Mc1020

Mr. T. Mayooran,
Department of Interdisciplinary Studies,
Faculty of Engineering,
University of Jaffna.
Email: mayooran@eng.jfn.ac.lk
Mathematics - MC1020
Curve Fitting

CURVE FITTING
Describes techniques to fit curves (curve fitting) to discrete data to
obtain intermediate estimates.
There are two general approaches for curve fitting:
• Least Squares regression: Data exhibit a significant degree of
scatter. The strategy is to derive a single curve that represents the
general trend of the data.
• Interpolation: Data is very precise. The strategy is to pass a curve
or a series of curves through each of the points.
Thursday, January 24, 2019 2MC1020-MathematicsMC1020-Mathematics

Introduction
In engineering, two types of applications are encountered:
–Trend analysis. Predicting values of dependent variable,
may include extrapolation beyond data points or
interpolation between data points.
–Hypothesis testing. Comparing existing mathematical
model with measured data.
Thursday, January 24, 2019 MC1020-Mathematics 3

Mathematical Background
• Arithmetic mean. The sum of the individual data points (yi)
divided by the number of points (n).
• Standard deviation. The most common measure of a spread
for a sample.
ni
n
y
y i
,,1, 

 

 2
)(,
1
yyS
n
S
S it
t
y

Mathematical Background (cont’d)
• Variance. Representation of spread by the square of the
standard deviation.
or
• Coefficient of variation. Has the utility to quantify the
spread of data.
 
1
/
22
2



 
n
nyy
S ii
y
1
)( 2
2




n
yy
S i
y
%100..
y
S
vc
y


Linear Regression
•Fitting a straight line to a set of paired
observations: (x1, y1), (x2, y2),…,(xn, yn)
yi = a0 + a1 xi + e
Where,
e = yi - a0 - a1 xi
a1 : slope
a0 : intercept
yi : measured value
e : error

Linear Regression: Residual
e Error
Line equation
y = a0 + a1 x

How to find a0 and
a1 so that the error
would be
minimum?

• Minimize the sum of the residual errors
for all available data?
Inadequate!
(see )
• Sum of the absolute values?
Inadequate!
(see )
• How about minimizing the distance that
an individual point falls from the line?
This does not work either! see 
 

n
i
ioi
n
i
i xaaye
1
1
1
)(
 

n
i
ii
n
i
i xaaye
1
10
1
Regression
line
Choosing Criteria For a “Best Fit”

11
• Best strategy is to minimize the sum of the
squares of the residuals between the measured-y
and the y calculated with the linear model:
• Yields a unique line for a given set of data
• Need to compute a0 and a1 such that Sr is
minimized!









n
i
iir
n
i
modelimeasuredi
n
i
ir
xaayS
yy
eS
1
2
10
1
2
1
2
)(
)( ,,
e Error
Thursday, January 24, 2019 MC1020-Mathematics

Linear Regression: Least Squares Fit
 

n
i
ii
n
i
ir xaayeS
1
2
10
1
2
)(min
   

n
i
n
i
iiii
n
i
ir xaayyyeS
1 1
2
10
2
1
2
)()model,measured,(
Yields a unique line for a given set of data.

Linear Regression: Least Squares Fit
 

n
i
ii
n
i
ir xaayeS
1
2
10
1
2
)(min
The coefficients a0 and a1 that minimize Sr must
satisfy the following conditions:













0
0
1
0
a
S
a
S
r
r

 
 
 










2
10
10
1
1
1
0
0
0)(2
0)(2
iiii
ii
iioi
r
ioi
o
r
xaxaxy
xaay
xxaay
a
S
xaay
a
S
Linear Regression:
Determination of ao and a1
 
 





2
10
10
00
iiii
ii
xaxaxy
yaxna
naa
2 equations with 2
unknowns, can be
solved simultaneously

Linear Regression:
Determination of ao and a1
  
  


 221
ii
iiii
xxn
yxyxn
a
xaya 10 

You can view/download
this whole slides from
following path:
Go to www.slideshare.net
Search MC1020

Error Quantification of Linear
Regression
• Total sum of the squares around the
mean for the dependent variable, y, is
St
• Sum of the squares of residuals
around the regression line is Sr


n
i
it yyS
1
2
)(
2
n
1i
i1oi
n
1i
2
ir xaayeS )( 


Regression
• St-Sr quantifies the improvement or
error reduction due to describing data
in terms of a straight line rather than
as an average value.
t
rt
S
SS
r

2
r2: coefficient of determination
r : correlation coefficient

Regression
For a perfect fit:
• Sr= 0 and r = r2 =1, signifying that the
line explains 100 percent of the
variability of the data.
• For r = r2 = 0, Sr = St, the fit
represents no improvement.

Least Squares Fit of a Straight
Line: Example
Fit a straight line to the x and y values
in the following Table:
5.119 ii yx
28 ix 0.24 iy
1402
 ix
4285.3
7
24
4
7
28
 yx
428571.3
7
24
4
7
28
 yx
xi yi xiyi xi
2
1 0.5 0.5 1
2 2.5 5 4
3 2 6 9
4 4 16 16
5 3.5 17.5 25
6 6 36 36
7 5.5 38.5 49
28 24 119.5 140

Least Squares Fit of a Straight Line:
Example (cont’d)
07142857.048392857.0428571.3
8392857.0
281407
24285.1197
)(
10
2
221









 
  
xaya
xxn
yxyxn
a
ii
iiii
Y = 0.07142857 + 0.8392857 x

Least Squares Fit of a Straight Line:
Example (Error Analysis)
9911.2
2
  ir eS
932.0868.02
 rr
xi yi
1 0.5
2 2.5
3 2.0
4 4.0
5 3.5
6 6.0
7 5.5
8.5765 0.1687
0.8622 0.5625
2.0408 0.3473
0.3265 0.3265
0.0051 0.5896
6.6122 0.7972
4.2908 0.1993
2
^
22
)( yye)y(y iii 
28 24.0 22.7143 2.9911
868.02



t
rt
S
SS
r
  7143.22
2
  yyS it

Least Squares Fit of a Straight
Line: Example (Error Analysis)
9457.1
17
7143.22
1





n
S
s t
y
7735.0
27
9911.2
2
/ 




n
S
s r
xy
yxy SS /
• The standard deviation (quantifies the spread around the
mean):
• The standard error of estimate (quantifies the spread
around the regression line)
Because , the linear regression model has good
fitness

Algorithm for linear regression
Regress(x,y,n,a1,a0,sxy,r2)
sumx=0; sumxy=0; st=0
sumy=0;sumx2=0; sr=0
DO i=1,n
sumx=sumx+xi
sumy=sumy+yi
sumxy=sumxy+xi*yi
sumx2=sumx2+xi*yi
END DO
xm=sumx/n
ym=sumy/n
a1=(n*sumxy-sumx*sumy)/(n*sumx2-sumx*sumx)
a0=ym-a1*xm
DO i=1,n
st=st+(yi-ym)^2
sr=sr+(yi-a1*xi-a0)^2
END DO
syx=(sr/(n-2))^0.5
r2=(st-sr)/st
END Regress

Matlab code for linear regression
function [y0, a0, a1, r2, r, k2] = lin_reg(x, y, x0)
% Number of known points
n = length(x);
% Initialization
j = 0; k = 0; l = 0; m = 0; r2 = 0;
% Accumulate intermediate sums
j = sum(x); k = sum(y);
l = sum(x.^2); m = sum(y.^2); r2 = sum(x.*y);
% Compute curve coefficients
a1 = (n*r2 - k*j)/(n*l - j^2); a0 = (k - a1*j)/n;
% Compute regression analysis
j = a1*(r2 - j*k/n);
m = m - k^2/n; k = m - j;
% Coefficient of determination
r2 = j/m;r = sqrt(r2);
% Std. error of estimate
k2 = sqrt(k/(n-2));
% Interpolation value
y0 = a0 + a1*x0;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[y0, a0, a1, r2, r, k2] = lin_reg(x, y, x0)
[y0] = lin_reg(x, y, x0)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Linearization of Nonlinear
Relationships
• The relationship between the dependent and
independent variables is linear.
• However, a few types of nonlinear functions
can be transformed into linear regression
problems.
The exponential equation.
The power equation.
The saturation-growth-rate equation.

Linearization of Nonlinear Relationships
1. The exponential equation.
 xb
eay 1
1
xbay 11lnln 
y* = ao + a1 x

2. The power equation
 2
2
b
xay
xbay logloglog 22 
y* = ao + a1 x*Thursday, January 24, 2019 MC1020-Mathematics 30

3. The saturation-growth-rate equation



xb
x
ay
3
3







xa
b
ay
111
3
3
3
y* = 1/y
ao = 1/a3
a1 = b3/a3
x* = 1/x

Example
Fit the following Equation:
2
2
b
xay 
to the data in the following table:
xi yi
1 0.5
2 1.7
3 3.4
4 5.7
5 8.4
15 19.7
X*=log xi Y*=logyi
0 -0.301
0.301 0.226
0.477 0.534
0.602 0.753
0.699 0.922
2.079 2.141
)log(log 2
2
b
xay 
2120
**
log
logloglet
b, aaa
x,y, XY


xbay logloglog 22 
*
10
*
XaaY 

Example
Xi Yi X*i=Log(X) Y*i=Log(Y) X*Y* X*^2
1 0.5 0.0000 -0.3010 0.0000 0.0000
2 1.7 0.3010 0.2304 0.0694 0.0906
3 3.4 0.4771 0.5315 0.2536 0.2276
4 5.7 0.6021 0.7559 0.4551 0.3625
5 8.4 0.6990 0.9243 0.6460 0.4886
Sum 15 19.700 2.079 2.141 1.424 1.169
1 2 22
0 1
5 1.424 2.079 2.141
1.75
5 1.169 2.079( )
0.4282 1.75 0.41584 0.334
i i i i
i i
n x y x y
a
n x x
a y a x
    
  
 

      
  
 

Linearization of Nonlinear
Functions: Example
log y = -0.334+1.75log x
1.75
0.46y x

Polynomial Regression
• Some engineering data is poorly
represented by a straight line.
• For these cases a curve is better
suited to fit the data.
• The least squares method can readily
be extended to fit the data to higher
order polynomials.

Polynomial Regression (cont’d)
A parabola is preferable

• A 2nd order polynomial (quadratic) is defined by:
• The residuals between the model and the data:
• The sum of squares of the residual:
exaxaay o  2
21
2
21 iioii xaxaaye 
  
22
21
2
iioiir xaxaayeS

0xxaxaay2
a
S
0xxaxaay2
a
S
0xaxaay2
a
S
2
i
2
i2i1oi
2
r
i
2
i2i1oi
1
r
2
i2i1oi
o
r












)(
)(
)(
 
 




4
i2
3
i1
2
ioi
2
i
3
i2
2
i1ioii
2
i2i1oi
xaxaxayx
xaxaxayx
xaxaany 3 linear equations
with 3 unknowns
(ao,a1,a2), can be
solved

• A system of 3x3 equations needs to be solved to determine the
coefficients of the polynomial.
• The standard error & the coefficient of determination
3
/


n
S
s r
xy t
rt
S
SS
r

2





































ii
ii
i
iii
iii
ii
yx
yx
y
a
a
a
xxx
xxx
xxn
2
2
1
0
432
32
2
*

General:
The mth-order polynomial:
• A system of (m+1)x(m+1) linear equations must be solved for
determining the coefficients of the mth-order polynomial.
• The standard error:
• The coefficient of determination:
exaxaxaay m
mo  .....2
21
 1
/


mn
S
s r
xy
t
rt
S
SS
r

2

Polynomial Regression- Example
Fit a second order polynomial to data:
2253
 ix
9794
 ix
xi yi xi
2 xi
3 xi
4 xiyi xi
2yi
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
15 152.6 55 225 979 585.6 2489
6.585 ii yx
15 ix
6.152 iy
552
 ix
433.25
6
6.152
,5.2
6
15
 yx 8.2488
2
 ii yx

(cont’d)
• The system of simultaneous linear equations:
2
210
86071.135929.247857.2
86071.1,35929.2,47857.2
xxy
aaa

































8.2488
6.585
6.152
*
97922555
2255515
55156
2
1
0
a
a
a
74657.3
2
  ir eS  39.2513
2
  yyS it

(cont’d)
xi yi ymodel ei
2 (yi-y`)2
0 2.1 2.4786 0.14332 544.42889
1 7.7 6.6986 1.00286 314.45929
2 13.6 14.64 1.08158 140.01989
3 27.2 26.303 0.80491 3.12229
4 40.9 41.687 0.61951 239.22809
5 61.1 60.793 0.09439 1272.13489
15 152.6 3.74657 2513.39333
•The standard error of estimate:
•The coefficient of determination:
12.1
36
74657.3
/ 

xys
99925.0,99851.0
39.2513
74657.339.2513 22


 rrr

Learning outcomes
Draw sketch graphs of standard curves
Fit graphs to data using straight-line forms
Fit graphs to data using the method of least
squares
Calculate measures of correlation

Practice Example 1
As machines are used over long periods of time, the
output product can get off target. Below is the
average value of how much off target a product is
getting manufactured as a function of machine use.
Table : Off target value as a function of machine
use.
Regress the data to ℎ = 𝑎0 + 𝑎1 𝑡. Find the amount
of off target after 50 hours of operation.
30 33 34 35 39 44 45
1.10 1.21 1.25 1.23 1.30 1.40 1.42

We will discuss some
additional Examples in
Tutorial session….

Mc1020

Recommended

Recommended

More Related Content

What's hot

What's hot (16)

Similar to Mc1020

Similar to Mc1020 (20)

Recently uploaded

Recently uploaded (20)

Mc1020