Problem with function @(t) [ (1/2)*a*t.^2]

 function eulers_for_mv
  a = .5 ;

t = linspace(0,2) ;
dt = t(2) - t(1) ;
x = zeros(1,length(t));
v = zeros(1,length(t));
for i=1:length(t)-1
    v(i+1) = v(i) + dt*a 
    x(i+1) = x(i) + dt*v(i) ;
end
original_eq = @(t) [x(1) + v(1)*t +(1/2)*a*t.^2] ;

original_eq(t) ---> returns to me a 1x200 matrix because I left some room between t and 1/2 it shouldnt matter no?? if I take this away writing ...t+(1/2) ... it`ll give me the correct 1x100 matrix. why??

Answer 1

[Star Strider]

MATLAB interprets spaces in matrices as separators, so it does matter in that situation. In matrices, put parentheses around expressions you want MATLAB to consider as single expressions.

If you want ‘v(1)*t +(1/2)*a*t.^2’ to be a single expression, write it as:

original_eq = @(t) [x(1) + (v(1)*t +(1/2)*a*t.^2)] ;

[Douglas Alves]

But what about x(1)? there shoud be an error because it`s separated from the other part. As you said, it sees spaces as separators so it should separate x(1) from the other part. no?

Thank you Star Strider...

[John D'Errico]

No. In the form you posed,

original_eq = @(t) [x(1) + v(1)*t +(1/2)*a*t.^2] ;

the interpreter will see

x(1) + v(1)*t

as one piece, since the separate + sign can only be interpreted one way.

When you had the + next to another term, MATLAB construes it as the unary plus operator, where we have plus(x)==x.

Think of it like this. Matlab takes each piece and evaluates it, then combines the pieces. Where there is an operator between them, matlab uses that operator. Where there is only space, MATLAB combines them into an array.

Essentially, you wrote this:

[A + B +C]

which the interpreter in MATLAB first takes the +c terms and converts it to C, so then the interpreter can only see the expression as:

[(A+B) , C]

Had you written it as

[A + B + C]

MATLAB would simply perform the sum of all three elements, and the [] would be superfluous.

[Douglas Alves]

I understand it perfectly. Thanks a lot!!

[Star Strider]

When I run:

z = original_eq(t);  
fznz = find(z ~= 0);

after the other elements of your code (I didn’t create a function file for it), the first 101 elements of ‘z’ (a (1x200) vector) are all zero. Only the last 99 are non-zero.

Note that x(1) and v(1) are both zero and do not change. It is adding x(1) to v(1)*t, because otherwise I would expect the vector to be 201 elements long. Because t(1)=0, the first term of (1/2)*a*t.^2 would also be zero, accounting for the first 101 elements all being zero. So it is concatenating the elements v(1)*t, and (1/2)*a*t.^2 as separate elements of a single vector, creating your (1x200) element vector.

In this example, note the difference between:

t = 1:5;
qq = @(t) [0 + t +t.^2];
w = qq(t)

and:

t = 1:5;
qq = @(t) [0 + t + t.^2];
w = qq(t)

So the spaces (or lack of them) matter in how MATLAB interprets the vector calculation. In the first one, the ‘+’ before t.^2 does nothing (although a minus sign would produce the negative of t.^2) because MATLAB is not interpreting it as an arithemtic operator, while in the second, MATLAB interprets all the spaces and the operators as signifying a single expression.