Function variable not being treated as a vector?

[javila]

Hi, @Proektsoftbg!

I got another question for you...

Why the macro approach in #640 works and the following one doesn't?

[Proektsoftbg]

Hi! The problem is that for historical reasons dots are allowed in variables, so p.k can be a variable with this name. You can actually define p.k = 1. When the function is created, it is compiled for performance and p.k is expected to be a variable later. At this moment, the program does not know if you will pass a variable or vector to p. We cannot change this behavior now because old worksheets may stop working. So, to explicitly specify that p will be a vector, define p=[0] before creating the function. Then p.k will be compiled as the k-th index of vector p.

However, the best and most efficient way is not to use indexing at all, and entirely by vectorized operators functions:

Please, check this example:

a = [4; 5; 6; 7; 8]
n = range(1; len(a); 1)
f(x) = sum(a*x^n)
f(1)
f(2)
g(x) = dot(a; x^n)
g(1)
g(2)

[javila]

You just made my tricks (#640) useless with those vector operations... 😢

Thanks for the explanation, @Proektsoftbg.

[javila]

Is this also considered efficient???

roots = [-1; -1; 2; 2; 2; 4]
gg(x) = -(1/8)*product(fill(vector(len(roots)); x) - roots)

[Proektsoftbg]

Not quite. This is equivalent to:

roots = [-1; -1; 2; 2; 2; 4]
gg(x) = -(1/8)*product(x - roots)

You do not need to put the scalar in a vector before the operation. The scalar operator is automatically applied to each element and the result is stored a new vector. Try this code:

a = [2; 3; 4]
b = a - 1

[Proektsoftbg]

For max. efficiency if needed always use the hp versions of vectors and matrices always when you do not have units or all element have equal units. Then my code will become:

a = hp([4; 5; 6; 7; 8])
n = range_hp(1; len(a); 1)
f(x) = sum(a*x^n)
f(1)