Iteration Methods

Consider the function

and the iteration

started at

initial value. Compute the first 10 iterates.

f = @(x) 1.8*(1-x).*x; %iteration function

x(1) = 0.9; %initial value

numIter=50;

for i = 1:numIter %number of iterates

x(i+1) = f(x(i));

end

%plot iterations points

X = x;

Y = zeros(1, length(x));

scatter(X,Y)

% Hint: Try this other representation

% szPoints = linspace(300,1,length(x)); %size of the points

% colorPoints = linspace(1,10,length(x)); %colors for each point

% scatter(X,Y,szPoints,colorPoints)

Exercise 0:

Compute the fixed point for the iterate function

and determine how many iterates are needed to achieve a precision of

when approximate numerically the fixed point, that is:

Hint: You must use a while loop instead of a for loop.

Sol: n=13

Exercise 1:

Compute the number of terms needed to approximate the sum of the Harmonic serie

with a precision of

Sol: n = 1.0e+6, suma = 1.644933066848770e+00

Exercise 2:

Do the Fibonacci iteration

starting from the initial values

. Compute the ratio

between two consecutive values and verify that it converges to the Golden ratio, Gr

. How many iterations you need to have

Sol: n = 25

Aitken acceleration

One possibility of accelerating the convergence of an iteration procedure is to use the Aitken method. The idea is to create a new succesion from the present iteration which converges faster:

Exercise 3:

Using the iteration function

. Compute the succcesive values

and also the Aitken succession. Use x0=2, maxIter=100, tol= 1.e-10 .

Sol:

iter = 0, x0 = 2.0000000000000000e+00

iter = 1, xAitk = 1.0191337034629475e+00

iter = 2, xAitk = 1.0000257973776343e+00

iter = 3, xAitk = 1.0000000000481342e+00

iter = 4, xAitk = 9.9999999999999989e-01

Graphical iteration: Logistic function (example of Chaos)

From a geometrical point of view a fixed point is the intersection of the line y = x with the plot of the funtion y = f(x).

Graphically the iteration procedure consist in two steps:

Projection on the function: make a line from to
Change x-y values: make a line from to the diagonal line

L = 2.8;

logist = @(x,L) L.*(1-x).*x;

x = -0.2:0.05:1.2;

y = logist(x,L);

%----------------------------------

% Plot function and coordinate axis

%----------------------------------

figure()

% Plot iteration function

plot (x,y,'Color','b','LineWidth',2);

hold on;

axis([-0.2 1.2 -0.2 1.1])

grid

% Plot coordinate axis

plot([-0.2 1.2],[0 0],'Color','k','LineWidth',2)

plot([0 0],[-0.2 2],'Color','k','LineWidth',2)

% Plot Diagonal line

plot([-0.2 2],[-0.2, 2]);

%--------------------------------

% Start and plot iterates:

%--------------------------------

% Initial point

x0 = 0.9;

y0 = 0.0;

% Iteration loop

for i = 1:10

y1 = logist(x0,L);

plot([x0 x0],[y0 y1],'Color','g','LineWidth',1); %go to the function

hold on;

pause(0.3)

x1 = y1;

plot([x0 x1],[y1 y1],'Color','g','LineWidth',1); %go to the diagonal

pause(0.3)

% update iteration

y0 = y1;

x0 = x1;

drawnow;

end

text(0.85,0.85,['Param = ' num2str(L,'%.4f')])

Exercise 4:

Plot the continuous evolution of the logistic iteration when going form

(start a new plot for each L). Choose always the initial point x0= 0.9 and do 100 iterates for each L.

Exercise 5:

Plot the Biffurcation Diagram of Feigenbaum. It consists in plotting the logistic iterates (n=200) for different values of

. To avoid initial inestabilities we will take always the initial point x0= 0.9 but we will plot only the points from the 50th iterate to the 200th. The final plot is shown below.