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Numerical Techniques

Numerical Techniques. Solving a Set of Linear Equations. Example:. + 2 = 1 2 - + 4 = -1 -2 + + 3 = 2. and. where. MATLAB can solve for using matrix left division: x = AB. % Solving a set of linear equations. A = [ 1 2 0 2 -1 4

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Numerical Techniques

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  1. Numerical Techniques

  2. Solving a Set of Linear Equations Example: + 2 = 1 2 - + 4 = -1 -2 + + 3 = 2 and where MATLAB can solve for using matrix left division: x = A\B

  3. % Solving a set of linear equations A = [ 1 2 0 2 -1 4 -2 1 3 ]; B = [ 1 -1 2]'; % t + 2u = 1 % 2t - u + 4v = -1 % -2t + u + 3v = 2 % A*x = B x = A\B; % matrix left division disp(x) disp(['transpose of A*x : ',num2str((A*x)')]) -0.4286 0.7143 0.1429 transpose of A*x : 1 -1 2

  4. Exercise Solve the following set of linear equations: 3- + = -2 - - 2 = 1 2- 2 + 3 = -3

  5. % Linear interpolation x = 0:5; y = [0.15 1.01 1.22 2.54 3.98 4.23]; % data X = linspace(0,5,101); Y = interp1(x,y,X,'linear'); figure(1) plot(x,y,'ko','MarkerFaceColor','k') hold on plot(X,Y,'-r') axis([0 5 0 5])

  6. % Cubic spline interpolation x = 0:5; y = [0.15 1.01 1.22 2.54 3.98 4.23]; % data X = linspace(0,5,101); Y = interp1(x,y,X,'spline'); figure(2) plot(x,y,'ko','MarkerFaceColor','k') hold on plot(X,Y,'-r') axis([0 5 0 5])

  7. Exercise Given the data appearing in the table at the bottom of this slide, do a linear interpolation to find matching y-values for the following set of x-values: 0.5, 1.5, 2.5, 3.5, 4.5

  8. % Curve fitting: 1st-order polynomial x = -3:3; y = [-3.59,-1.71,-0.93,-0.15, 1.27, 1.66, 2.98]; % data coef = polyfit(x,y,1); % 1st-order fit X = linspace(-3,3,61); Y = polyval(coef,X); figure(3) plot(x,y,'ko','MarkerFaceColor','k') hold on plot(X,Y,'-r') axis([-3 3 -4 4])

  9. % Curve fitting: 2nd-order polynomial x = -3:3; y = [9.79, 3.60, 1.35, 0.42, 0.88, 4.11, 8.42]; % data coef = polyfit(x,y,2); % 2nd-order fit X = linspace(-3,3,61); Y = polyval(coef,X); figure(4) plot(x,y,'ko','MarkerFaceColor','k') hold on plot(X,Y,'-r') axis([-3 3 0 10])

  10. Exercise Fit a first-order polynomial (that is, a straight line) to the data in the table at the bottom of this slide. Then prepare a plot with the following: the original data plotted as discrete points and the straight line defined by polyfit. You’ll need to use polyval in order to generate the straight line from the polynomial coefficients.

  11. % Roots of a polynomial coef = [1,-1,-4,4]; % p(x) = x^3 - x^2 - 4x + 4 x = roots(coef); % values of x for which p(x) = 0 disp(x) -2.0000 2.0000 1.0000

  12. Exercise Find the values of that satisfy the following equation:

  13. Function Handle A function handle is a numerical “tag” that identifies a function. It is sometimes more convenient to identify a function by its handle (which is numeric) than by its name (which is a string). A function handle can, for example, be passed as a numerical input argument to another function. An anonymous function is a function that is defined without ever being assigned a name. Every anonymous function will receive a (function) handle, and that is how the function is identified by other parts of the code.

  14. function A = areaCircle(r) • % areaCircle calculates the area of a circle. • % input: r = radius • % output: A = area • A = pi*(r.^2); • end % areaCircle

  15. % Create a function handle • h1 = @areaCircle; % function handle • fprintf('%8.6f\n',h1(1)) • 3.141593

  16. function plotFunction(x,h) • % plotFunction creates a plot, given a function handle. • % Inputs: • % x = abscissa vector • % h = function handle • figure(1) • plot(x,h(x)) • set(gca,'FontSize',20) • xlabel('x’) • ylabel('y’) • title(func2str(h)) • set(findobj(1,'LineWidth',0.5),'LineWidth',2) • saveas(1,func2str(h),'png') • end % plotFunction

  17. plotFunction(0:10,h1)

  18. Exercise

  19. % Create an anonymous function having a (function) handle • h2 = @(r) pi*(r.^2); • fprintf('%8.6f\n',h2(1)) • 3.141593

  20. Exercise

  21. Find zero of function

  22. % Find zeros with function fzero NonlinearFunction = @(x) exp(-x) - 0.5; x0 = fzero(NonlinearFunction,[0,1]); fprintf('zero of NonlinearFunction: %6.4f\n',x0) zero of NonlinearFunction: 0.6931

  23. Exercise Find the zero of the function for

  24. Find area under curve for

  25. % Find area under curve with function integral sinc = @(x) sin(pi*x)./(pi*x); area = integral(sinc,0,4); fprintf('area: %6.4f\n',area) area: 0.4750

  26. Exercise Find the area under the curve for

  27. Find volume under surface for

  28. % Find volume under surface with integral2 Gaussian = @(x,y) exp(-(x.^2 + y.^2)); ymin = @(x) -sqrt(1-x.^2); ymax = @(x) sqrt(1-x.^2); vol = integral2(Gaussian,-1,1,ymin,ymax); fprintf('volume: %6.4f\n',vol) volume: 1.9859

  29. Exercise Find the volume under the surface for (circle of radius 2)

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