Reputation: 3431
I am new to Matlab and I have to use fixed point iteration to find the x
value for the intersection between y = x
and y = sqrt(10/x+4)
, which after graphing it, looks to be around 1.4. I'm using an initial guess of x1 = 0
. This is my current Matlab code:
f = @(x)sqrt(10./(x+4));
x1 = 0;
xArray(10) = [];
for i = 1:10
x2 = f(x1);
xArray(i) = x2;
x1 = x1 + 1;
end
plot(xArray);
fprintf('%15.8e\n',xArray);
Now when I run this it seems like my x
is approaching 0.8. Can anyone tell me what I am doing wrong?
Upvotes: 1
Views: 11613
Reputation:
Well done. You've made a decent start at this.
Lets look at the graphical solution. BTW, this is how I'd have done the graphical part:
ezplot(@(x) x,[-1 3])
hold on
ezplot(@(x) sqrt(10./(x+4)),[-1 3])
grid on
Or, I might subtract the two functions, then looking for a zero of the difference, so where it crosses the x axis.
This is what the fixed point iteration does anyway, trying to solve for x, such that
x = sqrt(10/(x+4))
So how would I change your code to fix it? First of all, I'd want to use more descriptive names for the variables. You don't get charged by the character, and making your code easier to read & follow will pay off greatly in the future for you.
There were a couple of code issues. To initialize a vector, use a form like one of these:
xArray = zeros(1,10);
xArray(1,10) = 0;
Note that if xArray was ALREADY defined because you have been working on this problem, the latter form will only zero out that single element. So the first form is best by a large margin. It affirmatively creates an array, or overwrites an existing array if it is already present in your workspace.
Finally, I like to initialize an array like this with something special, rather than zero, so we can see when an element was overwritten. NaNs are good for this.
Next, there was no need to add one to x1 in your code. Again, I'd strongly suggest using better variable names. It is also a good idea to use comments. Be liberal.
I'd suggest the idea of a convergence tolerance. You can also have an iteration counter.
f = @(x)sqrt(10./(x+4));
% starting value
xcurrent = 0;
% count the iterations, setting a maximum in maxiter, here 25
iter = 0;
maxiter = 25;
% initialize the array to store our iterations
xArray = NaN(1,maxiter);
% convergence tolerance
xtol = 1e-8;
% before we start, the error is set to be BIG. this
% just lets our while loop get through that first iteration
xerr = inf;
% the while will stop if either criterion fails
while (iter < maxiter) && (xerr > xtol)
iter = iter + 1;
xnew = f(xcurrent);
% save each iteration
xArray(iter) = xnew;
% compute the difference between successive iterations
xerr = abs(xnew - xcurrent);
xcurrent = xnew;
end
% retain only the elements of xArray that we actually generated
xArray = xArray(1:iter);
plot(xArray);
fprintf('%15.8e\n',xArray);
What was the result?
1.58113883e+00
1.33856229e+00
1.36863563e+00
1.36479692e+00
1.36528512e+00
1.36522300e+00
1.36523091e+00
1.36522990e+00
1.36523003e+00
1.36523001e+00
1.36523001e+00
For a little more accuracy to see how well we did...
format long g
xcurrent
xcurrent =
1.36523001364783
f(xcurrent)
ans =
1.36523001338436
By the way, it is a good idea to know why the loop terminated. Did it stop for insufficient iterations?
The point of my response here was NOT to do your homework, since you were close to getting it right anyway. The point is to show some considerations on how you might improve your code for future work.
Upvotes: 2
Reputation: 2437
There is no need to add 1 to x1
. your output from each iteration is input for next iteration. So, x2
from output of f(x1)
should be the new x1
. The corrected code would be
for i = 1:10
x2 = f(x1);
xArray(i) = x2;
x1 = x2;
end
Upvotes: 1