my newton code is not working for fing root of f = 2.5- 1/x

hi sir, this is my code, I am tring to use my newton code to find the root of f = 2.5 - 1/x but my error keeps = 1 and runs in to error

f = @(x) 2.5 - 1/x
df = @(x) x.^-2
error = 100
xi = 10
iter = 1
while error > 1
xii = xi - (f(xi)/df(xi))
error = ((xii-xi)/xii)*100
error = abs(error)
xi=xii
iter = iter+1
end

I knot the root is at 2.5 but how can I get it ?

Answer 1

[DGM]

Well, it is working, but you're operating near a singularity with particularly problematic symmetry.

If you pick a large initial estimate, you're going to end up on the other side of the singularity, with the next step size being even larger. The solution will diverge away from the singularity location. Try a smaller initial estimate, something in the range of (0 0.8) (non-inclusive).

Alternatively, if you want to keep the solver from wandering into bad places, you can try something like this:

f = @(x) 2.5 - 1/x
df = @(x) x.^-2
error = 100
xi = 5 % normally this would cause the solver to diverge
constraint=[0 10]; % but keep the solution constrained

iter = 1
while error > 0.001
	xii = xi - (f(xi)/df(xi));
	xii = min(max(xii,constraint(1)+eps),constraint(2)-eps)
	error = ((xii-xi)/xii)*100;
	error = abs(error);
	xi=xii;
	iter = iter+1;
end

though bear in mind that a crude constraint like this might cause other problems in certain cases.

[wenchong chen]

the root i got is 0.4, now I am tring to plot root vs iter and error vs iter, this is my code but I got more root and error number with iter numbers

f = @(x) 2.5 - 1/x

df = @(x) x.^-2

xmvec=[];

errvec=[];

error = 100

xi = 5 % normally this would cause the solver to diverge

constraint=[0 10]; % but keep the solution constrained

iter = 1

while error > 0.001

xii = xi - (f(xi)/df(xi));

xii = min(max(xii,constraint(1)+eps),constraint(2)-eps)

error = ((xii-xi)/xii)*100;

error = abs(error);

xi=xii;

xmvec=[xmvec xi]

errvec=[errvec error];

xmvec=[xmvec xi]

errvec=[errvec error];

iter = iter+1;

end

subplot(2,1,1)

plot(1:iter-1,xmvec)

subplot(2,1,2)

plot(1:iter-1,errvec)

Error using plot

Vectors must be the same length.

[DGM]

These two lines are duplicated twice.

xmvec=[xmvec xi]
errvec=[errvec error];

once you remove those, it plots.

For the assignment, you might want to pick a small value for xi (e.g. 0.001 or something), otherwise the error plot might not be very interesting due to the extreme scale.

[wenchong chen]

thank you very much! one more question, how can I keep guesing it until the error=0, find the number?

[DGM]

I'm not sure what you're asking. I was going to say that the error will approach zero, but I see now that you're multiplying it by 100. I'm not sure why you're doing that.

f = @(x) 2.5 - 1/x
df = @(x) x.^-2
error = 1
xi = -5
constraint=[0 10];

iter = 1
while error > 1E-8 % we can pick whatever we want here
	xii = xi - (f(xi)/df(xi));
	xii = min(max(xii,constraint(1)+eps),constraint(2)-eps);
	error = ((xii-xi)/xii);
	error = abs(error);
	xi=xii;
	iter = iter+1;
end

fprintf('final error is %2.4e\n',error(end))

either way, the error can be made arbitrarily small. For this exit condition, it's 2E-9. For an exit condition of 1E-9, it's zero to within the datatype precision.

If you're asking something different, please clarify.