Handout 4 - Interpolation Examples

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1 Handout 4 - Interpolation Examples Middle East Technical University Example 1: Obtaining the n th Degree Newton s Interpolating Polynomial Passing through (n+1) Data Points Obtain the 4 th degree Newton s interpolating polynomial that passes through the following 5 data points. Interpolate for f(1) and extrapolate for f(6). x : f : Solution: f n (x) = f(x 0 ) + (x x 0 ) f[x 1, x 0 ] + This is value A of the table This is value E of the table (x x 0 )(x x 1 ) f[x 2, x 1, x 0 ] + This is value H of the table (x x 0 )(x x 1 )(x x 2 ) f[x 3, x 2, x 1, x 0 ] + This is value J of the table (x x 0 )(x x 1 )(x x 2 )(x x 3 ) f[x 4, x 3, x 2, x 1, x 0 ] Let s construct the FDD table x f(x) f [, ] f [,, ] f [,,, ] f [,,,, ] x 0 = A E H J x 1 = B F I x 2 = C G x 3 = D x 4 = A = f[x 1, x 0 ] = f(x 1) f(x 0 ) = = x 1 x B = f[x 2, x 1 ] = f(x 2) f(x 1 ) = = x 2 x C = f[x 3, x 2 ] = f(x 3) f(x 2 ) = = x 3 x D = f[x 4, x 3 ] = f(x 4) f(x 3 ) = = x 4 x E = f[x 2, x 1, x 0 ] = F = f[x 3, x 2, x 1 ] = B A x 2 x 0 = C B x 3 x 1 =

2 G = f[x 4, x 3, x 2 ] = H = f[x 3, x 2, x 1, x 0 ] = I = f[x 4, x 3, x 2, x 1 ] = J = f[x 4, x 3, x 2, x 1, x 0 ] = D C x 4 x 2 = F E x 3 x 0 = G F x 4 x 1 = I H = x 4 x 0 The 4 th degree interpolating polynomial is constructed using the 1 st row of the table as follows f 4 (x) = f(x 0 ) + (x x 0 ) A + (x x 0 )(x x 1 ) E + (x x 0 )(x x 1 )(x x 2 ) H + which simplifies to (x x 0 )(x x 1 )(x x 2 )(x x 3 ) J f 4 (x) = x x x x Following plot shows the 5 data points and the 4 th order interpolating polynomial passing through them. Using the obtained f 4 (x), the required interpolation and extrapolation can be done as f 4 (1) = (1) (1) (1) (1) = f 4 (6) = (6) (6) (6) (6) =

3 Important: At this point we can reveal the fact that the provided 5 data points were actually obtained using f = sin(x) function. The following plot compares sin(x) function with the obtained f 4 (x). As seen, in the range [0, 3.5], which is the range of the provided x values, the two functions are almost identical. However, outside this range, where we do extrapolation, the functions deviate from each other considerably. So the conclusion is that extrapolation might yield unexpected results and requires special caution. Example 2: Performing Different Order Interpolations using an Already Constructed FDD Table The following is the FDD table constructed in the previous example. x f f[,] f[,,] f[,,,] f[,,,,] a) Interpolate f(1) using all 5 points, i.e. using a 4 th order polynomial. b) Interpolate f(0.7) using 2 points, i.e. using a 1 st order polynomial. b) Interpolate f(2.8) using 3 points, i.e. using a 2 nd order polynomial. c) Interpolate f(3.2) using 4 points, i.e. using a 3 rd order polynomial. 3

4 Solution: a) This was already done in the previous example. Here x 0 = 0 and the values of the 1 st row of the FDD table is used as shown below. f 4 (1) = (1 x 0 )( ) + (1 x 0 )(1 x 1 )( ) + = (1 x 0 )(1 x 1 )(1 x 2 )( ) + (1 x 0 )(1 x 1 )(1 x 2 )(1 x 3 )( ) b) Two points that surround x = 0.7 are x 0 = 0.5 and x 1 = 1.5. Important: The first point we use for the interpolation is always called x 0. To perform interpolation with these two points, the second row of the table is used. Two-point interpolation only needs the values shown below in the blue box and to get those values only the part of the table inside the red border needs to be constructed. With two points, the following 1 st order polynomial can be used to interpolate f(0.7) f 1 (0.7) = (0.7 x 0 )( ) = Exercise: Show that if we had used all 5 points to interpolate f(0.7) the result would be c) Three points that are closest to x = 2.8 are x 0 = 1.5, x 1 = 3 and x 2 = 3.5. Once again the first point used in the interpolation is called x 0. This time we use the third row of the table. Three-point interpolation only needs the values shown below in the blue box and to get those values only the part of the table inside the red border needs to be constructed. 4

5 With three points, the following 2 nd order polynomial can be used to interpolate f(2.8) f 2 (2.8) = (2.8 x 0 )( ) + (2.8 x 0 )(2.8 x 1 )( ) = Exercise: Show that if we had used all 5 points to interpolate f(2.8) the result would be d) 4 points are the closest to x = 3.2 are x 0 = 0.5, x 1 = 1.5, x 2 = 3.0, x 3 = 3.5. Second row of the table will be used. Four-point interpolation only needs the values shown below in the blue box and to get those values only the part of the table inside the red border needs to be constructed. With four points, the following 3 rd order polynomial can be used to interpolate f(3.2) f 3 (3.2) = (3.2 x 0 )( ) + (3.2 x 0 )(3.2 x 1 )( ) + + (3.2 x 0 )(3.2 x 1 )(3.2 x 2 )( ) = Exercise: Show that if we had used all 5 points to interpolate f(3.2) the result would be Example 3: Lagrange Type Interpolating Functions The following data is given. It is not important here, but the f values are obtained by the function f = ln(x). x : f : a) Interpolate f(4.5) using 1 st order Lagrange type polynomial b) Interpolate f(8) using 2 nd order Lagrange type polynomial 5

6 Solution: a) 1 st order polynomial needs two points. Points that surround x = 4.5 are x 0 = 3 and x 1 = 5. f 1 (4.5) = 4.5 x 1 x 0 x 1 f(x 0 ) x 0 x 1 x 0 f(x 1 ) f 1 (4.5) = ( ) + ( ) = b) We need 3 points that surround x = 8. x 0 = 5, x 1 = 7 and x 2 = 10 can be used f 2 (8) = (8 x 1)(8 x 2 ) (x 0 x 1 )(x 0 x 2 ) f(x 0) + (8 x 0)(8 x 2 ) (x 1 x 0 )(x 1 x 2 ) f(x 1) + (8 x 0)(8 x 1 ) (x 2 x 0 )(x 2 x 1 ) f(x 2) f 2 (8) = (8 7)(8 10) (8 5)(8 10) (8 5)(8 7) ( ) + ( ) + (5 7)(5 10) (7 5)(7 10) (10 5)(10 7) ( ) = Exercise: Perform these interpolations by constructing FDD tables and show that identical results can be obtained. Example 4: Spline Interpolation Following data is given x : f : a) Interpolate f(13) and f(17) using linear splines. b) Interpolate f(13) and f(17) using quadratic splines. Solution: a) Linear splines are just straight lines joining the sequential points of the data set. To interpolate f(13), we need to use x 0 = 11 and x 1 = 15 f(13) = f(x 0 ) + f(x 1) f(x 0 ) (13 x x 1 x 0 ) = (13 11) = To interpolate f(17), we need to use x 0 = 15 and x 1 = 18 f(17) = f(x 0 ) + f(x 1) f(x 0 ) (17 x x 1 x 0 ) = (17 15) = Graphically these interpolations look like the following. Red dots are the data points and the blue ones are the interpolations. 6

7 b) There are n + 1 = 4 data points. In between there are n = 3 intervals with one quadratic polynomial in each. Each quadratic polynomial has 3 unknown coefficients to be determined. Totally there are 3n = 9 unknown coefficients. To find them we need to setup a system of 9 equations. Polynomials over the intervals are Interval 1 (8 x 11) : a 1 x 2 + b 1 x + c 1 Interval 2 (11 x 15) : a 2 x 2 + b 2 x + c 2 Interval 3 (15 x 18) : a 3 x 2 + b 3 x + c 3 At the knots (x = 8, 11, 15, 18), the splines must take the provided f values. At knot 0 (x = 8) : a 1 (8 2 ) + b 1 (8) + c 1 = 5 Eqn (1) At knot 1 (x = 11) : a 1 (11 2 ) + b 1 (11) + c 1 = 9 Eqn (2) a 2 (11 2 ) + b 2 (11) + c 2 = 9 Eqn (3) At knot 2 (x = 15) : a 2 (15 2 ) + b 2 (15) + c 2 = 10 Eqn (4) a 3 (15 2 ) + b 3 (15) + c 3 = 10 Eqn (5) At knot 3 (x = 18) : a 3 (18 2 ) + b 3 (18) + c 3 = 9 Eqn (6) At the interior knots (x = 11, 15), first derivatives of the right and left splines must be equal. At knot 1 (x = 11) : 2a 1 (11) + b 1 = 2a 2 (11) + b 2 Eqn (7) At knot 2 (x = 15) : 2a 2 (15) + b 2 = 2a 3 (15) + b 3 Eqn (8) For the last equation we set a 1 = 0, i.e. we force the first spline to be a straight line. Therefore, actually there are not 9 but 8 unknowns and 8 equations. The equations system that needs to be solved and the solution are shown below 7

8 b 1 c 1 a 2 b 2 c 2 a b 3 [ ] { c 3 } = { 0 } Middle East Technical University b 1 c 1 a 2 b 2 c 2 a 3 b 3 { c 3 } = { } With this solution, the equations of the splines in each interval are Spline 1 (8 x 11) : x Spline 2 (11 x 15) : x x Spline 3 (15 x 18) : x x + 60 To interpolate f(13) we need to use the 2 nd spline f(13) = (13) (13) = To interpolate f(17) we need to use the 3 rd spline f(17) = (17) (17) + 60 = The splines and the interpolated values are as follows. Red dots are the given data points and the blue ones are the interpolations. 8

9 Exercise: Repeat the solution with cubic splines and obtain the following results f(13) = , f(17) = Middle East Technical University The cubic splines that you are asked to determine are shown below. The overshoot of the quadratic splines in the 2 nd interval seen before does not exist here. Also this one behaves considerably different in the 3 rd interval. In general, it oscillates less. Exercise: Obtain the 3 rd order interpolating polynomial and draw it. Also calculate f(13) and f(17) with it. 9