# 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. (c) Numerical Factory - 2020