graphing sequences

Graphing SequencesSec. 9.4b

But first, we start with…The fifth and ninth terms of an arithmetic sequence are –5 and–17, respectively. Find the first term and a recursive rule forthe n-th term.

The general explicit rule for an arithmetic sequence:

1 1na a n d Plug in the given data:

5 1 5 1 5a a d 9 1 9 1 17a a d

1 4 5a d

1 8 17a d A system to solve!!!

But first, we start with…

1 4 5a d

1 8 17a d Subtract theequations: 4 12d

3d

First Term:

1 7a Recursive Rule:

1 3n na a

The fifth and ninth terms of an arithmetic sequence are –5 and–17, respectively. Find the first term and a recursive rule forthe n-th term.

Another similar problem:The third and sixth terms of a geometric sequence are –75 and–9375, respectively. Find the first term, common ratio, and anexplicit rule for the n-th term.

The general explicit rule for a geometric sequence:1

1n

na a r Plug in the given data:

23 1 75a a r

56 1 9375a a r

21 75a r 5

1 9375a r Another system to solve – Woo Hoo!!!

Another similar problem:

21 75a r

51 9375a r

Divide theequations:

3 125r 5r

First Term:

1 3a

Explicit Rule:13 5nna

The third and sixth terms of a geometric sequence are –75 and–9375, respectively. Find the first term, common ratio, and anexplicit rule for the n-th term.

Graphing SequencesWe can represent sequences graphically in two ways: (a) Asa scatter plot, and (b) Using the sequence graphing mode.

Ex: Produce on a calculator a graph of the sequence ka2 1ka k in which

METHOD 1 (Scatter Plot)

The command seq(K, K, 1, 10) L puts the first 10 naturalnumbers in list one.

1

The command L – 1 L puts the corresponding terms ofthe sequence in list two.

12

2

Now graph the scatter plot in window [–1, 15] by [–10, 100]!!!

Graphing SequencesWe can represent sequences graphically in two ways: (a) Asa scatter plot, and (b) Using the sequence graphing mode.

Ex: Produce on a calculator a graph of the sequence ka2 1ka k in which

METHOD 2 (Sequence Mode)

Put your calculator into Seq mode, then enter

into the Y = list, with nMin = 1 and nMax = 10.

Now graph the sequence in the same window!!!

2 1u n n

Graphing SequencesUsing a graphing calculator, generate the specific terms of thefollowing sequences:

1. (Explicit) 3 5ka k for k = 1, 2, 3,…

First Command: 0 K

Second Command: K + 1 K:3K – 5

Then press ENTER repeatedly!!!

Graphing SequencesUsing a graphing calculator, generate the specific terms of thefollowing sequences:

2. (Recursive) 1 2,a

First Command: –2

Second Command: ANS + 3

Then press ENTER repeatedly!!!

1 3n na a for n = 2, 3, 4,…

A Famous Recursive Sequence

The Fibonacci SequenceThe Fibonacci SequenceThe Fibonacci sequence can be defined recursively by

1 1,a 2 1,a 2 1n n na a a for all positive integers 3n

First Command: 0 A:1 B

Second Command: A + B C:A B:C A

To get this sequence on your calculator:

Then press ENTER repeatedly!!!

Sums of Sequences

Definition:Summation Notation

In summation notation, the sum of the terms ofthe sequence is denoted 1 2, , , na a a

1

n

kk

a

which is read “the sum of from k = 1 to n.”kaThe variable k is called the index of summation.

Our First “Exploration”Determine the number represented by each of the followingexpressions.

1.

5

1

3k

k 45

2.

82

5k

k 174

3. 12

0

cosn

n 1

4. 1

sinn

n

0

5.1

3

10kk

1

3

Gauss’s InsightFirst, read the second-to-last paragraph on page 739…

Your challenge is to find the sum of the natural numbersfrom 1 to 100 without a calculator.

1. Write the sum

1 2 3 98 99 100 2. Underneath this sum, write the sum

100 99 98 3 2 1 3. Add the numbers in each vertical column. You should get the same identical sum 100 times – what is it?

101 101 101 101 101 101

Gauss’s InsightFirst, read the second-to-last paragraph on page 739…

Your challenge is to find the sum of the natural numbersfrom 1 to 100 without a calculator.

4. What is the sum of the 100 identical numbers referred to in part 3? 100 101 10,1005. Explain why half the answer in part 4 is the answer to the challenge. Can you find it without a calculator?

The sum in part 4 involves two copies of the same progression,so it doubles the sum of the progression. The answer thatGauss gave was 5050.

1 1 1 11

2 1n

kk

a a a d a d a n d

Two ways to write an arithmetic sum:

Let’s Prove the General TheoremLet’s Prove the General Theorem

1

2 1n

k n n n nk

a a a d a d a n d

1 1 11

2n

k n n nk

a a a a a a a

Sum these two expressions vertically:

11

2n

k nk

a n a a

1

1 2

nn

kk

a aa n

1 1a n d Substitute

Let’s Prove the General TheoremLet’s Prove the General Theorem

na1

1 2

nn

kk

a aa n

for :

11

2 12

n

kk

na a n d

Theorem:Sum of a Finite Arithmetic Sequence

Let be a finite arithmeticsequence with common difference d. Then thesum of the terms of the sequence is

1 2 3, , , , na a a a

1 21

n

k nk

a a a a

1

2na a

n

12 1

2

na n d

Applying our New TheoremA corner section of a stadium has 8 seats along the frontrow. Each successive row has two more seats than therow preceding it. If the top row has 24 seats, how manyseats are in the entire section?

The number of seats in the rows forms an arithmetic sequence:

1 8a 24na 2d Let’s solve for n: 1 1na a n d

24 8 1 2n 9n

Applying our New TheoremA corner section of a stadium has 8 seats along the frontrow. Each successive row has two more seats than therow preceding it. If the top row has 24 seats, how manyseats are in the entire section?

Use the new formula:

9

8 249

2S

144 seats

To find with your calculator:

sum(seq(8 + (N – 1)2, N, 1, 9) = 144 seats

2 11 1 1 1

1

nn

kk

a a a r a r a r

The general notation:

Finding the Sum of a Geometric SequenceFinding the Sum of a Geometric Sequence

2 11 1 1 1

1

nn n

kk

r a a r a r a r a r

Multiply both sides by r :

1 11 1

n nn

k kk k

a r a a a r

Subtract the two summations:

Finding the Sum of a Geometric SequenceFinding the Sum of a Geometric Sequence

1 11 1

n nn

k kk k

a r a a a r

Factor out common terms:

11

1 1n

nk

k

a r a r

Solve for the summation:

1

1

1

1

nn

kk

a ra

r

Theorem:Sum of a Finite Geometric Sequence

Let be a finite geometricsequence with common ratio r = 1.

1 2 3, , , , na a a a

1 21

n

k nk

a a a a

1 1

1

na r

r

Then the sum of the terms in the sequence is

Applying our Second New Theorem

Find the sum of the geometric sequence given below.

104, 4 3,4 9, 4 27, , 4 1 3 Identify terms: 1 4a 1 3r 11n

The new formula:111

1

14

3

n

n

11

11

4 1 1 3

1 1 3S

3.000016935

Applying our Second New Theorem

Find the sum of the geometric sequence given below.

104, 4 3,4 9, 4 27, , 4 1 3 111

1

14

3

n

n

3.000016935

Support with a calculator:

sum(seq(4(–1/3)^(N – 1), N, 1, 11) = 3.000016935

Infinite Series

First, let’s return to an example from last class…

We found this sum:111

1

14 3.000016935

3

n

n

Now, we explore what happens when we change the “11” toINFINITY!!! 1

1

1lim 4

3

kn

nk

First, let’s return to an example from last class…

1

1

1lim 4

3

kn

nk

4 1 1 3lim

1 1 3

n

n

4 1 0

4 3

3

This is our first example of an infinite series, which isan expression where an infinite number of terms areadded together………(duh?)

Definition: Infinite Series

1 21

n nn

a a a a

An infinite series is an expression of the form

Note: An infinite series is not a true sum…

Definition: Infinite Series

1 21

lim limn

k nn nk

a a a a S

Sometimes a sequence of partial sums (all of which are truesums) approaches a finite limit S:

In this case we say that the series converges to S, and wedefine S as the sum of the infinite series. In sigma notation,

1 1

limn

k knk k

a a S

If the limit of the partial sums does not exist, then the seriesdiverges and has no sum.

Guided PracticeFor each of the following series, find the first five terms in thesequence of partial sums. Which of the series appear toconverge?

1) 0.1 + 0.01 + 0.001 + 0.0001 + …

First five partial sums: {0.1, 0.11, 0.111, 0.1111, 0.11111}

These appear to approach a limit of 1/9 So the seriesconverges to a sum of 1/9!!!

2) 10 + 20 + 30 + 40 + …

First five partial sums: {10, 30, 60, 100, 150}

These numbers approach no limit The series diverges

Guided PracticeFor each of the following series, find the first five terms in thesequence of partial sums. Which of the series appear toconverge?

3) 1 – 1 + 1 – 1 + …

First five partial sums: {1, 0, 1, 0, 1}

These numbers oscillate and do not approach a limit The series diverges

NOTE: You cannot apply certain rules (such as theassociate property of addition) to infinite series!!!

Theorem:Sum of an Infinite Geometric Series

A geometric series1

1

n

k

a r

converges if and only if .1r

If it does converge, the sum is .1

aS

r

Guided PracticeDetermine whether the given series converges. If it converges,give the sum.

1) 1

1

3 0.75k

k

The series converges

0.75 1r

First term:

03 0.75 3a

Sum:3

121 1 0.75

aS

r

Guided PracticeDetermine whether the given series converges. If it converges,give the sum.

2)

0

4

5

n

n

The series converges

4 5 1r

First term:

04 5 1 Sum:

1 5

1 4 5 9S

Guided PracticeDetermine whether the given series converges. If it converges,give the sum.

3)

1 2

n

n

The series diverges

2 1r

Guided PracticeDetermine whether the given series converges. If it converges,give the sum.

4)1 1 1

12 4 8

The series converges

1 2 1r

First term:

1a Sum:

12

1 1 2S

Guided PracticeExpress the given decimal in fraction form.

0.234 0.234234234...We can write this number as a sum:

0.234 0.000234 0.000000234 This is a convergent geometric series with a = 0.234 andr = 0.001. The sum is:

0.234

1 1 0.001

aS

r

0.234

0.999

234

999

26

111

Guided PracticeThe table below shows the December balance in a simpleinterest savings account each year from 1996 to 2000.

Year

Balance

1996

$18,000

1997

$20,016

1998

$22,032

1999

$24,048

2000

$26,064

(a) The balances form an arithmetic sequence. What is d ?

Find the difference between any two balances d = 2016

(b) Write a formula for the balance in the account n years after December 1996.

$18,000 $2016n

Guided Practice

(c) Find the sum of the December balances from 1996 to 2006, inclusive.

Sum of the eleven terms of the arithmetic sequence:

11

112 18000 10 2016

2S $308,880

The table below shows the December balance in a simpleinterest savings account each year from 1996 to 2000.

Year

Balance

1996

$18,000

1997

$20,016

1998

$22,032

1999

$24,048

2000

$26,064