ACCEPTED SOLUTION
Accepted Solutions
Its not clear to me what you are looking for or what result you expect
Is it something like this:
or do you expect a vector of values as result?
You should also know that the first index of a vector is by default 0, not 1. This behaviour can be changed by setting ORIGIN to a value other than zero.
What you calculated simply was
Note that du.1*du.1 is a scalar vector product!
Its not clear to me what you are looking for or what result you expect
Is it something like this:
or do you expect a vector of values as result?
You should also know that the first index of a vector is by default 0, not 1. This behaviour can be changed by setting ORIGIN to a value other than zero.
What you calculated simply was
Note that du.1*du.1 is a scalar vector product!
Did you compute this in Prime or in Mathcad 15?
I do not get the right result in Prime (should be 12)
@Sergey wrote:
Did you compute this in Prime or in Mathcad 15?
No, it was done in Prime.
leave away the "n=1" and upper limit "6". Leave the placeholders simply empty.
du.1^2 is squaring a vector, Actually this is an invalid operation and Prime could (should?) throw an error. Instead it applied implicit vectorization, doing the squaring element by element and the result is a vector. The sum (without lower and upper limit) simply sums up the elements of this vector.
What you wrote add this vector six times and the result again is a vector. To can use this running index from 1 to 6 but first you would have to set ORIGIN to 1 and then you would have to use a vector index du.1[n.
Doing it the way I showed is easier, though.
