%In response to questions in class, here
%is how you could use Matlab to check that five functions are
%linearly independent by taking the Wronskian.
%My comments begin with %, all other lines are matlab input or output.


>> syms t
  
>> f = [cos(t) sin(t) cos(2*t) sin(t)^2 cos(t)^2]
 
f =
 
[   cos(t),   sin(t), cos(2*t), sin(t)^2, cos(t)^2]


% I check first if somehow matlab will do this for me, no luck
 
>> help wronskian

wronskian.m not found.

>> W = [ f; diff(f); diff(diff(f)); diff(diff(diff(f))); diff(diff(diff(diff(f)))) ]
 
W =
 
[                 cos(t),                 sin(t),               cos(2*t),               sin(t)^2,               cos(t)^2]
[                -sin(t),                 cos(t),            -2*sin(2*t),        2*sin(t)*cos(t),       -2*sin(t)*cos(t)]
[                -cos(t),                -sin(t),            -4*cos(2*t),  2*cos(t)^2-2*sin(t)^2, -2*cos(t)^2+2*sin(t)^2]
[                 sin(t),                -cos(t),             8*sin(2*t),       -8*sin(t)*cos(t),        8*sin(t)*cos(t)]
[                 cos(t),                 sin(t),            16*cos(2*t), -8*cos(t)^2+8*sin(t)^2,  8*cos(t)^2-8*sin(t)^2]
 
>> det(W)
 
ans =
 
36*sin(2*t)*sin(t)^6-36*sin(t)^2*sin(2*t)*cos(t)^4+72*cos(2*t)*sin(t)^5*cos(t)+144*cos(2*t)*sin(t)^3*cos(t)^3+36*cos(t)^2*sin(2*t)*sin(t)^4-36*sin(2*t)*cos(t)^6+72*cos(2*t)*sin(t)*cos(t)^5
 
% All right, it can take the derivative symbolically

% now plug in t=0


>> subs(ans,{t},{0})

ans =

     0


% Hmm, it is singular at t=0, so try t=1

>> subs(det(W),{t},{1})

ans =

  -4.4409e-15
  
% close enough to zero that it probably is.
% At this point I notice that I took the wrong five functions.
%  They are linearly dependent since  cos(2*t) = cos(t)^2 - sin(t)^2
% So we should get zero for the Wronskian.

% While I have these linearly dependent functions, let's use Matlab to find the dependence relation.
% The null command finds a basis for the null space of a matrix

>> null(W)
 
ans =
 
[ empty sym ]
 
% Well, matlab found nothing in the nullspace of the symbolic matrix,
%  so let us evaluate the matrix at t=0 and find its null space.

>> null(subs(W,{t},{0}))

ans =

   -0.0000
   -0.0000
    0.5774
    0.5774
   -0.5774

% So a basis for the null space of W(0) is .5774 * [0 0 1 1 -1]
%  (The .5774 is 1/sqrt(3) to make the vector length one.
%  this is a strong hint that 1*cos(2t) + 1*sin^2(t) -1*cos^2(t) = 0.


------------


% okay, now let us use the right five functions.

>> f = [cos(t) sin(t) sin(2*t) sin(t)^2, cos(t)^2]
 
f =
 
[   cos(t),   sin(t), sin(2*t), sin(t)^2, cos(t)^2]
 
>> W = [ f; diff(f); diff(diff(f)); diff(diff(diff(f))); diff(diff(diff(diff(f)))) ]
 
W =
 
[                 cos(t),                 sin(t),               sin(2*t),               sin(t)^2,               cos(t)^2]
[                -sin(t),                 cos(t),             2*cos(2*t),        2*sin(t)*cos(t),       -2*sin(t)*cos(t)]
[                -cos(t),                -sin(t),            -4*sin(2*t),  2*cos(t)^2-2*sin(t)^2, -2*cos(t)^2+2*sin(t)^2]
[                 sin(t),                -cos(t),            -8*cos(2*t),       -8*sin(t)*cos(t),        8*sin(t)*cos(t)]
[                 cos(t),                 sin(t),            16*sin(2*t), -8*cos(t)^2+8*sin(t)^2,  8*cos(t)^2-8*sin(t)^2]
 
>> det(W)
 
ans =
 
-36*cos(2*t)*sin(t)^6+36*cos(2*t)*sin(t)^2*cos(t)^4+72*cos(t)*sin(2*t)*sin(t)^5+144*sin(2*t)*sin(t)^3*cos(t)^3-36*cos(2*t)*sin(t)^4*cos(t)^2+36*cos(2*t)*cos(t)^6+72*sin(t)*sin(2*t)*cos(t)^5
 
>> subs(det(W),{t},{0})

ans =

    36

% So the Wronskian is nonzero at t=0
% In fact it is always 36 as the following shows:

>> simplify(det(W))
 
ans =
 
36
 
% We could have used a loop to construct W as follows

% First make a matrix of the right size with the right first row

>> W = [f; f; f; f; f]
 
W =
 
[   cos(t),   sin(t), sin(2*t), sin(t)^2, cos(t)^2]
[   cos(t),   sin(t), sin(2*t), sin(t)^2, cos(t)^2]
[   cos(t),   sin(t), sin(2*t), sin(t)^2, cos(t)^2]
[   cos(t),   sin(t), sin(2*t), sin(t)^2, cos(t)^2]
[   cos(t),   sin(t), sin(2*t), sin(t)^2, cos(t)^2]


% Now make sure each row is the derivative of the preceeding row
 
>> for n=2:5 W(n,:) = diff(W(n-1,:)); end
>> W
 
W =
 
[                 cos(t),                 sin(t),               sin(2*t),               sin(t)^2,               cos(t)^2]
[                -sin(t),                 cos(t),             2*cos(2*t),        2*sin(t)*cos(t),       -2*sin(t)*cos(t)]
[                -cos(t),                -sin(t),            -4*sin(2*t),  2*cos(t)^2-2*sin(t)^2, -2*cos(t)^2+2*sin(t)^2]
[                 sin(t),                -cos(t),            -8*cos(2*t),       -8*sin(t)*cos(t),        8*sin(t)*cos(t)]
[                 cos(t),                 sin(t),            16*sin(2*t), -8*cos(t)^2+8*sin(t)^2,  8*cos(t)^2-8*sin(t)^2]
 
% Matlab note:
%  W(n,:) is the n-th row.  Likewise W(:,n) is the n-th column
% The :  tells you to take all entries.
% If you wanted the 2x3 matrix of the third and fifth rows and columns 1,2, and 4
%  you could write W([3 5],[1 2 4])

----------


% Okay, let us show that 1, e^t, ln(t), and t^2 are linearly independent functions

>> f = [1 exp(t) log(t) t^2]
 
f =
 
[      1, exp(t), log(t),    t^2]
 
>> W = [f;f;f;f];
>> for n=2:4 W(n,:) = diff(W(n-1,:)); end
>> W
 
W =
 
[      1, exp(t), log(t),    t^2]
[      0, exp(t),    1/t,    2*t]
[      0, exp(t), -1/t^2,      2]
[      0, exp(t),  2/t^3,      0]
 
>> det(W)
 
ans =
 
4*exp(t)*(-1+t+t^2)/t^3

% this is clearly nonzero for almost all t, for example at t=1 it is 4e

% It is amusing to have matlab find all t where the Wronskian is zero
 
>> solve(det(W))
 
ans =
 
[  1/2*5^(1/2)-1/2]
[ -1/2-1/2*5^(1/2)]
 
% So at (-1 +- sqrt(5) )/2 it is zero.  
% This has nothing to do with the linear independence of the functions however.