"I would have expected that x2 would be the equivalent of x1(1:end-1)"
Nope. Those two values include floating point error and are calculated from different end values. Read this:
Quoting that answer:
"To counteract such error accumulation, the algorithm of the COLON operator dictates that:
1. The first half of the output vector (in this example ‘v’) is calculated by adding integer multiples of the step ‘d’ to the left-hand endpoint ‘a’.
2. The second half is calculated by subtracting multiples of the step ‘d’ from the right-hand endpoint."
For your values, we can see that the first vector contains:
>> fs = 50000;
>> x1 = 0:1/fs:1;
>> sprintf('%.30f',x1(25001))
ans =
0.500000000000000000000000000000
but the second vector contains:
>> x2 = 0:1/fs:(1-1/fs);
>> sprintf('%.30f',x2(25001))
ans =
0.499999999999999940000000000000
This is, according the above explanation, because the lengths of the two vectors are different and the "0.5" value is therefore calculated from different end values. This is easy to calculate ourselves, based on the explanation above. From the left-hand end for the first vector:
>> sprintf('%.30f',0+25000*(1/fs))
ans =
0.500000000000000000000000000000
and from the right-hand end for the second vector:
>> sprintf('%.30f',(1-1/fs)-24999*(1/fs))
ans =
0.499999999999999940000000000000
which matches exactly the values that you are getting.
Of course comparing floating point numbers for equality is never a good idea: always use a tolerance:
If you really want to see what the floating point values really are, try this:
