Solution 1

> n=1;
>> for i=0:n-1
f=[3;4];b=[cos(pi/(4*n))];A=[cos((2*i+1)*pi/(4*n)) sin((2*i+1)*pi/(4*n))];
end
>> [x,fopt]=linprog(f,A,b)
Exiting: One or more of the residuals, duality gap, or total relative error
has stalled:
      the dual appears to be infeasible (and the primal unbounded).
      (The primal residual < TolFun=1.00e-008.)

x =

1.0e+012 *

-0.0000
-3.3538

fopt =

-1.3415e+013

>> n=2;
>> f=[3;4];b=[cos(pi/(4*n))];A=[cos((2*i+1)*pi/(4*n)) sin((2*i+1)*pi/(4*n))];
end
??? end
|
Error: Illegal use of reserved keyword "end".

>> for i=0:n-1
f=[3;4];b=[cos(pi/(4*n))];A=[cos((2*i+1)*pi/(4*n)) sin((2*i+1)*pi/(4*n))];
end
>> [x,fopt]=linprog(f,A,b)
Exiting: One or more of the residuals, duality gap, or total relative error
has stalled:
      the dual appears to be infeasible (and the primal unbounded).
      (The primal residual < TolFun=1.00e-008.)

x =

1.0e+010 *

-2.3918
0.0001

fopt =

-7.1750e+010

>> n=3;
>> for i=0:n-1
f=[3;4];b=[cos(pi/(4*n))];A=[cos((2*i+1)*pi/(4*n)) sin((2*i+1)*pi/(4*n))];
end
>> [x,fopt]=linprog(f,A,b)
Exiting: One or more of the residuals, duality gap, or total relative error
has stalled:
      the dual appears to be infeasible (and the primal unbounded).
      (The primal residual < TolFun=1.00e-008.)

x =

1.0e+008 *

-0.0019
-1.4561

fopt =

-5.8301e+008

>> n=4;
>> for i=0:n-1
f=[3;4];b=[cos(pi/(4*n))];A=[cos((2*i+1)*pi/(4*n)) sin((2*i+1)*pi/(4*n))];
end
>> [x,fopt]=linprog(f,A,b)
Exiting: One or more of the residuals, duality gap, or total relative error
has stalled:
      the dual appears to be infeasible (and the primal unbounded).
      (The primal residual < TolFun=1.00e-008.)

x =

1.0e+008 *

-3.0227
-0.6135

fopt =

-1.1522e+009

Notation. Let Rd[x] denote the vector space of polynomials of degre...