# 3.5 Spline interpolation

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 3.5 Spline interpolation Given a tabulated function f k = f(x k ), k = 0,... N, a spline is a polynomial between each pair of tabulated points, but one whose coefficients are determined slightly non-locally. The non-locality is designed to guarantee global smoothness in the interpolated function up to some order of derivative. Cubic splines are the most popular. They produce an interpolated function that is continuous through to the second derivative. Splines tend to be stabler than fitting a polynomial through the N + points, with less possibility of wild oscillations between the tabulated points. We shall explain how spline interpolation works by first going through the theory and then applying it to interpolate the function below: 0.5 f 5 f f f 4 f 7 f f 8 f 2 f Figure 3.: The function f(x) = +x 2 sin(x) between [ 4, 4]. The values of f(x k ) are given for i = 0,..., 8, x k = 4, 3, 2,, 0,, 2, 3, 4. Figure 3. shows the function f(x) = sin x in the region x [ 4, 4]. We are given the value of the +x 2 function at the points f i = f(x k ), k = 0,..., 8, where x k = 4, 3,..., 0,..., 3, 4. These are what is called the interpolating nodes Linear Spline Let s focus attention on one particular interval (x k, x k+ ). Linear interpolation in that interval gives the interpolation formula f = Af k + Bf k+ (3.) where A x k+ x x k+ x k, B A = x x k x k+ x k (3.2)

2 This is just like the piecewise Lagrange polynomial interpolation we looked at earlier. Figure 3.2 shows the piecewise interpolated function over the full range [ 4, 4]. 0.5 f 5 f f f 4 f 7 f f 8 f 2 f Figure 3.2: Piecewise linear interpolation with 9 nodes (solid) for the function +x 2 sin x between [ 4, 4] (dotted). We can see that linear interpolation works quite well for larger values of x but does particularly badly in (, ) as it fails to capture the curvature of the function. The accuracy can be improved by using more interpolating nodes, but an important issue is that the first derivatives of the interpolating function are discontinuous at the nodes Cubic spline interpolation The goal of cubic spline interpolation is to get an interpolation formula that is continuous in both the first and second derivatives, both within the intervals and at the interpolating nodes. This will give us a smoother interpolating function. In general, if the function you want to approximate is smooth, then cubic splines will do better than piecewise linear interpolation. Before you read on, I d like you to clear your mind a little. The following is a derivation of the cubic interpolation formula from Chapter 3 of Numerical Recipes by Press et al., which is a slight variation on the B 0 splines we discussed in lectures, but in my opinion illustrates the fundamental idea of spline interpolation more clearly. If you understand this, then you will find working with the B 0 splines much easier. Let s restate the problem. We have a function f(x) that is tabulated at the N + points f k = f(x k ), k = 0,..., N. In each interval (x k, x k+ ), we can fit a straight line through the points (x k, f k ) and (x k+, f k+ ) using the formula given by (3.). The problem is that with a linear function, the first derivative is not contin- 2

3 uous at the boundary between two adjacent intervals, while we want the second derivative to be continuous, even at the boundary! Now suppose, contrary to fact, that in addition to the tabulated values of f i, we also have tabulated values for the function s second derivatives, that is, a set of numbers f i. Then within each interval (x k, x k+ ), we can add to the right-hand side of equation (3.) a cubic polynomial whose second derivative varies linearly from a value f k on the left to a value f k+ on the right. Doing so, we will have the desired continuous second derivative. If we also construct the cubic polynomial to have zero values at x k and x k+, then adding it in will not spoil the agreement with the tabulated functional values f k and f k+ at the endpoints x k and x k+. A little side calculation shows that there is only one way to arrange this construction, namely replacing equation (3.) by where A and B are defined as before and f = Af k + Bf k+ + C f k + D f k+ (3.3) C (A3 A)(x k+ x k ) 2, D (B3 B)(x k+ x k ) 2 (3.4) Note that since A and B are linearly dependent on x, C and D (through A and B) have cubic x-dependence. We can readily check that f is in fact the second derivative of the new interpolating polynomial. We take derivatives of equation (3.3) with respect to x, using the definitions of A, B, C and D to compute da/dx, db/dx, dc/dx and dd/dx. The result is for the first derivative, and d f dx = f k+ f k 3A2 (x k+ x k ) f k x k+ x k + 3B2 (x k+ x k ) f k+ (3.5) d 2 f dx 2 = Af k + Bf k+ (3.) for the second derivative. Since A = at x k and A = 0 at x k+, and B = 0 at x k and B = at x k+, (3.) shows that f is just the tabulated second derivative, and also that the second derivative will be continuous across the boundary between two intervals, say (x k, x k ) and (x k, x k+ ). The only problem now is that we supposed the f k s to be known, when actually, they are not. However, we have not yet required that the first derivative, computed from (3.5), be continuous across the boundary between two intervals. The key idea of a cubic spline is to require this continuity and to use it to get equations for the second derivatives f k. The required equations are obtained by setting equation (3.5) evaluated for x = x k in the interval (x k, x k ) equal to the same equation evaluated for x = x k but in the interval (x k, x k+ ). With some rearrangement, this give (for k =,..., N ) x k x k f k + x k+ x k f k 3 + x k+ x k f k+ = f k+ f k f k f k (3.7) x k+ x k x k x k These are N linear equations in the N + unknowns f i, i = 0,..., N. Therefore there is a two-parameter family of possible solutions. For a unique solution, we need to specify further conditions, typically taken as boundary conditions at x 0 and x N. The most common ways of doing this is to set both f 0 and f N equal to zero, giving the so-called 3

4 natural cubic spline, which has zero second derivatives on both of its boundaries. Now we have the solution for f k, k = 0,..., N, we can substitute back into equation (3.3) to give the cubic interpolation formula in each interval (x k, x k+ ). This is probably the easiest approach to use if you want to write a numerical code to calculate cubic spline interpolation. So what are B-splines? In the above approach, we started with a functional form for the interpolation formula ( f = Af k + Bf k+ + C f k + D f k+ ), and had to use constraints ( f continuous at interval boundaries) to solve for f k. As mathematicians, we like to build things up from fundamental units. In this case, we can use a set of piecewise cubic polynomials defined on some sub-interval of [x 0, x N ], which are by construction continuous through to second derivative at the boundaries of intervals. They would form a set of basis functions, since linear combinations of these functions would also satisfy the continuity properties at the boundaries between adjacent intervals. To construct the cubic spline over the whole range [x 0, x N ], we would then simply need match the sum of the basis functions with tabulated values of f i at the interpolating nodes x i, i = 0,..., N. The B-splines are the basis functions that satisfy our continuity conditions. If the interpolation is over the region [x 0, x N ], then B 0 is defined by B 0 (x) = 0 x x 0 2h (2h + (x x 0)) 3 x 0 2h x x 0 h 2h (x x 0) 2 (2h + (x x 0 )) x 0 h x x 0 2h (x x 0) 2 (2h (x x 0 )) x 0 x x 0 + h (2h (x x 0)) 3 x 0 + h x x 0 + 2h 0 x x 0 + 2h (3.8) where h = x k+ x k = x N x 0 N is the width between interpolating nodes (assumed to be equal). Note that B 0 has non-zero values over four intervals. We can easily check that B 0 satisfy the continuity conditions at the boundaries of the intervals, namely, B 0, B 0 and B 0 are continuous at 2h, h, 0, h and 2h. B k- (x) B k (x) B k+ (x) B k+2 (x) x k-2 x k- x k x k+ x k+2 Figure 3.3: Cubic B-spline 4

5 We define B k (x) = B 0 (x kh + x 0 ) (i.e. the B 0 functions shifted to the right by k nodes). This is illustrated in Figure 3.3, which shows B k, B k, B k+ and B k+2. Note that all four of these basis functions are non-zero in the interval (x k, x k+ ). The cubic spline function, S 3 (x) in [x 0, x N ] is then written as the linear combination of the B k s: S 3 (x) = N+ a k B k (x) (3.9) k= The sum is from to N +, since B is nonzero in the interval (x 0, x ), and B N+ is nonzero in the interval (x N, x N ), so we need to take into account these functions. Is the problem now completely specified? We know that we need 4 conditions (coefficients) to uniquely specify a cubic, and in [x 0, x N ] there are N intervals, so altogether a total of 4N conditions are needed. The continuity conditions are automatically satisfied in the N interior points, since the B k s satisfy the continuity conditions (that s 3(N ) conditions). The other requirement is that S 3 must match the tabulated points, i.e. S 3 (x k ) = f(x k ), for k = 0,..., N (N + conditions). So there are two unspecified conditions, and like before we can take S 3 (x 0 ) = S 3 (x N ) = 0 (i.e., curvature equals zero at the boundaries, natural spline ). Now evaluating S 3 (x) at x k : S 3 (x k ) = a k B k (x k ) + a k B k (x k ) + a k+ B k+ (x k ) + a k+2 B k+2 (x k ) = f(x k ) (3.0) (all other B k s are zero) From the definitions of B k (x) and B 0 (x), we find B k (x k ) = B 0 (x 0 ) = 2h3 3 B k (x k ) = B 0 (x 0 + h) = h3 B k+ (x k ) = B 0 (x 0 h) = h3 B k+2 (x k ) = B 0 (x 0 2h) = 0 Substituting into (3.0), we get the recurrence relation for the coefficients a k a k + 4a k + a k+ = h 3 f(x k) (3.) for k = 0,..., N. What happens at the boundaries? Remember we set S 3 (x 0 ) = S 3 (x N ) = 0. Differentiating S 3 (x) gives So we need to find the second derivatives of B k. S 3 N+ (x) = a k B k (x) (3.2) k= 5

6 Differentiating B 0 twice, we get 0 (x) = B 0 x x 0 2h 2h + (x x 0 ) x 0 2h x x 0 h 2h 3(x x 0 ) x 0 h x x 0 2h + 3(x x 0 ) x 0 x x 0 + h 2h (x x 0 ) x 0 + h x x 0 + 2h 0 (3.3) So as before, we have 0 = S 3 (x 0 ) = a B (x 0 ) + a 0 B 0 (x 0 ) + a B (x 0 ) + a 2 B 2 (x 0 ) = a B 0 (x 0 + h) + a 0 B 0 (x 0 ) + a B (x 0 h) = a h 2ha 0 + a h = a 2a 0 + a (3.4) From (3.0), we have a + 4a 0 + a = h 3 f(x 0) (3.5) Subtract (3.4) from (3.5), we obtain Similarly, we find We now can write a matrix equation to solve for the coefficients a 0,..., a N : a 0 = h 3 f(x 0) (3.) a N = h 3 f(x N) (3.7) a 0 a a 2 a N a N = h 3 f(x 0 ) f(x ) f(x 2 ) f(x N ) f(x N ) (3.8) This set of equations is tridiagonal and can be solved in O(N) operations by the tridiagonal algorithm. The final two coefficients to completely determine S 3 are a = 2a 0 a a N+ = 2a N a N (3.9)

7 Worked Example Now we can try to apply cubic spline interpolation to the function f(x) = +x 2 sin x, 4 x 4, with the 9 nodes x k = 4, 3, 2,, 0,, 2, 3, 4. Here x 0 = 4, x N = 4 and h =. So from equation (3.8), the basis function B 0 (x) is given by B 0 (x) = 0 x (x + )3 x (x + 4)2 (x + ) 5 x (x + 4)2 (x + 2) 4 x 3 (x + 2)3 3 x 2 0 x 2 (3.20) The cubic spline S 3 (x) = 9 k= a k B k (x), and we can find the coefficients a k, k = 0,,..., 8 by solving the linear system a 0 a a 2 a 3 a 4 a 5 a a 7 a 8 f( 4) f( 3) f( 2) f( ) = f(0) f() f(2) f(3) f(4) (3.2) This can be readily solved to give a 0,..., a 8. Additionally, a = 2a 0 a, a 9 = 2a 8 a 7. The numerical values are a = , a 9 = a 0 = a = a 2 = a 3 = a 4 = a 5 = a = a 7 = a 8 = We shall just show the construction of S 3 (x) in the interval [ 4, 3]: (3.22) S 3 (x) = a B (x) + a 0 B 0 (x) + a B (x) + a 2 B 2 (x) = a B 0 (x + ) + a 0 B 0 (x) + a B 0 (x ) + a 2 B 0 (x 2) = a ((x + ) + 2)3 + a 0 ( (x + 4)2 (x + 2)) +a ( ((x ) + 4)2 ((x ) + )) + a 2 ( ((x 2) + )3 ) So need to work out the functional form of S 3 (x) for each interval. The whole process is somewhat tedious, but as shown in the next figure, it does give the right result! 7

8 Figure 3.4: Here s the function f(x) = +x 2 sin x again (dotted), and now with the cubic spline interpolation for [ 4, 3] superimposed on top. 8

### a = x 1 < x 2 < < x n = b, and a low degree polynomial is used to approximate f(x) on each subinterval. Example: piecewise linear approximation S(x)

Spline Background SPLINE INTERPOLATION Problem: high degree interpolating polynomials often have extra oscillations. Example: Runge function f(x) = 1 1+4x 2, x [ 1, 1]. 1 1/(1+4x 2 ) and P 8 (x) and P

### Piecewise Cubic Splines

280 CHAP. 5 CURVE FITTING Piecewise Cubic Splines The fitting of a polynomial curve to a set of data points has applications in CAD (computer-assisted design), CAM (computer-assisted manufacturing), and

### i = 0, 1,, N p(x i ) = f i,

Chapter 4 Curve Fitting We consider two commonly used methods for curve fitting, namely interpolation and least squares For interpolation, we use first polynomials then splines Then, we introduce least

### 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

### since by using a computer we are limited to the use of elementary arithmetic operations

> 4. Interpolation and Approximation Most functions cannot be evaluated exactly: x, e x, ln x, trigonometric functions since by using a computer we are limited to the use of elementary arithmetic operations

### the points are called control points approximating curve

Chapter 4 Spline Curves A spline curve is a mathematical representation for which it is easy to build an interface that will allow a user to design and control the shape of complex curves and surfaces.

### 3.4. Solving Simultaneous Linear Equations. Introduction. Prerequisites. Learning Outcomes

Solving Simultaneous Linear Equations 3.4 Introduction Equations often arise in which there is more than one unknown quantity. When this is the case there will usually be more than one equation involved.

### 3.4. Solving simultaneous linear equations. Introduction. Prerequisites. Learning Outcomes

Solving simultaneous linear equations 3.4 Introduction Equations often arise in which there is more than one unknown quantity. When this is the case there will usually be more than one equation involved.

### 1 Cubic Hermite Spline Interpolation

cs412: introduction to numerical analysis 10/26/10 Lecture 13: Cubic Hermite Spline Interpolation II Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mark Cowlishaw, Nathanael Fillmore 1 Cubic Hermite

### November 16, 2015. Interpolation, Extrapolation & Polynomial Approximation

Interpolation, Extrapolation & Polynomial Approximation November 16, 2015 Introduction In many cases we know the values of a function f (x) at a set of points x 1, x 2,..., x N, but we don t have the analytic

### 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

### Polynomials and Vieta s Formulas

Polynomials and Vieta s Formulas Misha Lavrov ARML Practice 2/9/2014 Review problems 1 If a 0 = 0 and a n = 3a n 1 + 2, find a 100. 2 If b 0 = 0 and b n = n 2 b n 1, find b 100. Review problems 1 If a

### Solving simultaneous equations. Jackie Nicholas

Mathematics Learning Centre Solving simultaneous equations Jackie Nicholas c 2005 University of Sydney Mathematics Learning Centre, University of Sydney 1 1 Simultaneous linear equations We will introduce

### We want to define smooth curves: - for defining paths of cameras or objects. - for defining 1D shapes of objects

lecture 10 - cubic curves - cubic splines - bicubic surfaces We want to define smooth curves: - for defining paths of cameras or objects - for defining 1D shapes of objects We want to define smooth surfaces

### The KaleidaGraph Guide to Curve Fitting

The KaleidaGraph Guide to Curve Fitting Contents Chapter 1 Curve Fitting Overview 1.1 Purpose of Curve Fitting... 5 1.2 Types of Curve Fits... 5 Least Squares Curve Fits... 5 Nonlinear Curve Fits... 6

### 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

### Introduction to the Finite Element Method (FEM)

Introduction to the Finite Element Method (FEM) ecture First and Second Order One Dimensional Shape Functions Dr. J. Dean Discretisation Consider the temperature distribution along the one-dimensional

### 4.5 Chebyshev Polynomials

230 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION 4.5 Chebyshev Polynomials We now turn our attention to polynomial interpolation for f (x) over [ 1, 1] based on the nodes 1 x 0 < x 1 < < x N 1. Both

Section 5.4 The Quadratic Formula 481 5.4 The Quadratic Formula Consider the general quadratic function f(x) = ax + bx + c. In the previous section, we learned that we can find the zeros of this function

### Polynomials can be added or subtracted simply by adding or subtracting the corresponding terms, e.g., if

1. Polynomials 1.1. Definitions A polynomial in x is an expression obtained by taking powers of x, multiplying them by constants, and adding them. It can be written in the form c 0 x n + c 1 x n 1 + c

### 1 Gaussian Elimination

Contents 1 Gaussian Elimination 1.1 Elementary Row Operations 1.2 Some matrices whose associated system of equations are easy to solve 1.3 Gaussian Elimination 1.4 Gauss-Jordan reduction and the Reduced

### Curve Fitting. Next: Numerical Differentiation and Integration Up: Numerical Analysis for Chemical Previous: Optimization.

Next: Numerical Differentiation and Integration Up: Numerical Analysis for Chemical Previous: Optimization Subsections Least-Squares Regression Linear Regression General Linear Least-Squares Nonlinear

### HIGHER ORDER DIFFERENTIAL EQUATIONS

HIGHER ORDER DIFFERENTIAL EQUATIONS 1 Higher Order Equations Consider the differential equation (1) y (n) (x) f(x, y(x), y (x),, y (n 1) (x)) 11 The Existence and Uniqueness Theorem 12 The general solution

### Linear Dependence Tests

Linear Dependence Tests The book omits a few key tests for checking the linear dependence of vectors. These short notes discuss these tests, as well as the reasoning behind them. Our first test checks

### We can display an object on a monitor screen in three different computer-model forms: Wireframe model Surface Model Solid model

CHAPTER 4 CURVES 4.1 Introduction In order to understand the significance of curves, we should look into the types of model representations that are used in geometric modeling. Curves play a very significant

### (Refer Slide Time: 1:42)

Introduction to Computer Graphics Dr. Prem Kalra Department of Computer Science and Engineering Indian Institute of Technology, Delhi Lecture - 10 Curves So today we are going to have a new topic. So far

### Computer Graphics. Geometric Modeling. Page 1. Copyright Gotsman, Elber, Barequet, Karni, Sheffer Computer Science - Technion. An Example.

An Example 2 3 4 Outline Objective: Develop methods and algorithms to mathematically model shape of real world objects Categories: Wire-Frame Representation Object is represented as as a set of points

### is identically equal to x 2 +3x +2

Partial fractions 3.6 Introduction It is often helpful to break down a complicated algebraic fraction into a sum of simpler fractions. 4x+7 For example it can be shown that has the same value as 1 + 3

### Math 407A: Linear Optimization

Math 407A: Linear Optimization Lecture 4: LP Standard Form 1 1 Author: James Burke, University of Washington LPs in Standard Form Minimization maximization Linear equations to linear inequalities Lower

### Basic Terminology for Systems of Equations in a Nutshell. E. L. Lady. 3x 1 7x 2 +4x 3 =0 5x 1 +8x 2 12x 3 =0.

Basic Terminology for Systems of Equations in a Nutshell E L Lady A system of linear equations is something like the following: x 7x +4x =0 5x +8x x = Note that the number of equations is not required

### 1 Review of Least Squares Solutions to Overdetermined Systems

cs4: introduction to numerical analysis /9/0 Lecture 7: Rectangular Systems and Numerical Integration Instructor: Professor Amos Ron Scribes: Mark Cowlishaw, Nathanael Fillmore Review of Least Squares

### 3.1. Solving linear equations. Introduction. Prerequisites. Learning Outcomes. Learning Style

Solving linear equations 3.1 Introduction Many problems in engineering reduce to the solution of an equation or a set of equations. An equation is a type of mathematical expression which contains one or

### 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

### Natural cubic splines

Natural cubic splines Arne Morten Kvarving Department of Mathematical Sciences Norwegian University of Science and Technology October 21 2008 Motivation We are given a large dataset, i.e. a function sampled

### Algebra Revision Sheet Questions 2 and 3 of Paper 1

Algebra Revision Sheet Questions and of Paper Simple Equations Step Get rid of brackets or fractions Step Take the x s to one side of the equals sign and the numbers to the other (remember to change the

### Zeros of Polynomial Functions

Zeros of Polynomial Functions The Rational Zero Theorem If f (x) = a n x n + a n-1 x n-1 + + a 1 x + a 0 has integer coefficients and p/q (where p/q is reduced) is a rational zero, then p is a factor of

### Finite Element Analysis Prof. Dr. B. N. Rao Department of Civil Engineering Indian Institute of Technology, Madras. Lecture - 28

Finite Element Analysis Prof. Dr. B. N. Rao Department of Civil Engineering Indian Institute of Technology, Madras Lecture - 28 In the earlier lectures, we have seen formulation for 3 node linear triangular

### EXAM. Practice Questions for Exam #2. Math 3350, Spring April 3, 2004 ANSWERS

EXAM Practice Questions for Exam #2 Math 3350, Spring 2004 April 3, 2004 ANSWERS i Problem 1. Find the general solution. A. D 3 (D 2)(D 3) 2 y = 0. The characteristic polynomial is λ 3 (λ 2)(λ 3) 2. Thus,

### Høgskolen i Narvik Sivilingeniørutdanningen STE6237 ELEMENTMETODER. Oppgaver

Høgskolen i Narvik Sivilingeniørutdanningen STE637 ELEMENTMETODER Oppgaver Klasse: 4.ID, 4.IT Ekstern Professor: Gregory A. Chechkin e-mail: chechkin@mech.math.msu.su Narvik 6 PART I Task. Consider two-point

### 3.6. Partial Fractions. Introduction. Prerequisites. Learning Outcomes

Partial Fractions 3.6 Introduction It is often helpful to break down a complicated algebraic fraction into a sum of simpler fractions. For 4x + 7 example it can be shown that x 2 + 3x + 2 has the same

### Equations, Inequalities & Partial Fractions

Contents Equations, Inequalities & Partial Fractions.1 Solving Linear Equations 2.2 Solving Quadratic Equations 1. Solving Polynomial Equations 1.4 Solving Simultaneous Linear Equations 42.5 Solving Inequalities

### Homework set 1: posted on website

Homework set 1: posted on website 1 Last time: Interpolation, fitting and smoothing Interpolated curve passes through all data points Fitted curve passes closest (by some criterion) to all data points

### CHAPTER 13. Definite Integrals. Since integration can be used in a practical sense in many applications it is often

7 CHAPTER Definite Integrals Since integration can be used in a practical sense in many applications it is often useful to have integrals evaluated for different values of the variable of integration.

### UNIT - I LESSON - 1 The Solution of Numerical Algebraic and Transcendental Equations

UNIT - I LESSON - 1 The Solution of Numerical Algebraic and Transcendental Equations Contents: 1.0 Aims and Objectives 1.1 Introduction 1.2 Bisection Method 1.2.1 Definition 1.2.2 Computation of real root

### Chapter 20. Vector Spaces and Bases

Chapter 20. Vector Spaces and Bases In this course, we have proceeded step-by-step through low-dimensional Linear Algebra. We have looked at lines, planes, hyperplanes, and have seen that there is no limit

### 4 Solving Systems of Equations by Reducing Matrices

Math 15 Sec S0601/S060 4 Solving Systems of Equations by Reducing Matrices 4.1 Introduction One of the main applications of matrix methods is the solution of systems of linear equations. Consider for example

### Numerical integration of a function known only through data points

Numerical integration of a function known only through data points Suppose you are working on a project to determine the total amount of some quantity based on measurements of a rate. For example, you

### Finite Element Analysis Prof. Dr. B. N. Rao Department of Civil Engineering Indian Institute of Technology, Madras. Lecture - 01

Finite Element Analysis Prof. Dr. B. N. Rao Department of Civil Engineering Indian Institute of Technology, Madras Lecture - 01 Welcome to the series of lectures, on finite element analysis. Before I start,

### Math 181 Spring 2007 HW 1 Corrected

Math 181 Spring 2007 HW 1 Corrected February 1, 2007 Sec. 1.1 # 2 The graphs of f and g are given (see the graph in the book). (a) State the values of f( 4) and g(3). Find 4 on the x-axis (horizontal axis)

### Module M1.4 Solving equations

F L E X I B L E L E A R N I N G A P P R O A C H T O P H Y S I C S Module M.4 Opening items. Module introduction.2 Fast track questions. Ready to study? 2 Solving linear equations 2. Simple linear equations

### 3 Polynomial Interpolation

3 Polynomial Interpolation Read sections 7., 7., 7.3. 7.3.3 (up to p. 39), 7.3.5. Review questions 7. 7.4, 7.8 7.0, 7. 7.4, 7.7, 7.8. All methods for solving ordinary differential equations which we considered

### SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING. Self Study Course

SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Study Course MODULE 17 MATRICES II Module Topics 1. Inverse of matrix using cofactors 2. Sets of linear equations 3. Solution of sets of linear

### 5 Numerical Differentiation

D. Levy 5 Numerical Differentiation 5. Basic Concepts This chapter deals with numerical approximations of derivatives. The first questions that comes up to mind is: why do we need to approximate derivatives

### 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

### 5.1. Systems of Linear Equations. Linear Systems Substitution Method Elimination Method Special Systems

5.1 Systems of Linear Equations Linear Systems Substitution Method Elimination Method Special Systems 5.1-1 Linear Systems The possible graphs of a linear system in two unknowns are as follows. 1. The

### Interpolating Polynomials Handout March 7, 2012

Interpolating Polynomials Handout March 7, 212 Again we work over our favorite field F (such as R, Q, C or F p ) We wish to find a polynomial y = f(x) passing through n specified data points (x 1,y 1 ),

### TOPIC 4: DERIVATIVES

TOPIC 4: DERIVATIVES 1. The derivative of a function. Differentiation rules 1.1. The slope of a curve. The slope of a curve at a point P is a measure of the steepness of the curve. If Q is a point on the

### Further Maths Matrix Summary

Further Maths Matrix Summary A matrix is a rectangular array of numbers arranged in rows and columns. The numbers in a matrix are called the elements of the matrix. The order of a matrix is the number

### Chapter 4. Shape Functions

Chapter 4 Shape Functions In the finite element method, continuous models are approximated using information at a finite number of discrete locations. Dividing the structure into discrete elements is called

### Solving Systems of Linear Equations. Substitution

Solving Systems of Linear Equations There are two basic methods we will use to solve systems of linear equations: Substitution Elimination We will describe each for a system of two equations in two unknowns,

### arxiv:cs/ v1 [cs.gr] 22 Mar 2005

arxiv:cs/0503054v1 [cs.gr] 22 Mar 2005 ANALYTIC DEFINITION OF CURVES AND SURFACES BY PARABOLIC BLENDING by A.W. Overhauser Mathematical and Theoretical Sciences Department Scientific Laboratory, Ford Motor

### Appendix E: Graphing Data

You will often make scatter diagrams and line graphs to illustrate the data that you collect. Scatter diagrams are often used to show the relationship between two variables. For example, in an absorbance

### 3.2. Solving quadratic equations. Introduction. Prerequisites. Learning Outcomes. Learning Style

Solving quadratic equations 3.2 Introduction A quadratic equation is one which can be written in the form ax 2 + bx + c = 0 where a, b and c are numbers and x is the unknown whose value(s) we wish to find.

### THEORY OF SIMPLEX METHOD

Chapter THEORY OF SIMPLEX METHOD Mathematical Programming Problems A mathematical programming problem is an optimization problem of finding the values of the unknown variables x, x,, x n that maximize

### Mth 95 Module 2 Spring 2014

Mth 95 Module Spring 014 Section 5.3 Polynomials and Polynomial Functions Vocabulary of Polynomials A term is a number, a variable, or a product of numbers and variables raised to powers. Terms in an expression

### CHAPTER 1 Splines and B-splines an Introduction

CHAPTER 1 Splines and B-splines an Introduction In this first chapter, we consider the following fundamental problem: Given a set of points in the plane, determine a smooth curve that approximates the

### An Introduction to Separation of Variables with Fourier Series Math 391w, Spring 2010 Tim McCrossen Professor Haessig

An Introduction to Separation of Variables with Fourier Series Math 391w, Spring 2010 Tim McCrossen Professor Haessig Abstract: This paper aims to give students who have not yet taken a course in partial

### 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

### is identically equal to x 2 +3x +2

Partial fractions.6 Introduction It is often helpful to break down a complicated algebraic fraction into a sum of simpler fractions. 4x+7 For example it can be shown that has the same value as + for any

### 3. INNER PRODUCT SPACES

. INNER PRODUCT SPACES.. Definition So far we have studied abstract vector spaces. These are a generalisation of the geometric spaces R and R. But these have more structure than just that of a vector space.

### An Insight into Division Algorithm, Remainder and Factor Theorem

An Insight into Division Algorithm, Remainder and Factor Theorem Division Algorithm Recall division of a positive integer by another positive integer For eample, 78 7, we get and remainder We confine the

### 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

### Review Sheet for Third Midterm Mathematics 1300, Calculus 1

Review Sheet for Third Midterm Mathematics 1300, Calculus 1 1. For f(x) = x 3 3x 2 on 1 x 3, find the critical points of f, the inflection points, the values of f at all these points and the endpoints,

### Inverses. Stephen Boyd. EE103 Stanford University. October 27, 2015

Inverses Stephen Boyd EE103 Stanford University October 27, 2015 Outline Left and right inverses Inverse Solving linear equations Examples Pseudo-inverse Left and right inverses 2 Left inverses a number

### with "a", "b" and "c" representing real numbers, and "a" is not equal to zero.

3.1 SOLVING QUADRATIC EQUATIONS: * A QUADRATIC is a polynomial whose highest exponent is. * The "standard form" of a quadratic equation is: ax + bx + c = 0 with "a", "b" and "c" representing real numbers,

### Sample Problems. Lecture Notes Equations with Parameters page 1

Lecture Notes Equations with Parameters page Sample Problems. In each of the parametric equations given, nd the value of the parameter m so that the equation has exactly one real solution. a) x + mx m

### Ordered Pairs. Graphing Lines and Linear Inequalities, Solving System of Linear Equations. Cartesian Coordinates System.

Ordered Pairs Graphing Lines and Linear Inequalities, Solving System of Linear Equations Peter Lo All equations in two variables, such as y = mx + c, is satisfied only if we find a value of x and a value

### ( ) FACTORING. x In this polynomial the only variable in common to all is x.

FACTORING Factoring is similar to breaking up a number into its multiples. For example, 10=5*. The multiples are 5 and. In a polynomial it is the same way, however, the procedure is somewhat more complicated

### { f (x)e sx if s = r. f(x)e sx if s r. (D r) n y = e rx f(x),

1 Recipes for Solving Constant-Coefficient Linear Nonhomogeneous Ordinary Differential Equations Whose Right Hand Sides are Sums of Products of Polynomials times Exponentials Let D = d dx. We begin by

### Computational Geometry Lab: FEM BASIS FUNCTIONS FOR A TETRAHEDRON

Computational Geometry Lab: FEM BASIS FUNCTIONS FOR A TETRAHEDRON John Burkardt Information Technology Department Virginia Tech http://people.sc.fsu.edu/ jburkardt/presentations/cg lab fem basis tetrahedron.pdf

### 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.

Introduction to Calculus for Business and Economics by Stephen J. Silver Department of Business Administration The Citadel I. Functions Introduction to Calculus for Business and Economics y = f(x) is a

### More Zeroes of Polynomials. Elementary Functions. The Rational Root Test. The Rational Root Test

More Zeroes of Polynomials In this lecture we look more carefully at zeroes of polynomials. (Recall: a zero of a polynomial is sometimes called a root.) Our goal in the next few presentations is to set

### 3. Interpolation. Closing the Gaps of Discretization... Beyond Polynomials

3. Interpolation Closing the Gaps of Discretization... Beyond Polynomials Closing the Gaps of Discretization... Beyond Polynomials, December 19, 2012 1 3.3. Polynomial Splines Idea of Polynomial Splines

### Systems of Linear Equations

Systems of Linear Equations Beifang Chen Systems of linear equations Linear systems A linear equation in variables x, x,, x n is an equation of the form a x + a x + + a n x n = b, where a, a,, a n and

### Inner Product Spaces

Math 571 Inner Product Spaces 1. Preliminaries An inner product space is a vector space V along with a function, called an inner product which associates each pair of vectors u, v with a scalar u, v, and

### Introduction to Finite Systems: Z 6 and Z 7

Introduction to : Z 6 and Z 7 The main objective of this discussion is to learn more about solving linear and quadratic equations. The reader is no doubt familiar with techniques for solving these equations

### REVISED GCSE Scheme of Work Mathematics Higher Unit 6. For First Teaching September 2010 For First Examination Summer 2011 This Unit Summer 2012

REVISED GCSE Scheme of Work Mathematics Higher Unit 6 For First Teaching September 2010 For First Examination Summer 2011 This Unit Summer 2012 Version 1: 28 April 10 Version 1: 28 April 10 Unit T6 Unit

### Numerical Analysis and Computing

Joe Mahaffy, mahaffy@math.sdsu.edu Spring 2010 #4: Solutions of Equations in One Variable (1/58) Numerical Analysis and Computing Lecture Notes #04 Solutions of Equations in One Variable, Interpolation

### 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

### 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

### Lesson 3. Numerical Integration

Lesson 3 Numerical Integration Last Week Defined the definite integral as limit of Riemann sums. The definite integral of f(t) from t = a to t = b. LHS: RHS: Last Time Estimate using left and right hand

### Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given.

Polynomials (Ch.1) Study Guide by BS, JL, AZ, CC, SH, HL Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Sasha s method

### 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

### 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

### The distance of a curve to its control polygon J. M. Carnicer, M. S. Floater and J. M. Peña

The distance of a curve to its control polygon J. M. Carnicer, M. S. Floater and J. M. Peña Abstract. Recently, Nairn, Peters, and Lutterkort bounded the distance between a Bézier curve and its control

### MATH SOLUTIONS TO PRACTICE FINAL EXAM. (x 2)(x + 2) (x 2)(x 3) = x + 2. x 2 x 2 5x + 6 = = 4.

MATH 55 SOLUTIONS TO PRACTICE FINAL EXAM x 2 4.Compute x 2 x 2 5x + 6. When x 2, So x 2 4 x 2 5x + 6 = (x 2)(x + 2) (x 2)(x 3) = x + 2 x 3. x 2 4 x 2 x 2 5x + 6 = 2 + 2 2 3 = 4. x 2 9 2. Compute x + sin

### CHAPTER 2 FOURIER SERIES

CHAPTER 2 FOURIER SERIES PERIODIC FUNCTIONS A function is said to have a period T if for all x,, where T is a positive constant. The least value of T>0 is called the period of. EXAMPLES We know that =