KFUPM SE301: Numerical Methods Topic 5: Interpolation Lectures 20-22:

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Presentation transcript:

KFUPM SE301: Numerical Methods Topic 5: Interpolation Lectures 20-22: Read Chapter 18, Sections 1-5 CISE301_Topic5

Lecture 20 Introduction to Interpolation Interpolation Problem Existence and Uniqueness Linear and Quadratic Interpolation Newton’s Divided Difference Method Properties of Divided Differences CISE301_Topic5

Introduction Interpolation was used for long time to provide an estimate of a tabulated function at values that are not available in the table. What is sin (0.15)? x sin(x) 0.0000 0.1 0.0998 0.2 0.1987 0.3 0.2955 0.4 0.3894 Using Linear Interpolation sin (0.15) ≈ 0.1493 True value (4 decimal digits) sin (0.15) = 0.1494 CISE301_Topic5

The Interpolation Problem Given a set of n+1 points, Find an nth order polynomial that passes through all points, such that: CISE301_Topic5

Example Temperature (degree) Viscosity 1.792 5 1.519 10 1.308 15 1.140 An experiment is used to determine the viscosity of water as a function of temperature. The following table is generated: Problem: Estimate the viscosity when the temperature is 8 degrees. CISE301_Topic5

Interpolation Problem Find a polynomial that fits the data points exactly. Linear Interpolation: V(T)= 1.73 − 0.0422 T V(8)= 1.3924 CISE301_Topic5

Existence and Uniqueness Given a set of n+1 points: Assumption: are distinct Theorem: There is a unique polynomial fn(x) of order ≤ n such that: CISE301_Topic5

Examples of Polynomial Interpolation Linear Interpolation Given any two points, there is one polynomial of order ≤ 1 that passes through the two points. Quadratic Interpolation Given any three points there is one polynomial of order ≤ 2 that passes through the three points. CISE301_Topic5

Linear Interpolation Given any two points, The line that interpolates the two points is: Example : Find a polynomial that interpolates (1,2) and (2,4). CISE301_Topic5

Quadratic Interpolation Given any three points: The polynomial that interpolates the three points is: CISE301_Topic5

General nth Order Interpolation Given any n+1 points: The polynomial that interpolates all points is: CISE301_Topic5

Divided Differences CISE301_Topic5

Divided Difference Table x F[ ] F[ , ] F[ , , ] F[ , , ,] x0 F[x0] F[x0,x1] F[x0,x1,x2] F[x0,x1,x2,x3] x1 F[x1] F[x1,x2] F[x1,x2,x3] x2 F[x2] F[x2,x3] x3 F[x3] CISE301_Topic5

Divided Difference Table x F[ ] F[ , ] F[ , , ] -5 2 -4 1 -3 6 -1 -15 f(xi) -5 1 -3 -1 -15 Entries of the divided difference table are obtained from the data table using simple operations. CISE301_Topic5

Divided Difference Table x F[ ] F[ , ] F[ , , ] -5 2 -4 1 -3 6 -1 -15 f(xi) -5 1 -3 -1 -15 The first two column of the table are the data columns. Third column: First order differences. Fourth column: Second order differences. CISE301_Topic5

Divided Difference Table -5 1 -3 -1 -15 x F[ ] F[ , ] F[ , , ] -5 2 -4 1 -3 6 -1 -15 CISE301_Topic5

Divided Difference Table -5 1 -3 -1 -15 x F[ ] F[ , ] F[ , , ] -5 2 -4 1 -3 6 -1 -15 CISE301_Topic5

Divided Difference Table -5 1 -3 -1 -15 x F[ ] F[ , ] F[ , , ] -5 2 -4 1 -3 6 -1 -15 CISE301_Topic5

Divided Difference Table -5 1 -3 -1 -15 x F[ ] F[ , ] F[ , , ] -5 2 -4 1 -3 6 -1 -15 f2(x)= F[x0]+F[x0,x1] (x-x0)+F[x0,x1,x2] (x-x0)(x-x1) CISE301_Topic5

Two Examples x y 1 2 3 8 x y 2 3 1 8 What do you observe? Obtain the interpolating polynomials for the two examples: x y 1 2 3 8 x y 2 3 1 8 What do you observe? CISE301_Topic5

Two Examples x Y 1 3 2 5 8 x Y 2 3 1 4 8 Ordering the points should not affect the interpolating polynomial. CISE301_Topic5

Properties of Divided Difference Ordering the points should not affect the divided difference: CISE301_Topic5

Example x f(x) 2 3 4 5 1 6 7 9 Find a polynomial to interpolate the data. CISE301_Topic5

Example x f(x) f[ , ] f[ , , ] f[ , , , ] f[ , , , , ] 2 3 1 -1.6667 1.5417 -0.6750 4 5 -4 4.5 -1.8333 -1 6 7 9 CISE301_Topic5

Summary CISE301_Topic5

Lecture 21 Lagrange Interpolation CISE301_Topic5

The Interpolation Problem Given a set of n+1 points: Find an nth order polynomial: that passes through all points, such that: CISE301_Topic5

Lagrange Interpolation Problem: Given Find the polynomial of least order such that: Lagrange Interpolation Formula: …. CISE301_Topic5

Lagrange Interpolation CISE301_Topic5

Lagrange Interpolation Example 1/3 1/4 1 y 2 -1 7 CISE301_Topic5

Example Find a polynomial to interpolate: Both Newton’s interpolation method and Lagrange interpolation method must give the same answer. x y 1 3 2 5 4 CISE301_Topic5

Newton’s Interpolation Method 1 2 -3/2 7/6 -5/8 3 -1 -4/3 -2 5 4 CISE301_Topic5

Interpolating Polynomial CISE301_Topic5

Interpolating Polynomial Using Lagrange Interpolation Method CISE301_Topic5

Lecture 22 Inverse Interpolation Error in Polynomial Interpolation CISE301_Topic5

Inverse Interpolation …. One approach: Use polynomial interpolation to obtain fn(x) to interpolate the data then use Newton’s method to find a solution to x CISE301_Topic5

Inverse Interpolation 1. Exchange the roles of x and y. 2. Perform polynomial Interpolation on the new table. 3. Evaluate …. …. CISE301_Topic5

Inverse Interpolation x y x y CISE301_Topic5

Inverse Interpolation Question: What is the limitation of inverse interpolation? The original function has an inverse. y1, y2, …, yn must be distinct. CISE301_Topic5

Inverse Interpolation Example 1 2 3 y 3.2 2.0 1.6 3.2 1 -.8333 1.0417 2.0 2 -2.5 1.6 3 CISE301_Topic5

Errors in polynomial Interpolation Polynomial interpolation may lead to large errors (especially for high order polynomials). BE CAREFUL When an nth order interpolating polynomial is used, the error is related to the (n+1)th order derivative. CISE301_Topic5

10th Order Polynomial Interpolation CISE301_Topic5

Errors in polynomial Interpolation Theorem CISE301_Topic5

Example CISE301_Topic5

Summary The interpolating polynomial is unique. Different methods can be used to obtain it. Newton’s divided difference Lagrange interpolation Others Polynomial interpolation can be sensitive to data. BE CAREFUL when high order polynomials are used. CISE301_Topic5