫xdx – ∫ dx 3! x3 + ∫ dx5! x5 –… The antiderivative of the power series of sin(x) is
- 14. Example: Given the Mac series of [sin(x)]' = [ Differentiation and Integration of Power Series Σ k=0 (2k+1)! (-1)kx2k+1∞ x – 3! x3 + 5! x5 + .. =7! x7 – Σ k=0 (2k)! (-1)kx2k∞ + 4! x4 6! x6 8! x8 +1 – – – .. =2! x2 sin(x) = cos(x) = x – 3! x3 + 5! x5 + .. ]'7! x7 – + 4! x4 6! x6 + …= 1 – –2! x2 = cos(x) The derivative of the power series of sin(x) is ∫sin(x)dx = ∫xdx – ∫ dx 3! x3 + ∫ dx5! x5 –… The antiderivative of the power series of sin(x) is 4! x4 6! x6 = + 2! x2 – – … + c
- 15. Example: Given the Mac series of [sin(x)]' = [ Differentiation and Integration of Power Series Σ k=0 (2k+1)! (-1)kx2k+1∞ x – 3! x3 + 5! x5 + .. =7! x7 – Σ k=0 (2k)! (-1)kx2k∞ + 4! x4 6! x6 8! x8 +1 – – – .. =2! x2 sin(x) = cos(x) = x – 3! x3 + 5! x5 + .. ]'7! x7 – + 4! x4 6! x6 + …= 1 – –2! x2 = cos(x) The derivative of the power series of sin(x) is ∫sin(x)dx = ∫xdx – ∫ dx 3! x3 + ∫ dx5! x5 –… The antiderivative of the power series of sin(x) is 4! x4 6! x6 = + 2! x2 = -cos(x) + c– – … + c
- 16. Example: Given the Mac series of [sin(x)]' = [ Differentiation and Integration of Power Series Σ k=0 (2k+1)! (-1)kx2k+1∞ x – 3! x3 + 5! x5 + .. =7! x7 – Σ k=0 (2k)! (-1)kx2k∞ + 4! x4 6! x6 8! x8 +1 – – – .. =2! x2 sin(x) = cos(x) = x – 3! x3 + 5! x5 + .. ]'7! x7 – + 4! x4 6! x6 + …= 1 – –2! x2 = cos(x) The derivative of the power series of sin(x) is ∫sin(x)dx = ∫xdx – ∫ dx 3! x3 + ∫ dx5! x5 –… The antiderivative of the power series of sin(x) is 4! x4 6! x6 = + 2! x2 = cos(x) + c– – … + c They all have infinite radius of convergence.
- 17. Example: Use the fact [ ]' = Differentiation and Integration of Power Series 1 – x 1 (1 – x)2 1 to find the Mac series of . (1 – x)2 1
- 18. Example: Use the fact [ ]' = Differentiation and Integration of Power Series 1 – x 1 (1 – x)2 1 to find the Mac series of . (1 – x)2 1 The Mac series of = 1 – x 1 Σ k=0 ∞ xk = 1 + x + x2 + x3 + ..
- 19. Example: Use the fact [ ]' = Differentiation and Integration of Power Series 1 – x 1 (1 – x)2 1 to find the Mac series of . (1 – x)2 1 The Mac series of = 1 – x 1 Σ k=0 ∞ xk = 1 + x + x2 + x3 + .. Hence the Mac series of it's derivative is 1 – x 1 Σk=0 ∞ xk = 1 + x + x2 + x3 + ..=
- 20. Example: Use the fact [ ]' = Differentiation and Integration of Power Series 1 – x 1 (1 – x)2 1 to find the Mac series of . (1 – x)2 1 The Mac series of = 1 – x 1 Σ k=0 ∞ xk = 1 + x + x2 + x3 + .. Hence the Mac series of it's derivative is 1 – x 1 Σk=0 ∞ xk = 1 + x + x2 + x3 + ..= (1 – x)2 1 d dx
- 21. Example: Use the fact [ ]' = Differentiation and Integration of Power Series 1 – x 1 (1 – x)2 1 to find the Mac series of . (1 – x)2 1 The Mac series of = 1 – x 1 Σ k=0 ∞ xk = 1 + x + x2 + x3 + .. Hence the Mac series of it's derivative is Σk=0 xk[ ]' ∞ 1 – x 1 Σk=0 ∞ xk = 1 + x + x2 + x3 + ..= (1 – x)2 1 d dx d dx =
- 22. Example: Use the fact [ ]' = Differentiation and Integration of Power Series 1 – x 1 (1 – x)2 1 to find the Mac series of . (1 – x)2 1 The Mac series of = 1 – x 1 Σ k=0 ∞ xk = 1 + x + x2 + x3 + .. Hence the Mac series of it's derivative is Σk=0 xk[ ]' = ∞ Σk=1 kxk-1 ∞ 1 – x 1 Σk=0 ∞ xk = 1 + x + x2 + x3 + ..= (1 – x)2 1 d dx d dx =
- 23. Example: Use the fact [ ]' = Differentiation and Integration of Power Series 1 – x 1 (1 – x)2 1 to find the Mac series of . (1 – x)2 1 The Mac series of = 1 – x 1 Σ k=0 ∞ xk = 1 + x + x2 + x3 + .. Hence the Mac series of it's derivative is Σk=0 xk[ ]' = ∞ Σk=1 kxk-1 = 1 + 2x + 3x2 + 4x3 +… ∞ 1 – x 1 Σk=0 ∞ xk = 1 + x + x2 + x3 + ..= (1 – x)2 1 d dx d dx =
- 24. Example: Use the fact [ ]' = Differentiation and Integration of Power Series 1 – x 1 (1 – x)2 1 to find the Mac series of . (1 – x)2 1 The Mac series of = 1 – x 1 Σ k=0 ∞ xk = 1 + x + x2 + x3 + .. Hence the Mac series of it's derivative is Σk=0 xk The radius of convergence of 1/(x – 1) is R = 1, [ ]' = ∞ Σk=1 kxk-1 = 1 + 2x + 3x2 + 4x3 +… ∞ 1 – x 1 Σk=0 ∞ xk = 1 + x + x2 + x3 + ..= (1 – x)2 1 d dx d dx =
- 25. Example: Use the fact [ ]' = Differentiation and Integration of Power Series 1 – x 1 (1 – x)2 1 to find the Mac series of . (1 – x)2 1 The Mac series of = 1 – x 1 Σ k=0 ∞ xk = 1 + x + x2 + x3 + .. Hence the Mac series of it's derivative is Σk=0 xk The radius of convergence of 1/(x – 1) is R = 1, hence R =1 is the radius of convergence of [ ]' = ∞ Σk=1 kxk-1 = 1 + 2x + 3x2 + 4x3 +… ∞ Σk=1 ∞ kxk-1, 1 – x 1 Σk=0 ∞ xk = 1 + x + x2 + x3 + ..= (1 – x)2 1 d dx d dx =
- 26. Example: Use the fact [ ]' = Differentiation and Integration of Power Series 1 – x 1 (1 – x)2 1 to find the Mac series of . (1 – x)2 1 The Mac series of = 1 – x 1 Σ k=0 ∞ xk = 1 + x + x2 + x3 + .. Hence the Mac series of it's derivative is Σk=0 xk The radius of convergence of 1/(x – 1) is R = 1, hence R =1 is the radius of convergence of and 1/(1 – x)2 = for all x in (-1, 1). [ ]' = ∞ Σk=1 kxk-1 = 1 + 2x + 3x2 + 4x3 +… ∞ Σk=1 ∞ kxk-1, Σk=1 ∞ kxk-1 1 – x 1 Σk=0 ∞ xk = 1 + x + x2 + x3 + ..= (1 – x)2 1 d dx d dx =
- 27. We may use the integral formula to find the power series of some non-elementary functions. Differentiation and Integration of Power Series
- 28. We may use the integral formula to find the power series of some non-elementary functions. Differentiation and Integration of Power Series Example: Find the Mac series of ∫ dxx sin(x)
- 29. We may use the integral formula to find the power series of some non-elementary functions. Differentiation and Integration of Power Series Example: Find the Mac series of ∫ dxx sin(x) The Mac series of isx sin(x) x – 3! x3 + 5! x5 + ..7! x7 –( ) / x
- 30. We may use the integral formula to find the power series of some non-elementary functions. Differentiation and Integration of Power Series Example: Find the Mac series of ∫ dxx sin(x) The Mac series of isx sin(x) x – 3! x3 + 5! x5 + ..7! x7 –( ) / x = 1 – 3! x2 + 5! x4 + ..7! x6 –
- 31. We may use the integral formula to find the power series of some non-elementary functions. Differentiation and Integration of Power Series Example: Find the Mac series of ∫ dxx sin(x) The Mac series of isx sin(x) x – 3! x3 + 5! x5 + ..7! x7 –( ) / x = 1 – 3! x2 + 5! x4 + ..7! x6 – Σ k=0 (2k+1)! (-1)kx2k∞ =
- 32. We may use the integral formula to find the power series of some non-elementary functions. Differentiation and Integration of Power Series Example: Find the Mac series of ∫ dxx sin(x) The Mac series of isx sin(x) x – 3! x3 + 5! x5 + ..7! x7 –( ) / x = 1 – 3! x2 + 5! x4 + ..7! x6 – Σ k=0 (2k+1)! (-1)kx2k∞ = So the Mac series of ∫ dx = x sin(x) Σ ∫ k=0 (2k+1)! (-1)kx2k∞ dx
- 33. We may use the integral formula to find the power series of some non-elementary functions. Differentiation and Integration of Power Series Example: Find the Mac series of ∫ dxx sin(x) The Mac series of isx sin(x) x – 3! x3 + 5! x5 + ..7! x7 –( ) / x = 1 – 3! x2 + 5! x4 + ..7! x6 – Σ k=0 (2k+1)! (-1)kx2k∞ = So the Mac series of ∫ dx = x sin(x) Σ ∫ k=0 (2k+1)! (-1)kx2k∞ dx = Σ k=0 (2k+1)(2k+1)! (-1)kx2k+1∞
- 34. We may use the integral formula to find the power series of some non-elementary functions. Differentiation and Integration of Power Series Example: Find the Mac series of ∫ dxx sin(x) The Mac series of isx sin(x) x – 3! x3 + 5! x5 + ..7! x7 –( ) / x = 1 – 3! x2 + 5! x4 + ..7! x6 – Σ k=0 (2k+1)! (-1)kx2k∞ = So the Mac series of ∫ dx = x sin(x) Σ ∫ k=0 (2k+1)! (-1)kx2k∞ dx = Σ k=0 (2k+1)(2k+1)! (-1)kx2k+1∞ = x – 3*3! x3 + + ..– 5*5! x5 7*7! x7
- 35. Differentiation and Integration of Power Series Example: Find ∫ e-x dx accurate to 3 decimal places. x=0 1 2
- 36. Differentiation and Integration of Power Series Example: Find ∫ e-x dx accurate to 3 decimal places. x=0 1 ∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want. 2 2
- 37. Differentiation and Integration of Power Series Example: Find ∫ e-x dx accurate to 3 decimal places. ex = n! xn x=0 1 ∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want. 2 2 Σk=0 ∞
- 38. Differentiation and Integration of Power Series Example: Find ∫ e-x dx accurate to 3 decimal places. ex = n! xn x=0 1 ∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want. 2 2 Σk=0 ∞ e-x = n! (-x2)n Σk=0 ∞2 =
- 39. Differentiation and Integration of Power Series Example: Find ∫ e-x dx accurate to 3 decimal places. ex = n! xn x=0 1 ∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want. 2 2 Σk=0 ∞ e-x = n! (-x2)n Σk=0 ∞2 x2 + 2! x4 3! x6 4! x8 += 1 – – – ..
- 40. Differentiation and Integration of Power Series Example: Find ∫ e-x dx accurate to 3 decimal places. ex = n! xn x=0 1 ∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want. 2 2 Σk=0 ∞ e-x = n! (-x2)n Σk=0 ∞2 x2 + 2! x4 3! x6 4! x8 += 1 Hence ∫ e-x dx =2 n! (-x)2n Σ ∫ dx = k=0 ∞ – – – ..
- 41. Differentiation and Integration of Power Series Example: Find ∫ e-x dx accurate to 3 decimal places. ex = n! xn x=0 1 ∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want. 2 2 Σk=0 ∞ e-x = n! (-x2)n Σk=0 ∞2 x2 + 2! x4 3! x6 4! x8 += 1 Hence ∫ e-x dx =2 n! (-x2)n Σ ∫ dx =k=0 ∞ (2n+1)n!Σk=0 ∞ (-1)nx2n+1 – – – ..
- 42. Differentiation and Integration of Power Series Example: Find ∫ e-x dx accurate to 3 decimal places. ex = n! xn x=0 1 ∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want. 2 2 Σk=0 ∞ e-x = n! (-x2)n Σk=0 ∞2 x2 + 2! x4 3! x6 4! x8 += 1 Hence ∫ e-x dx =2 n! (-x2)n Σ ∫ dx =k=0 ∞ (2n+1)n!Σk=0 ∞ = c + x (-1)nx2n+1 3 x3 + 5*2! x5 + ..7*3! x7 – – – .. – –
- 43. Differentiation and Integration of Power Series Example: Find ∫ e-x dx accurate to 3 decimal places. ex = n! xn x=0 1 ∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want. 2 2 Σk=0 ∞ e-x = n! (-x2)n Σk=0 ∞2 x2 + 2! x4 3! x6 4! x8 += 1 Hence ∫ e-x dx =2 n! (-x2)n Σ ∫ dx =k=0 ∞ (2n+1)n!Σk=0 ∞ = c + x (-1)nx2n+1 3 x3 + 5*2! x5 + ..7*3! x7 = P(x) – – – .. – –
- 44. Differentiation and Integration of Power Series Example: Find ∫ e-x dx accurate to 3 decimal places. ex = n! xn x=0 1 ∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want. 2 2 Σk=0 ∞ e-x = n! (-x2)n Σk=0 ∞2 x2 + 2! x4 3! x6 4! x8 += 1 Hence ∫ e-x dx =2 n! (-x2)n Σ ∫ dx =k=0 ∞ (2n+1)n!Σk=0 ∞ = c + x (-1)nx2n+1 3 x3 + 5*2! x5 + ..7*3! x7 = P(x) ∫ e-x dx = P(1) – P(0)x=0 1 2 – – – .. – – Therefore
- 45. Differentiation and Integration of Power Series Example: Find ∫ e-x dx accurate to 3 decimal places. ex = n! xn x=0 1 ∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want. 2 2 Σk=0 ∞ e-x = n! (-x2)n Σk=0 ∞2 x2 + 2! x4 3! x6 4! x8 += 1 Hence ∫ e-x dx =2 n! (-x2)n Σ ∫ dx =k=0 ∞ (2n+1)n!Σk=0 ∞ = c + x (-1)nx2n+1 3 x3 + 5*2! x5 + ..7*3! x7 = P(x) ∫ e-x dx = P(1) – P(0)x=0 1 = 1 3 1 + 5*2! 1 + ..7*3! 1 2 – – – .. – – – – Therefore which is a "decreasing" alternating series.
- 46. Differentiation and Integration of Power Series ∫ e-x dxx=0 1 = 12 Hence is a convergent alternating series. 3 1 + 5*2! 1 +7*3! 1 – – 1 9*4! – 1 11*5! + 13*6! … 1
- 47. Differentiation and Integration of Power Series ∫ e-x dxx=0 1 = 12 Hence is a convergent alternating series. From theorem of alternaing series, any "decreasing" convergent alternating series, it's sum S satisfies |S| < |a1=1st term|. 3 1 + 5*2! 1 +7*3! 1 – – 1 9*4! – 1 11*5! + 13*6! … 1
- 48. Differentiation and Integration of Power Series ∫ e-x dxx=0 1 = 12 Hence is a convergent alternating series. From theorem of alternaing series, any "decreasing" convergent alternating series, it's sum S satisfies |S| < |a1=1st term|. By trial and error, we find that < 0.0005.13*6! 1 3 1 + 5*2! 1 +7*3! 1 – – 1 9*4! – 1 11*5! + 13*6! … 1
- 49. Differentiation and Integration of Power Series ∫ e-x dxx=0 1 = 12 Hence is a convergent alternating series. From theorem of alternaing series, any "decreasing" convergent alternating series, it's sum S satisfies |S| < |a1=1st term|. By trial and error, we find that < 0.0005.13*6! 1 So the tail series 1 – .. < 0.0005+ 13*6! 1 15*7! 1 17*8! – 3 1 + 5*2! 1 +7*3! 1 – – 1 9*4! – 1 11*5! + 13*6! … 1
- 50. Differentiation and Integration of Power Series ∫ e-x dxx=0 1 = 1 3 1 + 5*2! 1 +7*3! 12 – –Hence is a convergent alternating series. From theorem of alternaing series, any "decreasing" convergent alternating series, it's sum S satisfies |S| < |a1=1st term|. By trial and error, we find that < 0.0005.13*6! 1 So the tail series 1 – .. < 0.0005+ 13*6! 1 15*7! 1 17*8! This implies the sum of the front terms 1 3 1 + 5*2! 1 +7*3! 1 – – – 1 9*4! – 1 11*5! 0.74673 is accurate to 3 decimal places. 1 9*4! – 1 11*5! + 13*6! … 1

No public clipboards found for this slide

Be the first to comment