When I run the below code, why dont I get value of Q=[ 2 3]

In the equation y=sin(2x + 3) , a=2 and b=3 (general form y=sin(a*x + b)). When I run the following code why don't I get Q=[2 3] .

 x=(0:0.01:2*pi);
  y=sin(2*x + 3); 
   for i=1:62

       P=[x(i)   1;
          x(i+1) 1];

       R=[asin(y(i));
         asin(y(i+1))];

       Q=R\P    % %Q=[a b];
    end

Answer 1

[James Tursa]

Your fundamental problem is that asin( ) is multi-valued and you are treating it as single-valued. To recover the 2 and 3, you would need to know ahead of time exactly which of these multi-values that you want, and you don't. And which of these multi-values you want will of course change as you go through the sin( ) cycles. E.g., to recover the 2 and 3 from the first pair you would need to do this:

x = (0:0.01:2*pi);
y = sin(2*x + 3);
P = [x(1) 1; x(2) 1];
Q = P \ [pi-asin(y(1));pi-asin(y(2))]
Q =
   2.000000000000002
   3.000000000000000

But the form I chose for the "R" part was specific to the range of the result that I wanted to get.

Picking your formulation we get this:

 Q = P \ [asin(y(1));asin(y(2))]
 Q =
  -2.000000000000000
   0.141592653589793

But these are legitimate results also, just in a different range. E.g., using this Q to see if they work:

>> sin(Q(1)*x(1)+Q(2)) - y(1)
ans =
     0
>> sin(Q(1)*x(2)+Q(2)) - y(2)
ans =
     0

Yep, they work.

Bottom line is you can't recover your original a and b constants using this method because you don't have enough information about which asin( ) result is needed.

(Plus, you had a typo anyway ... your Q = R\P should have been Q = P\R)