General Theory of Differential Equations Sections 2.8, , 4.1

Size: px
Start display at page:

Download "General Theory of Differential Equations Sections 2.8, 3.1-3.2, 4.1"

Transcription

1 A B I L E N E C H R I S T I A N U N I V E R S I T Y Department of Mathematics General Theory of Differential Equations Sections 2.8, , 4.1 Dr. John Ehrke Department of Mathematics Fall 2012

2 Questions Of Existence and Uniqueness Part of the difficulty in working with differential equations is understanding what constitutes a solution and knowing when a solution even exists. In this lecture, we will try to address questions of existence and uniqueness as they relate to solutions of linear differential equations. This will allow us to build up a general theory supporting our study of differential equations throughout the semester. We will begin with a small example to illustrate what can go wrong. Example Solve the differential equation dy ( y ) dx = 2. x Solution: This equation is separable and so we proceed as follows: Slide 2/27 Dr. John Ehrke Lecture 3 Fall 2012 dy y = 2 dx = ln y = 2 ln x + c x = e ln y 2 ln x+c = e = y = e c (e ln x) 2 = y = A x 2

3 Specifying a Solution The solutions of this equation are parabolas of the form y = A x 2, but we have a couple of questions to still be resolved. Are there some initial conditions for which no solutions exist? If a solution does exist, over what interval(s) are they uniquely defined? In answering the second question, consider what would happen if I asked you to solve the IVP corresponding to y(0) = 0. This is bad for two reasons: There is no rate of change defined at the origin, since f (x, y) is undefined there. A solution could come in on one branch and leave on a completely different branch, i.e. a solution of the IVP corresponding to y(1) = 2 is the parabola y = 2x 2, but this parabola passes through the origin and so cannot be extended uniquely for negative x. The solution is unique only from [0, ). Even for the first question above, no solution exists for any initial value along the y-axis other than the origin, because all the parabolas must pass through the origin. This example highlights just a few of the things that can go wrong when solving differential equations. Slide 3/27 Dr. John Ehrke Lecture 3 Fall 2012

4 Existence Uniqueness Theorem At this point one might wonder if there is any way to predict the behavior of solutions of a given differential equation without solving the ODE. The answer to this question is yes, and is summarized by the existence uniqueness theorem (EUT). First Order Existence Uniqueness Theorem Suppose that both the function f (x, y) and its partial derivative f / y are continuous on some rectangle R in the xy-plane containing the point (a, b) in its interior. Then, for some open interval I R containing the point a, the initial value problem dy = f (x, y), y(a) = b (1) dx has one and only one solution that is defined on the interval I. We will not do a complete proof of this result, but will introduce the machinery behind the proof and a make a few remarks. Our text does not have a complete proof of this result but many undergraduate texts include such a proof in the appendices. If you are interested in reading through the complete proof let me know and I would be happy to provide the proof for you. Slide 4/27 Dr. John Ehrke Lecture 3 Fall 2012

5 Method of Successive Approximations The approach we employ to establish the existence of solutions is called the method of successive approximations and utilizes a series of approximations called Picard iterates. This method is based on the fact that a function y(x) satisfies the initial value problem, (1) on the open interval I containing x = a if and only if it satisfies the integral equation y(x) = b + x a f (t, y(t)) dt (2) for all x I. In particular, if y(x) satisfies (2), then clearly y(a) = b, and differentiating both sides gives y (x) = f (x, y(x)) as desired. It is important to note that we traded solving a differential equation for an integral equation. This is almost always the case when one studies differential equations, as we will see. Slide 5/27 Dr. John Ehrke Lecture 3 Fall 2012

6 Picard Iterates (Demonstrating Existence) To attempt to solve (2), we begin with the initial function y 0 (x) b, and then define iteratively a sequence y 1, y 2, y 3,... of functions that we hope will converge to the solution. Specifically, we let y 1 (x) = b + y 2 (x) = b + y n+1 (x) = b +. x a x a x a f (t, y 0 (t)) dt f (t, y 1 (t)) dt f (t, y n (t)) dt Suppose we know that each of these functions {y n (x)} 0 is defined on some open interval (the same for each n) containing x = a, and that the limit y(x) = lim y n (x) exists at each point of this interval. n Slide 6/27 Dr. John Ehrke Lecture 3 Fall 2012

7 Continuing... Continuing from the previous slide, it follows that y(x) = lim y n+1 (x) n [ = lim b + n = b + lim = b + = b + x a x n a x a x a f ] f (t, y n (t)) dt f (t, y n (t)) dt ( t, lim n y n (t) f (t, y(t)) dt This solution method works provided that: ) dt we can validate the interchange of limit operations above. we can show there exists only this one solution. Slide 7/27 Dr. John Ehrke Lecture 3 Fall 2012

8 Demonstrating Uniqueness If we assume at this point that we have established the existence of a solution via successive approximations (there still is some work to be done to make that argument solid), then we only need to show that the hypotheses of the theorem lead to a unique solution. When proving uniqueness you generally assume there exists another solution to the IVP different from y(x), say φ(x) and look at the difference y(x) φ(x) and try to reach a contradiction by showing this difference must be zero. In this case, that is done by proving that y(x) φ(x) A x 0 y(t) φ(t) dt. To do this requires the use of some heavy duty analysis and the determination of a Lipschitz condition. A Lipschitz condition for two functions exists if there is a positive constant A such that f (x, y 1 ) f (x, y 2) A y 1 y 2. Arguments of this type are generally considered in a first semester ODE course in graduate school. We will not pursue this problem beyond this point. Slide 8/27 Dr. John Ehrke Lecture 3 Fall 2012

9 Applying the Existence Uniqueness Theorem Example Consider the first order ODE, xy = y 1. Identify for what values of (a, b), the initial value problem xy = y 1, y(a) = b has a unique solution, no solution, or more than one solution. Solution: Consider the point(s), (0, b) for all b 1. At these points, x = 0, and so f (x, y) = (y 1)/x fails to be continuous at x = 0 and so by the EUT there can be no solution to the IVP, dy dx = y 1, y(0) = b. x Consider the point, (0, 1). When we consider the point (0, 1), f (x, y) = 0/0 which is indeterminate. So the EUT says nothing about this point in regards to existence. We may have solutions, or we may not have solutions. In the case, a solution exists might it be unique? We will check the partial derivative, ( ) y 1 = 1 y x x which is discontinuous at x = 0. This means, that solutions to the IVP y(0) = a when they exist are not guaranteed to be unique and in particular we see this for (0, 1). Slide 9/27 Dr. John Ehrke Lecture 3 Fall 2012

10 Working With Complex Numbers To achieve our goals this semester we will need to learn how to handle complex numbers and operations involving complex variables. A complex number is of the form z = a + bi where a, b R. The complex conjugate of z, denoted z is given by z = a bi. Remember, that multiplying a number by its complex conjugate turns the number to a real number. In particular, z z = a 2 + b 2. The absolute value of z, denoted z, involves conjugation since Moreover, using z and z we obtain z = z z = a 2 + b 2. a = R(z) = z + z and b = I(z) = z z. 2 2i Recall that when working with complex numbers you will often need to multiply or divide by i. Recall that i = 1 and so i 2 = 1, i 3 = i and i 4 = 1. Slide 10/27 Dr. John Ehrke Lecture 3 Fall 2012

11 Polar Representation of a Complex Number z = a + bi = r cos θ + i(r sin θ) = r (cos θ + i sin θ) = re iθ Euler s Formula: e iθ = cos θ + i sin θ The form z = re iθ is called the polar form of the complex number a + bi. Under this description, a is called the real part, b the imaginary part when z is written in rectangular form. In polar form, r is called the modulus of z (denoted z ), and θ is called the argument of z (denoted arg(z)). Slide 11/27 Dr. John Ehrke Lecture 3 Fall 2012

12 The Complex Exponential Typically, when we call something an exponential in mathematics, we have in mind two very specific properties that it must satisfy. An exponential should satisfy the exponential rule. That is, if e is an exponential, then e x e y = e x+y. An exponential should satisfy the differential equation dy dt = ay. This says that an exponential has the property that it is self-replicating under differentiation (and integration). Question? Does our definition of the complex exponential, e iθ as described by Euler s formula define an exponential as stated above? e iθ = cos θ + i sin θ is called a complex-valued function of a real variable Slide 12/27 Dr. John Ehrke Lecture 3 Fall 2012

13 Power Series Representations The final motivation for our description of cos θ + i sin θ as being the complex exponential is provided by our knowledge of the Taylor series for the exponential, which is given by Inserting x = iθ we obtain e x = 1 + x 1! + x2 2! + x3 3! + x4 4! +. e iθ = 1 + iθ 1! + (iθ)2 + (iθ)3 + (iθ)4 + 2! 3! 4! ) = (1 θ2 2! + θ4 4! + + i (θ θ3 3! + ). This points to the fact that the real part of e iθ should be cos θ and the imaginary part is the Taylor series for sin θ. Slide 13/27 Dr. John Ehrke Lecture 3 Fall 2012

14 The Art of Complexifying To see how truly amazing and wondrous operations with complex variables can be we will demonstrate two examples which take advantage of the art of complexifying (yes that s a word). Example Obtain a solution to the integral e x cos x dx by complexifying the integral. Solution: Looking at the integrand, we can rewrite it as the real part of a complex variable, e x cos(x) = R ( e x e ix). By substituting this in for the real integrand we complexify the integral, ( ) ( ) e e x cos(x) dx = R e ( 1+i)x ( 1+i)x dx = R. 1 + i Taking the real part of this expression gives ( ) ( ) 1 1 i R 1 + i e( 1+i)x = R e x (cos x + i sin x) = e x (sin x cos x). 2 2 Slide 14/27 Dr. John Ehrke Lecture 3 Fall 2012

15 Constant Coefficient Equations In this example, we will introduce a technique that we will lean on later in the semester called exponential response. Example Obtain a general solution to the first order linear equation with constant coefficients given by x (t) + 2x(t) = cos(t). Solution: Complexifying the ODE leads to z (t) + 2z(t) = e it. We change to z(t) to represent the fact that we expect the solution to such an equation to be complex. We guess that the solution to such an equation is of the form z(t) = Ae it (i.e. matches the exponential response term up to a constant) where A is complex valued. Substituting our guess into the ODE gives A = 1/(2 + i) so z(t) = i eit R(z(t)) = x(t) = 2 5 cos t + 1 sin t. 5 Slide 15/27 Dr. John Ehrke Lecture 3 Fall 2012

16 The Exponential Response Theorem This method is so useful that we will establish the general case, that is, solve x + kx = Ae rt. The results of this are summarized in the theorem below: Exponential Response Theorem The exponential response to the first order linear constant coefficient ODE x + kx = Ae rt is given by as long as r + k 0. x(t) = A r + k ert (3) This formula is valid for complex values of r, but depending on the specific nature of the original oscillation you might have to obtain the real or imaginary part of (3). Slide 16/27 Dr. John Ehrke Lecture 3 Fall 2012

17 Second Order Linear Equations The general form of a second order differential equation is given by y = f (t, y, y ). In this lecture, we will consider the second order, linear homogeneous ODE with constant coefficients, given by ay + by + cy = 0 (4) We are going to make some assumptions about solutions to (4) which we will come back and verify later. 1 A general solution (4) is of the form y(x) = c 1 y 1 + c 2 y 2. 2 y 1 and y 2 are linearly independent solutions of (4). (y 1 cy 2, for some constant c 0) 3 Initial conditions involve both y and y which help to determine c 1 and c 2 above. 4 The solutions, y 1 and y 2 are called the fundamental set of solutions for the equation (4). Slide 17/27 Dr. John Ehrke Lecture 3 Fall 2012

18 Making a guess... A natural question to ask at this point is, What is the nature of the solutions y 1 and y 2? This question is best answered by making a guess (as we saw in the last lecture) that such solutions are exponential in nature. So, to try and solve the equation ay + by + cy = 0, (5) we guess that y(t) = e rt is a solution. Substituting into (2) gives ar 2 e rt + bre rt + ce rt = 0 = (r 2 + pr + q)e rt = 0 where p = b/a and q = c/a and a 0. This implies that y(t) = e rt is a solution if and only if r 2 + pr + q = 0. Slide 18/27 Dr. John Ehrke Lecture 3 Fall 2012

19 The Characteristic Equation So we have determined that y = e rt if and only if r 2 + pr + q = 0. This equation is called the characteristic equation for the ODE, (2). There are three cases to consider, each of which leads to different solution behaviors. 1 distinct real roots r 1, r 2 y 1 = e r 1t, y 2 = e r 2t 2 complex conjugate pair r = a ± bi y 1 = e (a+bi)t, y 2 = e (a bi)t 3 repeated real root, r gives you y 1 = e rt, but how do we find y 2? We will explore the physical interpretations of these cases, as well as solution techniques in the next lecture, but for the remainder of this lecture, we will discuss the general theory which underpins these techniques. The questions we need to answer are: 1 Why is c 1 y 1 + c 2 y 2 a solution to the equation, (2)? 2 Why are all solutions of this form? Slide 19/27 Dr. John Ehrke Lecture 3 Fall 2012

20 Superposition Principle Superposition Principle If y 1 and y 2 are solutions to a linear homogeneous ODE, then y = c 1 y 1 + c 2 y 2 is also a solution. To make our lives easier in manipulating differential equations we often want to think about differential equations as linear operators acting on y. We define the linear operator L corresponding to the ODE y + py + qy = 0 by y + py + qy = 0 = D 2 y + pdy + qy = 0 = (D 2 + pd + q)y = 0 = Ly = 0 We denote by D, the differential operator, D = d/dy. L = D 2 + pd + q is a linear operator if it satisfies the following properties: L(u 1 + u 2 ) = L(u 1 ) + L(u 2 ) L(αu) = αl(u), for some constant α. Slide 20/27 Dr. John Ehrke Lecture 3 Fall 2012

21 Linear Operators Under the definition of a linear operator on the previous slide, we show that L = D 2 + pd + q is linear. L(u 1 + u 2) = (D 2 + pd + q)(u 1 + u 2) = D 2 (u 1 + u 2) + pd(u 1 + u 2) + q(u 1 + u 2) = u 1 + u 2 + p(u 1 + u 2) + qu 1 + qu 2 = (u 1 + pu 1 + qu 1 ) + (u 2 + pu 2 + qu 2) = Lu 1 + Lu 2 We claim it is obvious that L(αu) = αl(u) by properties of the derivative. We are now prepared to answer our first question: If we suppose that y 1 and y 2 are solutions to y + py + qy = 0, does that automatically mean c 1 y 1 + c 2y 2 is a solution? L(c 1 y 1 + c 2y 2) = L(c 1 y 1 ) + L(c 2y 2) = c 1 L(y 1 ) + c 2L(y 2) = 0 (since y 1 solves the ODE, then L(y 1 ) = 0) Thus, c 1 y 1 + c 2y 2 solves y + py + qy = 0 if y 1 and y 2 do. Slide 21/27 Dr. John Ehrke Lecture 3 Fall 2012

22 Two solutions are better than one... In answering the second of our questions for this lecture, Why do all solutions look like c 1 y 1 + c 2y 2? we will show that this solution form is necessary to satisfy IVPs. Theorem The set of solutions {c 1 y 1 + c 2y 2} is enough to satisfy any initial values imposed on the equation y + py + qy = 0. This means there does not exist some initial condition y(x 0) = a, y (x 0) = b such that a solution not of the form c 1 y 1 + c 2y 2 is the only solution that satisfies this IVP. Proof: Let y(x 0) = a, y (x 0) = b. Then applying these initial conditions to the solution c 1 y 2 + c 2y 2 we obtain the system of linear equations with unknowns c 1, c 2 given by c 1 y 1 (x 0) + c 2y 2(x 0) = a c 2y 1(x 0) + c 2y 2(x 0) = b This system of equations is solvable if and only if the determinant, y 1 y 2 y 1 y 0. 2 Slide 22/27 Dr. John Ehrke Lecture 3 Fall 2012

23 The Wronskian Given y 1, y 2, the Wronskian of y 1, y 2, denoted W(y 1, y 2 ) is given by W(y 1, y 2 ) = y 1 y 2 y 1 y = y 1y 2 y 2 y 1 2 The Wronskian explains why linearly independent solutions are required since if y 2 = cy 1, then W(y 1, y 2 ) = y 1 y 2 y 1 y = y 1 cy 1 2 y 1 cy = cy 1y 1 cy 1 y 1 = 0 1 In this case, because W(y 1, y 2 ) = 0, then the resulting system of equations from the initial conditions is not solvable. This leads to a very important theorem, Abel s Theorem If y 1, y 2 are solutions to the ODE y + py + qy = 0, then either 1 W(y 1, y 2) 0 for all x 2 or W(y 1, y 2) 0 for all x Slide 23/27 Dr. John Ehrke Lecture 3 Fall 2012

24 Abel s Theorem Proof: To prove Abel s theorem we note that y 1 and y 2 satisfy y 1 + p(t)y 1 + q(t)y 1 = 0 (6) y 2 + p(t)y 2 + q(t)y 2 = 0 (7) If we multiply the first equation by y 2, the second by y 1 and add the results, we obtain (y 1 y 2 y 1 y 2) + p(t)(y 1 y 2 y 1y 2) = 0. (8) Next, we let W(t) = W(y 1, y 2)(t) and observe that W (t) = y 1 y 2 y 1 y 2. Then we can write (3) in the form W + p(t)w = 0 (9) which is a first order linear equation in W, and has solution ( ) W(t) = c exp p(t) dt where c is a constant. The value of c depends on the fundamental solutions, y 1 and y 2. However, since the exponential function is never 0, W(t) is not 0 unless c = 0, in which case W(t) 0 for all t, which completes the proof. (10) Slide 24/27 Dr. John Ehrke Lecture 3 Fall 2012

25 Normalized Solutions Not all solutions y 1 and y 2 are created equal. There are in fact, best solution, called normalized solutions which we will introduce below. Normalized Solutions Two solutions Y 1 and Y 2 of the second order linear ODE, y + py + qy = 0 are called normalized if they satisfy the initial conditions, Y 1 (0) = 1, Y 1 (0) = 0 Y 2 (0) = 0, Y 2 (0) = 1 Example Using the above definition, find the normalized solutions for y + y = 0 and y y = 0. Compare your results. Remember to use the characteristic equation to determine your fundamental set of solutions. Slide 25/27 Dr. John Ehrke Lecture 3 Fall 2012

26 Why do we care about normalized solutions? If Y 1 and Y 2 are normalized at x = 0, then the solution to the IVP y + py + qy = 0 with initial conditions y(0) = a, y (0) = b is y(x) = ay 1 + by 2. In example, we can simply read the solutions off straight from the ODE. This is highly desired by those in the engineering fields, and is generally a nicer version of linearly independent solutions than what we have been working with thus far. The general solution y(x) = c 1 Y 1 + c 2 Y 2 is called the normalized general solution. Slide 26/27 Dr. John Ehrke Lecture 3 Fall 2012

27 Wrapping up the discussion... We are now ready to finish answering the second of our two questions, Why all solutions look like c 1 y 1 + c 2y 2? First, we state the second order existence uniqueness theorem. Second Order Existence Uniqueness Theorem Given the second order linear differential equation y + p(x)y + q(x)y = 0, (11) where p and q are continuous for all x in a rectangle about the initial point x 0, there is exactly one solution satisfying the initial conditions, y(x 0) = a and y (x 0) = b. How does this help answer our question? Well, given a solution u(x) satisfying u(0) = a, u (0) = b, then by the existence uniqueness theorem above, there can only be one such solution, and we know that ay 1 + by 2 is a solution, where Y 1 and Y 2 are normalized solutions of (11). This says that u(x) is of the form c 1 y 1 + c 2y 2 for some fundamental set of solutions, y 1, y 2, which is what we needed to show. Slide 27/27 Dr. John Ehrke Lecture 3 Fall 2012

tegrals as General & Particular Solutions

tegrals as General & Particular Solutions tegrals as General & Particular Solutions dy dx = f(x) General Solution: y(x) = f(x) dx + C Particular Solution: dy dx = f(x), y(x 0) = y 0 Examples: 1) dy dx = (x 2)2 ;y(2) = 1; 2) dy ;y(0) = 0; 3) dx

More information

Second-Order Linear Differential Equations

Second-Order Linear Differential Equations Second-Order Linear Differential Equations A second-order linear differential equation has the form 1 Px d 2 y dx 2 dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. We saw in Section 7.1

More information

9. Particular Solutions of Non-homogeneous second order equations Undetermined Coefficients

9. Particular Solutions of Non-homogeneous second order equations Undetermined Coefficients September 29, 201 9-1 9. Particular Solutions of Non-homogeneous second order equations Undetermined Coefficients We have seen that in order to find the general solution to the second order differential

More information

So far, we have looked at homogeneous equations

So far, we have looked at homogeneous equations Chapter 3.6: equations Non-homogeneous So far, we have looked at homogeneous equations L[y] = y + p(t)y + q(t)y = 0. Homogeneous means that the right side is zero. Linear homogeneous equations satisfy

More information

1. First-order Ordinary Differential Equations

1. First-order Ordinary Differential Equations Advanced Engineering Mathematics 1. First-order ODEs 1 1. First-order Ordinary Differential Equations 1.1 Basic concept and ideas 1.2 Geometrical meaning of direction fields 1.3 Separable differential

More information

First Order Non-Linear Equations

First Order Non-Linear Equations First Order Non-Linear Equations We will briefly consider non-linear equations. In general, these may be much more difficult to solve than linear equations, but in some cases we will still be able to solve

More information

a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)

a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4) ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x

More information

2 Integrating Both Sides

2 Integrating Both Sides 2 Integrating Both Sides So far, the only general method we have for solving differential equations involves equations of the form y = f(x), where f(x) is any function of x. The solution to such an equation

More information

Lecture Notes on Polynomials

Lecture Notes on Polynomials Lecture Notes on Polynomials Arne Jensen Department of Mathematical Sciences Aalborg University c 008 Introduction These lecture notes give a very short introduction to polynomials with real and complex

More information

Nonhomogeneous Linear Equations

Nonhomogeneous Linear Equations Nonhomogeneous Linear Equations In this section we learn how to solve second-order nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form ay by cy G x where

More information

A First Course in Elementary Differential Equations. Marcel B. Finan Arkansas Tech University c All Rights Reserved

A First Course in Elementary Differential Equations. Marcel B. Finan Arkansas Tech University c All Rights Reserved A First Course in Elementary Differential Equations Marcel B. Finan Arkansas Tech University c All Rights Reserved 1 Contents 1 Basic Terminology 4 2 Qualitative Analysis: Direction Field of y = f(t, y)

More information

COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS

COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS BORIS HASSELBLATT CONTENTS. Introduction. Why complex numbers were introduced 3. Complex numbers, Euler s formula 3 4. Homogeneous differential equations 8 5.

More information

Lectures 5-6: Taylor Series

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,

More information

The integrating factor method (Sect. 2.1).

The integrating factor method (Sect. 2.1). The integrating factor method (Sect. 2.1). Overview of differential equations. Linear Ordinary Differential Equations. The integrating factor method. Constant coefficients. The Initial Value Problem. Variable

More information

THE COMPLEX EXPONENTIAL FUNCTION

THE COMPLEX EXPONENTIAL FUNCTION Math 307 THE COMPLEX EXPONENTIAL FUNCTION (These notes assume you are already familiar with the basic properties of complex numbers.) We make the following definition e iθ = cos θ + i sin θ. (1) This formula

More information

Solutions to Homework 10

Solutions to Homework 10 Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x

More information

ENCOURAGING THE INTEGRATION OF COMPLEX NUMBERS IN UNDERGRADUATE ORDINARY DIFFERENTIAL EQUATIONS

ENCOURAGING THE INTEGRATION OF COMPLEX NUMBERS IN UNDERGRADUATE ORDINARY DIFFERENTIAL EQUATIONS Texas College Mathematics Journal Volume 6, Number 2, Pages 18 24 S applied for(xx)0000-0 Article electronically published on September 23, 2009 ENCOURAGING THE INTEGRATION OF COMPLEX NUMBERS IN UNDERGRADUATE

More information

Differentiation and Integration

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

More information

Taylor and Maclaurin Series

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

More information

A Brief Review of Elementary Ordinary Differential Equations

A Brief Review of Elementary Ordinary Differential Equations 1 A Brief Review of Elementary Ordinary Differential Equations At various points in the material we will be covering, we will need to recall and use material normally covered in an elementary course on

More information

A Second Course in Elementary Differential Equations: Problems and Solutions. Marcel B. Finan Arkansas Tech University c All Rights Reserved

A Second Course in Elementary Differential Equations: Problems and Solutions. Marcel B. Finan Arkansas Tech University c All Rights Reserved A Second Course in Elementary Differential Equations: Problems and Solutions Marcel B. Finan Arkansas Tech University c All Rights Reserved Contents 8 Calculus of Matrix-Valued Functions of a Real Variable

More information

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS A second-order linear differential equation has the form 1 Px d y dx dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. Equations of this type arise

More information

Some Notes on Taylor Polynomials and Taylor Series

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

More information

Methods of Solution of Selected Differential Equations Carol A. Edwards Chandler-Gilbert Community College

Methods of Solution of Selected Differential Equations Carol A. Edwards Chandler-Gilbert Community College Methods of Solution of Selected Differential Equations Carol A. Edwards Chandler-Gilbert Community College Equations of Order One: Mdx + Ndy = 0 1. Separate variables. 2. M, N homogeneous of same degree:

More information

The Heat Equation. Lectures INF2320 p. 1/88

The Heat Equation. Lectures INF2320 p. 1/88 The Heat Equation Lectures INF232 p. 1/88 Lectures INF232 p. 2/88 The Heat Equation We study the heat equation: u t = u xx for x (,1), t >, (1) u(,t) = u(1,t) = for t >, (2) u(x,) = f(x) for x (,1), (3)

More information

19.6. Finding a Particular Integral. Introduction. Prerequisites. Learning Outcomes. Learning Style

19.6. Finding a Particular Integral. Introduction. Prerequisites. Learning Outcomes. Learning Style Finding a Particular Integral 19.6 Introduction We stated in Block 19.5 that the general solution of an inhomogeneous equation is the sum of the complementary function and a particular integral. We have

More information

Lecture 14: Section 3.3

Lecture 14: Section 3.3 Lecture 14: Section 3.3 Shuanglin Shao October 23, 2013 Definition. Two nonzero vectors u and v in R n are said to be orthogonal (or perpendicular) if u v = 0. We will also agree that the zero vector in

More information

Homework #2 Solutions

Homework #2 Solutions MAT Spring Problems Section.:, 8,, 4, 8 Section.5:,,, 4,, 6 Extra Problem # Homework # Solutions... Sketch likely solution curves through the given slope field for dy dx = x + y...8. Sketch likely solution

More information

3.8 Finding Antiderivatives; Divergence and Curl of a Vector Field

3.8 Finding Antiderivatives; Divergence and Curl of a Vector Field 3.8 Finding Antiderivatives; Divergence and Curl of a Vector Field 77 3.8 Finding Antiderivatives; Divergence and Curl of a Vector Field Overview: The antiderivative in one variable calculus is an important

More information

Derive 5: The Easiest... Just Got Better!

Derive 5: The Easiest... Just Got Better! Liverpool John Moores University, 1-15 July 000 Derive 5: The Easiest... Just Got Better! Michel Beaudin École de Technologie Supérieure, Canada Email; mbeaudin@seg.etsmtl.ca 1. Introduction Engineering

More information

The Fourth International DERIVE-TI92/89 Conference Liverpool, U.K., 12-15 July 2000. Derive 5: The Easiest... Just Got Better!

The Fourth International DERIVE-TI92/89 Conference Liverpool, U.K., 12-15 July 2000. Derive 5: The Easiest... Just Got Better! The Fourth International DERIVE-TI9/89 Conference Liverpool, U.K., -5 July 000 Derive 5: The Easiest... Just Got Better! Michel Beaudin École de technologie supérieure 00, rue Notre-Dame Ouest Montréal

More information

Math 201 Lecture 23: Power Series Method for Equations with Polynomial

Math 201 Lecture 23: Power Series Method for Equations with Polynomial Math 201 Lecture 23: Power Series Method for Equations with Polynomial Coefficients Mar. 07, 2012 Many examples here are taken from the textbook. The first number in () refers to the problem number in

More information

Finding y p in Constant-Coefficient Nonhomogenous Linear DEs

Finding y p in Constant-Coefficient Nonhomogenous Linear DEs Finding y p in Constant-Coefficient Nonhomogenous Linear DEs Introduction and procedure When solving DEs of the form the solution looks like ay + by + cy =, y = y c + y p where y c is the complementary

More information

r (t) = 2r(t) + sin t θ (t) = r(t) θ(t) + 1 = 1 1 θ(t) 1 9.4.4 Write the given system in matrix form x = Ax + f ( ) sin(t) x y 1 0 5 z = dy cos(t)

r (t) = 2r(t) + sin t θ (t) = r(t) θ(t) + 1 = 1 1 θ(t) 1 9.4.4 Write the given system in matrix form x = Ax + f ( ) sin(t) x y 1 0 5 z = dy cos(t) Solutions HW 9.4.2 Write the given system in matrix form x = Ax + f r (t) = 2r(t) + sin t θ (t) = r(t) θ(t) + We write this as ( ) r (t) θ (t) = ( ) ( ) 2 r(t) θ(t) + ( ) sin(t) 9.4.4 Write the given system

More information

CITY UNIVERSITY LONDON. BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION

CITY UNIVERSITY LONDON. BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION No: CITY UNIVERSITY LONDON BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION ENGINEERING MATHEMATICS 2 (resit) EX2005 Date: August

More information

MATH 425, PRACTICE FINAL EXAM SOLUTIONS.

MATH 425, PRACTICE FINAL EXAM SOLUTIONS. MATH 45, PRACTICE FINAL EXAM SOLUTIONS. Exercise. a Is the operator L defined on smooth functions of x, y by L u := u xx + cosu linear? b Does the answer change if we replace the operator L by the operator

More information

Solving DEs by Separation of Variables.

Solving DEs by Separation of Variables. Solving DEs by Separation of Variables. Introduction and procedure Separation of variables allows us to solve differential equations of the form The steps to solving such DEs are as follows: dx = gx).

More information

LS.6 Solution Matrices

LS.6 Solution Matrices LS.6 Solution Matrices In the literature, solutions to linear systems often are expressed using square matrices rather than vectors. You need to get used to the terminology. As before, we state the definitions

More information

Solutions for Review Problems

Solutions for Review Problems olutions for Review Problems 1. Let be the triangle with vertices A (,, ), B (4,, 1) and C (,, 1). (a) Find the cosine of the angle BAC at vertex A. (b) Find the area of the triangle ABC. (c) Find a vector

More information

Solving Cubic Polynomials

Solving Cubic Polynomials Solving Cubic Polynomials 1.1 The general solution to the quadratic equation There are four steps to finding the zeroes of a quadratic polynomial. 1. First divide by the leading term, making the polynomial

More information

Solutions to Linear First Order ODE s

Solutions to Linear First Order ODE s First Order Linear Equations In the previous session we learned that a first order linear inhomogeneous ODE for the unknown function x = x(t), has the standard form x + p(t)x = q(t) () (To be precise we

More information

6. Define log(z) so that π < I log(z) π. Discuss the identities e log(z) = z and log(e w ) = w.

6. Define log(z) so that π < I log(z) π. Discuss the identities e log(z) = z and log(e w ) = w. hapter omplex integration. omplex number quiz. Simplify 3+4i. 2. Simplify 3+4i. 3. Find the cube roots of. 4. Here are some identities for complex conjugate. Which ones need correction? z + w = z + w,

More information

3 Contour integrals and Cauchy s Theorem

3 Contour integrals and Cauchy s Theorem 3 ontour integrals and auchy s Theorem 3. Line integrals of complex functions Our goal here will be to discuss integration of complex functions = u + iv, with particular regard to analytic functions. Of

More information

correct-choice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were:

correct-choice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were: Topic 1 2.1 mode MultipleSelection text How can we approximate the slope of the tangent line to f(x) at a point x = a? This is a Multiple selection question, so you need to check all of the answers that

More information

Higher Order Equations

Higher Order Equations Higher Order Equations We briefly consider how what we have done with order two equations generalizes to higher order linear equations. Fortunately, the generalization is very straightforward: 1. Theory.

More information

System of First Order Differential Equations

System of First Order Differential Equations CHAPTER System of First Order Differential Equations In this chapter, we will discuss system of first order differential equations. There are many applications that involving find several unknown functions

More information

ARBITRAGE-FREE OPTION PRICING MODELS. Denis Bell. University of North Florida

ARBITRAGE-FREE OPTION PRICING MODELS. Denis Bell. University of North Florida ARBITRAGE-FREE OPTION PRICING MODELS Denis Bell University of North Florida Modelling Stock Prices Example American Express In mathematical finance, it is customary to model a stock price by an (Ito) stochatic

More information

Problem 1 (10 pts) Find the radius of convergence and interval of convergence of the series

Problem 1 (10 pts) Find the radius of convergence and interval of convergence of the series 1 Problem 1 (10 pts) Find the radius of convergence and interval of convergence of the series a n n=1 n(x + 2) n 5 n 1. n(x + 2)n Solution: Do the ratio test for the absolute convergence. Let a n =. Then,

More information

COMPLEX NUMBERS AND SERIES. Contents

COMPLEX NUMBERS AND SERIES. Contents COMPLEX NUMBERS AND SERIES MIKE BOYLE Contents 1. Complex Numbers Definition 1.1. A complex number is a number z of the form z = x + iy, where x and y are real numbers, and i is another number such that

More information

College of the Holy Cross, Spring 2009 Math 373, Partial Differential Equations Midterm 1 Practice Questions

College of the Holy Cross, Spring 2009 Math 373, Partial Differential Equations Midterm 1 Practice Questions College of the Holy Cross, Spring 29 Math 373, Partial Differential Equations Midterm 1 Practice Questions 1. (a) Find a solution of u x + u y + u = xy. Hint: Try a polynomial of degree 2. Solution. Use

More information

Differentiation of vectors

Differentiation of vectors Chapter 4 Differentiation of vectors 4.1 Vector-valued functions In the previous chapters we have considered real functions of several (usually two) variables f : D R, where D is a subset of R n, where

More information

Limits and Continuity

Limits and Continuity Math 20C Multivariable Calculus Lecture Limits and Continuity Slide Review of Limit. Side limits and squeeze theorem. Continuous functions of 2,3 variables. Review: Limits Slide 2 Definition Given a function

More information

MATH 381 HOMEWORK 2 SOLUTIONS

MATH 381 HOMEWORK 2 SOLUTIONS MATH 38 HOMEWORK SOLUTIONS Question (p.86 #8). If g(x)[e y e y ] is harmonic, g() =,g () =, find g(x). Let f(x, y) = g(x)[e y e y ].Then Since f(x, y) is harmonic, f + f = and we require x y f x = g (x)[e

More information

The Method of Partial Fractions Math 121 Calculus II Spring 2015

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

More information

INTERPOLATION. Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x, y).

INTERPOLATION. Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x, y). INTERPOLATION Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x, y). As an example, consider defining and x 0 =0, x 1 = π 4, x

More information

Computing divisors and common multiples of quasi-linear ordinary differential equations

Computing divisors and common multiples of quasi-linear ordinary differential equations Computing divisors and common multiples of quasi-linear ordinary differential equations Dima Grigoriev CNRS, Mathématiques, Université de Lille Villeneuve d Ascq, 59655, France Dmitry.Grigoryev@math.univ-lille1.fr

More information

ORDINARY DIFFERENTIAL EQUATIONS

ORDINARY DIFFERENTIAL EQUATIONS ORDINARY DIFFERENTIAL EQUATIONS GABRIEL NAGY Mathematics Department, Michigan State University, East Lansing, MI, 48824. SEPTEMBER 4, 25 Summary. This is an introduction to ordinary differential equations.

More information

Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: f (x) =

Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: f (x) = Vertical Asymptotes Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: lim f (x) = x a lim f (x) = lim x a lim f (x) = x a

More information

Introduction to Complex Numbers in Physics/Engineering

Introduction to Complex Numbers in Physics/Engineering Introduction to Complex Numbers in Physics/Engineering ference: Mary L. Boas, Mathematical Methods in the Physical Sciences Chapter 2 & 14 George Arfken, Mathematical Methods for Physicists Chapter 6 The

More information

COMPLEX NUMBERS. a bi c di a c b d i. a bi c di a c b d i For instance, 1 i 4 7i 1 4 1 7 i 5 6i

COMPLEX NUMBERS. a bi c di a c b d i. a bi c di a c b d i For instance, 1 i 4 7i 1 4 1 7 i 5 6i COMPLEX NUMBERS _4+i _-i FIGURE Complex numbers as points in the Arg plane i _i +i -i A complex number can be represented by an expression of the form a bi, where a b are real numbers i is a symbol with

More information

Matrix Methods for Linear Systems of Differential Equations

Matrix Methods for Linear Systems of Differential Equations Matrix Methods for Linear Systems of Differential Equations We now present an application of matrix methods to linear systems of differential equations. We shall follow the development given in Chapter

More information

Second Order Linear Nonhomogeneous Differential Equations; Method of Undetermined Coefficients. y + p(t) y + q(t) y = g(t), g(t) 0.

Second Order Linear Nonhomogeneous Differential Equations; Method of Undetermined Coefficients. y + p(t) y + q(t) y = g(t), g(t) 0. Second Order Linear Nonhomogeneous Differential Equations; Method of Undetermined Coefficients We will now turn our attention to nonhomogeneous second order linear equations, equations with the standard

More information

Rolle s Theorem. q( x) = 1

Rolle s Theorem. q( x) = 1 Lecture 1 :The Mean Value Theorem We know that constant functions have derivative zero. Is it possible for a more complicated function to have derivative zero? In this section we will answer this question

More information

Linear algebra and the geometry of quadratic equations. Similarity transformations and orthogonal matrices

Linear algebra and the geometry of quadratic equations. Similarity transformations and orthogonal matrices MATH 30 Differential Equations Spring 006 Linear algebra and the geometry of quadratic equations Similarity transformations and orthogonal matrices First, some things to recall from linear algebra Two

More information

Reference: Introduction to Partial Differential Equations by G. Folland, 1995, Chap. 3.

Reference: Introduction to Partial Differential Equations by G. Folland, 1995, Chap. 3. 5 Potential Theory Reference: Introduction to Partial Differential Equations by G. Folland, 995, Chap. 3. 5. Problems of Interest. In what follows, we consider Ω an open, bounded subset of R n with C 2

More information

Taylor Polynomials and Taylor Series Math 126

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

More information

Systems with Persistent Memory: the Observation Inequality Problems and Solutions

Systems with Persistent Memory: the Observation Inequality Problems and Solutions Chapter 6 Systems with Persistent Memory: the Observation Inequality Problems and Solutions Facts that are recalled in the problems wt) = ut) + 1 c A 1 s ] R c t s)) hws) + Ks r)wr)dr ds. 6.1) w = w +

More information

Zeros of Polynomial Functions

Zeros of Polynomial Functions Review: Synthetic Division Find (x 2-5x - 5x 3 + x 4 ) (5 + x). Factor Theorem Solve 2x 3-5x 2 + x + 2 =0 given that 2 is a zero of f(x) = 2x 3-5x 2 + x + 2. Zeros of Polynomial Functions Introduction

More information

DEFINITION 5.1.1 A complex number is a matrix of the form. x y. , y x

DEFINITION 5.1.1 A complex number is a matrix of the form. x y. , y x Chapter 5 COMPLEX NUMBERS 5.1 Constructing the complex numbers One way of introducing the field C of complex numbers is via the arithmetic of matrices. DEFINITION 5.1.1 A complex number is a matrix of

More information

Student name: Earlham College. Fall 2011 December 15, 2011

Student name: Earlham College. Fall 2011 December 15, 2011 Student name: Earlham College MATH 320: Differential Equations Final exam - In class part Fall 2011 December 15, 2011 Instructions: This is a regular closed-book test, and is to be taken without the use

More information

Chapter 5 Polar Coordinates; Vectors 5.1 Polar coordinates 1. Pole and polar axis

Chapter 5 Polar Coordinates; Vectors 5.1 Polar coordinates 1. Pole and polar axis Chapter 5 Polar Coordinates; Vectors 5.1 Polar coordinates 1. Pole and polar axis 2. Polar coordinates A point P in a polar coordinate system is represented by an ordered pair of numbers (r, θ). If r >

More information

22 Matrix exponent. Equal eigenvalues

22 Matrix exponent. Equal eigenvalues 22 Matrix exponent. Equal eigenvalues 22. Matrix exponent Consider a first order differential equation of the form y = ay, a R, with the initial condition y) = y. Of course, we know that the solution to

More information

FIXED POINT ITERATION

FIXED POINT ITERATION FIXED POINT ITERATION We begin with a computational example. solving the two equations Consider E1: x =1+.5sinx E2: x =3+2sinx Graphs of these two equations are shown on accompanying graphs, with the solutions

More information

3.4 Complex Zeros and the Fundamental Theorem of Algebra

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

More information

Some Polynomial Theorems. John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.

Some Polynomial Theorems. John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom. Some Polynomial Theorems by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.com This paper contains a collection of 31 theorems, lemmas,

More information

LINE INTEGRALS OF VECTOR FUNCTIONS: GREEN S THEOREM. Contents. 2. Green s Theorem 3

LINE INTEGRALS OF VECTOR FUNCTIONS: GREEN S THEOREM. Contents. 2. Green s Theorem 3 LINE INTEGRALS OF VETOR FUNTIONS: GREEN S THEOREM ontents 1. A differential criterion for conservative vector fields 1 2. Green s Theorem 3 1. A differential criterion for conservative vector fields We

More information

2 Complex Functions and the Cauchy-Riemann Equations

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)

More information

Chapter 4. Linear Second Order Equations. ay + by + cy = 0, (1) where a, b, c are constants. The associated auxiliary equation is., r 2 = b b 2 4ac 2a

Chapter 4. Linear Second Order Equations. ay + by + cy = 0, (1) where a, b, c are constants. The associated auxiliary equation is., r 2 = b b 2 4ac 2a Chapter 4. Linear Second Order Equations ay + by + cy = 0, (1) where a, b, c are constants. ar 2 + br + c = 0. (2) Consequently, y = e rx is a solution to (1) if an only if r satisfies (2). So, the equation

More information

11.7 Polar Form of Complex Numbers

11.7 Polar Form of Complex Numbers 11.7 Polar Form of Complex Numbers 989 11.7 Polar Form of Complex Numbers In this section, we return to our study of complex numbers which were first introduced in Section.. Recall that a complex number

More information

TOPIC 3: CONTINUITY OF FUNCTIONS

TOPIC 3: CONTINUITY OF FUNCTIONS TOPIC 3: CONTINUITY OF FUNCTIONS. Absolute value We work in the field of real numbers, R. For the study of the properties of functions we need the concept of absolute value of a number. Definition.. Let

More information

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization

More information

ECG590I Asset Pricing. Lecture 2: Present Value 1

ECG590I Asset Pricing. Lecture 2: Present Value 1 ECG59I Asset Pricing. Lecture 2: Present Value 1 2 Present Value If you have to decide between receiving 1$ now or 1$ one year from now, then you would rather have your money now. If you have to decide

More information

Problem Set 5 Due: In class Thursday, Oct. 18 Late papers will be accepted until 1:00 PM Friday.

Problem Set 5 Due: In class Thursday, Oct. 18 Late papers will be accepted until 1:00 PM Friday. Math 312, Fall 2012 Jerry L. Kazdan Problem Set 5 Due: In class Thursday, Oct. 18 Late papers will be accepted until 1:00 PM Friday. In addition to the problems below, you should also know how to solve

More information

6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives

6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives 6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise

More information

Separable First Order Differential Equations

Separable First Order Differential Equations Separable First Order Differential Equations Form of Separable Equations which take the form = gx hy or These are differential equations = gxĥy, where gx is a continuous function of x and hy is a continuously

More information

Copyrighted Material. Chapter 1 DEGREE OF A CURVE

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

More information

Sequences and Series

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

More information

Elementary Functions

Elementary Functions Chapter Three Elementary Functions 31 Introduction Complex functions are, of course, quite easy to come by they are simply ordered pairs of real-valued functions of two variables We have, however, already

More information

Calculus. Contents. Paul Sutcliffe. Office: CM212a.

Calculus. Contents. Paul Sutcliffe. Office: CM212a. Calculus Paul Sutcliffe Office: CM212a. www.maths.dur.ac.uk/~dma0pms/calc/calc.html Books One and several variables calculus, Salas, Hille & Etgen. Calculus, Spivak. Mathematical methods in the physical

More information

Math 2280 - Assignment 6

Math 2280 - Assignment 6 Math 2280 - Assignment 6 Dylan Zwick Spring 2014 Section 3.8-1, 3, 5, 8, 13 Section 4.1-1, 2, 13, 15, 22 Section 4.2-1, 10, 19, 28 1 Section 3.8 - Endpoint Problems and Eigenvalues 3.8.1 For the eigenvalue

More information

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

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

More information

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

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

More information

1 Inner Products and Norms on Real Vector Spaces

1 Inner Products and Norms on Real Vector Spaces Math 373: Principles Techniques of Applied Mathematics Spring 29 The 2 Inner Product 1 Inner Products Norms on Real Vector Spaces Recall that an inner product on a real vector space V is a function from

More information

On closed-form solutions to a class of ordinary differential equations

On closed-form solutions to a class of ordinary differential equations International Journal of Advanced Mathematical Sciences, 2 (1 (2014 57-70 c Science Publishing Corporation www.sciencepubco.com/index.php/ijams doi: 10.14419/ijams.v2i1.1556 Research Paper On closed-form

More information

Reducibility of Second Order Differential Operators with Rational Coefficients

Reducibility of Second Order Differential Operators with Rational Coefficients Reducibility of Second Order Differential Operators with Rational Coefficients Joseph Geisbauer University of Arkansas-Fort Smith Advisor: Dr. Jill Guerra May 10, 2007 1. INTRODUCTION In this paper we

More information

3.2 Sources, Sinks, Saddles, and Spirals

3.2 Sources, Sinks, Saddles, and Spirals 3.2. Sources, Sinks, Saddles, and Spirals 6 3.2 Sources, Sinks, Saddles, and Spirals The pictures in this section show solutions to Ay 00 C By 0 C Cy D 0. These are linear equations with constant coefficients

More information

Figure 2.1: Center of mass of four points.

Figure 2.1: Center of mass of four points. Chapter 2 Bézier curves are named after their inventor, Dr. Pierre Bézier. Bézier was an engineer with the Renault car company and set out in the early 196 s to develop a curve formulation which would

More information

Lecture 13 Linear quadratic Lyapunov theory

Lecture 13 Linear quadratic Lyapunov theory EE363 Winter 28-9 Lecture 13 Linear quadratic Lyapunov theory the Lyapunov equation Lyapunov stability conditions the Lyapunov operator and integral evaluating quadratic integrals analysis of ARE discrete-time

More information

REVIEW EXERCISES DAVID J LOWRY

REVIEW EXERCISES DAVID J LOWRY REVIEW EXERCISES DAVID J LOWRY Contents 1. Introduction 1 2. Elementary Functions 1 2.1. Factoring and Solving Quadratics 1 2.2. Polynomial Inequalities 3 2.3. Rational Functions 4 2.4. Exponentials and

More information

1 if 1 x 0 1 if 0 x 1

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

More information