% question 6
% an example of a signal with frequency of 10 oscillations per second
%  f(t) = cos(10*2*pi*t), where t is expressed in seconds
t = 0:0.1:3;  % a first trial
s1 = cos(10*2*pi*t);
plot(t,s1)
t = 0:1/20:3;  % try two times the frequency
s2 = cos(10*2*pi*t);
plot(t,s2)
% we need to sample at least two times the frequency for every second,
% here we sample for three seconds at a rate of 20 samples per second,
% thus a total of 60 samples as absolute lower bound
% # 7 answer for n = 3
p3 = [ 0 0 1; 0 1 0; 1 0 0]

p3 =

     0     0     1
     0     1     0
     1     0     0

p3*[1:3]'

ans =

     3
     2
     1

% for any n, say n = 37
n = 37;
rows = [1:n];  % there is a one on every row
cols = [n:-1:1] % this defines where the ones in the columns come

cols =

  Columns 1 through 12 

    37    36    35    34    33    32    31    30    29    28    27    26

  Columns 13 through 24 

    25    24    23    22    21    20    19    18    17    16    15    14

  Columns 25 through 36 

    13    12    11    10     9     8     7     6     5     4     3     2

  Column 37 

     1

p37 = sparse(rows,cols,1);
p37*[1:n]'

ans =

    37
    36
    35
    34
    33
    32
    31
    30
    29
    28
    27
    26
    25
    24
    23
    22
    21
    20
    19
    18
    17
    16
    15
    14
    13
    12
    11
    10
     9
     8
     7
     6
     5
     4
     3
     2
     1

% # 4 We have seen polyfit, fft, and spline.  
%  polyfit fits the approximate data with a polynomial, preferably of low degree
% we use the fft to remove noise from signals, because with fft we can get the
% essential characteristics (amplitudes and frequencies) from a signal
% splines are used in modeling, to design objects
% # 5, see http://www.math.uic.edu/~jan/MCS320f01.html, last lecture
% # 8
% Let xA, xB, and xC be the amount of shares of type A, B, and C
% max 4*xA + 6*xB + 9*xC    the objective is to maximize the interest
% subject to the following constraints:
%     xA + xB + xC <= 100,000
%                      xC <=   20,000
%   - xA                 <=  -10,000
%    xA >= 0,  xB >= 0,  xC >= 0    do not forget these constraints!
% the answer to part (b):
f = [-4   -6   -9];   % linprog deals with minimization
A = [1 1 1;  0 0 1;  -1 0 0;  -1 0 0;  0 -1 0; 0 0 -1]

A =

     1     1     1
     0     0     1
    -1     0     0
    -1     0     0
     0    -1     0
     0     0    -1

b = [100000;  20000;  -10000;  0;0;0];
linprog(f,A,b)
Optimization terminated successfully.

ans =

  1.0e+004 *

    1.0000
    7.0000
    2.0000

format bank
ans

ans =

      10000.00
      70000.00
      20000.00

diary off