CHAPTER 5 CURVE FITTING
Presenter: Dr. Zalilah Sharer © 2018 School of Chemical and Energy Engineering Universiti Teknologi Malaysia 23 September 2018 TOPICS
Linear regression (exponential model, power equation and saturation growth rate equation)
Polynomial Regression
Polynomial Interpolation (Linear interpolation, Quadratic Interpolation, Newton DD)
Lagrange Interpolation Curve Fitting
• Curve fitting describes techniques to fit curves at points between the discrete values to obtain intermediate estimates. • Two general approaches for curve fitting: a) Least –Squares Regression - to fits the shape or general trend by sketch a best line of the data without necessarily matching the individual points (figure PT5.1, pg 426). - 2 types of fitting: i) Linear Regression ii) Polynomial Regression Figure shows sketches developed from same set of data by 3 engineers. a) least-squares regression - did not attempt to connect the point, but characterized the general upward trend of the data with a straight line b) Linear interpolation - Used straight- line segments or linear interpolation to connect the points. Very common practice in engineering. If the values are close to being linear , such approximation provides estimates that are adequate for many engineering calculations. However, if the data is widely spaced, significant errors can be introduced by such linear interpolation. c) Curvilinear interpolation - Used curves to try to capture suggested by the data. ∑ Our goal here to develop systematic and objective method deriving such curves. a) Least-square Regression : i) Linear Regression • Is used to minimize the discrepancy/differences between the data points and the curve plotted . Sometimes, polynomial interpolation is inappropriate and may yield unsatisfactory results when used to predict intermediate values (see Fig. 17.1, pg 455).
Fig. 17.1 a) : shows 7 experimentally derived data points exhibiting significant variability. Data exhibiting significant error. Curve Fitting
Linear Regression is fitting a ‘ best’ straight line through the points. The mathematical expression for the straight line is:
y = a 0+a 1x+e Eq 17.1
where, a1- slope
a0 - intercept e - error, or residual, between the model and the observations Rearranging the eq. above as:
e = y - a0 - a1x
Thus, the error or residual, is the discrepancy between the true value y and the approximate value, a0+a 1x, predicted by the linear equation. Criteria for a ‘best’ Fit • To know how a “best” fit line through the data is by minimize the sum of residual error , given by ;
n n
∑ei =∑(yi − a0 − a1xi ) ----- Eq 17.2 i=1 i=1
where; n : total number of points
• A strategy to overcome the shortcomings: The sum of the squares of the errors between the measured y and the y calculated with the linear model is shown in Eq 17.3;
n n n S = e2 = (y − y )2 = (y − a − a x )2 r ∑ i ∑i,measured i,mod el ∑ i 0 1 i ----- Eq 17.3 i=1 i=1 i=1 Least-squares fit for a straight line
• To determine values for ao and a 1, i) differentiate equation 17.3 with respect to each coefficient, ii) setting the
derivations equal to zero (minimize Sr), iii) set ΣΣΣao = n.a o to give equations 17.4 and 17.5, called as normal equations ,
(refer text book) which can be solved simultaneously for a 1 and ao;
n ∑ x i y i − ∑ x i ∑ y i a = ----- Eq 17.6 1 2 2 n ∑x i − () ∑ x i
a 0 = y − a 1 x ----- Eq 17.7 Example 1
Use least-squares regression to fit a straight line to:
x 1 2 3 4 5 6 7
y 0.5 2.5 2.0 4.0 3.5 6.0 5.5 • Two criteria for least-square regression will provide the best estimates
of ao and a1 called maximum likelihood principle in statistics: i. The spread of the points around the line of similar magnitude along the entire range of the data . ii. The distribution of these points about the line is normal .
• If these criteria are met, a “ standard deviation ” for the regression line is given by equation:
------Eq. 17.9
S r S y = x n − 2 sy/x : standard error of estimate “y/x” : predicted value of y corresponding to a particular value of x n -2 : two data derived estimates ao and a1 were used to compute Sr (we have lost 2 degree of freedom) • Equation 17.9 is derived from Standard Deviation (Sy) about the mean :
S S = t ------(PT5.2, pg 442 ) y n − 1
2 St = ∑ ( yi − y) ------(PT5.3, pg 442 )
St : total sum of squares of the residuals between data points and the mean.
• Just as the case with the standard deviation, the standard error of the estimate quantifies the spread of the data. Estimation of error in summary 1. Standard Deviation
S ----- (PT5.2, pg 442 ) S = t y n −1 2 St = ∑(yi − y) ----- (PT5.3, pg 442 )
2. Standard error of the estimate
n n 2 2 Sr = ∑ei = ∑(yi − a0 − a1x) i=1i = 1 ----- Eq 17.8 S r Sy = x n − 2 ----- Eq 17.9
where, y/x designates that the error is for a predict value of y corresponding to a particular value of x. 3. Determination coefficient
S − S r 2 = t r ----- Eq 17.10 S t
4. Correlation coefficient
S t − S r n∑ xi y i − (∑ xi )( ∑ y i ) r = @ r = ----- Eq 17.11 S t 2 2 2 2 n∑xi − ( ∑xi ) n ∑y i − ( ∑ y i ) Example 2 Use least-squares regression to fit a straight line to:
x 1 2 3 4 5 6 7
y 0.5 2.5 2.0 4.0 3.5 6.0 5.5
Compute the standard deviation (Sy), the standard error of estimate (Sy/x ) and the correlation coefficient (r) for data above (use Example 1 result) Work with your buddy and lets do Quiz 1
Use least-squares regression to fit a straight line to:
x 1 2 3 4 5 6 7 8 9
y 1 1.5 2 3 4 5 8 10 13
Compute the standard error of estimate (Sy/x ) and the correlation coefficient (r) Quiz 2
Compute the standard error of the estimate and the correlation coefficient.
x 0.25 0.75 1.25 1.50 2.00
y -0.45 -0.60 0.70 1.88 6.00 Linearization of Nonlinear Relationships
• Linear regression provides a powerful technique for fitting the best line to data, where the relationship between the dependent and independent variables is linear . • But, this is not always the case, thus first step in any regression analysis should be to plot and visually inspect whether the data is a linear model or not. Figure 17.8: a) data is ill-suited for linear regression, b) parabola is preferable. Nonlinear Relationships
• Linear regression is predicated on the fact that the relationship between the dependent and independent variables is linear - this is not always the case. • Three common examples are:
β1x exponential: y = α1e
β2 power : y = α2 x x saturation-growth-rate: y = α3 β3 + x Linearization of Nonlinear Relationships
• One option for finding the coefficients for a nonlinear fit is to linearize it. For the three common models, this may involve taking logarithms or inversion:
Model Nonlinear Linearized
β1x exponential: y = α1e ln y = ln α1 + β1x
β2 power : y = α2 x log y = log α2 + β2 log x
x 1 1 β3 1 saturation-growth-rate: y = α3 = + β3 + x y α3 α3 x Linearization of Nonlinear Relationships
• After linearization, Linear regression can be applied to determine the linear relation. • For example, the linearized exponential equation:
ln y = lnα1 + β1x
y a0 a1x Figure 17.9: Type of polynomial equations and their linearized versions, respectively. • Fig. 17.9, pg 453 shows population growth of radioactive decay behavior.
Fig. 17.9 (a) : the exponential model
β1x ------(17.12) y =α1e
α1 , β1 : constants, β1 ≠ 0
This model is used in many fields of engineering to characterize quantities.
Quantities increase : β1 positive
Quantities decrease : β1 negative Example 2
Fit an exponential model y = a ebx to:
x 0.4 0.8 1.2 1.6 2.0 2.3
y 750 1000 1400 2000 2700 3750
Solution • Linearized the model into; ln y = ln a + bx
y = a 0 + a 1x ----- (Eq. 17.1)
• Build the table for the parameters used in eqs 17.6 and 17.7, as in example 17.1, pg 444. 2 xi yi ln yi xi (x i)(ln yi) 0.4 750 6.620073 0.16 2.648029 0.8 1000 6.900775 0.64 5.520620 1.2 1400 7.244228 1.44 8.693074 1.6 2000 7.600902 2.56 12.161443 2.0 2700 7.901007 4.00 15.802014 2.3 3750 8.229511 5.29 18.927875 ΣΣΣ 8.38.38.3 44.496496 14.09 63.753055 n = 6 n n
∑ x i = 8 3. ∑ ln y i = 44 .496496 i =1 i =1 n n 2 ∑x = 14 09. ∑ ( x i )(ln y i ) = 63 .753055 i i =1 i =1 8 3. 44 .496496 x = = 1.383333 ln y = = 7.416083 6 6 a0 = ln a = ln y − b x = 7.416083 − (0.843 )( 1.383333 ) ln a = 6.25
n Σ ( x i )(ln y i ) − Σ x i Σ (ln y i ) a 1 = b = 2 2 n Σ x i − ( Σ x i ) ( 6 )( 63 .753055 ) − (8 .3 )( 44 .496496 ) b = = 0 .843 ( 6 )( 14 .09 ) − (8 .3 ) 2 Straight-line: ln y = ln a + bx ∴ ln y = 6.25 + 0.843 x
Exponential: y = a ebx
ln a = 6.25 ⇒ a = e 25.6 = 518 ∴ y = a ebx = 518 e 843.0 x Figure 17.9: Type of polynomial equations and their linearized versions, respectively. Power Equation
• Equation (17.13 ) can be linearized by taking base-10 logarithm to yield:
β 2 ------(17.13) y = α 2 x
log y = log α 2 + β 2 log x ------(17.16)
• A plot of log y versus log x will yield a straight line with
slope of βββ2 and an intercept of log ααα2. Example 4
Linearization of a Power equation and fit equation (17.13) to the data in table below using a logarithmic transformation of the data.
x 1 2 3 4 5
y 0.5 1.7 3.4 5.7 8.4 2 xi yi log xi log yi (log xi) (log xi)( log yi)
1 0.5 0 -0.301 0 0 2 1.7 0.301 0.226 0.090601 0.068026 3 3.4 0.477 0.534 0.227529 0.254718 4 5.7 0.602 0.753 0.362404 0.453306 5 8.4 0.699 0.922 0.488601 0.644478 ΣΣΣ 2.079 2.134 1.169135 1.420528 n = 5 n n
∑log xi = .2 079 ∑log yi = .2 134 i=1 i=1 n n 2 ∑(log xi ) = .1 169135 ∑(log xi )(log yi ) = .1 420528 i=1 i=1 .2 079 .2 134 log x = = .0 4158 log y = = .0 4268 5 5
nΣ(log x i )(log y i ) − (Σ log x i )( Σ log y i ) b = 2 2 nΣ(log x i ) − (Σ log x i ) (5)(1.420528 ) − (2.079 )(2.134 ) b = = 1.75 (5)(1.169135 ) − (2.079 ) 2 log a = log y − b (log x ) = 0.4268 − (1.75 )( 0.4158 ) ln a = −0.3
Straight-line:
log y = log a + b log x ∴ log y = − 0.3 + 1.75 log x
Power: y = a x b
log a = − 3.0 ⇒ a =10 − 3.0 = 5.0
∴ y = a xb = 0.5 x 75.1 • Fig. 17.10 a), pg 455, is a plot of the original data in its untransformed state, while fig. 17.10 b) is a plot of the transformed data.
• The intercept, log α2 = -0.300, and by taking
the antilogarithm, α2 = 10 -0.3 = 0.5.
• The slope is β2 = 1.75, consequently, the power equation is : y = 0.5x 1.75 Figure 17.9: Type of polynomial equations and their linearized versions, respectively. Saturation growth rate equation • Equation (17.14) can be linearized by inverting it to yield: x y = α 3 ------(17.14) β 3 + x
1 β 1 1 = 3 + y α 3 x α 3 ------(17.17)
• A plot of 1/y versus 1/x will yield a straight line with slope of
βββ3/ααα3 and an intercept of 1/ ααα3 • In their transformed forms, these models are fit using linear regression in order to evaluate the constant coefficients. • This model well-suited for characterizing population growth under limiting conditions. Example 5
Linearization of a saturation-growth rate equation to the data in table below.
x 0.75 2 2.5 4 6 8 8.5 y 0.8 1.3 1.2 1.6 1.7 1.8 1.7 n = 7 n 1 n 1 ∑ = 2.8926 ∑ = 5.2094 i=1 xi i=1 yi 2 n 1 n 1 1 ∑ = 2.3074 ∑ = 2.8127 i=1 xi i=1 xi yi 1 2.8926 1 5.2094 = = 0.4132 = = 0.7442 x 7 y 7 2 xi yi 1/ x i 1/ yi (1/x i) (1/ x i)(1/yi)
0.75 0.8 1.33333 1.25000 1.7777 1.6666 2 1.3 0.50000 0.76923 0.2500 0.3846 2.5 1.2 0.40000 0.83333 0.1600 0.3333 4 1.6 0.25000 0.62500 0.0625 0.1562 6 1.7 0.16667 0.58823 0.0278 0.0981 8 1.8 0.12500 0.55555 0.0156 0.0694 8.5 1.7 0.11765 0.58823 0.0138 0.1045 ΣΣΣ 2.89260 5.20940 2.3074 2.8127
1 1 1 1 n Σ − Σ Σ b x y x y = i i i i a 2 1 1 2 n Σ − ( Σ ) x i x i b ( 7 )( 2 .8127 ) − ( 2 .8926 )( 5 .2094 ) = = 0 .5935 a ( 7 )( 2 .3074 ) − ( 2 .8926 ) 2 1 1 b 1 = − = .0 7442 − ( .0 5935 )( .0 4132 ) a y a x 1 = .0 4990 a 1 1 b 1 Straight-line: = + y a a x
1 1 ∴ = .0 4990 + .0 5935 y x
x Saturation-growth: y = a b + x 1 = .0 4990 ⇒ ∴ a = 2 a b = .0 5935 ⇒ ∴ b = ( .0 5935 )( 2) = .1 187 a x ∴ y = 2 1.187 + x Lets do Quiz 3
Fit a power equation and saturation growth rate equation to: x 1 2 3 4 5 6 7 y 2.1 2.2 2.3 2.4 2.5 2.6 2.7 Figure 17.8: a) data is ill-suited for linear regression, b) parabola is preferable. Polynomial Regression
• Another alternative is to fit polynomials to the data using polynomial regression.
• The least-squares procedure can be readily extended to fit the data to a higher-order polynomial.
• For example, to fit a second–order polynomial or quadratic:
2 y = a0 +a1x + a2 x + e • The sum of the squares of the residual is: n 2 2 Sr = ∑ ( yi − a0 − a1xi − a2 xi ) i=1 where n= total number of points • Then, taking the derivative of equation (17.18) with respect to
each of the unknown coefficients, ao , a1 ,and, a2 of the polynomial, as in:
δ S r 2 = −2∑ ( yi − a0 − a1 xi − a 2 xi ) δ a0
δ S r 2 = −2∑ xi ( yi − a0 − a1 xi − a 2 xi ) δ a1
δ S r 2 2 = −2∑ xi ( yi − a0 − a1 xi − a 2 xi ) δ a 2 • Setting the equations equal to zero and rearrange to develop set of normal equations and by setting ΣΣΣao = n.ao
2 (n)a0 + (∑ xi )a1 + (∑ xi )a2 = ∑ yi 2 3 (∑xi )a0 + ( ∑xi )a1 + ( ∑xi )a2 = ∑ xi yi ----- 17.19 2 3 4 2 (∑xi )a0 + ( ∑xi )a1 + ( ∑xi )a2 = ∑ xi yi • The above 3 equations are linear with 3 unknowns coefficients (ao , a1 ,and, a2) which can be calculated directly from observed data. • In matrix form:
2 n Σxi Σxi a0 Σyi x x 2 x3 × a = Σx y Σ i Σ i Σ i 1 i i 2 3 4 2 Σxi Σxi Σxi a2 Σxi yi • The two -dimensional case can be easily extended to an mth -order polynomial as:
2 m y = a0 + a1 x + a2 x + ... + am x + e • Thus, standard error for mth -order polynomial :
S S = r ----- 17.20 y / x n − (m + )1 Example 6 Fit a second order polynomial to the data in the first 2 columns of table 17.4:
2 3 4 2 xi yi xi xi xi xi yi xi yi 0 2.1 0 0 0 0 0 1 7.7 1 1 1 7.7 7.7 2 13.6 4 8 16 27.2 54.4 3 27.2 9 27 81 81.6 244.8 4 40.9 16 64 256 163.6 654.4 5 61.1 25 125 625 305.5 1527.5 ΣΣΣ 15 152.6 55 225 979 585.6 2488.8
• From the given data: 4 m = 2 Σxi = 15 Σ xi = 979 y = 25.433 3 n = 6 Σ yi = 152.6 Σ xiyi = 585.6 Σ xi = 225 2 2 x = 2.5 Σ xi = 55 Σ xi yi = 2488.8 2 n Σxi Σxi a0 Σyi 2 3 Σxi Σxi Σxi × a1 = Σxi yi 2 3 4 2 Σxi Σxi Σxi a3 Σxi yi • Therefore, the simultaneous linear equations are:
6 15 55 ao 152 6. 15 55 225 a1 = 585 6. 55 225 979 a2 2488 8. • Solving these equations through a technique such as Gauss elimination gives:
ao = 2.47857, a 1 = 2.35929, and a 2 = 1.86071 • Therefore, the least-squares quadratic equation for this case is: y = 2.47857 + 2.35929x + 1.86071x 2
• To calculate st and sr , build table 17.4 for columns 3 and 4. 2 2 2 xi yi (yi- y ) (yi-ao-a1xi-a2xi ) 0 2.1 544.44 0.14332 1 7.7 314.47 1.00286 2 13.6 140.03 1.08158 3 27.2 3.12 0.80491 4 40.9 239.22 0.61951 5 61.1 1272.11 0.09439 Σ 152.6 2513.39 3.74657
St = Σ(yi − y) = 2513 39. 2 2 Sr = Σ(yi − a0 − a1xi − a2 xi ) = .3 74657 The standard error (regression polynomial):
Sr .3 74657 S y = = = 12.1 x n − (m + )1 6 − (2 + )1 • The correlation coefficient can be calculated by using equations 17.10 and 17.11, respectively: S − S r 2 = t r S t
S − S n∑ xi y i − (∑ xi )( ∑ y i ) r = t r @ r = S t 2 2 2 2 n∑xi − ( ∑xi ) n ∑y i − ( ∑ y i )
2 Therefore, r = (S t – Sr) / S t = (2513.39 – 3.74657) / 2513.39 r2 = 0.99851 ∴The correlation coefficient is, r = 0.99925
• The results indicate that 99.851% of the original uncertainty has been explained by the model. This result supports the conclusion that the quadratic equation represents an excellent fit, as evident from Fig.17.11. Figure 17.11: fit of a second-order polynomial TOPICS
Polynomial Interpolation (Linear interpolation, Quadratic Interpolation, Newton DD)
Lagrange Interpolation Spline Interpolation Interpolation • Polynomial Interpolation is a common method to determine intermediate values between data points. • General equation for nth order polynomial is:
2 n f (x) = a0 + a1 x + a2 x ..... + a n x ----- 18.1 • Polynomial interpolation consists of determining the unique nth -order polynomial that fits n+1 data point. • For n+1 data points, there is only one polynomial of order n that passes through all the points. • For example, there is only one straight line (first-order polynomial) that connects two points (Fig. 18.1a) and only one parabola connects a set of three points (Fig.18.1b). • Two popular alternative mathematical formats used to express an interpolating polynomial: a. Newton polynomial b. Lagrange polynomial
18.1 Newton’s Divided-Difference Interpolating Polynomials • The most popular and useful in polynomial forms. • Consists of the first- and second-order versions.
18.1.1 Linear Interpolation • The simplest form of interpolation is to connect two data points with a straight line . Figure 18.2: graphical depiction of linear interpolation. • This linear interpolation technique can be depicted graphically as shown in fig 18.2, in which, the similar triangles can be rearranged to yield a linear-interpolation formula ;
f (x1 ) − f (x0 ) f1 (x) = f (x0 ) + (x − x0 ) ----- 18.2 x1 − x0
- f1(x) is refer to first order interpolation polynomial
f (x ) − f (x ) -The term 1 0 is a finite-divided-difference x1 − x0
approximation of the first derivative.
• In general, the smaller the interval between the data points, the better the approximation. Figure 18.2: graphical depiction of linear interpolation. Example 7
Estimate the natural logarithm of 2 using linear interpolation. First, perform the computation by interpolating between ln 1 = 0 and ln 6 = 1.791759. Then, repeat the procedure, but use a smaller interval from ln 1 to ln 4 (1.386294 ). Note that the true value of ln 2 is 0.6931472. Solution
By using equation (18.2), a linear interpolation for ln 2
from xo = 1 to x1 = 6 to give; x f(x)
x0=1 f(x 0) =0 f (x1) − f (x0 ) x=2 f (x)=? f1(x) = f (x0 ) + (x − x0 ) 1 x1 − x0 x1=6 f(x1)=1.791759 .1 791759 − 0 f (2) = 0 + (2 − )1 = .0 3583519 1 6 − 1
.0 6931472 − .0 3583519 ε = *100 = 48 3. % t .0 6931472 • Then, using the smaller interval from xo = 1 to x1 = 4 yields;
x f(x)
x0=1 f(x 0) =0
x=2 f1(x)=?
x1=4 f(x1)=1.386294
1.386294 − 0 f (2) = 0 + (2 −1) = 0.4620981 1 4 −1 .0 6931472 − 0.4620981 ε = *100 = 33 3. % t .0 6931472
• Thus, using the shorter interval reduces the percent
relative error to εt = 33.3%. • Both interpolations are shown in Fig.18.3, along with true function. Figure 18.3: Comparison of two linear interpolations with different intervals. Quiz 4
Estimate the logarithm of 5 to the base 10 (log5) using linear interpolation.
a) Interpolate between log 4=0.60206 and log6=0.7781513 b) Interpolate between log4.5=0.6532125 and log 5.5=0.7403627
For each of the interpolations, compute the percent relative error based on the true value 18.1.2 Quadratic Interpolation The error in example 18.1 (linear interpolation) resulted from approximation of a curve with a straight line. With 3 data points, the estimation can be improved with a second-Order Polynomial (quadratic polynomial or parabola). Thus;
f2 (x) = b0 + b1(x − x0 ) + b2 (x − x0 )( x − x1) ----- (18.3)
Although equation (18.3) seem to differ from the general polynomial (equation 18.1), the two equations are equivalent, by multiplying the terms in equation (18.3) to yield;
2 f2(x) = bo + b 1x – b1xo + b 2x + b 2xox1 – b2xx o – b2xx 1 or in collecting terms, 2 f2(x) = ao + a1x + a2x where;
ao = bo – b1xo + b2xox1
a1 = b1 – b2xo – b2x1
a2 = b2
• Thus, equations 18.1 and 18.3 are alternative, equivalent formulations of the unique second-order polynomial joining three points.
• To determine the values of coefficient ( bo, b1, and, b2), rearrange and use Eq 18.3, substitute Eq 18.4 into 18.3, and substitute Eq 18.4 and 18.5 into Eq 18.3 to yield Eq 18.6.
b0 = f (x0 ) ----- 18.4
f (x1) − f (x0 ) b1 = x1 − x0 ----- 18.5 f (x ) − f (x ) f (x ) − f (x ) 2 1 − 1 0 x − x x − x b = 2 1 1 0 2 ----- 18.6 x2 − x0 Example 8
Fit a second-order polynomial to the three points used in linear interpolation example
xo = 1 f(x 0) = 0 x1 = 4 f(x 1) =1.386294 x2 = 6 f(x 2) =1.791759
Use the polynomial to evaluate ln 2 Solution
b0 = f (x0 ) = 0
f (x1) − f (x0 ) .1 386294 − 0 b1 = = = .0 4620981 x − x0 4 −1
f (x2 ) − f (x1) f (x1) − f (x0 ) 1.791759 −1.386294 − − 0.4620981 x2 − x1 x1 − x0 6 − 4 b2 = = = − .0 0518731 x2 − x0 6 −1 Substituting these values into equation 18.3 yields the quadratic formula:
f2(x) = bo + b1(x – xo) + b2(x – xo)(x – x1)
f2(x) = 0 + 0.462098(x – 1) – 0.0518731(x – 1)(x – 4) which can be evaluated at x = 2 for;
f2(2) = 0.5658444 which represents a relative error of εt = 18.4%.
• Thus, the curvature introduced by the quadratic formula (Fig.18.4) improves the interpolation compared with the result obtained using straight lines in example 18.1, Fig.18.3. Figure 18.4: The use of quadratic formula to estimate ln 2 improves the interpolation. Example 9 Given the data, calculate f(3.4) using Newton’s polynomials of order 1 to 2. 3.4
x 1 2 2.5 3 4 5
f(x) 1 5 7 8 2 1
Solution ⇒ x 0 = 3 f ( x 0 ) = 8 ⇒ x1 = 4 f ( x1 ) = 2 ⇒ x 2 = 2.5 f ( x 2 ) = 7 or ⇒ x 2 = 5 f ( x 2 ) = 1 Use equations 18.4-18.6 to find b 0, b 1 and b 2.
st 1 order need f [x 0, x 1] ) From eq 18.4 ;
b0 = f (x0 ) = 8 From eq 18.5;
f (x1 ) − f (x0 ) 2 − 8 b1 = = = −6 x1 − x0 4 − 3
From equation 18.3,
f1 (x) = b0 + b1 (x − x0 )
∴ f1 (x) = 8 + (−6)( x − 3) ⇒ f1 (3.4) = 8 + (−6)( 3.4 − 3) = 5.6 nd 2 order (need f [x 0, x 1, x 2] ) From equation 18.6;
f (x2 ) − f (x1) f (x1) − f (x0 ) 7 − 2 − −(−6) x2 − x1 x1 − x0 2.5− 4 b2 = = = −5.75 x2 − x0 2.5−3 From eq 18.3 for 2 nd order f2 (x) = b0 +b1(x − x0 ) +b2 (x − x0 )( x − x1)
∴ f2 (x) =8+ (−6)( x −3) + (−5.75 )( x −3)( x − 4) ⇒ f2 (3.4) =8+ (−6)( 3.4−3) + (−5.75 )( 3.4−3)( 3.4− 4)
f2 (3.4) = 6.98 Quiz 5
Given the data, calculate f(4) using Newton’s polynomial of order 1 to 2
x 1 2 3 5 6 f(x) 4.75 4 5.25 19.75 36 18.1.3 General Form of Newton’s Interpolating Polynomials • The n th order polynomial is:
fn(x) = b o + b 1(x – xo) + b 2(x – xo)(x – x1)+ . . .
+ b n(x – xo)(x – x1). . . (x – xn-1) ----- (18.7)
• For n’th-order polynomial, n + 1 data points are required:
[ xo, f(x o) ], [ x1,f(x 1) ] . . . . [ xn, f(x n) ]
• Then, we used these data points and following equation’s to evaluate the coefficients ----- 18.8 b 0 = f ( x 0 )
b1 = f [ x1 , x 0 ] ----- 18.9
b 2 = f [ x 2 , x1 , x 0 ] ----- 18.10 M L ----- 18.11 b n = f [ x n , x n −1 , , x 1 , x 0 ] • Where the bracketed [ ] function evaluations are finite divided differences (FDD) ; • For example, the first finite divided difference is represented generally as;
f (xi ) − f (x j ) f [xi , x j ] = ----- 18.12 xi − x j
Second finite divided difference
f [xi , x j ] − f [x j , xk ] ----- 18.13 f [xi , x j , xk ] = xi − xk
• Similarly, the n’th finite divided difference is f [x , x L x ] − f [x , x L x ] L n n−1 1 n−1 n−2 0 ----- 18.14 f [xn , xn−1 x1, x0 ] = xn − x0 • These differences can be used to evaluate the coefficients in equations (18.8) through (18.11), which can then substituted into equation (18.7) to yield the interpolating polynomial, called as Newton’s divided-difference interpolating polynomial ;
f (x) = f (x ) + (x − x ) f [x , x ]+ (x − x )( x − x ) f [x , x , x ] n 0 0 1 0 0 1 2 1 0 ----- 18.15 + .... + (x − x0 )( x − x1 )... ()x − xn−1 f []xn , xn−1,...., x0 Example 10
From example 18.2, data points at x 0=1, x 1=4 and x 2=6 were used to estimate ln 2 with a parabola. Now adding a fourth point (x 3=5; f(x 3)=1.609438), Estimate ln 2 with a third order Newton’s interpolating Polynomial .
x0=1 x1=4 x2=6 x3=5
f(x0)=0 f(x1)=1.386294 f(x2)=1.791759 f(x3)=1.609438 Solution
The third-order polynomial with n = 3 is, f3(x) = bo + b 1(x – xo) + b 2(x – xo)(x – x1) + b 3(x – xo)(x – x1)(x-x2)
The first divided differences are (use eq 18.12):
f ( x1 ) − f ( x0 ) 1.386294 − 0 f [ x1 , x 0 ] = = = 0.4620981 x1 − x 0 4 − 1 f (x2 ) − f (x1) .1 791759 − .1 386294 f [x2 , x1] = = = .0 2027326 x2 − x1 6 − 4
f (x3 ) − f (x2 ) .1 609438 − .1 386294 f [x3 , x2 ] = = = .0 1823216 x3 − x2 5 − 6
The second divided differences (use equation 18.13):
f [x2 , x1]− f [x1, x0 ] 0.2027326 − 0.4620981 f [x2 , x1, x0 ] = = = −0.05187311 x2 − x0 6 −1
f [x3,x2 ]− f [x2 ,x1] 0.1823216 −0.2027326 f [x3,x2 ,x1] = = = −0.02041100 x3 − x1 5− 4 The third divided differences:
f [x3 , x2 , x1 ] − f [x2 , x1, x0 ] − .0 02041100 − (− .0 05187311 ) f [x3 , x2 , x1, x0 ] = = x3 − x0 5 −1
f [x3 , x2 , x1, x0 ] = .0 007865529 Finally, using eqs 18.8-18.11:
b0 = f ( x 0 ) = 0
b1 = f[ x1, x 0 ] = 0.4620981
b2 = f[] x 2 , x1, x 0 = −0.05187311
b3 = f[] x 3, x 2 , x1, x 0 = 0.007865529 Insert all values into eqs 18.7;
f3 (x) = 0 + .0 4620981 (x − )1 − .0 05187311 (x − )(1 x − )4 + .0 007865529 (x − )(1 x − 4)( x − 6)
f3 (2) = .0 6287686 , which represents a relative error of εt = 9.3%. Quiz 6
Calculate f (4) with a third and fourth order Newton’s interpolating polynomial.
x 1 2 3 5 6 f(x) 4.75 4 5.25 19.75 36 Solution 1st FDD;
f ( x1 ) − f ( x 0 ) 19 .75 − 5.25 f [ x1 , x 0 ] = = = 7.25 x1 − x 0 5 − 3
f (x2 ) − f (x1) 4 −19 .75 f [x2 , x1] = = = 5.25 x2 − x1 2 − 5
f (x3 ) − f (x2 ) 36 − 4 f [x3 , x2 ] = = = 8 x3 − x2 6 − 2 2nd FDD
f [x2 , x1]− f [x1, x0 ] 5.25 − 7.25 f [x2 , x1, x0 ] = = = 2 x2 − x0 2 − 3
f [x3, x2 ]− f [x2 , x1] 8 − 5.25 f [x3, x2 , x1] = = = 2.75 x3 − x1 6 − 5
3rd FDD
f [x3 , x2 , x1 ] − f [x2 , x1 , x0 ] 2.75 − 2 f [x3 , x2 , x1, x0 ] = = x3 − x0 6 − 3
f [x3 , x2 , x1, x0 ] = 0.25 Finally,
b0 = f ( x 0 ) = 5.25
b1 = f []x1 , x 0 = 7.25
b2 = f []x 2 , x1 , x 0 = 2
b3 = f []x ,3 x 2 , x1 , x 0 = 0.25
f 3 (x) = 5.25 + 7.25 (x − 3) + 2(x − 3)( x − 5) + 0.25 (x − 3)( x − 5)( x − 2)
f3(x) = bo + b 1(x – xo) + b 2(x – xo)(x – x1) + b 3(x – xo)(x – x1)(x-x2)
∴ f 3 (4) = 10 Lagrange Interpolating Polynomials • The Lagrange interpolating polynomial is simply a reformulation of the Newton’s polynomial that avoids the computation of divided differences: n
f n ( x) = ∑ Li ( x) f ( xi ) ----- 18.20 i = 0
n x − x j where Li ( x) = ∏ ----- 18.21 j = 0 xi − x j ≠ ij
• Where Π designates the “product of”. For example the linear version (n =1) is
x − x1 x − x 0 f1 ( x) = f ( x 0 ) + f ( x1 ) ----- 18.22 x 0 − x1 x1 − x 0 • And the second-order version (n=2) is
(x − x1 )(x − x 2 ) (x − x 0 )(x − x 2 ) f 2 ( x) = f ( x 0 ) + f ( x1 ) (x 0 − x1 )(x 0 − x 2 ) (x1 − x 0 )(x1 − x 2 ) (x − x )(x − x ) + 0 1 f ( x ) 2 ----- 18.23 (x 2 − x 0 )(x 2 − x 1 )
• For n=3
(x − x1)(x − x2 )(x − x3 ) (x − x0 )(x − x2 )(x − x3 ) f2 (x) = f (x0 ) + f (x1) (x0 − x1 x0 )(−x2 x0 − )(x3 )(x1 − x0 x1 )(−x2 x1 − )(x3 )
(x − x0 x )(− x1 x − x )(3 ) (x − x0 x )(− x1 x − x )(2 ) + f (x2 ) + f (x3) (x2 − x0 x2 )(−x1 x2 − )(x3 )(x3 − x0 x3 )(−x1 x3 − )(x2 ) Example 11 Use a Lagrange interpolating polynomial of the first and second order to evaluate ln 2 based on the data given.
x0 = 1 f(x 0) = 0
x1 = 4 f(x 1) = 1.386294
x2 = 6 f(x 2) = 1.791760 Solution: First-order polynomial at x = 2, use eq. 18.22; 2 − 4 2 −1 f (2) = (0) + .1( 386294 ) = .0 4620981 1 1− 4 4 −1 Second-order polynomial at x = 2, use eq 18.23; (2 − 4)( 2 − 6) (2 − )(1 2 − 6) f ( x ) = (0) + .1( 386294 ) + 2 1( − 4)( 1 − 6) (4 − )(1 4 − 6) (2 − )(1 2 − 4) .1( 791760 ) = .0 5658444 (6 − )(1 6 − 4) Working with your buddy to do Quiz 7 Quiz 7
Given the data, calculate f(4) using the Lagrange polynomials of order 1 to 2.
x 1 2 3 5 6 f( x) 4.75 4 5.25 19.75 36 Chapter 5 Lets do past year questions
Q1 Q3 Q6 Q8 Assignment (please use excell)
Given the data below, use least squares regression to fit a) a straight line b) a power equation c) a saturation-growth-rate equation and d) a parabola. Find the r 2 value and justify which is the best equation to represent the data.
x 5 10 15 20 25 30 35 40 45 50
y 17 24 31 33 37 37 40 40 42 51 Lets do past year question - Q3 and Q6
Develop a MATHLAB program which can be written in a command window to solve this problem which will give the linear regression equation Lets try mathlab to solve linear regression problem
T( oC) 4 8 12 16 20 24 28
V (10 -2cm 2/s) 1.88 1.67 1.49 1.34 1.22 1.11 1.02 Question? THE END
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