Why am i getting this error message? "Array indices must be positive integers or logical values."

x = 120;
for n=1:10
f= (x(cosh(600/x) -12)-108);
x_1 = x-(f/diff(f))
x=x_1;
end

Array indices must be positive integers or logical values.

Answer 1

[Bjorn Gustavsson]

In matlab (and most other programming-languages) you will have to write out the multiplication-operator (* for matrix multiplication or .* for elementwise multiplication). The way you write:

x(cosh(600/x)-12)

Will be interpreted as the cosh(600/x)-12th element of the matrix x. Note here that 600/x would be a right-matrix-division, for elementwise division you will have to use ./.

Most likely you want:

f= (x*(cosh(600/x) -12)-108);

Then you will have a problem with the next line too, because f is now a scalar which make the function diff return an empty array, [], so you will have to do something about that.

If you are very new to matlab take a trip through the on-ramp courses on the Matlab web-site.

HTH

[Hiruni Adithya Rajapakshe]

wow sir, thank you very much!

[Bjorn Gustavsson]

Good, if the answer was satisfactory, then accept it.

[Hiruni Adithya Rajapakshe]

This is the answer i get now, can you please show me where i went wrong.. My ambision here to find the value for the equation using Newton Resphan method. It supposed to be an iterative same answer.

x = 120;
for n=1:10
f= (x*(cosh(600./x) -12)-108);
x_1 = x-(f./diff(f))
x=x_1;
end

x_1 =

     []


x_1 =

     []


x_1 =

     []


x_1 =

     []


x_1 =

     []


x_1 =

     []


x_1 =

     []


x_1 =

     []


x_1 =

     []


x_1 =

     []

[Bjorn Gustavsson]

Yes, your f is a scalar - and if you check the help and documentation of diff you will see:

help diff
 DIFF Difference and approximate derivative.
    DIFF(X), for a vector X, is [X(2)-X(1)  X(3)-X(2) ... X(n)-X(n-1)].
    DIFF(X), for a matrix X, is the matrix of row differences,
       [X(2:n,:) - X(1:n-1,:)].

The diff-function expects an array as input. You call it with a scalar, which causes diff to return an empty array. To my eyes it seems as if you want to implement a Newton-solver, for that you want to divide with the derivative of your function. I suggest you introduce a new variable, perhaps named dfdx that you assign the derivative of the function whos value you assign to f. That should be straightforward for you with pen and paper. Alternatively you could calculate a numerical estimate of the derivative, but that is more programming.

[Hiruni Adithya Rajapakshe]

yeah that sounds great! again thank you very much sir!