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11.5 = Recursion & Iteration. Arithmetic = adding (positive or negative)

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Presentation on theme: "11.5 = Recursion & Iteration. Arithmetic = adding (positive or negative)"— Presentation transcript:

1 11.5 = Recursion & Iteration

2 Arithmetic = adding (positive or negative)

3 3, 6, 9, 12, …

4 Arithmetic = adding (positive or negative) 3, 6, 9, 12, … d = 3

5 Arithmetic = adding (positive or negative) 3, 6, 9, 12, … d = 3 *Formula for the n th term based on a 1 and d. a n = a 1 +(n–1)d

6 Arithmetic = adding (positive or negative) 3, 6, 9, 12, … d = 3 *Formula for the n th term based on a 1 and d. a n = a 1 +(n–1)d Geometric = multiplying (#’s > 1 or #’s < 1)

7 Arithmetic = adding (positive or negative) 3, 6, 9, 12, … d = 3 *Formula for the n th term based on a 1 and d. a n = a 1 +(n–1)d Geometric = multiplying (#’s > 1 or #’s < 1) 2, 10, 50, 250, …

8 Arithmetic = adding (positive or negative) 3, 6, 9, 12, … d = 3 *Formula for the n th term based on a 1 and d. a n = a 1 +(n–1)d Geometric = multiplying (#’s > 1 or #’s < 1) 2, 10, 50, 250, … r = 5

9 Arithmetic = adding (positive or negative) 3, 6, 9, 12, … d = 3 *Formula for the n th term based on a 1 and d. a n = a 1 +(n–1)d Geometric = multiplying (#’s > 1 or #’s < 1) 2, 10, 50, 250, … r = 5 *Formula for the n th term based on a 1 and r. a n = a 1 r (n – 1)

10 Arithmetic = adding (positive or negative) 3, 6, 9, 12, … d = 3 *Formula for the n th term based on a 1 and d. a n = a 1 +(n–1)d Geometric = multiplying (#’s > 1 or #’s < 1) 2, 10, 50, 250, … r = 5 *Formula for the n th term based on a 1 and r. a n = a 1 r (n – 1) Recursion = formula-based (“neither”)

11 Arithmetic = adding (positive or negative) 3, 6, 9, 12, … d = 3 *Formula for the n th term based on a 1 and d. a n = a 1 +(n–1)d Geometric = multiplying (#’s > 1 or #’s < 1) 2, 10, 50, 250, … r = 5 *Formula for the n th term based on a 1 and r. a n = a 1 r (n – 1) Recursion = formula-based (“neither”) 2, 4, 16, 256, …

12 Arithmetic = adding (positive or negative) 3, 6, 9, 12, … d = 3 *Formula for the n th term based on a 1 and d. a n = a 1 +(n–1)d Geometric = multiplying (#’s > 1 or #’s < 1) 2, 10, 50, 250, … r = 5 *Formula for the n th term based on a 1 and r. a n = a 1 r (n – 1) Recursion = formula-based (“neither”) 2, 4, 16, 256, … -The pattern is that you’re squaring each previous term.

13 Arithmetic = adding (positive or negative) 3, 6, 9, 12, … d = 3 *Formula for the n th term based on a 1 and d. a n = a 1 +(n–1)d Geometric = multiplying (#’s > 1 or #’s < 1) 2, 10, 50, 250, … r = 5 *Formula for the n th term based on a 1 and r. a n = a 1 r (n – 1) Recursion = formula-based (“neither”) 2, 4, 16, 256, … -The pattern is that you’re squaring each previous term. a n+1 = (a n ) 2

14 Recursion = formula-based (“neither”) 2, 4, 16, 256, … -The pattern is that you’re squaring each previous term. a n+1 = (a n ) 2 *Note that this formula only applies to this particular example!!!

15 Recursion = formula-based (“neither”) 2, 4, 16, 256, … -The pattern is that you’re squaring each previous term. a n+1 = (a n ) 2 *Note that this formula only applies to this particular example!!! Basic Formula: next term = #(1 st term) # ± #

16 Recursion = formula-based (“neither”) 2, 4, 16, 256, … -The pattern is that you’re squaring each previous term. a n+1 = (a n ) 2 *Note that this formula only applies to this particular example!!! Basic Formula: next term = #(1 st term) # ± # **The #’s are possibilities, but not requirements.

17 Recursion = formula-based (“neither”) 2, 4, 16, 256, … -The pattern is that you’re squaring each previous term. a n+1 = (a n ) 2 *Note that this formula only applies to this particular example!!! Basic Formula: next term = #(1 st term) # ± # **The #’s are possibilities, but not requirements. Exs. a n = 3a n-1 + 4 a n+1 = (a n ) 2 – 9 a n+2 = 2a n – a n+1

18 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1

19 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10

20 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1

21 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1

22 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41

23 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41

24 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1

25 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1 = 4(41) + 1

26 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1 = 4(41) + 1 = 165

27 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1 = 4(41) + 1 = 165 a 3 = 165

28 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1 = 4(41) + 1 = 165 a 3 = 165 a 3+1 = 4a 3 + 1

29 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1 = 4(41) + 1 = 165 a 3 = 165 a 3+1 = 4a 3 + 1 = 4(165) + 1

30 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1 = 4(41) + 1 = 165 a 3 = 165 a 3+1 = 4a 3 + 1 = 4(165) + 1 = 661

31 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1 = 4(41) + 1 = 165 a 3 = 165 a 3+1 = 4a 3 + 1 = 4(165) + 1 = 661 a 4 = 661

32 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1 = 4(41) + 1 = 165 a 3 = 165 a 3+1 = 4a 3 + 1 = 4(165) + 1 = 661 a 4 = 661 a 4+1 = 4a 4 + 1

33 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1 = 4(41) + 1 = 165 a 3 = 165 a 3+1 = 4a 3 + 1 = 4(165) + 1 = 661 a 4 = 661 a 4+1 = 4a 4 + 1 = 4(661) + 1

34 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1 = 4(41) + 1 = 165 a 3 = 165 a 3+1 = 4a 3 + 1 = 4(165) + 1 = 661 a 4 = 661 a 4+1 = 4a 4 + 1 = 4(661) + 1 = 2645

35 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1 = 4(41) + 1 = 165 a 3 = 165 a 3+1 = 4a 3 + 1 = 4(165) + 1 = 661 a 4 = 661 a 4+1 = 4a 4 + 1 = 4(661) + 1 = 2645 a 5 = 2645

36 Ex. 1 Find the first five terms of each sequence. a 1 = 10, a n+1 = 4a n + 1 a 1 = 10 a 1+1 = 4a 1 + 1 = 4(10) + 1 = 41 a 2 = 41 a 2+1 = 4a 2 + 1 = 4(41) + 1 = 165 a 3 = 165 a 3+1 = 4a 3 + 1 = 4(165) + 1 = 661 a 4 = 661 a 4+1 = 4a 4 + 1 = 4(661) + 1 = 2645 a 5 = 2645

37 Ex. 2 Write a recursive formula for the sequence. 16, 10, 7, 5.5, 4.75

38 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75

39 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference!

40 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference! a 1 = 16

41 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference! a 1 = 16 a 2 = 0.5(16) ± ? = 10

42 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference! a 1 = 16 a 2 = 0.5(16) ± ? = 10 8 ± ? = 10

43 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference! a 1 = 16 a 2 = 0.5(16) ± ? = 10 8 ± ? = 10 a 3 = 0.5(10) ± ? = 7

44 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference! a 1 = 16 a 2 = 0.5(16) ± ? = 10 8 ± ? = 10 a 3 = 0.5(10) ± ? = 7 5 ± ? = 7

45 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference! a 1 = 16 a 2 = 0.5(16) ± ? = 10 8 ± ? = 10 a 3 = 0.5(10) ± ? = 7 5 ± ? = 7 a 4 = 0.5(7) ± ? = 5.5

46 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference! a 1 = 16 a 2 = 0.5(16) ± ? = 10 8 ± ? = 10 a 3 = 0.5(10) ± ? = 7 5 ± ? = 7 a 4 = 0.5(7) ± ? = 5.5 3.5 ± ? = 5.5

47 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference! a 1 = 16 a 2 = 0.5(16) ± ? = 10 8 ± ? = 10 a 3 = 0.5(10) ± ? = 7 5 ± ? = 7 a 4 = 0.5(7) ± ? = 5.5 3.5 ± ? = 5.5 a 5 = 0.5(5.5) ± ? = 4.75

48 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference! a 1 = 16 a 2 = 0.5(16) ± ? = 10 8 ± ? = 10 a 3 = 0.5(10) ± ? = 7 5 ± ? = 7 a 4 = 0.5(7) ± ? = 5.5 3.5 ± ? = 5.5 a 5 = 0.5(5.5) ± ? = 4.75 2.75 ± ? = 4.75

49 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference! a 1 = 16 a 2 = 0.5(16) ± ? = 10 8 ± 2 = 10 a 3 = 0.5(10) ± ? = 7 5 ± 2 = 7 a 4 = 0.5(7) ± ? = 5.5 3.5 ± 2 = 5.5 a 5 = 0.5(5.5) ± ? = 4.75 2.75 ± 2 = 4.75

50 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference! a 1 = 16 a 2 = 0.5(16) ± ? = 10 8 ± 2 = 10 a 3 = 0.5(10) ± ? = 7 5 ± 2 = 7 a 4 = 0.5(7) ± ? = 5.5 3.5 ± 2 = 5.5 a 5 = 0.5(5.5) ± ? = 4.75 2.75 ± 2 = 4.75So a n+1 = 0.5a n + 2

51 Ex. 2 Write a recursive formula for the sequence. 16 10 7 5.5 4.75 -6 -3 -1.5 -0.75 *Each difference is half the previous difference! a 1 = 16 a 2 = 0.5(16) ± ? = 10 8 ± 2 = 10 a 3 = 0.5(10) ± ? = 7 5 ± 2 = 7 a 4 = 0.5(7) ± ? = 5.5 3.5 ± 2 = 5.5 a 5 = 0.5(5.5) ± ? = 4.75 2.75 ± 2 = 4.75So a n+1 = 0.5a n + 2

52 Iteration = proceeding terms of a recursive sequence

53 Iteration = proceeding terms of a recursive sequence Ex. 3 Find the first three iterates of the function for the given initial value. f(x) = -6x + 3x 0 = 8

54 Iteration = proceeding terms of a recursive sequence Ex. 3 Find the first three iterates of the function for the given initial value. f(x) = -6x + 3x 0 = 8 a 1 = -6(8) + 3

55 Iteration = proceeding terms of a recursive sequence Ex. 3 Find the first three iterates of the function for the given initial value. f(x) = -6x + 3x 0 = 8 a 1 = -6(8) + 3 = -45

56 Iteration = proceeding terms of a recursive sequence Ex. 3 Find the first three iterates of the function for the given initial value. f(x) = -6x + 3x 0 = 8 a 1 = -6(8) + 3 = -45 a 2 = -6(-45) + 3

57 Iteration = proceeding terms of a recursive sequence Ex. 3 Find the first three iterates of the function for the given initial value. f(x) = -6x + 3x 0 = 8 a 1 = -6(8) + 3 = -45 a 2 = -6(-45) + 3 = 273

58 Iteration = proceeding terms of a recursive sequence Ex. 3 Find the first three iterates of the function for the given initial value. f(x) = -6x + 3x 0 = 8 a 1 = -6(8) + 3 = -45 a 2 = -6(-45) + 3 = 273 a 3 = -6(273) + 3

59 Iteration = proceeding terms of a recursive sequence Ex. 3 Find the first three iterates of the function for the given initial value. f(x) = -6x + 3x 0 = 8 a 1 = -6(8) + 3 = -45 a 2 = -6(-45) + 3 = 273 a 3 = -6(273) + 3 = -1635

60 Iteration = proceeding terms of a recursive sequence Ex. 3 Find the first three iterates of the function for the given initial value. f(x) = -6x + 3x 0 = 8 a 1 = -6(8) + 3 = -45 a 2 = -6(-45) + 3 = 273 a 3 = -6(273) + 3 = -1635


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