Question

(20) Consider the function f(x) = e* - 3x. Using only and exactly the four points on the graph off with x-coordinates -1,0, 1 and 2, use MATLAB's polyfit function to determine a 3' degree polynomial that approximates f on the interval (-1, 2]. Plot the function f(x) and the 360 degree polynomial you have determined on the same set of axes. f must be blue and have a dashed line style, while the polynomial must be a red solid line with a 'thickness' that is larger than the default. Include a legend. Now: Use MATLAB's roots function to determine all roots of the 3rd degree polynomial and plot these roots as green triangles on the same axes as before. b. Use MATLAB's fzero function with a starting point of zero to determine a root of f(x). Plot this root as a rather big yellow triangle on the same set of axes as before. Use the Newton-Raphson algorithm for root finding and start with an initial guess (x) of 0. Only calculate up till Xz. Write as a comment in your script file, a message interpreting the results you got by using the Newton-Raphson algorithm as was instructed here. a. C.

Please MATLAB for all coding with good commenting.

Answer 1

All the explanation is in the code comments. Hope this helps!

Code:

% given function

f = @(x) exp(x) - 3*x;

% given x values

x = [-1 0 1 2];

% f(x) for x

y = f(x);

% use polyfit to approximate

p = polyfit(x, y, 3);

figure

hold on;

% plot the fuction and polynomial for range [-1, 2]

x = -1:0.1:2;

% f has blue dashed line

plot(x, f(x), 'b--')

% polynomial has red thick line

plot(x, polyval(p, x), 'r', 'LineWidth', 2)

xlabel('x')

% added legend at end to support all the marked points

legend('f(x)', 'polynomial')

% (a) use roots function for roots of polynomial

r = roots(p);

% plot as green triangles

plot(r, polyval(p, r), 'g^', 'MarkerFaceColor', 'g', 'DisplayName', 'roots of polynomial')

% (b)

r = fzero(f, 0);

% plot as big yellow triangle

plot(r, f(r), 'y^', 'MarkerSize', 10, 'DisplayName', 'root of function')

% (c) use newton raphson method to find root of f

x0 = 0;

% df/dx = e^x - 3

df = @(x) exp(x) - 3;

% loop till x3

for i=1:3

% next x

xn = x0 - f(x0)/df(x0);

% update x0 for next iteration

x0 = xn;

end

% report the x3, x3 = 0.6190 almost equal to r (= 0.6191)

xn

Sample output:

3.5 3 -f(x) polynomial roots of polynomial root of function 2.5 2 1.5 1 0.5 0 -0.5 -3 -2 0 1 2 -1 X

Note - the yellow coloured triangle is quite faint, it overlaps with the middle green triangle

xn = 0.6198

Code screenshots:

% given function f = @(x) exp(x) - 3*X; % given x values X = (-1 0 1 2]; % f(x) for x y = f(x); % use polyfit to approximate D = polyfit(x, y, 3); figure hold on; % plot the fuction and polynomial for range (-1, 2] x = -1:0.1:2; % f has blue dashed line plot(x, f(x), 'b--') % polynomial has red thick line plot(x, polyval(p, x), 'r', 'LineWidth', 2) xlabel('x') % added legend at end to support all the marked points legend('f(x)', 'polynomial') % (a) use roots function for roots of polynomial r = roots(p); % plot as green triangles plot(r, polyval(P, r), 'a', 'MarkerFacecolor', 's', 'DisplayName', 'roots of polynomial') % (6) r = fzero(f, 0); % plot as big yellow triangle plot(r, f(r), 'y', 'Markersize', ie, 'DisplayName', 'root of function)

% (C) use newton raphson method to find root of f x@ = e; % df/dx = e^x - 3 df = @(x) exp(x) - 3; % loop till x3 for i=1:3 % next x xn = x3 - f(xa)/df(xa); % update xe for next iteration x = xn; end % report the X3, X3 = 0.6190 almost equal to r (= 0.6191) xn