FINITE DIFFERENCES AND INTERPOLATION FINITE DIFFERENCES • Finite Difference Operators • Relation between the operators • Newton’s Forward Difference Interpolation Formula • Newton’s Backward Difference Interpolation Formula • Lagrange’s Interpolation Formula • Divided Differences • Newton’s divided difference formula Polynomial Interpolation Using Simple Operators Shift Operator Ef(x) = f(x + h) Forward Difference Op. Df(x) = f(x + h) - f(x) Backward Difference Op. f(x) = f(x) - f(x - h) Central Difference Op. f(x) = f(x + h/2) - f(x - h/2) Average Operator f(x) = {f(x + h/2) + f(x - h/2)}/2 Derivative Operator Df(x) = df(x)/dx Compound Operators • EDf(x) = E{Df(x)} = E{f(x+h) - f(x)} = Ef(x+h) - Ef(x) = f(x + 2h) - f(x + h) • Now try DEf(x) and see if the result is the same. Powers of operators • EEf(x) = E2f(x) = f(x + 2h) • EEE ... Ef(x) = Enf(x) = f(x + nh) E-1f(x) = f(x - h) E-nf(x) = f(x - nh) E1/2f(x) = f(x + h/2) E0f(x) = f(x) Relationships between operators Df(x) = f(x + h) - f(x) = Ef(x) - f(x) = (E - 1)f(x) Therefore, D = E - 1 Verify these relationships: = 1 - E-1 = E1/2 - E-1/2 = (E1/2 + E-1/2)/2 2 = D - Relationship between E and D • Ef(x) = f(x + h) = f(x) + hf (x) + h2 f (x)/2! + h3 f (x)/3! + . = f(x) + hD f(x) + h2/2! D2f(x) + h3/3! D3f(x) + . = (1 + hD + h2D2/2! + h3D3/3! + . .)f(x) ehD = f(x) Therefore, E = ehD • You can try to derive the following: = 2sinh(hD/2) = cosh(hD/2) Definition • Interpolation is the process of finding equations to approximate straight lines and curves that best fit given sets of data. 9 WHAT IS INTERPOLATION? Given (x0,y0), (x1,y1), …, (xn,yn), finding the value of ‘y’ at a value of ‘x’ in (x0, xn) is called interpolation. 10 NEWTON GREGORY FORWARD INTERPOLATION x - x 0 For convenience we put p = h and f0 = y0. Then we have p(p–1) p(p-- 1)(p 2) P(x + ph) = y + pDDD y +23 y + y + + 0 o 02! 0 3! 0 p(p--- 1)(p 2)L(p n+1) Dny n! 0 11 NEWTON GREGORY BACKWARD INTERPOLATION FORMULA xx Taking p = n , h we get the interpolation formula as: P(x +ph) = y + py + p(p 1) 2y + p(p 1)(p 2) 3y + ….. + n 0 n 2! n 3! n p (p 1) (p 2) (p n 1) n y n! n 12 Difference tables: An easy way to compute powers of either the forward or backward difference operator is to construct a difference table using a spread sheet. Here is the forward difference table for the data from the example. X F(X) D D2 D3 D4 0 0.00 0.21 0.09 0.06 0.04 1 0.21 0.30 0.15 0.10 -0.03 2 0.51 0.45 0.25 0.07 0.01 3 0.96 0.70 0.32 0.08 -0.02 4 1.66 1.02 0.40 0.06 0.02 5 2.68 1.42 0.46 0.08 6 4.10 1.88 0.54 7 5.98 2.42 8 8.40 Example Estimate f (3.17)from the data using Newton Forward Interpolation. x: 3.1 3.2 3.3 3.4 3.5 f(x): 0 0.6 1.0 1.2 1.3 14 Solution First let us form the difference table x y Dy D2y D3y D4y 3.1 0 0.6 3.2 0.6 - 0.2 0.4 0 3.3 1.0 - 0.2 0.1 0.2 0.1 3.4 1.2 -0.1 0.1 3.5 1.3 Here x0 = 3.1, x = 3.17, h = 0.1. 15 Example Estimate f(42) from the following data using newton backward interpolation. x: 20 25 30 35 40 45 f(x): 354 332 291 260 231 204 16 Solution The difference table is: x f f 2f 3f 4f 5f 20 354 - 22 25 332 - 19 - 41 29 30 291 10 -37 - 31 - 8 45 35 260 2 8 - 29 0 40 231 2 - 27 45 204 and p = - 0.6 Here xn = 45, h = 5, x = 42 17 Solution Newton backward formula is: p(p +1) 2 p(p+1)(p+2) 3 p(p +1)(p + 2)(p + 3) 4 P(x) = yn+pyn+ yn+ yn+ yn + 2! 3! 4! p(p +1)(p + 2)(p + 3)(p + 4) 5 yn 5! P(42)=204+(-0.6)(-27)+ (-0.6)(0.4) 2+ (-0.6)(0.40(1.4) 0+ (-0.6)(0.4)(1.4)(2.4) 8 + 2 6 24 (-0.6)(0.4)(1.4)(2.4)(3.4) 45 = 219.1430 120 Thus, f(42) = 219.143 18 NEWTONS DIVIDED DIFFERENCE • What is divided difference? f[x ] – f[x ] f[x ,x ] = 1 0 0 1 x1 – x0 f[x ,x ] – f[x ,x ] f[x0,x1,x2] = 1 2 0 1 x2 – x1 f[x1 ,x 2 - x k ]- f[x 0 ,...,x k-1 ] f[x0, x1, …, xk-1,xk] = xk0 - x for k = 3, 4, ….. n. These Ist, IInd... and kth order differences are denoted by Df, D2f, …, Dkf. • The divided difference interpolation polynomial is: P(x) = f(x0) + (x – x0) f [x0, x1] + (x –x0) (x – x1) f [x0, x1 , x2] + ……+ (x – x0)…..(x –xn-1) f[x0, x1, …, xn] 19 Example • For the data x: –1 0 2 5 f(x) :7 10 22 235 • Find the divided difference polynomial and estimate f(1). 20 Solution X f Df D2f D3f -1 7 3 0 10 1 6 2 2 22 13 71 5 235 P(x) = f(x0) + (x – x0) f[x0, x1] + (x – x0) ( x – x1) f [x0, x1, x2] + (x – x0) (x – x1) (x – x2) f [x0, x1, x2, x3] = 7 + (x +1) 3 + (x +1) (x – 0) 1 + (x +1) (x –0) (x – 2) 2 = 2x3 – x2 + 10 P (1) = 11 21 Use Newton’s divided-difference method to compute f(2) from the experimental data shown in the following table: 22 23 Lagrange Interpolation Lagrange Interpolating takes the following general formula: f N (x) 24 Linear Interpolation 25 Example Use a Lagrange interpolating polynomial of the first and second order to evaluate f(2) on the basis of the data: x0 = 1 f(x0) = 0 x1 = 4 f(x1) = 1.386294 x2 = 6 f(x2) = 1.791760 26 Quadratic Interpolation For 3 points of data 27 Example – cont. 28 Example – cont. 29.