# MATH 105: PRACTICE PROBLEMS FOR SERIES: SPRING Problems

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 MATH 05: PRACTICE PROBLEMS FOR SERIES: SPRING 20 INSTRUCTOR: STEVEN MILLER Sequences and Series.. Problems Question : Let a n =. Does the series +n+n 2 n= a n converge or diverge? Prove your Question 2: Let a n = n +2 n +( 2) n. Does the series n= a n converge or diverge? Prove your Question 3: Let a n = n 6 n. Does the series Question : Let a n = n n!. Does the series Question 5: Let a n = n2n 3 n. Does the series Question 6: Let a n = n 3 /2 n/2. Does the series n= a n converge or diverge? Prove your.2. Taylor Series (one variable). Question.. Find the first five terms of the Taylor series for f(x) = x 8 + x + 3 at x = 0. Question.2. Find the first three terms of the Taylor series for f(x) = x 8 + x + 3 at x =. Question.3. Find the first three terms of the Taylor series for f(x) = cos(5x) at x = 0. Question.. Find the first five terms of the Taylor series for f(x) = cos 3 (5x) at x = 0. Question.5. Find the first two terms of the Taylor series for f(x) = e x at x = 0. Question.6. Find the first six terms of the Taylor series for f(x) = e x8 = exp(x 8 ) at x = 0. Question.7. Find the first four terms of the Taylor series for f(x) = 2π e x2 /2 = exp( x 2 /2)/ 2π at x = 0. Question.8. Find the first three terms of the Taylor series for f(x) = x at x = 3. Question.9. Find the first three terms of the Taylor series for f(x) = ( + x) /3 at x = 2. Date: May 3, 20.

2 2 INSTRUCTOR: STEVEN MILLER Question.0. Find the first three terms of the Taylor series for f(x) = x log x at x =. Question.. Find the first three terms of the Taylor series for f(x) = log(+x) at x = 0. Question.2. Find the first three terms of the Taylor series for f(x) = log( x) at x =. Question.3. Find the first two terms of the Taylor series for f(x) = log(( x) e x ) = log(( x) exp(x)) at x = 0. Question.. Find the first three terms of the Taylor series for f(x) = cos(x) log( + x) at x = 0. Question.5. Find the first two terms of the Taylor series for f(x) = log(+2x) at x = 0.

3 PRACTICE PROBLEMS Sequences and Series. 2. Solutions Question : Let a n =. Does the series +n+n 2 n= a n converge or diverge? Prove your Solution: This series converges. Notice that for all n, +n+n 2 > n 2, so /(+n+n 2 ) < /n 2, meaning that each term of this series is strictly less than /n 2. Since n= /n2 converges, this series converges as well. Question 2: Let a n = n +2 n +( 2) n. Does the series n= a n converge or diverge? Prove your Solution: We claim that this series diverges. Remember that a necessary (but not sufficient!) condition for a series to converge is that lim n a n = 0. However, notice that the odd terms of this series are all growing! That is, if n = 2k + for some integer k, then we have () a 2k+ = (2k + ) = (2k k+ ) + ( 2) 2k+ because 2 2k+ + ( 2) 2k+ = 0. Therefore lim k a 2k+ =, so we cannot possibly have lim n a n = 0, so the series cannot converge. Question 3: Let a n = n 6 n. Does the series Solution: This series also diverges. Notice that n /(6n) = n 3 /6, so n= n /(6n) = 6 n= n3. But n 3 n for all integer n, so this series diverges by comparison to the sequence a n = n. If instead we had n /6 n, then it would converge by the ratio test. Question : Let a n = n n!. Does the series Solution: This series converges. Notice that n/n! = /(n )!, so shifting our index by we have n (2) n! = n!, n= recalling that 0! =. We claim that for n, n! > n 2. Notice that when n =, n! = 2, while n 2 = 6. We prove this claim using induction (notice we ve already shown the base case). Suppose that for some k, k! > k 2. We show that (k + )! > (k + ) 2. We see (k + )! = (k + )k! > (k + )k 2 by our inductive assumption. Thus if we can show k 2 > k +, we ll be done. Consider the polynomial f(x) = x 2 x. We see that f (x) = 2x > 0 for all x > /2. Therefore, since k 2 k > 0 when k =, k 2 k > 0 for all k. Therefore (k + )! = (k + )k! > (k + )k 2 > (k + ) 2, as we wanted to show. Therefore /n! < /n 2, and the series converges by comparison to the sequence a n = /n 2. n=0

4 INSTRUCTOR: STEVEN MILLER We can also solve this by using the ratio test. ρ a n+ n a n (n + )/(n + )! n n/n! n n n n n n! n + (n + )! n n! n + (n + )n! n n + n! n n + lim n n! = 0 = 0. Question 5: Let a n = n2n 3 n. Does the series Solution: This series converges. Again, we claim that for sufficiently large n, n(2/3) n < /n 2, which is equivalent to showing that (3/2) n /n > n 2. Taking logs, we see this is equivalent to n log(3/2) log(n) > 2 log(n), or n log(3/2) > 3 log(n), or n/ log(n) > 3/ log(3/2) (assuming n > so log(n) > 0). Notice that the function f(x) = x/ log(x) has derivative (log(x) )/(log(x) 2 ), which is positive for all n e. Therefore, since f(e 0 ) = e 0 /0 > 3/ log(3/2), (3/2) n /n > n 2 for all n e 0. Thus the series converges by comparison with the sequence a n = /n 2. Question 6: Let a n = n 3 /2 n/2. Does the series n= a n converge or diverge? Prove your Solution: Let s use the Ratio test. ρ a n+ n a n n (n + ) 3 /2 (n+)/2 n 3 /2 n/2 (n + ) 3 n n 3 2 n/2 2 (n+)/2 = n + n + n + 2 n/2 lim n n n n 2 n/2 2 /2 = n + n + n + lim n n n n 2 /2 = n + n + n + lim lim lim lim = n n n n n n n 2 /2 2 /2 < ; as this is less than, the series converges. Note the hardest part of this problem is doing the algebra well and canceling correctly.

5 2.2. Taylor Series (one variable). PRACTICE PROBLEMS 5 Recall that the Taylor series of degree n for a function f at a point x 0 is given by f(x 0 ) + f (x 0 )(x x 0 ) + f (x 0 ) (x x 0 ) 2 2! + f (x 0 ) (x x 0 ) f (n) (x 0 ) (x x 0 ) n, 3! n! where f (k) denotes the k th derivative of f. We can write this more compactly with summation notation as k=0 n k=0 f (k) (x 0 ) (x x 0 ) k, k! where f (0) is just f. In many cases the point x 0 is 0, and the formulas simplify a bit to n f (k) (0) x k = f(0) + f (0)x + f (0) x f (n) (0) x n. k! 2! n! The reason Taylor series are so useful is that they allow us to understand the behavior of a complicated function near a point by understanding the behavior of a related polynomial near that point; the higher the degree of our approximating polynomial, the smaller the error in our approximation. Fortunately, for many applications a first order Taylor series (ie, just using the first derivative) does a very good job. This is also called the tangent line method, as we are replacing a complicated function with its tangent line. One thing which can be a little confusing is that there are n + terms in a Taylor series of degree n; the problem is we start with the zeroth term, the value of the function at the point of interest. You should never be impressed if someone tells you the Taylor series at x 0 agrees with the function at x 0 this is forced to hold from the definition! The reason is all the (x x 0 ) k terms vanish, and we are left with f(x 0 ), so of course the two will agree. Taylor series are only useful when they are close to the original function for x close to x 0. Question 2.. Find the first five terms of the Taylor series for f(x) = x 8 + x + 3 at x = 0. Solution: To find the first five terms requires evaluating the function and its first four derivatives: f(0) = 3 f (x) = 8x 7 + x 3 f (0) = 0 f (x) = 56x 6 + 2x 2 f (0) = 0 f (x) = 336x 5 + 2x f (0) = 0 f () (x) = 680x + 2 f () = 2. Therefore the first five terms of the Taylor series are f(0) + f (0)x + + f () (0) x = 3 + 2!! x = 3 + x. This answer shouldn t be surprising as we can view our function as f(x) = 3 + x + x 8 ; thus our function is presented in such a way that it s easy to see its Taylor series about 0. If we wanted the first six terms of its Taylor series expansion about 0, the answer would be the

6 6 INSTRUCTOR: STEVEN MILLER same. We won t see anything new until we look at the degree 8 Taylor series (ie, the first nine terms), at which point the x 8 term appears. Question 2.2. Find the first three terms of the Taylor series for f(x) = x 8 + x + 3 at x =. Solution: We can find the expansion by taking the derivatives and evaluating at and not 0. We have Therefore the first three terms gives f(x) = x 8 + x + 3 f() = 5 f (x) = 8x 7 + x 3 f () = 2 f (x) = 56x 6 + 2x 2 f () = 68. f() + f ()(x ) + f () (x ) 2 = 5 + 2(x ) + 3(x ) 2. 2! Important Note: Another way to do this problem is one of my favorite tricks, namely converting a Taylor expansion about one point to another. We write x as (x ) + ; we have just added zero, which is one of the most powerful tricks in mathematics. We then have x 8 + x + 3 = ((x ) + ) 8 + ((x ) + ) + 3; we can expand each term by using the Binomial Theorem, and after some algebra we ll find the same answer as before. For example, ((x ) + ) equals ( ) ( ) (x ) 0 + (x ) 3 0 ( ) ( ) ( ) + (x ) (x ) 3 + (x ) In this instance, it is not a good idea to use this trick, as this makes the problem more complicated rather than easier; however, there are situations where this trick does make life easier, and thus it is worth seeing. We ll see another trick in the next problem (and this time it will simplify things). Question 2.3. Find the first three terms of the Taylor series for f(x) = cos(5x) at x = 0. Solution: The standard way to solve this is to take derivatives and evaluate. We have Thus the answer is f(x) = cos(5x) f(0) = f (x) = 5 sin(5x) f (0) = 0 f (x) = 25 cos(5x) f (0) = 25. f(0) + f (0)x + f (0) x 2 = x2. Important Note: We discuss a faster way of doing this problem. This method assumes we know the Taylor series expansion of a related function, g(u) = cos(u). This is one of the three

7 PRACTICE PROBLEMS 7 standard Taylor series expansions one sees in calculus (the others being the expansions for sin(u) and exp(u); a good course also does log( ± u)). Recall cos(u) = u2 2! + u! u6 6! + = k=0 ( ) k u 2k. (2k)! If we replace u with 5x, we get the Taylor series expansion for cos(5x): cos(5x) = (5x)2 2! + (5x)! (5x)6 6! +. As we only want the first three terms, we stop at the x 2 term, and find it is 25x 2 /2. The answer is the same as before, but this seems much faster. Is it? At first it seems like we avoided having to take derivatives. We haven t; the point is we took the derivatives years ago in Calculus when we found the Taylor series expansion for cos(u). We now use that. We see the advantage of being able to recall previous results we can frequently modify them (with very little effort) to cover a new situation; however, we can of course only do this if we remember the old results! Question 2.. Find the first five terms of the Taylor series for f(x) = cos 3 (5x) at x = 0. Solution: Doing (a lot of!) differentiation and algebra leads to 75 2 x x we calculated more terms than needed because of the comment below. Note that f (x) = 5 cos 2 (5x) sin(5x). To calculate f (x) involves a product and a power rule, and we can see that it gets worse and worse the higher derivative we need! It is worth doing all these derivatives to appreciate the alternate approach given below. Important Note: There is a faster way to do this problem. From the previous exercise, we know cos(5x) = 25 2 x2 + terms of size x 3 or higher. Thus to find the first five terms is equivalent to just finding the coefficients up to x. Unfortunately our expansion is just a tad too crude; we only kept up to x 2, and we need to have up to x. So, let s spend a little more time and compute the Taylor series of cos(5x) of degree : that is 25 2 x x. If we cube this, we ll get the first six terms in the Taylor series of cos 3 (5x). In other words, we ll have the degree 5 expansion, and all our terms will be correct up to the x 6 term. The reason is when we cube, the only way we can get a term of degree 5 or less is covered. Thus we need to compute ( 25 2 x ) 3 2 x ; however, as we only care about the terms of x 5 or lower, we can drop a lot of terms in the product. For instance, one of the factors is the x term; if it hits another x term or an x 2 it x 6 ;

8 8 INSTRUCTOR: STEVEN MILLER will give an x 6 or higher term, which we don t care about. Thus, taking the cube but only keeping terms like x 5 or lower degree, we get ( ( 3 + ) 2 25 ) ( ) ( 3 2 x ) 2 ( ) ( ) x x. After doing a little algebra, we find the same answer as before. So, was it worth it? To each his own, but again the advantage of this method is we reduce much our problem to something we ve already done. If we wanted to do the first seven terms of the Taylor series, we would just have to keep a bit more, and expand the original function cos(5x) a bit further. As mentioned above, to truly appreciate the power of this method you should do the problem the long way (ie, the standard way). Question 2.5. Find the first two terms of the Taylor series for f(x) = e x at x = 0. Solution: This is merely the first two terms of one of the most important Taylor series of all, the Taylor series of e x. As f (x) = e x, we see f (n) (x) = e x for all n. Thus the answer is f(0) + f (0)x = + x. More generally, the full Taylor series is e x = + x + x2 2! + x3 3! + x! + = Question 2.6. Find the first six terms of the Taylor series for f(x) = e x8 x = 0. n=0 x n n!. = exp(x 8 ) at Solution: The first way to solve this is to keep taking derivatives using the chain rule. Very quickly we see how tedious this is, as f (x) = 8x 7 exp(x 8 ), f (x) = 6x exp(x 8 ) + 56x 6 exp(x 8 ), and of course the higher derivatives become even more complicated. We use the faster idea mentioned above. We know e u so replacing u with x 8 gives = + u + u2 2! + u3 3! + = n=0 e x8 = + x 8 + (x8 ) ! As we only want the first six terms, the highest term is x 5. Thus the answer is just we would only have the x 8 term if we wanted at least the first nine terms! For this problem, we see how much better this approach is; knowing the first two terms of the Taylor series expansion of e u suffice to get the first six terms of e x8. This is magnitudes easier than calculating all those derivatives. Again, we see the advantage of being able to recall previous results. Question 2.7. Find the first four terms of the Taylor series for f(x) = 2π e x2 /2 = exp( x 2 /2)/ 2π at x = 0. u n n!,

9 PRACTICE PROBLEMS 9 Solution: The answer is 2π x2 π. We can do this by the standard method of differentiating, or we can take the Taylor series expansion of e u and replace u with x 2 /2. Question 2.8. Find the first three terms of the Taylor series for f(x) = x at x = 3. Solution: If f(x) = x /2, f (x) = 2 x /2 and f (x) = x 3/2. Evaluating at /3 gives ( 3 + x ) 3 ( 3 x ) Question 2.9. Find the first three terms of the Taylor series for f(x) = ( + x) /3 at x = 2. Solution: Doing a lot of differentiation and algebra yields ( ) /3 3 + ( ) 2/3 ( 2 x ) 2 ( ) 2/3 ( 2 x ) Question 2.0. Find the first three terms of the Taylor series for f(x) = x log x at x =. Solution: One way is to take derivatives in the standard manner and evaluate; this gives (x )2 (x ) +. 2 Important Note: Another way to do this problem involves two tricks we ve mentioned before. The first is we need to know the series expansion of log(x) about x =. One of the the most important Taylor series expansions, which is often done in a Calculus class, is log( + u) = u u2 2 + u3 3 u + = ( ) k+ u k. k We then write x log x = ((x ) + ) log ( + (x )) ; we can now grab the Taylor series from ) (x )2 (x )2 ((x ) + ) ((x ) = (x ) n= +. Question 2.. Find the first three terms of the Taylor series for f(x) = log(+x) at x = 0. Question 2.2. Find the first three terms of the Taylor series for f(x) = log( x) at x =. Solution: The expansion for log( x) is often covered in a Calculus class; equivalently, it can be found from log( + u) by replacing u with x. We find log( x) = (x + x 2 + x ) 3 + x n = n. For this problem, we get x + x2 2. n=

10 0 INSTRUCTOR: STEVEN MILLER Question 2.3. Find the first two terms of the Taylor series for f(x) = log(( x) e x ) = log(( x) exp(x)) at x = 0. Solution: Taking derivatives and doing the algebra, we see the answer is just zero! The first term that has a non-zero coefficient is the x 2 term, which comes in as x 2 /2. A better way of doing this is to simplify the expression before taking the derivative. As the logarithm of a product is the sum of the logarithms, we have log(( x) e x ) equals log( x) + log e x. But log e x = x, and log( x) = x x 2 /2. Adding the two expansions gives x 2 /2, which means that the first two terms of the Taylor series vanish. Question 2.. Find the first three terms of the Taylor series for f(x) = cos(x) log( + x) at x = 0. Solution: Taking derivatives and doing the algebra gives x x 2 /2. Important Note: A better way of doing this is to take the Taylor series expansions of each piece and then multiply them together. We need only take enough terms of each piece so that we are sure that we get the terms of order x 2 and lower correct. Thus ) ) cos(x) log( + x) = ( x2 2 + (x x2 2 + = x x Question 2.5. Find the first two terms of the Taylor series for f(x) = log(+2x) at x = 0. Solution: The fastest way to do this is to take the Taylor series of log( + u) and replace u with 2x, giving 2x.

### Some Notes on Taylor Polynomials and Taylor Series

Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited

### Taylor Polynomials and Taylor Series Math 126

Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will

### Worksheet on induction Calculus I Fall 2006 First, let us explain the use of for summation. The notation

Worksheet on induction MA113 Calculus I Fall 2006 First, let us explain the use of for summation. The notation f(k) means to evaluate the function f(k) at k = 1, 2,..., n and add up the results. In other

### Taylor and Maclaurin Series

Taylor and Maclaurin Series In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions

### Course Notes for Math 162: Mathematical Statistics Approximation Methods in Statistics

Course Notes for Math 16: Mathematical Statistics Approximation Methods in Statistics Adam Merberg and Steven J. Miller August 18, 6 Abstract We introduce some of the approximation methods commonly used

### MITES Physics III Summer Introduction 1. 3 Π = Product 2. 4 Proofs by Induction 3. 5 Problems 5

MITES Physics III Summer 010 Sums Products and Proofs Contents 1 Introduction 1 Sum 1 3 Π Product 4 Proofs by Induction 3 5 Problems 5 1 Introduction These notes will introduce two topics: A notation which

### 0.8 Rational Expressions and Equations

96 Prerequisites 0.8 Rational Expressions and Equations We now turn our attention to rational expressions - that is, algebraic fractions - and equations which contain them. The reader is encouraged to

### Lectures 5-6: Taylor Series

Math 1d Instructor: Padraic Bartlett Lectures 5-: Taylor Series Weeks 5- Caltech 213 1 Taylor Polynomials and Series As we saw in week 4, power series are remarkably nice objects to work with. In particular,

### 9.2 Summation Notation

9. Summation Notation 66 9. Summation Notation In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with a

### MATH 2300 review problems for Exam 3 ANSWERS

MATH 300 review problems for Exam 3 ANSWERS. Check whether the following series converge or diverge. In each case, justify your answer by either computing the sum or by by showing which convergence test

### Series Convergence Tests Math 122 Calculus III D Joyce, Fall 2012

Some series converge, some diverge. Series Convergence Tests Math 22 Calculus III D Joyce, Fall 202 Geometric series. We ve already looked at these. We know when a geometric series converges and what it

### Limit processes are the basis of calculus. For example, the derivative. f f (x + h) f (x)

SEC. 4.1 TAYLOR SERIES AND CALCULATION OF FUNCTIONS 187 Taylor Series 4.1 Taylor Series and Calculation of Functions Limit processes are the basis of calculus. For example, the derivative f f (x + h) f

### Calculus I Notes. MATH /Richards/

Calculus I Notes MATH 52-154/Richards/10.1.07 The Derivative The derivative of a function f at p is denoted by f (p) and is informally defined by f (p)= the slope of the f-graph at p. Of course, the above

### 10.2 Series and Convergence

10.2 Series and Convergence Write sums using sigma notation Find the partial sums of series and determine convergence or divergence of infinite series Find the N th partial sums of geometric series and

### Math 55: Discrete Mathematics

Math 55: Discrete Mathematics UC Berkeley, Spring 2012 Homework # 9, due Wednesday, April 11 8.1.5 How many ways are there to pay a bill of 17 pesos using a currency with coins of values of 1 peso, 2 pesos,

### We can express this in decimal notation (in contrast to the underline notation we have been using) as follows: 9081 + 900b + 90c = 9001 + 100c + 10b

In this session, we ll learn how to solve problems related to place value. This is one of the fundamental concepts in arithmetic, something every elementary and middle school mathematics teacher should

### Introduction to Complex Fourier Series

Introduction to Complex Fourier Series Nathan Pflueger 1 December 2014 Fourier series come in two flavors. What we have studied so far are called real Fourier series: these decompose a given periodic function

### Approximating functions by Taylor Polynomials.

Chapter 4 Approximating functions by Taylor Polynomials. 4.1 Linear Approximations We have already seen how to approximate a function using its tangent line. This was the key idea in Euler s method. If

### Base Conversion written by Cathy Saxton

Base Conversion written by Cathy Saxton 1. Base 10 In base 10, the digits, from right to left, specify the 1 s, 10 s, 100 s, 1000 s, etc. These are powers of 10 (10 x ): 10 0 = 1, 10 1 = 10, 10 2 = 100,

### 6.8 Taylor and Maclaurin s Series

6.8. TAYLOR AND MACLAURIN S SERIES 357 6.8 Taylor and Maclaurin s Series 6.8.1 Introduction The previous section showed us how to find the series representation of some functions by using the series representation

### 4.3 Lagrange Approximation

206 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION Lagrange Polynomial Approximation 4.3 Lagrange Approximation Interpolation means to estimate a missing function value by taking a weighted average

### Chapter 31 out of 37 from Discrete Mathematics for Neophytes: Number Theory, Probability, Algorithms, and Other Stuff by J. M.

31 Geometric Series Motivation (I hope) Geometric series are a basic artifact of algebra that everyone should know. 1 I am teaching them here because they come up remarkably often with Markov chains. The

### Homework 5 Solutions

Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which

### Calculating Logarithms By Hand

Calculating Logarithms By Hand W. Blaine Dowler June 14, 010 Abstract This details methods by which we can calculate logarithms by hand. 1 Definition and Basic Properties A logarithm can be defined as

### Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00

18.781 Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00 Throughout this assignment, f(x) always denotes a polynomial with integer coefficients. 1. (a) Show that e 32 (3) = 8, and write down a list

### Differentiation and Integration

This material is a supplement to Appendix G of Stewart. You should read the appendix, except the last section on complex exponentials, before this material. Differentiation and Integration Suppose we have

### Math 4310 Handout - Quotient Vector Spaces

Math 4310 Handout - Quotient Vector Spaces Dan Collins The textbook defines a subspace of a vector space in Chapter 4, but it avoids ever discussing the notion of a quotient space. This is understandable

### APPLICATIONS OF THE ORDER FUNCTION

APPLICATIONS OF THE ORDER FUNCTION LECTURE NOTES: MATH 432, CSUSM, SPRING 2009. PROF. WAYNE AITKEN In this lecture we will explore several applications of order functions including formulas for GCDs and

### SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS

(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1 SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES Be able to identify polynomial, rational, and algebraic

### Lies My Calculator and Computer Told Me

Lies My Calculator and Computer Told Me 2 LIES MY CALCULATOR AND COMPUTER TOLD ME Lies My Calculator and Computer Told Me See Section.4 for a discussion of graphing calculators and computers with graphing

### Joint Probability Distributions and Random Samples (Devore Chapter Five)

Joint Probability Distributions and Random Samples (Devore Chapter Five) 1016-345-01 Probability and Statistics for Engineers Winter 2010-2011 Contents 1 Joint Probability Distributions 1 1.1 Two Discrete

### Review of Fundamental Mathematics

Review of Fundamental Mathematics As explained in the Preface and in Chapter 1 of your textbook, managerial economics applies microeconomic theory to business decision making. The decision-making tools

### Chapter 8. Exponential and Logarithmic Functions

Chapter 8 Exponential and Logarithmic Functions This unit defines and investigates exponential and logarithmic functions. We motivate exponential functions by their similarity to monomials as well as their

### Lecture 5 Rational functions and partial fraction expansion

S. Boyd EE102 Lecture 5 Rational functions and partial fraction expansion (review of) polynomials rational functions pole-zero plots partial fraction expansion repeated poles nonproper rational functions

### Derivatives Math 120 Calculus I D Joyce, Fall 2013

Derivatives Mat 20 Calculus I D Joyce, Fall 203 Since we ave a good understanding of its, we can develop derivatives very quickly. Recall tat we defined te derivative f x of a function f at x to be te

### 1.7 Graphs of Functions

64 Relations and Functions 1.7 Graphs of Functions In Section 1.4 we defined a function as a special type of relation; one in which each x-coordinate was matched with only one y-coordinate. We spent most

### Practice with Proofs

Practice with Proofs October 6, 2014 Recall the following Definition 0.1. A function f is increasing if for every x, y in the domain of f, x < y = f(x) < f(y) 1. Prove that h(x) = x 3 is increasing, using

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### Representation of functions as power series

Representation of functions as power series Dr. Philippe B. Laval Kennesaw State University November 9, 008 Abstract This document is a summary of the theory and techniques used to represent functions

### Vieta s Formulas and the Identity Theorem

Vieta s Formulas and the Identity Theorem This worksheet will work through the material from our class on 3/21/2013 with some examples that should help you with the homework The topic of our discussion

### Partial Fractions. Combining fractions over a common denominator is a familiar operation from algebra:

Partial Fractions Combining fractions over a common denominator is a familiar operation from algebra: From the standpoint of integration, the left side of Equation 1 would be much easier to work with than

### Session 7 Fractions and Decimals

Key Terms in This Session Session 7 Fractions and Decimals Previously Introduced prime number rational numbers New in This Session period repeating decimal terminating decimal Introduction In this session,

### Real Roots of Univariate Polynomials with Real Coefficients

Real Roots of Univariate Polynomials with Real Coefficients mostly written by Christina Hewitt March 22, 2012 1 Introduction Polynomial equations are used throughout mathematics. When solving polynomials

### To give it a definition, an implicit function of x and y is simply any relationship that takes the form:

2 Implicit function theorems and applications 21 Implicit functions The implicit function theorem is one of the most useful single tools you ll meet this year After a while, it will be second nature to

### Using a table of derivatives

Using a table of derivatives In this unit we construct a Table of Derivatives of commonly occurring functions. This is done using the knowledge gained in previous units on differentiation from first principles.

### Homework # 3 Solutions

Homework # 3 Solutions February, 200 Solution (2.3.5). Noting that and ( + 3 x) x 8 = + 3 x) by Equation (2.3.) x 8 x 8 = + 3 8 by Equations (2.3.7) and (2.3.0) =3 x 8 6x2 + x 3 ) = 2 + 6x 2 + x 3 x 8

### 3. Mathematical Induction

3. MATHEMATICAL INDUCTION 83 3. Mathematical Induction 3.1. First Principle of Mathematical Induction. Let P (n) be a predicate with domain of discourse (over) the natural numbers N = {0, 1,,...}. If (1)

### Microeconomic Theory: Basic Math Concepts

Microeconomic Theory: Basic Math Concepts Matt Van Essen University of Alabama Van Essen (U of A) Basic Math Concepts 1 / 66 Basic Math Concepts In this lecture we will review some basic mathematical concepts

### HOMEWORK 4 SOLUTIONS. All questions are from Vector Calculus, by Marsden and Tromba

HOMEWORK SOLUTIONS All questions are from Vector Calculus, by Marsden and Tromba Question :..6 Let w = f(x, y) be a function of two variables, and let x = u + v, y = u v. Show that Solution. By the chain

### Calculus C/Multivariate Calculus Advanced Placement G/T Essential Curriculum

Calculus C/Multivariate Calculus Advanced Placement G/T Essential Curriculum UNIT I: The Hyperbolic Functions basic calculus concepts, including techniques for curve sketching, exponential and logarithmic

### MATH 4330/5330, Fourier Analysis Section 11, The Discrete Fourier Transform

MATH 433/533, Fourier Analysis Section 11, The Discrete Fourier Transform Now, instead of considering functions defined on a continuous domain, like the interval [, 1) or the whole real line R, we wish

### HOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)!

Math 7 Fall 205 HOMEWORK 5 SOLUTIONS Problem. 2008 B2 Let F 0 x = ln x. For n 0 and x > 0, let F n+ x = 0 F ntdt. Evaluate n!f n lim n ln n. By directly computing F n x for small n s, we obtain the following

### 1 if 1 x 0 1 if 0 x 1

Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or

### Squaring, Cubing, and Cube Rooting

Squaring, Cubing, and Cube Rooting Arthur T. Benjamin Harvey Mudd College Claremont, CA 91711 benjamin@math.hmc.edu I still recall my thrill and disappointment when I read Mathematical Carnival [4], by

### Tips for Solving Mathematical Problems

Tips for Solving Mathematical Problems Don Byrd Revised late April 2011 The tips below are based primarily on my experience teaching precalculus to high-school students, and to a lesser extent on my other

### Techniques of Integration

CHPTER 7 Techniques of Integration 7.. Substitution Integration, unlike differentiation, is more of an art-form than a collection of algorithms. Many problems in applied mathematics involve the integration

### Copyrighted Material. Chapter 1 DEGREE OF A CURVE

Chapter 1 DEGREE OF A CURVE Road Map The idea of degree is a fundamental concept, which will take us several chapters to explore in depth. We begin by explaining what an algebraic curve is, and offer two

### Section 6-2 Mathematical Induction

6- Mathematical Induction 457 In calculus, it can be shown that e x k0 x k k! x x x3!! 3!... xn n! where the larger n is, the better the approximation. Problems 6 and 6 refer to this series. Note that

### 3.4 Complex Zeros and the Fundamental Theorem of Algebra

86 Polynomial Functions.4 Complex Zeros and the Fundamental Theorem of Algebra In Section., we were focused on finding the real zeros of a polynomial function. In this section, we expand our horizons and

### Math 115 Spring 2011 Written Homework 5 Solutions

. Evaluate each series. a) 4 7 0... 55 Math 5 Spring 0 Written Homework 5 Solutions Solution: We note that the associated sequence, 4, 7, 0,..., 55 appears to be an arithmetic sequence. If the sequence

### (2 4 + 9)+( 7 4) + 4 + 2

5.2 Polynomial Operations At times we ll need to perform operations with polynomials. At this level we ll just be adding, subtracting, or multiplying polynomials. Dividing polynomials will happen in future

### CLASS 3, GIVEN ON 9/27/2010, FOR MATH 25, FALL 2010

CLASS 3, GIVEN ON 9/27/2010, FOR MATH 25, FALL 2010 1. Greatest common divisor Suppose a, b are two integers. If another integer d satisfies d a, d b, we call d a common divisor of a, b. Notice that as

### Lecture 1: Course overview, circuits, and formulas

Lecture 1: Course overview, circuits, and formulas Topics in Complexity Theory and Pseudorandomness (Spring 2013) Rutgers University Swastik Kopparty Scribes: John Kim, Ben Lund 1 Course Information Swastik

### General Framework for an Iterative Solution of Ax b. Jacobi s Method

2.6 Iterative Solutions of Linear Systems 143 2.6 Iterative Solutions of Linear Systems Consistent linear systems in real life are solved in one of two ways: by direct calculation (using a matrix factorization,

### Payment streams and variable interest rates

Chapter 4 Payment streams and variable interest rates In this chapter we consider two extensions of the theory Firstly, we look at payment streams A payment stream is a payment that occurs continuously,

### GREEN CHICKEN EXAM - NOVEMBER 2012

GREEN CHICKEN EXAM - NOVEMBER 2012 GREEN CHICKEN AND STEVEN J. MILLER Question 1: The Green Chicken is planning a surprise party for his grandfather and grandmother. The sum of the ages of the grandmother

### 2 Complex Functions and the Cauchy-Riemann Equations

2 Complex Functions and the Cauchy-Riemann Equations 2.1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. Likewise, in complex analysis, we study functions f(z)

### LECTURE 2. Part/ References Topic/Sections Notes/Speaker. 1 Enumeration 1 1.1 Generating functions... 1. Combinatorial. Asst #1 Due FS A.

Wee Date Sections f aculty of science MATH 895-4 Fall 0 Contents 1 Sept 7 I1, I, I3 Symbolic methods I A3/ C Introduction to Prob 4 A sampling 18 IX1 of counting sequences Limit Laws and and generating

### MATH 132: CALCULUS II SYLLABUS

MATH 32: CALCULUS II SYLLABUS Prerequisites: Successful completion of Math 3 (or its equivalent elsewhere). Math 27 is normally not a sufficient prerequisite for Math 32. Required Text: Calculus: Early

### Probability and Statistics Prof. Dr. Somesh Kumar Department of Mathematics Indian Institute of Technology, Kharagpur

Probability and Statistics Prof. Dr. Somesh Kumar Department of Mathematics Indian Institute of Technology, Kharagpur Module No. #01 Lecture No. #15 Special Distributions-VI Today, I am going to introduce

### Induction Problems. Tom Davis November 7, 2005

Induction Problems Tom Davis tomrdavis@earthlin.net http://www.geometer.org/mathcircles November 7, 2005 All of the following problems should be proved by mathematical induction. The problems are not necessarily

### Dr. Foote s Algebra & Calculus Tips or How to make life easier for yourself, make fewer errors, and get more points

Dr. Foote s Algebra & Calculus Tips or How to make life easier for yourself, make fewer errors, and get more points This handout includes several tips that will help you streamline your computations. Most

### Factoring Polynomials and Solving Quadratic Equations

Factoring Polynomials and Solving Quadratic Equations Math Tutorial Lab Special Topic Factoring Factoring Binomials Remember that a binomial is just a polynomial with two terms. Some examples include 2x+3

### GRE Prep: Precalculus

GRE Prep: Precalculus Franklin H.J. Kenter 1 Introduction These are the notes for the Precalculus section for the GRE Prep session held at UCSD in August 2011. These notes are in no way intended to teach

### GRE MATH REVIEW #5. 1. Variable: A letter that represents an unknown number.

GRE MATH REVIEW #5 Eponents and Radicals Many numbers can be epressed as the product of a number multiplied by itself a number of times. For eample, 16 can be epressed as. Another way to write this is

### The Method of Partial Fractions Math 121 Calculus II Spring 2015

Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method

### MBA Jump Start Program

MBA Jump Start Program Module 2: Mathematics Thomas Gilbert Mathematics Module Online Appendix: Basic Mathematical Concepts 2 1 The Number Spectrum Generally we depict numbers increasing from left to right

### Absolute Value of Reasoning

About Illustrations: Illustrations of the Standards for Mathematical Practice (SMP) consist of several pieces, including a mathematics task, student dialogue, mathematical overview, teacher reflection

### Sequences and Series

Sequences and Series Consider the following sum: 2 + 4 + 8 + 6 + + 2 i + The dots at the end indicate that the sum goes on forever. Does this make sense? Can we assign a numerical value to an infinite

### POLYNOMIAL FUNCTIONS

POLYNOMIAL FUNCTIONS Polynomial Division.. 314 The Rational Zero Test.....317 Descarte s Rule of Signs... 319 The Remainder Theorem.....31 Finding all Zeros of a Polynomial Function.......33 Writing a

### Induction. Margaret M. Fleck. 10 October These notes cover mathematical induction and recursive definition

Induction Margaret M. Fleck 10 October 011 These notes cover mathematical induction and recursive definition 1 Introduction to induction At the start of the term, we saw the following formula for computing

### Name: ID: Discussion Section:

Math 28 Midterm 3 Spring 2009 Name: ID: Discussion Section: This exam consists of 6 questions: 4 multiple choice questions worth 5 points each 2 hand-graded questions worth a total of 30 points. INSTRUCTIONS:

### 1 Lecture: Integration of rational functions by decomposition

Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.

### Notes on Factoring. MA 206 Kurt Bryan

The General Approach Notes on Factoring MA 26 Kurt Bryan Suppose I hand you n, a 2 digit integer and tell you that n is composite, with smallest prime factor around 5 digits. Finding a nontrivial factor

### Mathematical Induction. Mary Barnes Sue Gordon

Mathematics Learning Centre Mathematical Induction Mary Barnes Sue Gordon c 1987 University of Sydney Contents 1 Mathematical Induction 1 1.1 Why do we need proof by induction?.... 1 1. What is proof by

### Section 12.6: Directional Derivatives and the Gradient Vector

Section 26: Directional Derivatives and the Gradient Vector Recall that if f is a differentiable function of x and y and z = f(x, y), then the partial derivatives f x (x, y) and f y (x, y) give the rate

### Polynomial Invariants

Polynomial Invariants Dylan Wilson October 9, 2014 (1) Today we will be interested in the following Question 1.1. What are all the possible polynomials in two variables f(x, y) such that f(x, y) = f(y,

### Taylor Series and Asymptotic Expansions

Taylor Series and Asymptotic Epansions The importance of power series as a convenient representation, as an approimation tool, as a tool for solving differential equations and so on, is pretty obvious.

### Linear and quadratic Taylor polynomials for functions of several variables.

ams/econ 11b supplementary notes ucsc Linear quadratic Taylor polynomials for functions of several variables. c 010, Yonatan Katznelson Finding the extreme (minimum or maximum) values of a function, is

### Fractions and Decimals

Fractions and Decimals Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles December 1, 2005 1 Introduction If you divide 1 by 81, you will find that 1/81 =.012345679012345679... The first

### x if x 0, x if x < 0.

Chapter 3 Sequences In this chapter, we discuss sequences. We say what it means for a sequence to converge, and define the limit of a convergent sequence. We begin with some preliminary results about the

### DIFFERENTIABILITY OF COMPLEX FUNCTIONS. Contents

DIFFERENTIABILITY OF COMPLEX FUNCTIONS Contents 1. Limit definition of a derivative 1 2. Holomorphic functions, the Cauchy-Riemann equations 3 3. Differentiability of real functions 5 4. A sufficient condition

### LOGNORMAL MODEL FOR STOCK PRICES

LOGNORMAL MODEL FOR STOCK PRICES MICHAEL J. SHARPE MATHEMATICS DEPARTMENT, UCSD 1. INTRODUCTION What follows is a simple but important model that will be the basis for a later study of stock prices as

### Alex, I will take congruent numbers for one million dollars please

Alex, I will take congruent numbers for one million dollars please Jim L. Brown The Ohio State University Columbus, OH 4310 jimlb@math.ohio-state.edu One of the most alluring aspectives of number theory

### Mathematics 31 Pre-calculus and Limits

Mathematics 31 Pre-calculus and Limits Overview After completing this section, students will be epected to have acquired reliability and fluency in the algebraic skills of factoring, operations with radicals

### Continuous Compounding and Discounting

Continuous Compounding and Discounting Philip A. Viton October 5, 2011 Continuous October 5, 2011 1 / 19 Introduction Most real-world project analysis is carried out as we ve been doing it, with the present

### 106 Chapter 5 Curve Sketching. If f(x) has a local extremum at x = a and. THEOREM 5.1.1 Fermat s Theorem f is differentiable at a, then f (a) = 0.

5 Curve Sketching Whether we are interested in a function as a purely mathematical object or in connection with some application to the real world, it is often useful to know what the graph of the function