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Dhiraj Bhaskar
Dhiraj Bhaskar

Unit: Curve-Fitting Introduction: The document discusses using the least squares method to fit curves to data points by finding polynomial functions that minimize the errors between the fitted curve and data points. It provides examples of fitting first degree (linear) and second degree (parabolic) polynomial curves to data sets. For linear curves, it shows calculating the constants c0 and c1 using sums and determinants to find the best fit line Y=c0+c1X. For parabolic curves, it similarly calculates constants c0, c1, and c2 to find the best fit quadratic curve Y=c0+c1X+c2X^2. It also provides a worked example

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Unit: Curve-Fitting Introduction: In experiments we get lot of random given data (x,y). It is difficult to fit the curve passing through these data points. So we need to find some polynomial function by which approximate curve can be drawn. This curve may not pass through all the point but errors will be less. Least square method is a technique which gives best possible polynomial like Y c + c (x). by which errors will be less. ' 2 1 ( ) n n i i i Error y y ο€½ ο€½ ο€­ οƒ₯ οƒ₯ ο€½ n i i E 1 2 Curve fitting is the process of constructing a curve, or mathematical function that has the best fit to a series of data points, possibly subject to constraints. Fitted curves can be used as an aid for data visualization, to infer values of a function where no data are available, and to summarize the relationships among two or more variables
A first degree polynomial equation: Yn c + c (x). A first degree polynomial equation is an exact fit through any two points with distinct x coordinates. A second degree polynomial: Yn c + c (x) + c (x)2 . Fitting a Line (first degree polynomial) equation: n= number of data points βˆ‘ βˆ‘ βˆ‘ βˆ‘ . βˆ‘ OR Determinants: d=det d1= d2= Constants: c0=d1/d & c1=d2/d Fitting a Line : + (x). Problem: Using least square method Fit a line: Yn c + c (x) X 1 2 3 4 Y 2 6 12 20 Solution: n= number of data points=4 βˆ‘ βˆ‘ βˆ‘ βˆ‘ . βˆ‘ X Y 1 2 1 2 2 3 4 12 3 12 9 36 4 20 16 80 βˆ‘ =10 βˆ‘ . 40 βˆ‘ 30 βˆ‘ 130
OR Determinants: d=det 1 1 2 =20 d1= Dd2= =120 Constants: C0=d1/d = ‐ 5 and C1=d2/d=6 Best fitted line: + (X). ANSWER: + (X). Line Flowchart
% LINE FITTING Yf=C0+C1(X) clc clear all %-------------Direct input of Data----------- x=[1 2 3 4]; y=[2 6 12 20]; n=length(x) %--------------OR-User UInput of data----- % n=input('Enter n='); % for i=1:n % fprintf('Enter x(%d)=',i); % x(i)=input(''); % fprintf('Enter y(%d)=',i); % y(i)=input(''); % end %-------------------------------------------- s1=sum(x); s2=sum(x.^2); s3=sum(y); s4=sum(x.*y); %-------------------------------------------- d=[n s1; s1 s2]; d1=[s3 s1; s4 s2]; d2=[n s3; s1 s4]; d=det(d); d1=det(d1); d2=det(d2); %-------------------------------------------- c0=d1/d; c1=d2/d; fprintf('BEST FIT ,Yn=%0.3f+%0.3f(X)n',c0,c1) %-------------------------------------------- yn=c0+c1*x; plot(x,y,'*',x,yn) %--------------------------------------------
Solved Problems Fit the line yn=c0+c1*x x=[1 2 3 4]; y=[2 6 12 20]; ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ X Y X2 XY ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 1 2 1 2 2 6 4 12 3 12 9 36 4 20 16 80 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 10 40 30 130 Calculate d d1 d2 c 0 c1 by you BEST FIT ,Yn=‐5.000+6.000(X) Fit the line yn=c0+c1*x L=a0+a1T x=[ 20 30 40 50 60 70]; y=[800.3 800.4 800.6 800.7 800.9 801]; ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ X Y X2 X.*Y ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 20 800 400 16006 30 800 900 24012 40 801 1600 32024 50 801 2500 40035 60 801 3600 48054 70 801 4900 56070 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 270 4804 13900 216201 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ Calculate d d1 d2 c 0 c1 by you
BEST FIT ,Yn=799.994+0.015(X) Fitting a second degree polynomial: + (X) + (X)2 βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ 0 1 2 βˆ‘ βˆ‘ βˆ‘ . 1 2 1 2 3 2 3 4 0 1 2 5 6 7 D det 1 2 1 2 3 2 3 4 D1 det 5 1 2 6 2 3 7 3 4 D2 Det 5 2 1 6 3 2 7 4 D3 Det 1 5 1 2 6 2 3 7 Constants: C0=d1/d , C1=d2/d and C2=d3/d Problem: Using least square method fit a second degree polynomial + (X) + (X)2 X 1 2 3 4 5 6 7 8 9 Y 2 6 7 8 10 11 11 10 9 Solution: βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ = 0 1 2
Table calculatins to be done by you X Y X2 X3 X4 XY XY2 S1 S5 S2 S3 S4 S6 S7 . 45 74 286 2025 15333 421 2771 On making table and solving S1 45, S2 286, S3 2025, S4 15333, s5 74 , s6 421, s7 2771, D det 166320 D1 det ‐154440 D2 Det D3 Det ‐44460 Constants: C0=D1/D = ‐0.9286 C1=D2/D= 3. C2=D3/D= ‐0.2673 Best fitted parabola: . .+3.523(X) 0.267(X)2
Program % Quadratic FITTING Yf=C0+C1(X)+C2(x^2); % Quadratic FITTING Yf=C0+C1(X)+C2(x^2); clc clear all %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ x=[1 2 3 4 5 6 7 8 9]; y=[2 6 7 8 10 11 11 10 9]; n=9; %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ x2=x.^2; x3=x.^3; x4=x.^4; xy=x.*y; x2y=(x.^2).*y; s1=sum(x); s2=sum(x2); s3=sum(x3); s4=sum(x4); s5=sum(y); s6=sum(xy); s7=sum(x2y); %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ d=[n s1 s2; s1 s2 s3 s2 s3 s4]; d1=[s5 s1 s2; s6 s2 s3 s7 s3 s4]; d2=[n s5 s2; s1 s6 s3 s2 s7 s4]; d3=[n s1 s5; s1 s2 s6 s2 s3 s7]; %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ d=det(d); d1=det(d1); d2=det(d2); d3=det(d3); c0=d1/d; c1=d2/d; c2=d3/d; %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ fprintf('BEST FIT ,Yn=%0.3f+%0.3f(X)+%0.3f(x^2)n',c0,c1,c2) %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ yn=c0+c1*x+c2*x2; plot(x,y,'*',x,yn,'r') %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ Option for i=1:n fprintf('%0.0f %0.0f %0.0f %0.0f %0.0f %0.0f %0.0f n',x(i),y(i),x2(i),x3(i),x4(i),xy(i),x2y(i)) end fprintf('‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ n'); fprintf('%0.0f %0.0f %0.0f %0.0f %0.0f %0.0f %0.0f n',s1, s5,s2,s3,s4,s6,s7)
Solved Problem Fit the following quadratic/2nd degree/ parabolic curve + (X) + (X)2 x=[‐4 ‐3 ‐2 ‐1 0 1 2 3 4 5]; y=[21 12 4 1 2 7 15 30 45 67]; ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ X Y x2 x3 x4 xy x2y ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ‐4 21 16 ‐64 256 ‐84 336 ‐3 12 9 ‐27 81 ‐36 108 ‐2 4 4 ‐8 16 ‐8 16 ‐1 1 1 ‐1 1 ‐1 1 0 2 0 0 0 0 0 1 7 1 1 1 7 7 2 15 4 8 16 30 60 3 30 9 27 81 90 270 4 45 16 64 256 180 720 5 67 25 125 625 335 1675 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 5 204 85 125 1333 513 319 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ Remaining d d1 d2 d3 c0 c1 c3 …to be done by user
Flowchart: Curve Fitting (Parabola) s1=sum(x); s2=sum(x.^2); s3=sum(x.^3); s4=sum(x.^4); s5=sum(y); s6=sum(x.*y); s7=sum((x.^2).*y); Calculation of Determinents d, d1, d2 c0=d1/d; c1=d2/d; c2=d3/d; Start Break Read x(i), y(i) For ( i=1 to n) Read n yf=c0+c1*x+c2*x.^2 Print C0, C1, C2 plot(x,y)
Numerical Power For the power equation: PV^gamma =C P=[0.5 1 1.5 2 2.5 3]; V=[1.62 1 0.75 0.62 0.52 0.46]; Solution: Log(P)+ gamma*log(V)=log(C) Log(P)= log(C) - gamma *log(V) Y = C0 + C1 (X) C0=log(c) C1=-gamma Y=Log(P) X=Log(V) ------------------------------------------------------------------- V P X Y X^2 XY ------------------------------------------------------------------- 1.62 0.5 0.482426 -0.693147 0.232735 -0.334392 1.0 1.0 0. 0. 0. 0. 0.75 1.5 -0.287682 0.405465 0.082761 -0.116645 0.62 2. -0.478036 0.693147 0.228518 -0.331349 0.52 2.5 -0.653926 0.916291 0.427620 -0.599187 0.46 3. -0.776529 1.098612 0.602997 -0.853104 ----------------------------------------------------------- S1= -1.7137 S3=2.4204 S2=1.5746 S4=-2.2347 d = 6.5109 d1 =-0.0185 d2 = -9.2602 c0 =-0.0028 c1 = -1.4223 c = exp(c0)= 0.9972 gamma =-c1=1.4223 PV^1.422=0.997165
Program Power clc clear all %‐‐‐‐‐‐‐‐‐‐‐‐‐Power Equation‐‐‐‐‐‐‐‐‐‐‐ %PV^Gamma=C p=[0.5 1.0 1.5 2.0 2.5 3.0]; v=[1.62 1.0 0.75 0.62 0.52 0.46]; n=6; y=log10(p); x=log10(v); xy=x.*y; x2=x.^2; s1=sum(x); s2=sum(x2); s3=sum(y); s4=sum(xy); %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ d=n*s2‐s1*s1; d1=s3*s2‐s4*s1; d2=n*s4‐s1*s3; %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ c0=d1/d; c1=d2/d; gamma=‐c1; c=exp(c0); %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ fprintf('PV^%0.3f=%fn',gamma,c) %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ %plot(p,v); for i=1:n fprintf('%f %f %f %f %f %f n',v(i),p(i),x(i),y(i),x2(i),xy(i)); end
Numerical Exponential For the Exponential equation: y=aebx x=[2 4 6 8]; y=[25 38 56 84]; Solution y=aebx ln(y)=ln(a)+bx ln(e) Y =C0 +C1(x) yn=ln(y) C0=ln(a) C1=b ------------------------------------------------------ x y yn x^2 x*yn ------------------------------------------------------ 2. 25. 3.218876 4. 6.437752 4. 38. 3.637586 16. 14.550345 6. 56. 4.025352 36. 24.152110 8. 84. 4.430817 64. 35.446534 ----------------------------------------------------- S1=20 s3= 15.3126 s2= 120 s4=80.5867 d = 80 d1 = 225.7808 d2 = 16.0944 c0=2.8223 c1=0.2012 a=exp(c0)= 16.8148 b = c1=0.2012 y=16.815 e (0.2012x)
Program Exponential clc clear all %‐‐‐‐‐‐‐‐‐‐‐‐‐Exponential‐‐‐‐‐‐‐‐‐‐‐ %y=ae^bx %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ x=[2 4 6 8]; y=[25 38 56 84]; n=length(x); %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ y1=log(y); %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ x2=x.^2; xy1=x.*y1; s1=sum(x); s2=sum(x2); s3=sum(y1); s4=sum(xy1); %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ d=n*s2‐s1*s1; d1=s3*s2‐s4*s1; d2=n*s4‐s1*s3; %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ c0=d1/d; c1=d2/d; a=exp(c0); b=c1; %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ fprintf('y=%0.3f e^(%0.3fx)n',a,b) %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ for i=1:n fprintf('%f %f %f %f %f n',x(i),y(i),y1(i),x2(i),xy1(i)); end
Solver :% POLYFIT clear all x=[19 25 30 36 40 45 50]; y=[76 77 79 80 82 83 85]; d=1; %d=1 (line) d=2(parabola) c=polyfit(x,y,d) % yn = polyval(c, x); yn=c(2)+c(1)*x; plot(x,y,'*',x,yn)

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  • 1. Unit: Curve-Fitting Introduction: In experiments we get lot of random given data (x,y). It is difficult to fit the curve passing through these data points. So we need to find some polynomial function by which approximate curve can be drawn. This curve may not pass through all the point but errors will be less. Least square method is a technique which gives best possible polynomial like Y c + c (x). by which errors will be less. ' 2 1 ( ) n n i i i Error y y ο€½ ο€½ ο€­ οƒ₯ οƒ₯ ο€½ n i i E 1 2 Curve fitting is the process of constructing a curve, or mathematical function that has the best fit to a series of data points, possibly subject to constraints. Fitted curves can be used as an aid for data visualization, to infer values of a function where no data are available, and to summarize the relationships among two or more variables
  • 2. A first degree polynomial equation: Yn c + c (x). A first degree polynomial equation is an exact fit through any two points with distinct x coordinates. A second degree polynomial: Yn c + c (x) + c (x)2 . Fitting a Line (first degree polynomial) equation: n= number of data points βˆ‘ βˆ‘ βˆ‘ βˆ‘ . βˆ‘ OR Determinants: d=det d1= d2= Constants: c0=d1/d & c1=d2/d Fitting a Line : + (x). Problem: Using least square method Fit a line: Yn c + c (x) X 1 2 3 4 Y 2 6 12 20 Solution: n= number of data points=4 βˆ‘ βˆ‘ βˆ‘ βˆ‘ . βˆ‘ X Y 1 2 1 2 2 3 4 12 3 12 9 36 4 20 16 80 βˆ‘ =10 βˆ‘ . 40 βˆ‘ 30 βˆ‘ 130
  • 3. OR Determinants: d=det 1 1 2 =20 d1= Dd2= =120 Constants: C0=d1/d = ‐ 5 and C1=d2/d=6 Best fitted line: + (X). ANSWER: + (X). Line Flowchart
  • 4. % LINE FITTING Yf=C0+C1(X) clc clear all %-------------Direct input of Data----------- x=[1 2 3 4]; y=[2 6 12 20]; n=length(x) %--------------OR-User UInput of data----- % n=input('Enter n='); % for i=1:n % fprintf('Enter x(%d)=',i); % x(i)=input(''); % fprintf('Enter y(%d)=',i); % y(i)=input(''); % end %-------------------------------------------- s1=sum(x); s2=sum(x.^2); s3=sum(y); s4=sum(x.*y); %-------------------------------------------- d=[n s1; s1 s2]; d1=[s3 s1; s4 s2]; d2=[n s3; s1 s4]; d=det(d); d1=det(d1); d2=det(d2); %-------------------------------------------- c0=d1/d; c1=d2/d; fprintf('BEST FIT ,Yn=%0.3f+%0.3f(X)n',c0,c1) %-------------------------------------------- yn=c0+c1*x; plot(x,y,'*',x,yn) %--------------------------------------------
  • 5. Solved Problems Fit the line yn=c0+c1*x x=[1 2 3 4]; y=[2 6 12 20]; ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ X Y X2 XY ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 1 2 1 2 2 6 4 12 3 12 9 36 4 20 16 80 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 10 40 30 130 Calculate d d1 d2 c 0 c1 by you BEST FIT ,Yn=‐5.000+6.000(X) Fit the line yn=c0+c1*x L=a0+a1T x=[ 20 30 40 50 60 70]; y=[800.3 800.4 800.6 800.7 800.9 801]; ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ X Y X2 X.*Y ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 20 800 400 16006 30 800 900 24012 40 801 1600 32024 50 801 2500 40035 60 801 3600 48054 70 801 4900 56070 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 270 4804 13900 216201 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ Calculate d d1 d2 c 0 c1 by you
  • 6. BEST FIT ,Yn=799.994+0.015(X) Fitting a second degree polynomial: + (X) + (X)2 βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ 0 1 2 βˆ‘ βˆ‘ βˆ‘ . 1 2 1 2 3 2 3 4 0 1 2 5 6 7 D det 1 2 1 2 3 2 3 4 D1 det 5 1 2 6 2 3 7 3 4 D2 Det 5 2 1 6 3 2 7 4 D3 Det 1 5 1 2 6 2 3 7 Constants: C0=d1/d , C1=d2/d and C2=d3/d Problem: Using least square method fit a second degree polynomial + (X) + (X)2 X 1 2 3 4 5 6 7 8 9 Y 2 6 7 8 10 11 11 10 9 Solution: βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ βˆ‘ = 0 1 2
  • 7. Table calculatins to be done by you X Y X2 X3 X4 XY XY2 S1 S5 S2 S3 S4 S6 S7 . 45 74 286 2025 15333 421 2771 On making table and solving S1 45, S2 286, S3 2025, S4 15333, s5 74 , s6 421, s7 2771, D det 166320 D1 det ‐154440 D2 Det D3 Det ‐44460 Constants: C0=D1/D = ‐0.9286 C1=D2/D= 3. C2=D3/D= ‐0.2673 Best fitted parabola: . .+3.523(X) 0.267(X)2
  • 8. Program % Quadratic FITTING Yf=C0+C1(X)+C2(x^2); % Quadratic FITTING Yf=C0+C1(X)+C2(x^2); clc clear all %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ x=[1 2 3 4 5 6 7 8 9]; y=[2 6 7 8 10 11 11 10 9]; n=9; %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ x2=x.^2; x3=x.^3; x4=x.^4; xy=x.*y; x2y=(x.^2).*y; s1=sum(x); s2=sum(x2); s3=sum(x3); s4=sum(x4); s5=sum(y); s6=sum(xy); s7=sum(x2y); %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ d=[n s1 s2; s1 s2 s3 s2 s3 s4]; d1=[s5 s1 s2; s6 s2 s3 s7 s3 s4]; d2=[n s5 s2; s1 s6 s3 s2 s7 s4]; d3=[n s1 s5; s1 s2 s6 s2 s3 s7]; %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ d=det(d); d1=det(d1); d2=det(d2); d3=det(d3); c0=d1/d; c1=d2/d; c2=d3/d; %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ fprintf('BEST FIT ,Yn=%0.3f+%0.3f(X)+%0.3f(x^2)n',c0,c1,c2) %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ yn=c0+c1*x+c2*x2; plot(x,y,'*',x,yn,'r') %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ Option for i=1:n fprintf('%0.0f %0.0f %0.0f %0.0f %0.0f %0.0f %0.0f n',x(i),y(i),x2(i),x3(i),x4(i),xy(i),x2y(i)) end fprintf('‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ n'); fprintf('%0.0f %0.0f %0.0f %0.0f %0.0f %0.0f %0.0f n',s1, s5,s2,s3,s4,s6,s7)
  • 9. Solved Problem Fit the following quadratic/2nd degree/ parabolic curve + (X) + (X)2 x=[‐4 ‐3 ‐2 ‐1 0 1 2 3 4 5]; y=[21 12 4 1 2 7 15 30 45 67]; ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ X Y x2 x3 x4 xy x2y ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ‐4 21 16 ‐64 256 ‐84 336 ‐3 12 9 ‐27 81 ‐36 108 ‐2 4 4 ‐8 16 ‐8 16 ‐1 1 1 ‐1 1 ‐1 1 0 2 0 0 0 0 0 1 7 1 1 1 7 7 2 15 4 8 16 30 60 3 30 9 27 81 90 270 4 45 16 64 256 180 720 5 67 25 125 625 335 1675 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 5 204 85 125 1333 513 319 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ Remaining d d1 d2 d3 c0 c1 c3 …to be done by user
  • 10. Flowchart: Curve Fitting (Parabola) s1=sum(x); s2=sum(x.^2); s3=sum(x.^3); s4=sum(x.^4); s5=sum(y); s6=sum(x.*y); s7=sum((x.^2).*y); Calculation of Determinents d, d1, d2 c0=d1/d; c1=d2/d; c2=d3/d; Start Break Read x(i), y(i) For ( i=1 to n) Read n yf=c0+c1*x+c2*x.^2 Print C0, C1, C2 plot(x,y)
  • 11. Numerical Power For the power equation: PV^gamma =C P=[0.5 1 1.5 2 2.5 3]; V=[1.62 1 0.75 0.62 0.52 0.46]; Solution: Log(P)+ gamma*log(V)=log(C) Log(P)= log(C) - gamma *log(V) Y = C0 + C1 (X) C0=log(c) C1=-gamma Y=Log(P) X=Log(V) ------------------------------------------------------------------- V P X Y X^2 XY ------------------------------------------------------------------- 1.62 0.5 0.482426 -0.693147 0.232735 -0.334392 1.0 1.0 0. 0. 0. 0. 0.75 1.5 -0.287682 0.405465 0.082761 -0.116645 0.62 2. -0.478036 0.693147 0.228518 -0.331349 0.52 2.5 -0.653926 0.916291 0.427620 -0.599187 0.46 3. -0.776529 1.098612 0.602997 -0.853104 ----------------------------------------------------------- S1= -1.7137 S3=2.4204 S2=1.5746 S4=-2.2347 d = 6.5109 d1 =-0.0185 d2 = -9.2602 c0 =-0.0028 c1 = -1.4223 c = exp(c0)= 0.9972 gamma =-c1=1.4223 PV^1.422=0.997165
  • 12. Program Power clc clear all %‐‐‐‐‐‐‐‐‐‐‐‐‐Power Equation‐‐‐‐‐‐‐‐‐‐‐ %PV^Gamma=C p=[0.5 1.0 1.5 2.0 2.5 3.0]; v=[1.62 1.0 0.75 0.62 0.52 0.46]; n=6; y=log10(p); x=log10(v); xy=x.*y; x2=x.^2; s1=sum(x); s2=sum(x2); s3=sum(y); s4=sum(xy); %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ d=n*s2‐s1*s1; d1=s3*s2‐s4*s1; d2=n*s4‐s1*s3; %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ c0=d1/d; c1=d2/d; gamma=‐c1; c=exp(c0); %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ fprintf('PV^%0.3f=%fn',gamma,c) %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ %plot(p,v); for i=1:n fprintf('%f %f %f %f %f %f n',v(i),p(i),x(i),y(i),x2(i),xy(i)); end
  • 13. Numerical Exponential For the Exponential equation: y=aebx x=[2 4 6 8]; y=[25 38 56 84]; Solution y=aebx ln(y)=ln(a)+bx ln(e) Y =C0 +C1(x) yn=ln(y) C0=ln(a) C1=b ------------------------------------------------------ x y yn x^2 x*yn ------------------------------------------------------ 2. 25. 3.218876 4. 6.437752 4. 38. 3.637586 16. 14.550345 6. 56. 4.025352 36. 24.152110 8. 84. 4.430817 64. 35.446534 ----------------------------------------------------- S1=20 s3= 15.3126 s2= 120 s4=80.5867 d = 80 d1 = 225.7808 d2 = 16.0944 c0=2.8223 c1=0.2012 a=exp(c0)= 16.8148 b = c1=0.2012 y=16.815 e (0.2012x)
  • 14. Program Exponential clc clear all %‐‐‐‐‐‐‐‐‐‐‐‐‐Exponential‐‐‐‐‐‐‐‐‐‐‐ %y=ae^bx %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ x=[2 4 6 8]; y=[25 38 56 84]; n=length(x); %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ y1=log(y); %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ x2=x.^2; xy1=x.*y1; s1=sum(x); s2=sum(x2); s3=sum(y1); s4=sum(xy1); %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ d=n*s2‐s1*s1; d1=s3*s2‐s4*s1; d2=n*s4‐s1*s3; %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ c0=d1/d; c1=d2/d; a=exp(c0); b=c1; %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ fprintf('y=%0.3f e^(%0.3fx)n',a,b) %‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ for i=1:n fprintf('%f %f %f %f %f n',x(i),y(i),y1(i),x2(i),xy1(i)); end
  • 15. Solver :% POLYFIT clear all x=[19 25 30 36 40 45 50]; y=[76 77 79 80 82 83 85]; d=1; %d=1 (line) d=2(parabola) c=polyfit(x,y,d) % yn = polyval(c, x); yn=c(2)+c(1)*x; plot(x,y,'*',x,yn)