chapter 11 sec 6 recursion and special sequences

Chapter 11 Sec 6Chapter 11 Sec 6

Recursion and Special Recursion and Special SequencesSequences

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Algebra 2 Chapter 11 Sections 6 & 7

Fibonacci Sequence in NatureFibonacci Sequence in Nature• A shoot on a sneezewort plant must A shoot on a sneezewort plant must

grow for two months before it is strong grow for two months before it is strong enough to put out another shoot. After enough to put out another shoot. After that it puts out at least one shoot every that it puts out at least one shoot every month.month.

• Notice that the sequence 1, 1, 2, 3, 5, Notice that the sequence 1, 1, 2, 3, 5, 8, 13, ..has a pattern. Each term in the sequence is the sum of the 8, 13, ..has a pattern. Each term in the sequence is the sum of the two previous terms. This is called a two previous terms. This is called a Fibonacci Sequence.Fibonacci Sequence.

• First term First term aa11 11

• Second term Second term aa22 11

• Third term Third term aa33 aa11 + + aa22 1 + 1 = 21 + 1 = 2

• Nth term Nth term aann aan – n – 22 + a + an – n – 11

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Recursive FormulaRecursive Formula• The formula The formula aann = a = an – n – 22 + a + an – n – 11 is an example of a is an example of a

recursive formula.recursive formula. This means that each term is This means that each term is formulated from one or more previous terms. To be able formulated from one or more previous terms. To be able to use a recursive formula, you must be given the to use a recursive formula, you must be given the value(s) of the first term(s) so that you can start the value(s) of the first term(s) so that you can start the sequence and then use the formula to generate the rest of sequence and then use the formula to generate the rest of the terms. the terms.

• Ex. Find the first five terms of the sequence in which Ex. Find the first five terms of the sequence in which aa11 = 4 and = 4 and

aan + n + 11 = = 33aann – 2, – 2, n n >> 1. 1.

4, 10, 28, 82, 244

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Example 1Example 1Find the first five terms of the sequence in which Find the first five terms of the sequence in which

aa11 = = 5 and 5 and aan + n + 11 = = 22aann + 7, + 7, nn >> 1. 1.

5, 17, 41, 89, 185

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IterationIterationIteration is the process of composing a function Iteration is the process of composing a function

with itself. with itself.

For example start with For example start with xx00 = 3 and = 3 and ff((xx) = 2) = 2x + x + 22

Put Put xx00 in in ff((xx00) to get ) to get xx11, so , so xx1 1 = f= f((xx00) =2(3) + 2 = 8) =2(3) + 2 = 8

xx2 2 = f= f((xx11)) = f= f(8) = 18;(8) = 18;

xx3 3 = f= f((xx22)) = f= f(18) = 38; and …(18) = 38; and …

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Example 2Example 2

Find the first three iterates Find the first three iterates xx11, x, x2 2 , x, x3 3 of the of the

function function ff((xx)) = = 33x – x – 1 for an initial value 1 for an initial value

of of xx00 = 5. = 5.

xx11 = f = f((xx00)) = = 3(5)3(5) – – 1 = 14 1 = 14

xx22 = f = f((xx11)) = = 3(14)3(14) – – 1 = 411 = 41

xx33 = f = f((xx22)) = = 3(41)3(41) – – 1 = 122 1 = 122

Chapter 11 Sec 7Chapter 11 Sec 7

The Binomial The Binomial TheoremTheorem

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Pascal’s TrianglePascal’s TriangleUse this to find the coefficients of powers on binomials. Use this to find the coefficients of powers on binomials.

Before we start here are the ground rules for any Before we start here are the ground rules for any expansion of the form (expansion of the form (a + ba + b))nn..

1. There are 1. There are nn + 1 terms. + 1 terms.

2. The exponent 2. The exponent nn of ( of (a + ba + b))nn is the exponent of is the exponent of a a in the in the first term and of first term and of b b in the last term.in the last term.

3. In successive terms, the exponent of 3. In successive terms, the exponent of a a decreases by decreases by one and one and b b increases by one.increases by one.

4. The sum of the exponents in each term is 4. The sum of the exponents in each term is nn..

5. The coefficients are symmetric.5. The coefficients are symmetric.

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Pascal’s TrianglePascal’s Triangle

11

1 11 1

1 2 11 2 1

1 3 3 11 3 3 1

1 4 6 4 11 4 6 4 1

1 5 10 10 5 11 5 10 10 5 1

1 6 15 20 15 6 11 6 15 20 15 6 1

1 7 21 35 35 21 7 11 7 21 35 35 21 7 1

((a + ba + b))00

((a + ba + b))11

((a + ba + b))22

((a + ba + b))33

((a + ba + b))44

((a + ba + b))55

((a + ba + b))66

((a + ba + b))77

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Example 1Example 1

Expand (Expand (p + qp + q))55. .

((p + qp + q))5 5 = = pp55 + 5 + 5pp44q q + 10 + 10pp33qq22 + 10 + 10pp22 qq33 + 5 + 5pqpq44 + + qq55

1 5 10 10 5 11 5 10 10 5 1((a + ba + b))55

1 4 6 4 11 4 6 4 1((a + ba + b))44Expand (Expand (x – x – 3)3)44. .

((x – x – 3)3)44 = = xx44 + + 44xx33(–3)(–3) + + 66xx22(–3)(–3)22 + + 44xx (–3)(–3)33 + + (–3)(–3)44

= = xx44 – 12 – 12xx33 + 54 + 54xx2 2 – 108 – 108xx + 81 + 81

1 4 6 4 11 4 6 4 1((a + ba + b))44Expand (3Expand (3x + x + 2)2)44. .

(3(3x + x + 2)2)44 = = (3(3xx))44 + + 44(3(3xx))33(2)(2) + + 66(3(3xx))22(2)(2)22 + + 44(3(3xx)) (2)(2)33 + + (2)(2)44

= 81= 81xx44 + 216 + 216xx33 + 216 + 216xx2 2 + 96 + 96xx + 16 + 16

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The Binomial TheoremThe Binomial Theorem

Notice the rules still apply.Notice the rules still apply.

1. There are 1. There are nn + 1 terms. + 1 terms.

2. The exponent 2. The exponent nn of ( of (a + ba + b))nn is the exponent of is the exponent of a a in the first term in the first term and of and of b b in the last term.in the last term.

3. In successive terms, the exponent of 3. In successive terms, the exponent of a a decreases by one and decreases by one and b b increases by one.increases by one.

4.4. The sum of the exponents in each term is The sum of the exponents in each term is nn..

5.5. The coefficients are symmetric.The coefficients are symmetric.

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Using the B-TheoremUsing the B-Theorem

Expand (Expand (t – st – s))55..

The expansion will have six terms. Use The expansion will have six terms. Use

to find the first 3 terms, then use symmetry for the rest.to find the first 3 terms, then use symmetry for the rest.

((t – st – s))55 = =11tt55(–(–ss))00+5+5tt44(–(–ss))11+10+10tt33(–(–ss))22 +10 +10tt22 (–(–ss))33+5+5tt(–(–ss))44 + + (–(–ss))55

= = tt55 – 5 – 5tt44ss11 + 10+ 10tt33ss22 – 10 – 10tt22ss33 + 5 + 5tsts44 – – ss55

10 5, 1,or 21

45,

1

5,1

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FactorialsFactorialsThe factors in the coefficients of the binomial theorem The factors in the coefficients of the binomial theorem

involve special products called involve special products called factorials. factorials. For example the For example the

product is written 4! And is read product is written 4! And is read 4 factorial. 4 factorial. In In

general, if general, if nn is a positive integer, then is a positive integer, then

By definition 0! = 1.By definition 0! = 1.

Ex. Evaluate Ex. Evaluate

1234

12...321! nnnnn

!5!3

!8

12345123

12345678

!5!3

!8

!5678!8 :Note

56or 123

678

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Binomial Theorem in Factorial Binomial Theorem in Factorial FormForm

Sometimes you need to know a particular term of a Sometimes you need to know a particular term of a

binomial expansion. Note that when the Binomial binomial expansion. Note that when the Binomial

Theorem is written in Sigma notation, Theorem is written in Sigma notation, kk = 0 for the first = 0 for the first

term, term, kk = 1 for the second term, and so on. In general, the = 1 for the second term, and so on. In general, the

value of value of kk is always one less than the term you are finding. is always one less than the term you are finding.

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Example 4Example 4

Expand (3Expand (3xx – – yy))44 . .

Now replace Now replace kk with 0, 1, 2, 3, & 4 to get the five terms. with 0, 1, 2, 3, & 4 to get the five terms.

n

k

kkn bakkn

n

0 !!

!

4

0

44 3!!4

!43

k

kk yxkk

yx

432234 3!3!1

!43

!2!2

!43

!1!3

!43 yyxyxyxx

432234 349627481 yxyyxyxx

432234 125410881 yxyyxyxx

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Example 5Example 5Find the fourth term in the expansion of (Find the fourth term in the expansion of (a + a + 33bb))44..Use the Binomial Theorem to write the expansion in Use the Binomial Theorem to write the expansion in sigma notation.sigma notation.

n = n = 4, 4, kk = 3 = 3

((a + a + 33bb))44 = = kk

k

bakk

3!!4

!4 44

0

334 3!3!34

!4ba

33 108274 abba

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Daily AssignmentDaily Assignment

• Chapter 11 Sections 6 & 7 Chapter 11 Sections 6 & 7 • Study Guide (SG)Study Guide (SG)

• Pg 151 – 152 OddPg 151 – 152 Odd

• Pg 153 – 154 #1 – 5 on each pagePg 153 – 154 #1 – 5 on each page

chapter 11 sec 6 recursion and special sequences

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