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

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

11.5 = Recursion & Iteration

Arithmetic = adding (positive or negative)

3, 6, 9, 12, …

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

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

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)

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, …

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

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)

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”)

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, …

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.

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

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!!!

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) # ± #

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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!

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

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

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

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

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

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

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

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

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

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

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

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

Iteration = proceeding terms of a recursive sequence

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

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

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

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

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

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

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

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