Can someone tell me what is wrong with this?

x1=5;
tau=0.1;
k=1;
while x(k)-x(k+1)>0.01
    k=k+1;
    x(k+1)=x(k)-tau*(2*x(k)-2);
end

figure(1)
plot(x,'b*-.');
grid;

[Bhaskar R]

I don't know your context. Its working fine if I test x as a random 100 samples.

x1=5;
tau=0.1;
k=1;
x = rand(1, 100);
while x(k)-x(k+1)>0.01
    k=k+1;
    x(k+1)=x(k)-tau*(2*x(k)-2);
end
figure(1)
plot(x,'b*-.');
grid;

[M]

what is wrong with this

Can you give us more details ?

Do you get an error ? What do you expect the code to do ?

[Bhaskar R]

There is no syntax or runtime error in your code.

Code is doing as

In the vector if present and previous value diffence is greate than 0.01 then next value calculated as

x(k+1)=x(k)-tau*(2*x(k)-2);

present value assigned as previous value - tau*(2*previous value - 2)

If code ends when the condition met that means it won't check for the all values of the tha vector x

[Ewa Jablonska]

That's the task and I have it using for loop, but I can not figure out how to do it with while loop.

[M]

Why do you want it to work with a while loop ?

To see a comparison between for and while loops, you can have a look here:

https://mathworks.com/matlabcentral/answers/66814-for-loops-vs-while-loops

[Ewa Jablonska]

Because our lecturer said we should figure it out with while loop. :/

Answer 1

[Ewa Jablonska]

Thanks for effort, i mixed up + with -. :/ Correct:

%Value of step coefficient tau=0.1

x(1)=5;

k=1;

tau=0.1;

while x(k)>1.01

k=k+1;

x(k+1)=x(k)-tau*(2*x(k)-2);

end

plot(x,'b*-');

grid;

Answer 2

[Fangjun Jiang]

The problem is, vector x is not defined. When you do x1=5, I think you meant x(1)=5. But even with that, when k=1, x(2) is not defined yet.

The solution is to add these lines at the begining.

N=100;x=zeros(N,1);x(1)=5;

[Ewa Jablonska]

%Value of step coefficient tau=0.1

x(1)=5;

k=1;

tau=0.1;

while x(k)>0.1

k=k+1;

x(k+1)=x(k)+tau*(2*x(k)+2);

end

plot(x,'b*-')

error:

Index exceeds array bounds.

Error in zad12p (line 7)

x(k+1)=x(k)+tau*(2*x(k)+2);

Answer 3

[David Hill]

function x = Minimum(tau,xx)
x(1)=xx;
x(2)=x(1)-tau*(2*x(1)-2);
k=2;
while abs(x(k)-x(k-1))>0.01
    x(k+1)=x(k)-tau*(2*x(k)-2);
    k=k+1;
end
end

[Ewa Jablonska]

and how to see the result?

[David Hill]

you can see x in the workspace,

type 'disp(x)' and return,

or just type 'x' and return.

Answer 4

[Steven Lord]

I'm guessing from the fact that you define x1 but don't use it that the x variable may not exist. You may have been thinking / hoping that MATLAB would use x1, x2, etc. when it saw x(1), x(2), etc. but it does not.

The fact that you're using x(k+1) (not x(k-1)) in your while condition also smells a bit strange. You may be walking off the end of the x array.

But if correcting these does not resolve the problem, please state more clearly the specific difficulty you're experiencing.

Do you receive an error message (text displayed in red)? If so, tell us on which line the error occurs and show us the full and exact text of the error, all the text displayed in red exactly as it's displayed in the Command Window.

Do you receive a warning (text displayed in orange)? If so, do the same as for an error but show us all the text displayed in orange.

Do you get a different answer than you expected? If so, show us the answer you received and the answer you expected and explain why you expected what you expected.

Does something else happen? If so, what?

[Ewa Jablonska]

%Value of step coefficient tau=0.1

x(1)=5;

k=1;

tau=0.1;

while x(k)>0.1

k=k+1;

x(k+1)=x(k)-tau*(2*x(k)+2);

end

plot(x,'b*-')

error: Index exceeds array bounds.

Error in zad12p (line 7)

x(k+1)=x(k)-tau*(2*x(k)+2);

No oragne just red.

Tasks given and must use while loop, because we did for "for loop":

The ponit is that function should end at specific value but it does not even with higher go to value.