Section 3.1: Sequences and Series

Similar documents
Section 3.1: Sequences and Series

Subtracting Fractions

CS201 Discussion 10 DRAWTREE + TRIES

COMP 423 lecture 11 Jan. 28, 2008

Ray surface intersections

Unit #9 : Definite Integral Properties, Fundamental Theorem of Calculus

SIMPLIFYING ALGEBRA PASSPORT.

Integration. October 25, 2016

Introduction to Integration

10.5 Graphing Quadratic Functions

3.5.1 Single slit diffraction

3.5.1 Single slit diffraction

Unit 5 Vocabulary. A function is a special relationship where each input has a single output.

Integration. September 28, 2017

Section 10.4 Hyperbolas

12-B FRACTIONS AND DECIMALS

What do all those bits mean now? Number Systems and Arithmetic. Introduction to Binary Numbers. Questions About Numbers

50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines. can be calculated by the following formula:

Section 3.2: Arithmetic Sequences and Series

Engineer To Engineer Note

such that the S i cover S, or equivalently S

In the last lecture, we discussed how valid tokens may be specified by regular expressions.

a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1

MATH 25 CLASS 5 NOTES, SEP

SOME EXAMPLES OF SUBDIVISION OF SMALL CATEGORIES

Lists in Lisp and Scheme

CSCI 3130: Formal Languages and Automata Theory Lecture 12 The Chinese University of Hong Kong, Fall 2011

MIPS I/O and Interrupt

6.2 Volumes of Revolution: The Disk Method

Answer Key Lesson 6: Workshop: Angles and Lines

Matrices and Systems of Equations

A Tautology Checker loosely related to Stålmarck s Algorithm by Martin Richards

CS 241 Week 4 Tutorial Solutions

If f(x, y) is a surface that lies above r(t), we can think about the area between the surface and the curve.

EECS 281: Homework #4 Due: Thursday, October 7, 2004

Simplifying Algebra. Simplifying Algebra. Curriculum Ready.

ECE 468/573 Midterm 1 September 28, 2012

Definition of Regular Expression

9.1 apply the distance and midpoint formulas

INTRODUCTION TO SIMPLICIAL COMPLEXES

Representation of Numbers. Number Representation. Representation of Numbers. 32-bit Unsigned Integers 3/24/2014. Fixed point Integer Representation

Rational Numbers---Adding Fractions With Like Denominators.

The Fundamental Theorem of Calculus

Improper Integrals. October 4, 2017

2014 Haskell January Test Regular Expressions and Finite Automata

MA1008. Calculus and Linear Algebra for Engineers. Course Notes for Section B. Stephen Wills. Department of Mathematics. University College Cork

Physics 152. Diffraction. Difrraction Gratings. Announcements. Friday, February 2, 2007

ZZ - Advanced Math Review 2017

CS280 HW1 Solution Set Spring2002. now, we need to get rid of the n term. We know that:

COMBINATORIAL PATTERN MATCHING

binary trees, expression trees

Dr. D.M. Akbar Hussain

Study Guide for Exam 3

Dynamic Programming. Andreas Klappenecker. [partially based on slides by Prof. Welch] Monday, September 24, 2012

Fig.25: the Role of LEX

MATH 2530: WORKSHEET 7. x 2 y dz dy dx =

Fig.1. Let a source of monochromatic light be incident on a slit of finite width a, as shown in Fig. 1.

2 Computing all Intersections of a Set of Segments Line Segment Intersection

What are suffix trees?

Misrepresentation of Preferences

Stained Glass Design. Teaching Goals:

Mid-term exam. Scores. Fall term 2012 KAIST EE209 Programming Structures for EE. Thursday Oct 25, Student's name: Student ID:

1 Quad-Edge Construction Operators

Math 142, Exam 1 Information.

a(e, x) = x. Diagrammatically, this is encoded as the following commutative diagrams / X

COMPUTATIONAL INTELLIGENCE

Geometric transformations

CS321 Languages and Compiler Design I. Winter 2012 Lecture 5

CS481: Bioinformatics Algorithms

Lecture 7: Integration Techniques

Very sad code. Abstraction, List, & Cons. CS61A Lecture 7. Happier Code. Goals. Constructors. Constructors 6/29/2011. Selectors.

CS311H: Discrete Mathematics. Graph Theory IV. A Non-planar Graph. Regions of a Planar Graph. Euler s Formula. Instructor: Işıl Dillig

Before We Begin. Introduction to Spatial Domain Filtering. Introduction to Digital Image Processing. Overview (1): Administrative Details (1):

Homework. Context Free Languages III. Languages. Plan for today. Context Free Languages. CFLs and Regular Languages. Homework #5 (due 10/22)

1. SEQUENCES INVOLVING EXPONENTIAL GROWTH (GEOMETRIC SEQUENCES)

Angle Properties in Polygons. Part 1 Interior Angles

MTH 146 Conics Supplement

Math 464 Fall 2012 Notes on Marginal and Conditional Densities October 18, 2012

P(r)dr = probability of generating a random number in the interval dr near r. For this probability idea to make sense we must have

ASTs, Regex, Parsing, and Pretty Printing

TO REGULAR EXPRESSIONS

Today. Search Problems. Uninformed Search Methods. Depth-First Search Breadth-First Search Uniform-Cost Search

Questions About Numbers. Number Systems and Arithmetic. Introduction to Binary Numbers. Negative Numbers?

. (b) Evaluate the sum given by. Exercise #1: A sequence is defined by the equation a n 2n

9 4. CISC - Curriculum & Instruction Steering Committee. California County Superintendents Educational Services Association

What do all those bits mean now? Number Systems and Arithmetic. Introduction to Binary Numbers. Questions About Numbers

Fall 2017 Midterm Exam 1 October 19, You may not use any books, notes, or electronic devices during this exam.

ΕΠΛ323 - Θεωρία και Πρακτική Μεταγλωττιστών

4452 Mathematical Modeling Lecture 4: Lagrange Multipliers

6.3 Volumes. Just as area is always positive, so is volume and our attitudes towards finding it.

Graphs with at most two trees in a forest building process

COMPUTER SCIENCE 123. Foundations of Computer Science. 6. Tuples

Solutions to Math 41 Final Exam December 12, 2011

Creating Flexible Interfaces. Friday, 24 April 2015

If you are at the university, either physically or via the VPN, you can download the chapters of this book as PDFs.

An Efficient Divide and Conquer Algorithm for Exact Hazard Free Logic Minimization

Intermediate Information Structures

Quiz2 45mins. Personal Number: Problem 1. (20pts) Here is an Table of Perl Regular Ex

CS143 Handout 07 Summer 2011 June 24 th, 2011 Written Set 1: Lexical Analysis

Fall 2018 Midterm 1 October 11, ˆ You may not ask questions about the exam except for language clarifications.

Transcription:

Section.: Sequences d Series Sequences Let s strt out with the definition of sequence: sequence: ordered list of numbers, often with definite pttern Recll tht in set, order doesn t mtter so this is one wy tht sequence differs from set. Also, repetition doesn t mtter in set but does in sequence: if number is repeted in sequence, it isn t considered duplicte d cnot be removed without chging the sequence. Sequences, like sets, c be finite or infinite. If sequence is finite, then either the lst term or the number of terms must be specified so tht it s cler where the sequence stops. Which of the following sequences re infinite? Which re finite? ) 7,, 5, 9, b), 4, 9, 6, 5, 6, 00 c) 4,,,,,,,... 4 8 6 56 b) d c) re finite, becuse their lst terms re given. ), however, goes on forever so is infinite. To begin with, let s exmine some sequences in detil. We will begin by looking for ptterns in ech sequence. Wht is the pttern for the following sequences? Wht is the next term for ech sequence? ) 7,, 5, 9, b), 4, 9, 6, 5, 6, 00 c) 4,,,,,,,... 4 8 6 56

d), 6,, 4, e), 6, 5, 4, ) The pttern is tht you dd 4 to the previous term to get the next term. The next term is then. b) The pttern is tht if you sy tht is the first term d 4 is the second term, then n will be the nth term. So the next term fter 6 is 49. c) The pttern is to divide ech term by two (or multiply by hlf) to get the next term. So the term fter /6 will be /. d) The pttern is to multiply ech term by to get the next term. The next term is then 48. e) The pttern is to subtrct 9 from the previous term, so the next one is. Note tht in this previous exmple, the lst two sequences looked very similr for three of their first four terms. However, the third term is different so the pttern for the two sequences is not the sme d subsequent terms could look very different. Nottion We will use the nottion n for the nth term in sequence, where n is the index. For exmple, the first term would then be, the second term, d so on. The index n, then, is positive integer (or nturl number, if you like). Other nottions my strt their counting with o being the first term. For the purposes of this course, we ll stick to strting t n =. Defining Sequence There re three wys to define sequence: ) List ll of the terms, or enough terms to set up the pttern. If the sequence is finite, then either the lst term or the number of terms must be given. ) Give generl formul for the nth term. ) Give recursive formul for the nth term. Let s look t exmples of ech type. For instce, the sequences 7,, 5, 9, d, 4, 9, 6, 5, 6, 00 re exmples of sequences defined by listing the terms.

Generl Formul A generl formul is formul tht gives n s function of n only. Let s look t the following exmples to exmine some sequences defined in this wy. Give the first four terms of the sequence given by the generl formul 4n. n = 4n +, so = 4 + = 7 = 4 + = = 4 + = 5 4 = 4 4 + = 9 The first four terms re then 7,, 5, d 9. This is the sme sequence tht ws given s prt ) in the first two exmples of this section. n Give ll terms of the sequence given by the formul n for n5. This is finite sequence, since restrictions hve been plced on the vlues of n. The terms re then: 4 5 9 7 4 8 5 4

You c see from the previous exmples tht the generl formul llows you to clculte n for y vlue of n. The very useful thing bout the generl formul is tht you don t need to know the previous term to clculte prticulr term. For instce, if you wt to know the 50 th term of the sequence 7,, 5, 9,, you c determine tht the pttern is to dd 4 to the previous term to get the next term. However, to get the 50 th term, you d hve to clculte the 49 th first, but the 49 th requires the 48 th, d so on. But if you insted use the expression 4n, which gives the sme sequence, then the 50 th term is just 50 4n 450 0 d there s no need to clculte preceding terms. Hdy! Recursive Definition A recursive formul gives formul for the next term in terms of the previous one. For exmple, in our old friend 7,, 5, 9,, the next term is found by dding 4 to the previous term: 4. However, tht s not enough informtion to uniquely define the series becuse you don t know where to strt. A complete definition must include the first term lso. Therefore, the recursive definition for our old friend 7,, 5, 9, would be 7 Recursive definitions, then, must specify the first term or terms d lso the rule which llows you to clculte the next term from the previous term or terms. 4 Clculte the first four terms of the sequence given by 0 The first term is lredy given,. Then 4 0 04 4 0 0 79 79 0 78 0 694

Give recursive formul for the sequence, 6, 8, 54, The pttern is tht the next term equls the previous term times three. Therefore, Recursive definitions hve the sme drwbck tht we ve seen before: if we wt to know the 00 th term, we need to clculte the 99 th first, d so on. Only the generl formul llows us to clculte ech term directly without knowing the previous one. Fiboncci sequence The Fiboncci sequence is the most fmous exmple of recursive sequence:,,,, 5, 8,, The pttern c be quite difficult to spot you get the next term from the sum of the two previous terms. The recursive formul for this sequence is therefore n n n Here, the first two terms must be given to strt off with so tht you re then ble to clculte the third term from the previous two. Series A series is the sum of sequence, which c be finite or infinite. Here re two exmples: ) 6 + 0 + 4 + 8 + 64 b)... 9 7 Nottion The sum of the first n terms of sequence is denoted by S n (lso sometimes clled the nth prtil sum). If the series is finite, it could be the sum of ll of the terms. S is how we write the sum of infinite series, like the second exmple bove.

For the series 6 + 0 + 4 + 8 + 64, clculte S d S 4. S = 6 + 0 + 4 = 60 S 4 = 6 + 0 + 4 + 8 = 88 However, it s esy to see tht this method becomes very cumbersome for lrge vlues of n. We ll develop some more efficient methods in the next two sections. Sigm nottion It s esy to tke sequence in list form d trsform it into series by chging ll of the comms to + signs. However, wht if you re given the generl formul insted? For exmple, let s tke 7,, 5, 9, which we know to be 4n. Since the generl form is so useful for finding n when n is lrge, it would be nice if we could retin tht informtion while writing our sum. To do so, we ll introduce new nottion clled sigm nottion. It uses the Greek letter sigm (the uppercse one): Σ, which is commonly used to me sum of. Let s look t exmple of sigm nottion d discuss wht ll of the prts me. Consider the following 5 i (4i ) The letter i is index here, d it runs from the vlue given t the bottom of the sigm to the number t the top of the sigm in steps of. Here, i runs from to 5. We re summing, then, the vlue of 4i + for ech vlue of i s it runs from to 5: 5 i Let s look t more exmples. i i i i4 i5 (4i ) 4 4 4 44 45 7 5 9 75

Clculte i (i 5) i i i i (i 5) 5 5 5 Clculte 8 j 9 j6 9 j6 j 6 j 7 j 8 j 9 8 j 86 87 88 89 4 0 6 Clculte 6 k 6 j k k k 4 k 5 k 6 5 The tricky thing bout the lst one is deciding how my terms there re. You my, s is shown bove, write out ll of the possible vlues of the index. Or you my use the following nifty rule:

# terms = lst first + For instce, the lst exmple hd the index running from to 6. The number of terms, then, for tht series is 6 + = 5. Write the following series in sigm nottion: 4 9 6 5...00 Let s pick our index first. If we wt to be lzy, insted of strting our index t, we could strt t d our series would be 0 k Other cceptble swers would involve chging our strting point for the index 9 to give j or i or even l 55 j fvourite number. 8 i0 k 65 l57 if 57 hppens to be your Write the following sequence in sigm nottion:... 4 5 6 j To write infinite series in sigm nottion, you just replce the finl vlue of the index with. j