3.5 Spline interpolation


 Matthew May
 1 years ago
 Views:
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 nonlocally. The nonlocality 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 righthand 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 xdependence. 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 twoparameter 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 socalled 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 Bsplines? 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 subinterval 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 Bsplines 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 nonzero 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 k2 x k x k x k+ x k+2 Figure 3.3: Cubic Bspline 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 nonzero 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
More informationPiecewise 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 (computerassisted design), CAM (computerassisted manufacturing), and
More informationi = 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
More information4.3 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
More informationsince 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
More informationthe 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.
More information3.4. 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.
More information3.4. 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.
More information1 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
More informationNovember 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
More informationSome 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 informationPolynomials 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
More informationSolving 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
More informationWe 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
More informationThe 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
More informationTaylor 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 informationIntroduction 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 onedimensional
More information4.5 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
More information5.4 The Quadratic Formula
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
More informationPolynomials 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
More information1 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 GaussJordan reduction and the Reduced
More informationCurve 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 LeastSquares Regression Linear Regression General Linear LeastSquares Nonlinear
More informationHIGHER 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
More informationLinear 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
More informationWe can display an object on a monitor screen in three different computermodel 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
More information(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
More informationComputer 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: WireFrame Representation Object is represented as as a set of points
More informationis 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
More informationMath 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
More informationBasic 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
More information1 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
More information3.1. 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
More informationINTERPOLATION. 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 informationNatural 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
More informationAlgebra 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
More informationZeros of Polynomial Functions
Zeros of Polynomial Functions The Rational Zero Theorem If f (x) = a n x n + a n1 x n1 + + 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
More informationFinite 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
More informationEXAM. 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,
More informationHøgskolen i Narvik Sivilingeniørutdanningen STE6237 ELEMENTMETODER. Oppgaver
Høgskolen i Narvik Sivilingeniørutdanningen STE637 ELEMENTMETODER Oppgaver Klasse: 4.ID, 4.IT Ekstern Professor: Gregory A. Chechkin email: chechkin@mech.math.msu.su Narvik 6 PART I Task. Consider twopoint
More information3.6. 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
More informationEquations, 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
More informationHomework 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
More informationCHAPTER 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.
More informationUNIT  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
More informationChapter 20. Vector Spaces and Bases
Chapter 20. Vector Spaces and Bases In this course, we have proceeded stepbystep through lowdimensional Linear Algebra. We have looked at lines, planes, hyperplanes, and have seen that there is no limit
More information4 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
More informationNumerical 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
More informationFinite 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,
More informationMath 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 xaxis (horizontal axis)
More informationModule 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
More information3 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
More informationSCHOOL 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
More information5 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
More informationLimit 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
More information5.1. 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 5.11 Linear Systems The possible graphs of a linear system in two unknowns are as follows. 1. The
More informationInterpolating 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 ),
More informationTOPIC 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
More informationFurther 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
More informationChapter 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
More informationSolving 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,
More informationarxiv: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
More informationAppendix 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
More information3.2. 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.
More informationTHEORY 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
More informationMth 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
More informationCHAPTER 1 Splines and Bsplines an Introduction
CHAPTER 1 Splines and Bsplines 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
More informationAn 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
More informationReview 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 decisionmaking tools
More informationis 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
More information3. 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.
More informationAn 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
More informationcorrectchoice 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 informationReview 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,
More informationInverses. 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 Pseudoinverse Left and right inverses 2 Left inverses a number
More informationwith "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,
More informationSample 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
More informationOrdered 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
More information( ) 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
More information{ 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 ConstantCoefficient Linear Nonhomogeneous Ordinary Differential Equations Whose Right Hand Sides are Sums of Products of Polynomials times Exponentials Let D = d dx. We begin by
More informationComputational 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
More information1 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.
More informationIntroduction to Calculus for Business and Economics. by Stephen J. Silver Department of Business Administration The Citadel
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 informationMore 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
More information3. 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
More informationSystems 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
More informationInner 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
More informationIntroduction 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
More informationREVISED 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
More informationNumerical 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
More informationVieta 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
More informationLS.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 informationLesson 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
More informationLagrange 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
More informationLINE 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 informationTaylor 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 informationThe 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
More informationMATH 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
More informationCHAPTER 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 =
More information1.1 Solving a Linear Equation ax + b = 0
1.1 Solving a Linear Equation ax + b = 0 To solve an equation ax + b = 0 : (i) move b to the other side (subtract b from both sides) (ii) divide both sides by a Example: Solve x = 0 (i) x = 0 x = (ii)
More informationNumerical Methods Lecture 5  Curve Fitting Techniques
Numerical Methods Lecture 5  Curve Fitting Techniques Topics motivation interpolation linear regression higher order polynomial form exponential form Curve fitting  motivation For root finding, we used
More information