Interpolation • In interpolation or extrapolation we usually want to do the following - We have data points xi yi , i = 12} N - We want to know the y value at xxz i - In interpolation x1 xxN and in extrapolation xx 1 or xx! N - Extrapolation is dangerous; it is used e.g. in solving differential equations. - Data set may have noise: the interpolate should go smoothly through the data set not necessarily through all points. - One application is approximating (special) functions - In this case we have an infinite number of points available. - In some cases interpolation is done by using a few points in the neighborhood of x. - This may result in noncontinuous derivative of the interpolate. - In spline interpolation one condition is that also the derivative is continuous. - In polynomial interpolation one is not particularly interested in the polynomial coefficients only in its values. - Calculating coefficients is rather error prone. Scientific computing III 2011: 6. Interpolation 1 Interpolation • Interpolation vs. curve fitting: Scientific computing III 2011: 6. Interpolation 2 Interpolation • Degree of interpolation is (number of points used)-1. - Below an example of interpolation of a smooth function original function low degree interpolation high degree interpolation - When the original function has sharp corners an interpolation polynomial with a lower degree may work better Scientific computing III 2011: 6. Interpolation 3 Interpolation: polynomials • We have a data set xi yi , yi = fx()i , i = 1} N - We have to find a polynomial P N – 1 x that fulfills the condition PN – 1()xi = yi , i = 1} N - It is easy to show that the polynomial is at most of degree N – 1 and it is unique if all x i are different. - A straightforward way to determine the coefficients is to use the methods we have already learned - Let the polynomial be of the form 2 N–1 yc=1+c2xc+3x ++}cNx - From the above condition we get a group of linear equations ­ 2 N–1 °c1+c2x1+c3x1 ++}cNx1 =y1 ° °c+cx+cx2 ++}cxN–1=y ®12232 N2 2 ° } ° °c+cx+cx2 ++}cxN–1=y ¯12N3N NN N Scientific computing III 2011: 6. Interpolation 4 Interpolation: polynomials • Theorem: existence of interpolating polynomial: If points x1 x2 } xN are distinct, then for arbitrary real values y1 y2 } yN there is a unique polynomial P of degree d N – 1 such that Px i = yi for 1 d i d N . - Proof by induction: Suppose we have already a polynomial P that reproduces a part of the data set: Px i = yi , 1 ddik, (For example: this can be a constant polynomial for data point 1: Px = y1 .) Then we add another term to P so that it will go through the point xk + 1 yk + 1 : Qx = Px + cxx – 1 xx– 2 } xx– k Q reproduces data points 12 } k because P does and the added term is zero for all these points. Now we adjust constant c so that Q reproduces the data point k + 1: Qx k + 1 = Px k + 1 + cx k + 1 – x1 xk + 1 – x2 } xk + 1 – xk = yk + 1 . From this equation we can solve c because all x i are distinct. QED. Scientific computing III 2011: 6. Interpolation 5 Interpolation: polynomial - In matrix form: 1 x x2 } xN – 1 1 1 1 c1 y1 2 N – 1 1 x2 x2 } x2 c2 y2 2 N – 1 c = y 1 x3 x3 } x3 3 3 }}}} } } } 2 N–1 cN yN 1 xNxN}xN - This square matrix has a name of its own: Vandermonde matrix and its a little bad behaving: >> v=1:5 >> v=1:8 v = v = 1 2 3 4 5 1 2 3 4 5 6 7 8 >> m=vander(v) >> m=vander(v) m = m = 1 1 1 1 1 1 1 1 1 1 1 ... 16 8 4 2 1 128 64 32 16 8 4 ... 81 27 9 3 1 2187 729 243 81 27 9 ... 256 64 16 4 1 16384 4096 1024 256 64 16 ... 625 125 25 5 1 78125 15625 3125 625 125 25 ... >> rcond(m) 279936 46656 7776 1296 216 36 ... ans = 823543 117649 16807 2401 343 49 ... 2.2699e-05 2097152 262144 32768 4096 512 64 ... >> rcond(m) ans = 6.0171e-10 Scientific computing III 2011: 6. Interpolation 6 Interpolation: polynomial - A better way is to calculate P N – 1 by using so called Lagrange’s polynomials. - Let the polynomials l1 l2 } lN be defined as N§·xx– l ()x = ¨¸--------------j, i=1}N i ©¹x–x j=1i j jiz - These functions have the property li()xj= Gij - Now we can write P N – 1 x as N P ()x = fx()l()x N – 1 ¦ i i i=1 - It is easy to check that P N–1 x goes through all the data points xiyi . - Because functions l ihave degree less than N it follows that P N–1also has degree less than N. Scientific computing III 2011: 6. Interpolation 7 Interpolation: polynomial - Example: x : 13e 14e 1 43e y : 2 –1 7 2 1 4 §·x– - -- x–1§·x– - -- ©¹4 ©¹3 93 69 l x = ------------------------------------------------------= 1 8 x 3 – ------ x 2 + ------ x – 6 1 11 1 14 2 2 §·- -- – - -- §·---1– §·- -- – - -- ©¹34 ©¹3©¹33 1 4 §·x– - -- x–1§·x– - -- ©¹3 ©¹3 192 512 1216 256 l x = ------------------------------------------------------ = – --------- x 3 + --------- x 2 – ------------ x + --------- 2 11 1 14 13 13 39 39 §·- -- – - -- §·---1– §·- -- – - -- ©¹43 ©¹4©¹43 1 1 4 §·x– - -- §·x– - -- §·x– - -- ©¹3©¹4©¹3 23 31 2 l x = -----------------------------------------------------= – 6 x 3 + ------ x 2 – ------ x + - -- 3 1 1 4 2 6 3 §·1– - -- §·1– - -- §·1– - -- ©¹3©¹4©¹3 1 1 §·x– - -- §·x– - -- x–1 ©¹3©¹4 36 57 24 3 l x = ------------------------------------------------------ = ------ x 3 – ------ x 2 + ------ x – ------ 4 1 41 4 13 13 13 13 §·-4 -- – - -- §·- -- – - -- §·---1– ©¹33 ©¹34 ©¹3 186 1577 5281 560 P x = 2l x–l x+7l x+2l x = --------- x 3 – ------------ x 2 + ------------ x – --------- 3 1 2 3 4 13 26 78 39 Scientific computing III 2011: 6. Interpolation 8 Interpolation: polynomials - So we end up with the interpolation polynomial (let’s leave the subscript out) xx–2 xx–3 } xx–N Px()= ------------------------------------------------------------------------- y 1 x1–x2 x1–x3 } x1 – xN xx–1 xx–3 } xx–N + ------------------------------------------------------------------------- y 2 x2–x1 x2–x3 } x2 – xN + } xx–1 xx–2 } xx–N–1 + ----------------------------------------------------------------------------------- y N xN–x1 xN–x2 } xN – xN – 1 - The error estimation of the above polynomial can be given as follows: - Let x 1 x 2 } x N be distinct numbers in >@ab and let’s assume that f has N continuous derivatives in >@ab - Then for each x >@ ab [ x ab so that N f N [ x fx =Px +------------------------- xx– (1) N! i i=1 Scientific computing III 2011: 6. Interpolation 9 Interpolation: polynomials - Proof: - For xx= k , k = 12 } N fx k = Px k and any [ xk ab fulfills (1) - For x k z x we define function gt as tx– gt = ft –Pt – >@fx –Px N -----------------i(2) i = 1 xx–i - Since f has N continuous derivatives and P has all derivatives continuous and xx z k o gt has N continuous deriv- atives in >@ab - For tx= k Generalized Rolle’s theorem: N x–x gx = fx –Px – >@fx –Px --------------------ki= 0 Assume k k k 1. fx continuous on >@ab , xx–i i = 1 2. derivatives f 1 }x f N x exist in ab , 3. x x x ab, - For tx= 01 } N >@ 4. fx j = 0 , for j = 01} N . N xx– Then , , such that N . gx = fx –Px – >@fx –Px -----------------i= 0 c acb f c = 0 xx– i = 1 i - Thus g vanishes at N + 1 points xx 1 x 2 } x N in >@ab . - Generalized Rolle’s theorem says that [[ { x in ab for which g N [ =0. - From (2) we get dN­½ tx– 0=g N [ =f N [–P N [–>@fx –Px -------- N -----------------i (3) N®¾ i = 1 dt ¯¿ xx–i t=[ Scientific computing III 2011: 6. Interpolation 10 Interpolation: polynomials - Now P is at most of degree N – 1 o P N { 0 - The product term is a polynomial of degree N o N N –1 N tx– dN­½ tx– N! -----------------i= xx– tN+Ot N–1 o -------- -----------------i = ------------------------------------- i N®¾ N xx–i dt ¯¿ xx–i xx– i = 1 i=1 i = 1 i=1 i - (3) now becomes N! 0 = f N [– >@fx –Px ------------------------------------- N xx– i=1 i - And solving for fx we get f N [ fx =Px +----------------- N xx– N! i=1 i QED. - Note the analogy between the error formula of the Taylor series: f N [ ----------------- xx– N N! 0 and f N ----------------- [ xx– xx– } xx– N! 1 2 N Scientific computing III 2011: 6. Interpolation 11 Interpolation: polynomials - Example: - Prepare a table for function fx =ex, x>@01 . - Precision d decimals, step size h - What step size is needed for linear interpolation to give absolute error no more than 10–6 ? - Let x >@01 and xj d x d xj + 1 - Error is now f 2 f 2 f 2 fx –Px d---------------- [ xx– xx– =------------------- [ xx– xx– =------------------- [ x– jh xj– +1h 2! j j+1 2 j j+1 2 1 fx –Px d---max f 2 [ max x– jh xj– +1h 2 [ >@01 xj ddxxj+1 Maximum of x– jh xj– +1his at xj= +12ehwith value h2e4 eh2 o fx –Px d -------- 8 - We want the error to be less than 10–6 : eh2 8 –3 --------d 1 0 – 6 h 2 d ---10–6 h 1.72u10 8 e - So we could choose h=0.001 Scientific computing III 2011: 6.