How can i modify this code so x can = 0

function qw=fNR(x)
qw=zeros(100,1)
qw(2)=x
i=2

while abs(qw(i)-qw(i-1))>1e-6
    qw(i+1)=qw(i)-ff1(qw(i))/fdf1(qw(i))
    i=i+1
end

qw=qw(2:i-1)
[qw(end) i] 
plot(qw)
end 

how can i model this functio so it can take fNR(0)

fdf1

function y=fdf1(x)
y=3*x.^2-12*x+11;
end 

ff1

function y=ff1(x)
y=x.^3-6*x.^2+11*x-5;
end 

Answer 1

[Image Analyst]

Try this:

function qw=fNR(x)
qw=zeros(100,1)
qw(2)=x
i=2
maxIterations = 1000;
loopCounter = 0;
while (abs(qw(i)-qw(i-1))>1e-6) && loopCounter < maxIterations
    qw(i+1)=qw(i)-ff1(qw(i))/fdf1(qw(i))
    i=i+1
    loopCounter = loopCounter + 1;
end
qw=qw(2:i-1)
if isempty(qw)
    message = sprintf('qw is empty')
    uiwait(warndlg(message));
    return;
end
[qw(end) i]
plot(qw)
end

[Muhammad Choudhury]

when i call fNR(0) i get an error message:

Array indices must be positive integers or logical values.

Error in fNR (line 12)

[qw(end) i] %this displays on the screen the last value of the iteration

[DGM]

That's showing up because the loop never starts. Since the loop never starts, pruning the vector:

qw=qw(2:i-1);

results in a zero-length vector. Trying to index to qw(end) is the same as qw(0), which isn't a valid index.

You can at least get it to start by doing this.

i=2;
dqw = 1;
while dqw > 1e-6
    qw(i+1) = qw(i)-ff1(qw(i))/fdf1(qw(i));
    [qw(i) qw(i-1)]
    dqw = abs(qw(i)-qw(i-1))
    i = i+1;
end

[Muhammad Choudhury]

ah thanks makes sense

[DGM]

Oops. I didn't realize I was commenting on the answer instead of on the question.

[Image Analyst]

@Muhammad Choudhury I know that was your error with the original code. Obviously you never ran my code because my code throws up an alert that the array is empty rather than your error.