CE 30125 - Lecture 2 LECTURE 2 INTRODUCTION TO INTERPOLATION • Interpolation function: a function that passes exactly through a set of data points. • Interpolating functions to interpolate values in tables x sin(x) 0.0 0.000000 0.5 0.479426 1.0 0.841471 1.5 0.997495 2.0 0.909297 2.5 0.598472 • In tables, the function is only specified at a limited number or discrete set of indepen- dent variable values (as opposed to a continuum function). • We can use interpolation to find functional values at other values of the independent variable, e.g. sin(0.63253) p. 2.1 CE 30125 - Lecture 2 • In numerical methods, like tables, the values of the function are only specified at a discrete number of points! Using interpolation, we can describe or at least approximate the function at every point in space. • For numerical methods, we use interpolation to • Interpolate values from computations • Develop numerical integration schemes • Develop numerical differentiation schemes • Develop finite element methods • Interpolation is typically not used to obtain a functional description of measured data since errors in the data may lead to a poor representation. • Curve fitting to data is handled with a separate set of techniques p. 2.2 CE 30125 - Lecture 2 Linear Interpolation • Linear interpolation is obtained by passing a straight line between 2 data points y g(x) f(x) f(x1) f(x0) x x0 x1 fx = the exact function for which values are known only at a discrete set of data points gx = the interpolated approximation to fx x0 , x1 = the data points (also referred to as interpolation points or nodes) • In tabular form: xo fxo x gx x1 fx1 p. 2.3 CE 30125 - Lecture 2 • If gx is a linear function then gx= Ax+ B (1) where A and B are unknown coefficients • To pass through points xo fxo and x1 fx1 we must have: gxo = fxo Axo + B = fxo (2) gx1 = fx1 Ax1 + B = fx1 (3) • 2 unknowns and 2 equations solve for AB • Using (2) Bfx= o – Axo Substituting into (3) Ax1 + fxo – Axo = fx1 p. 2.4 CE 30125 - Lecture 2 fx– fx A = -----------------------------1 o- x1 – xo fxx – fxx B = ----------------------------------------o 1 1 o- x1 – xo • Substituting for A and B into equation (1) x1 – x xx– o gx= fxo --------------------- + fx1 --------------------- x1 – xo x1 – xo This is the formula for linear interpolation p. 2.5 CE 30125 - Lecture 2 Example 1 • Use values at xo and x1 to get an interpolated value at x = 0.632 using linear interpola- tion Table 1: x fx= sinx xo = 0.5 fxo = 0.47942554 0.632 g0.632 = ? x1 = 1.0 fx1 = 0.84147099 1.0– 0.632 0.632– 0.5 g0.632 = 0.479425-------------------------------+ 0 . 8 4147099------------------------------- 1.0– 0.5 1.0– 0.5 g0.632 = 0.57500 p. 2.6 CE 30125 - Lecture 2 Error for Linear Interpolating Functions • Error is defined as: ex fx– gx •ex represents the difference between the exact function fx and the interpolating or approximating function gx. • We note that at the interpolating points xo and x1 exo = 0 ex1 = 0 • This is because at the interpolating point we have by definition gxo = fxo gx1 = fx1 p. 2.7 CE 30125 - Lecture 2 Derivation of e(x) f(x) f(x) g(x) f(x1) f(x0) x x0 x1 ex fx– gx Step 1 • Expand fx in Taylor Series (T.S.) about xo 2 2 df xx– o d f fx= fx++xx– ------ ----------------------------- where x x (4) o o dx 2! 2 o xx= dx o x = • The third term is the actual remainder term and represents all other terms in the series since it is evaluated at x = ! p. 2.8 CE 30125 - Lecture 2 Step 2 df • Express ------ in terms of fx and fx dx o 1 xx= o • We can accomplish this by simply evaluating the T.S. in (4) at xx= 1 . 2 2 df x1 – xo d f fx= fx++x – x ------ ----------------------- -------- (5) 1 o 1 o dx 2! 2 xx= dx o x = 2 2 df fx1 fxo x1 – xo 1 d f ------ = --------------------- – --------------------- – ----------------------- ----------------------------- (6) dx x – x x – x 2! x – x 2 xx= 1 o 1 o 1 o dx o x = 2 df fx1 fxo x1 – xo d f ------ = --------------------- – --------------------- – ----------------------------- (7) dx x – x x – x 2 2 xx= 1 o 1 o dx o x = • We note that this is a discrete approximation to the first derivative (a F.D. Formula) p. 2.9 CE 30125 - Lecture 2 Step 3 • Substitute Equation 7 into T.S. form of fx, Equation (4). • This gives us an expression for fx in terms of the discrete values fxo and fx1 . 2 2 2 fx1 fxo x1 – xo d f xx– o d f fx = fx + xx– --------------------- – --------------------- – ----------------------------- + ----------------------------- (8) o o 2 2 x1 – xo x1 – xo 2 dx 2 dx x = x = 2 2 xx– o xx– o xx– o –x1 + xo xx– o d f fx = fx ++---------------------fx – ---------------------fx ------------------------------------------- + --------------------- -------- (9) o 1 o 2 x1 – xo x1 – xo 2 2 dx x = 2 fxo fx1 xx– o d f fx = x – x – x + x --------------------- ++xx– --------------------- – x ++x xx– ------------------ -------- (10) 1 o o o 1 o o 2 x1 – xo x1 – xo 2 dx x = 2 x1 – x xx– o xx– o xx– 1 d f fx = fx ---------------- ++fx ---------------- ----------------------------------------------- (11) o 1 2 x1 – xo x1 – xo 2 dx x = p. 2.10 CE 30125 - Lecture 2 • The first part of Equation (11) is simply the linear interpolation formula. The second part is in fact the error. Thus: ex fx– gx x – x xx– xx– xx– d2f ex fx ---------------1 - ++fx ---------------o- -------------------------------------o 1 -------- o 1 2 x1 – xo x1 – xo 2 dx x = x1 – x xx– o –fxo ---------------- – fx1 ---------------- x1 – xo x1 – xo 2 xx– o xx– 1 d f ex= ----------------------------------------------- xo x1 2 dx2 x = • If we assume that the interval xo x1 is small, then the second derivative won’t change dramatically in the interval! 2 2 2 2 d f d f d f d f xo + x1 -------- -------- -------- -------- where xm ---------------- dx2 dx2 dx2 dx2 2 x = xx= o xx= 1 xx= m p. 2.11 CE 30125 - Lecture 2 • Thus we typically evaluate the derivative term in the error expression using the midpoint in the interval 1 d2f ex ---xx– o xx– 1 -------- 2 dx2 xx= m • Another problem is that we typically don’t know the second derivative at the midpoint of the interval, xm • However using finite differencing formulae we can approximate this derivative knowing the functional values at the interpolating points • Maximum error occurs at the midpoint for linear interpolation (where xx– o xx– 1 is the largest) 1 d2f max exx xx ---xm – xo xm – x1 -------- 0 1 2 dx2 xx= m • However hx 1 – xo and h h --- = x – x and --- = x – x 2 m o 2 1 m p. 2.12 CE 30125 - Lecture 2 • Thus h2 d2f max exx xx= ----- -------- 0 1 8 dx2 xm • Notes on Error for linear interpolation • The error expression has a polynomial and a derivative portion. • Maximum error occurs approximately at the midpoint between xo and x1 • Error increases as the interval h increases 2 2 • Error increases as f x increases. Again note that f x can be approximated with finite difference (F.D.) formulae if at least 3 surrounding functional values are available. (We will discuss F.D. formulae later.) p. 2.13 CE 30125 - Lecture 2 Example 2 • Compute an error estimate for the problem in Example 1. • Recall we found that g0.632 = 0.57500 • Error is estimated as: 2 1 d f ex ---xx– o xx– 1 -------- 2 dx2 xx= m • Since x = 0.750 is the midpoint at the interval 0.5 1.0 , we have 2 1 d f e0.632 --- 0.632– 0.5 0.632– 1.0 -------- 2 dx2 x = 0.750 d2f e 0.632 –0.024288-------- 2 dx x = 0.75 p. 2.14 CE 30125 - Lecture 2 • Since we have not yet extensively discussed approximating derivatives using discrete 2 d f values, we will compute -------- using analytical methods: dx2 x = 0.75 2 d f -------- ==–sin0.750 – 0.68164 dx2 x = 0.75 2 d f • Substituting in the value for -------- , we obtain an estimate for the error: dx2 x = 0.75 e0.632 –0.024288 –0.68164 = 0.016555 • Computing the actual error (the actual solution - the estimated error): Ex= sinx – gx E0.632 ==sin0.632 – 0.57500 0.01576 • The estimated error, ex is a good approximation of the actual error Ex! p. 2.15.