Fixed-point Iteration in MATLAB

[Fatima Hasan]

Homework Statement

The question is attached below.

Relevant Equations

-

I used the same code , but changing the equation ( line 4). In part (D) , I got a complex number , does it mean that the function is divergence ?
Here's my code :
Part c :

Part d :

Could someone confirm my answer please ? Any help would be greatly appreciated !

 

[BvU]

x0=g(x0)

How can your program work at all ? Or don't I understand Matlab ?
C)
Did you check with the plot ? where is g(x) = x ? Not at 1.75 !

D)
1^anything = 1 so how come that solution is not found ?

 

[Fatima Hasan]

x0=g(x0)

How can your program work at all ? Or don't I understand Matlab ?
C)
Did you check with the plot ? where is g(x) = x ? Not at 1.75 !

D)
1^anything = 1 so how come that solution is not found ?

I set the initial conditions ( Lines : 4 - 8 ). The fixed point iteration in part C is x0=g(x0) and to compute the error ,we have to find the absolute value of the difference between x and xold. Once we have computed the error, the current value of x is stored in xold. x(1) is g(x0) when i =1 , x2 is g(x1) when i =2 , and so on ...

 

[BvU]

Does x0 = g(x0) change the value of x ?

 

[Fatima Hasan]

Does x0 = g(x0) change the value of x ?

No, I think I have to write x instead of x0 in line 10 ,so x is varying in each iteration. Therefore , the error will be changed as well. Am I right now ?

 

[Joshy]

Doesn't part d say "$\text{for} ~~ x > 0$"? I don't completely understand this, but is that right to start with $-1$? Can you start with a different arbitrarily small number like $1e-15$? I'm curious to see what happens and might even give it a go myself.

$x0 = g(x0)$ looks right to me.

 

[Fatima Hasan]

Doesn't part d say "$\text{for} ~~ x > 0$"? I don't completely understand this, but is that right to start with $-1$? Can you start with a different arbitrarily small number like $1e-15$? I'm curious to see what happens and might even give it a go myself.

$x0 = g(x0)$ looks right to me.

I wrote the same code but changing the initial value of x0 , from $-1$ to $1e-15$ . Here's what I got :

Partc

Part d

 

[Joshy]

Okay I see another thing (what others clearly noticed before me)... I was so concentrated on the function itself, but you do have another $x$ you're comparing $x0$ to when you check for your error; you're also plotting with respect to $x$, which your code does not update. Make sure the $x$ and $x0$ is consistent.

A recommendation... I think a more elegant approach requires a modest change in your code. Change your $x$ into an array just like in line 19, but do this in the beginning at line 7 instead. Make the array something like:

x = x0.*ones(1, 20); % can be arbitrary numbers you're going to overwrite it anyways make sure your first one x(1) is x0

In the for loop you can refer to xold as $x(i-1)$ and the current iteration of x as $x(i)$ .

for i = 2:20 % start with 2 instead since your initialized version is x(1)
   
    x(i) = g( x(i-1) ); % x(i) is your current x, and x(i-1) is xold
    err = abs ( x(i) - x(i-1) );
   
    if(tol > err)
        break;
    end
end

When you get to plotting... no need to overwrite x or to create a new array.

 

[Fatima Hasan]

Okay I see another thing (what others clearly noticed before me)... I was so concentrated on the function itself, but you do have another $x$ you're comparing $x0$ to when you check for your error; you're also plotting with respect to $x$, which your code does not update. Make sure the $x$ and $x0$ is consistent.

A recommendation... I think a more elegant approach requires a modest change in your code. Change your $x$ into an array just like in line 19, but do this in the beginning at line 7 instead. Make the array something like:

x = x0.*ones(1, 20); % can be arbitrary numbers you're going to overwrite it anyways make sure your first one x(1) is x0

In the for loop you can refer to xold as $x(i-1)$ and the current iteration of x as $x(i)$ .

for i = 2:20 % start with 2 instead since your initialized version is x(1)
  
    x(i) = g( x(i-1) ); % x(i) is your current x, and x(i-1) is xold
    err = abs ( x(i) - x(i-1) );
  
    if(tol > err)
        break;
    end
end

When you get to plotting... no need to overwrite x or to create a new array.

Thanks for your reply. Could you please tell me how to write a MATLAB code in PF ?

I changed the code and that's what I got:

 

[Joshy]

I made a mistake changing $x$ into an array- my apologies (you can revert to your older version I think it worked better). I think your purpose for plotting both were to see where they intersect... shows that it should have converge.

There's an icon just above the text box.

I have no further feedback I think I'm going to have to fidget with this if I want to give you a better response. I might, but won't promise anything.

edit:

Did a little more digging. I think those last two are suppose to be difficult on purpose :) (difficult as in it won't converge easily) but it was fun to explore. That starting value really matters a lot for those two.

If I were to look at (roughly) $x^2 - sin(x)$ and I knew that $x$ was a really tiny magnitude ie. $|x| << 1$... breaking it up into two parts I would know that $x^2$ is super tiny might be possible to neglect it and $sin(x)$ would be almost itself. If $x$ is large, then the sinusoidal is negligible and somehow $x = x^2$. Good luck.

Part d was really interesting. I should have looked at the formula... if you start with a tiny number like I suggested then it makes sense... $x$ is some small number to the power $x-cos(x)$ (probably a negative number), which makes it something like $1 \over x^\text{small number}$. Of course it'll become ginormous just as we saw. Our trick for part c doesn't work here.

(note the $10^{10}$ scale)

Your best bet is starting off with a bigger number, but how big? Funny enough... as you start close to 1 (+ or - minus a small fraction) then it's close to convergence. $1^x$... it's going to be $1$. You stray further away from that then ie $0.5^{0.5-cos(0.5)}$ and iterating it over and over again it's going to become some large number again.

Somewhere starting between 0.5 and 0.7 interesting things happen. I would recommend playing around there. I recommend plotting the error to see what's going on too ;) (this plot is when my starting value was 0.7).