How to access zero index in the following code?

I need to calculate value of tol for i=1:20

tol=abs(hvq(i))-abs(hvq(i-1))

Now in the first iteration as it is giving me error saying, "attempted to access hvq(0); index must be a positive integer or logical". I know that MATLAB is not based on zero indexing. Now how to resolve this issue? In the first iteration 'abs(hvq(i-1))' should take value =0.

[Stephen23]

This is unrelated to one/zero based indexing. The question is simply how to specify the index before the first index. The answer is simply: don't, because it does not exist (not matter whether one or zero based).

Answer 1

[John D'Errico]

Do you mean that you cannot use an if test? It is not possible to test for when i==1, and use a slightly different formula? Hmm. I thought that was possible. :)

The point being, just use a test.

if i == 1
  altered forula for tol
else
  usual formula
end

Since you will use a loop to do this anyway, this will work.

[Abhinav]

Thank you :) It worked!

Answer 2

[Steven Lord]

Compute the absolute value of the whole vector first. Concatenate a 0 at the start of the vector then use diff.

x = (1:10).^2;
dx = diff([0 abs(x)]);
result = [x; dx]

Each element in the second row of result is the difference between the element immediately above it in the first row and the element immediately before that one in the first row. The only exception is the first element in the second row, which is the same as the element above it in the first row (minus 0, which rarely makes a difference.)