Transcript 11.5 = Recursion & Iteration

11.5 = Recursion & Iteration
Arithmetic = adding (positive or negative)
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 nth term based on a1 and d.
an = a1+(n–1)d
Arithmetic = adding (positive or negative)
3, 6, 9, 12, …
d=3
*Formula for the nth term based on a1 and d.
an = a1+(n–1)d
Geometric = multiplying (#’s > 1 or #’s < 1)
Arithmetic = adding (positive or negative)
3, 6, 9, 12, …
d=3
*Formula for the nth term based on a1 and d.
an = a1+(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 nth term based on a1 and d.
an = a1+(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 nth term based on a1 and d.
an = a1+(n–1)d
Geometric = multiplying (#’s > 1 or #’s < 1)
2, 10, 50, 250, …
r=5
*Formula for the nth term based on a1 and r.
an = a1r(n – 1)
Arithmetic = adding (positive or negative)
3, 6, 9, 12, …
d=3
*Formula for the nth term based on a1 and d.
an = a1+(n–1)d
Geometric = multiplying (#’s > 1 or #’s < 1)
2, 10, 50, 250, …
r=5
*Formula for the nth term based on a1 and r.
an = a1r(n – 1)
Recursion = formula-based (“neither”)
Arithmetic = adding (positive or negative)
3, 6, 9, 12, …
d=3
*Formula for the nth term based on a1 and d. an = a1+(n–1)d
Geometric = multiplying (#’s > 1 or #’s < 1)
2, 10, 50, 250, …
r=5
*Formula for the nth term based on a1 and r.
Recursion = formula-based (“neither”)
2, 4, 16, 256, …
an = a1r(n – 1)
Arithmetic = adding (positive or negative)
3, 6, 9, 12, …
d=3
*Formula for the nth term based on a1 and d. an = a1+(n–1)d
Geometric = multiplying (#’s > 1 or #’s < 1)
2, 10, 50, 250, …
r=5
*Formula for the nth term based on a1 and r.
an = a1r(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 nth term based on a1 and d. an = a1+(n–1)d
Geometric = multiplying (#’s > 1 or #’s < 1)
2, 10, 50, 250, …
r=5
*Formula for the nth term based on a1 and r.
an = a1r(n – 1)
Recursion = formula-based (“neither”)
2, 4, 16, 256, …
-The pattern is that you’re squaring each previous term.
an+1 = (an)2
Recursion = formula-based (“neither”)
2, 4, 16, 256, …
-The pattern is that you’re squaring each previous term.
an+1 = (an)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.
an+1 = (an)2
*Note that this formula only applies to this particular
example!!!
Basic Formula: next term = #(1stterm)# ± #
Recursion = formula-based (“neither”)
2, 4, 16, 256, …
-The pattern is that you’re squaring each previous term.
an+1 = (an)2
*Note that this formula only applies to this particular
example!!!
Basic Formula: next term = #(1stterm)# ± #
**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.
an+1 = (an)2
*Note that this formula only applies to this particular
example!!!
Basic Formula: next term = #(1stterm)# ± #
**The #’s are possibilities, but not requirements.
Exs.
an = 3an-1 + 4
an+1 = (an)2 – 9
an+2 = 2an – an+1
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1 = 4(41) + 1
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1 = 4(41) + 1 = 165
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1 = 4(41) + 1 = 165
a3 = 165
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1 = 4(41) + 1 = 165
a3 = 165
a3+1 = 4a3 + 1
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1 = 4(41) + 1 = 165
a3 = 165
a3+1 = 4a3 + 1 = 4(165) + 1
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1 = 4(41) + 1 = 165
a3 = 165
a3+1 = 4a3 + 1 = 4(165) + 1 = 661
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1 = 4(41) + 1 = 165
a3 = 165
a3+1 = 4a3 + 1 = 4(165) + 1 = 661
a4 = 661
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1 = 4(41) + 1 = 165
a3 = 165
a3+1 = 4a3 + 1 = 4(165) + 1 = 661
a4 = 661
a4+1 = 4a4 + 1
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1 = 4(41) + 1 = 165
a3 = 165
a3+1 = 4a3 + 1 = 4(165) + 1 = 661
a4 = 661
a4+1 = 4a4 + 1 = 4(661) + 1
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1 = 4(41) + 1 = 165
a3 = 165
a3+1 = 4a3 + 1 = 4(165) + 1 = 661
a4 = 661
a4+1 = 4a4 + 1 = 4(661) + 1 = 2645
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1 = 4(41) + 1 = 165
a3 = 165
a3+1 = 4a3 + 1 = 4(165) + 1 = 661
a4 = 661
a4+1 = 4a4 + 1 = 4(661) + 1 = 2645
a5 = 2645
Ex. 1 Find the first five terms of each sequence.
a1 = 10, an+1 = 4an + 1
a1 = 10
a1+1 = 4a1 + 1 = 4(10) + 1 = 41
a2 = 41
a2+1 = 4a2 + 1 = 4(41) + 1 = 165
a3 = 165
a3+1 = 4a3 + 1 = 4(165) + 1 = 661
a4 = 661
a4+1 = 4a4 + 1 = 4(661) + 1 = 2645
a5 = 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!
a1 = 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!
a1 = 16
a2 = 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!
a1 = 16
a2 = 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!
a1 = 16
a2 = 0.5(16) ± ? = 10
8 ± ? = 10
a3 = 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!
a1 = 16
a2 = 0.5(16) ± ? = 10
8 ± ? = 10
a3 = 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!
a1 = 16
a2 = 0.5(16) ± ? = 10
8 ± ? = 10
a3 = 0.5(10) ± ? = 7
5±?=7
a4 = 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!
a1 = 16
a2 = 0.5(16) ± ? = 10
8 ± ? = 10
a3 = 0.5(10) ± ? = 7
5±?=7
a4 = 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!
a1 = 16
a2 = 0.5(16) ± ? = 10
8 ± ? = 10
a3 = 0.5(10) ± ? = 7
5±?=7
a4 = 0.5(7) ± ? = 5.5
3.5 ± ? = 5.5
a5 = 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!
a1 = 16
a2 = 0.5(16) ± ? = 10
8 ± ? = 10
a3 = 0.5(10) ± ? = 7
5±?=7
a4 = 0.5(7) ± ? = 5.5
3.5 ± ? = 5.5
a5 = 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!
a1 = 16
a2 = 0.5(16) ± ? = 10
8 ± 2 = 10
a3 = 0.5(10) ± ? = 7
5±2=7
a4 = 0.5(7) ± ? = 5.5
3.5 ± 2 = 5.5
a5 = 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!
a1 = 16
a2 = 0.5(16) ± ? = 10
8 ± 2 = 10
a3 = 0.5(10) ± ? = 7
5±2=7
a4 = 0.5(7) ± ? = 5.5
3.5 ± 2 = 5.5
a5 = 0.5(5.5) ± ? = 4.75
2.75 ± 2 = 4.75
So an+1 = 0.5an + 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!
a1 = 16
a2 = 0.5(16) ± ? = 10
8 ± 2 = 10
a3 = 0.5(10) ± ? = 7
5±2=7
a4 = 0.5(7) ± ? = 5.5
3.5 ± 2 = 5.5
a5 = 0.5(5.5) ± ? = 4.75
2.75 ± 2 = 4.75
So an+1 = 0.5an + 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 + 3
x0 = 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 + 3
x0 = 8
a1 = -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 + 3
x0 = 8
a1 = -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 + 3
x0 = 8
a1 = -6(8) + 3 = -45
a2 = -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 + 3
x0 = 8
a1 = -6(8) + 3 = -45
a2 = -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 + 3
x0 = 8
a1 = -6(8) + 3 = -45
a2 = -6(-45) + 3 = 273
a3 = -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 + 3
x0 = 8
a1 = -6(8) + 3 = -45
a2 = -6(-45) + 3 = 273
a3 = -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 + 3
x0 = 8
a1 = -6(8) + 3 = -45
a2 = -6(-45) + 3 = 273
a3 = -6(273) + 3 = -1635