# 1 curve-fitting interpolation. 2 curve fitting regression linear regression polynomial regression...

Click here to load reader

Post on 19-Dec-2015

320 views

Embed Size (px)

TRANSCRIPT

- Slide 1
- 1 Curve-Fitting Interpolation
- Slide 2
- 2 Curve Fitting Regression Linear Regression Polynomial Regression Multiple Linear Regression Non-linear Regression Interpolation Newton's Divided-Difference Interpolation Lagrange Interpolating Polynomials Spline Interpolation
- Slide 3
- 3 Polynomial Interpolation Objective: Given n+1 points, we want to find the polynomial of order n that passes through all the points.
- Slide 4
- 4 Polynomial Interpolation The n th -order polynomial that passes through n+1 points is unique, but it can be written in different mathematical formats: The Newton's Form The Lagrange Form The conventional form
- Slide 5
- 5 Linear Interpolation (Newton's Form) Objective: Connecting two points with a straight line. f 1 (x) represents the first-order interpolating polynomial.
- Slide 6
- 6 Two linear interpolations of f(x)=ln(x) on two different intervals.
- Slide 7
- 7 Quadratic Interpolation (Newton's Form) Connecting three points with a second-order polynomial or parabola. One way to form a 2 nd -order polynomial is The advantage is that b 0, b 1, and b 2 can be calculated conveniently. Only the format is different. There is till only one unique 2nd-order polynomial that passes through three points. Can be rewritten in the conventional form. i.e., as
- Slide 8
- 8 Quadratic Interpolation Finding b 0, b 1, b 2 Given three points (x 0, f(x 0 )), (x 1, f(x 1 )), and (x 2, f(x 2 )), we can create three equations with three unknowns b 0, b 1, and b 2 as which can be solved for b 0, b 1, and b 2
- Slide 9
- 9 Quadratic Interpolation Finding b 0, b 1, b 2 Alternatively, we can also calculate b 0, b 1, and b 2 as b 1 : Finite-divided difference for f'(x) b 2 : Finite-divided difference for f"(x)
- Slide 10
- 10 Comparing Linear and Quadratic Interpolation The quadratic interpolation formula includes an additional term which represents the 2 nd -order curvature.
- Slide 11
- 11 Fig 18.4 Linear vs. quadratic interpolation of ln(x)
- Slide 12
- 12 General Form of Newton's Interpolating Polynomials
- Slide 13
- 13 Graphical depiction of the recursive nature of finite divided differences.
- Slide 14
- 14 Cubic interpolation of ln(x)
- Slide 15
- 15 Lagrange Interpolating Polynomials Simply a reformulation of the Newtons polynomial that avoids the computation of divided differences: e.g.: 1 st and 2 nd -order polynomials in Lagrange form:
- Slide 16
- 16 Second order case of Lagrange polynomial. Each of the three terms is a 2 nd -order polynomial that passes through one of the data points and is zero at the other two. The summation of three terms must, therefore, be the unique 2 nd- order polynomial that passes exactly through three points.
- Slide 17
- 17 Coefficients of an Interpolating Polynomial Newton and Lagrange polynomials are well suited for determining intermediate values between points. However, they do not provide a polynomial in the conventional form: To calculate a 0, a 1, , a n, we can use simultaneous linear systems of equations.
- Slide 18
- 18 Given n+1 points, (x 0, f(x 0 )), (x 1, f(x 1 )), , (x n, f(x n )), we have n+1 equations which can be solved for n+1 unknowns: Coefficients of an Interpolating Polynomial Solve this system of linear equations for a 0, a 1, , a n.
- Slide 19
- 19 Coefficients of an Interpolating Polynomial Solving the system of linear equations directly is not the most efficient method. This system is typically ill-conditioned. The resulting coefficients can be highly inaccurate when n is large.
- Slide 20
- 20 Extrapolation Extrapolation is the process of estimating a value of f(x) that lies outside the range of the known base points, x 0, x 1, , x n. Extreme care should be exercised where one must extrapolate.
- Slide 21
- 21 Spline Interpolation For some cases, polynomials can lead to erroneous results because of round off error and overshoot. Alternative approach is to apply lower-order polynomials to subsets of data points. Such connecting polynomials are called spline functions.
- Slide 22
- 22
- Slide 23
- 23 (a)Linear spline Derivatives are not continuous Not smooth (b) Quadratic spline Continuous 1 st derivatives (c) Cubic spline Continuous 1 st & 2 nd derivatives Smoother
- Slide 24
- 24 Quadratic Spline
- Slide 25
- 25 Observations n+1 points n intervals Each interval is connected by a 2 nd -order polynomial f i (x) = a i x 2 +b i x+c i, i=1, , n. Each polynomial has 3 unknowns Altogether there are 3n unknowns Need 3n equations (or conditions) to solve for 3n unknowns Quadratic Interpolation
- Slide 26
- 26 1.The function values of adjacent polynomials must be equal at the interior knots. This condition can be represented as Quadratic Interpolation ( 3n conditions) Since there are n-1 interior knots, this condition yields 2 n-2 equations.
- Slide 27
- 27 2.The first and last functions must pass through the end points. This adds 2 more equations: Quadratic Interpolation ( 3n conditions) 3.The first derivatives at the interior knots must be equal. This adds n-1 more equations: We now have 2n - 2 + 2 + n - 1 = 3n - 1 equations. We need one more equation.
- Slide 28
- 28 4.Assume the 2 nd derivatives is zero at the first point. This gives us the last condition as Quadratic Interpolation ( 3n conditions) With this condition selected, the first two points are connected by a straight line. Note: This is not the only possible choice or assumption we can make.
- Slide 29
- 29 Cubic Spline 1.The function values must be equal at the interrior knots ( 2n-2 conditoins). 2.The 1 st and last functions must pass through the end points ( 2 conditions). 3.The 1 st derivatives at the interior knots must be equals ( n-1 conditions). 4.The 2 nd derivatives at the interior knots must be equals ( n-1 conditions). 5.Assume the 2 nd derivatives at the end points are zero ( 2 conditions). This condition makes the spline a "natural" spline.
- Slide 30
- 30 Efficient way to derive cubic spline The cubic equation on each interval can be expressed as There are only two unknowns in each equations the 2 nd derivatives at the end of each interval:
- Slide 31
- 31 Efficient way to derive cubic spline The unknowns can be evaluated using the following equation: If this equation is written for all the interior knots, n-1 simultaneous equations result with n-1 unknowns.
- Slide 32
- 32 Summary Polynomial interpolation for approximate complicated functions. (Data are exact) Newton's or Lagrange Polynomial interpolation are suitable for evaluating intermediate points. Cubic spline Overcome the problem of "overshoot" Easier to derive Smooth (continuous 2 nd -order derivatives)