CURVE FITTING LEAST SQUARE LINE. Consider the class of linear function of the form. = Ax+ B...(1)

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1 CURVE FITTIG LEAST SQUARE LIE Consider the class of linear function of the form y = f( x) = B...() In previous chapter we saw how to construct a polynomial that passes through a set of points. If all numerical values { x }, { y } are nown to several significant digits of accuracy, then the polynomial interpolation can be used successfully ; other wise it can not. y, it realize that the true value f ( x ) satisfies For values { } x, { } f ( x ) = y e.() where e is the measurement error How do find the best approximation of the form () that goes near (not always through) the points? To answer this question we need to discuss the errors e = f ( x ) y () There are several norms that can be used with the residuals in () to measure how far the curve y = f( x) lies from the data { } max { } MAXIMUM ERROR : E ( f ) = max f ( x ) y = e...() K= K= K = K = AVERAGE ERROR : E ( f ) = f ( x ) y = e...(5) ROOT MEA SQUARE ERROR ( RMS): E ( f ) = f ( x ) y = e...(6)

2 Example Compare the maximum error, average error and RMS error for the linear approximation f ( x) = 8.6.6x to the data points (-,), (,9), (,7), (,5), (,), (,), (5,) and (6,-). x y f ( x) = 8.6.6x e = f( x) y e =8 K = { } MAXIMUM ERROR : E ( f ) = max.,.,.,.,.8,.6, =.8 AVERAGE ERROR : E ( f ) = e = (.6) =.5 8 K = 8 ROOT MEA SQUARE ERROR ( RMS): E ( f ) = e = (.6) =.8 Best fitting line is found by minimizing one of the quantities in equation () through (6). Hence there are three best fitting lines that we could find. The third norm E ( f ) is the traditional choice because it is much easier to minimize computationally.

3 FIDIG THE LEAST SQUARES LIES { } = Let ( x, y ) be a set of points ( where { x } are distinct). The least square line y = f( x) = B is the line minimize the RMS error E ( f ). ( x, y ) ( x, y ) ( x, y ) ( x, y ) The quantity E ( f ) will be minimum if and only if the quantity = [ ] E ( ( f )) = ( B) y is minimum Theorem { } = Suppose that ( x, y ) are points. The coefficients of the least square line y = B are the solution of the following linear system x A x B= xy = = = x A B = y = =...(7)

4 Proof Where E( AB, ) = ( B y ) = The minimum value of E( AB, ) is determined by setting the partial derivatives E( AB, ) E( AB, ) and equal to zero and solving these equation for A and B. A B EAB (, ) A B y x Bx yx = = = ( ) = ( ) =...(8) EAB (, ) = ( B y ) = ( Bx y ) =...(9) B from (8) and (9) = = x A x B= xy = = = x A B = y = =

5 Example Find the least squares line y = B for the following data and evaluate E ( f ) x -6-6 y 7 5 f ( x ) According to the above table x y f ( x ) x x y e = f( x ) y ( e ) Where =5 then A B= -8 A 5B= 7 solve x system to find A= -.6 and B=. y = B y =.6x. and E( f) =. =.8 5 5

6 THE POWER FIT y= m Theorem { } = Suppose that ( x, y ) squares power curve points, where { x } are distinct. The coefficients A of the least m y = is given by Proof, where m x y = A = m x = = m ( ) EA ( ) = y m x y E m m = = ( y) x A = = A = m x = Example Find the power fits y= and y= for the following data and use ( ) determine which curve is best. x y xy xy = = for y = A = and for y = A = x x = = 6

7 Establish following table x y x x x 6 x x y x y for y = A = =.687 and for y = A = = f( x) = y=.687 x and g( x) = y=.59x Find best curve fit using E ( f ) x y f ( x ) g( x ) e = f( x ) y e g( x ) y = ( e ) ( e ) for y =.687 x E ( f ) = 8.9 =.97 and for y =.59 x E ( f ) =.8 = the best power fit is y =.59x 7

8 DATA LIEARIZATIO METHOD FOR y= Ce { } = Suppose that we are given points ( x, y ) y= Ce and we want to fit an exponential curve...() The first step is to tae logarithm of both sides ln y= ln( Ce ) = lnc ln e then ln y= ln C...() Then say Y = ln y, X = x, and B = lnc from () Y = AX B Then use X A X B= XY = = = X A B = Y = = to find A, B and C Example Use data linearization method and find the exponential fit y= Ce for the following data x y where, y = Ce then lny = lnc let Y = ln y, x = X and B = lnc 8

9 Establish table x = X y Y = ln y X Y X =5 AB=6.9 -x A 5B=6.98 A=.9 A=.9 B= lnc = B=.57 C = e =.579 y= Ce y=.579e.9x 9

10 CHAGE OF VARIABLES FOR DATA LIEARIZATIO Function y=f(x) y= A B x Linearized form Y=AXB y= A B x Change of variables and constants X =, Y = y x D y = = x C x C y D D C Y =, A=, B= y D D y = B B y = Y =, X = x y y = x B = A B y x Y =, X = y x y = Aln x B y = Aln x B X = ln x, Y = y y = Ce ln ln y = C Y = ln y, X = x, B= lnc y A = Cx ln ln ln y= A x C Y = ln y, X = ln x, B= lnc y= ( B) y = B X = x, Y = y y= Cxe Dx ln y = B x y Y = ln, X = x x L y = L ln lnc Ce = y L Y = ln, X = x, B= ln C y

11 Example : Fit a curve of the form y = A Cx to the following data x y 5 and determine E ( f ) A A y = then = Cx = Cx ln Alnx lnc A = Cx y y y where Y = ln, X = ln x, B = ln C y x y X = ln x Y = ln y X XY =.685 A.796 B = A B =.68 solve x system then Where -.85 A= A=.9889 and B= A=.989 and B= ln C C = e C = e =.77 y = f( x) = x ow find E ( f ) using f(x) and given data B

12 x y f ( x ) e f( x ) y = ( ) e E( f) =.85 =.5 EXERCISE : Let y = a x a x Find the curve fit for the function y using the following table and determine E ( f ) x.5. y EXERCISE : Find the curve fit y = f( x) = Cxe B with f() =, for the points (-,), (, 6) and (,9) using the least square technique and determine E ( f ).