The discussion centers on issues encountered while implementing fixed-point iteration in MATLAB. A user experienced divergence when changing the initial value, leading to complex results, and sought clarification on whether this indicated a divergence in the function. Participants emphasized the importance of ensuring that the variables x and x0 are consistent throughout the code and suggested using an array to track iterations more effectively. Additionally, they discussed how the choice of initial conditions significantly impacts convergence, recommending starting values close to 1 for better results. Overall, the conversation highlights the nuances of fixed-point iteration and the critical role of initial conditions in achieving convergence.
#1
Fatima Hasan
315
14
Homework Statement
The question is attached below.
Relevant Equations
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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 !
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 ?
#3
Fatima Hasan
315
14
BvU said:
Matlab:
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 ...
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 ?
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#6
Jody
Gold Member
453
230
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.
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#7
Fatima Hasan
315
14
Joshy said:
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
#8
Jody
Gold Member
453
230
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:
Matlab:
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)## .
Matlab:
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.
#9
Fatima Hasan
315
14
Joshy said:
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:
Matlab:
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)## .
Matlab:
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:
#10
Jody
Gold Member
453
230
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).
