I've had many request for solutions so here are some for the simpler exercises. Caveat: I hope the following solutions work but my typing and brain are sometimes not in sync. If you find errors, please let me know.
If your solution looks different - great! Differences are good. Do you know why? Finding out might be worthwhile and you are welcome to send a note to let me know what you've found.
1. x = 32:2:75
2. x = [2 5 1 6]
a = x + 16
b = x(1:2:end) + 3
c = sqrt(x) or c = x.^(0.5)
d = x.^2 or d = x.*x
3. x = [3 2 6 8]', y = [4 1 3 5]'
a = y + sum(x)
b = x.^y
c = x./y
z = x.*y
w = sum(z)
x'*y - w (same thing)
5. The function "rats" just displays the contents of a variable
a = 2:2:20 (or whatever max #)
b = 10:-2:-4
c1 = 1:5, c2 = 1./c1 , rats(c2)
d1 = 0:4, d2 = 1:5, d3 = d1./d2 , rats(d3)
6. n = 1:100;
x = ( (-1).^(n+1) ) ./ (2*n - 1);
y = sum(x)
8. t = 1:0.2:2
a = log(2 + t + t.^2)
b = exp(t).*(1 + cos(3*t))
c = cos(t).^2 + sin(t).^2 (all ones!)
d = atan(t)
e = cot(t)
f = sec(t).^2 + cot(t) - 1
12. t = 1790:2000;
term = 1 + exp(-0.0313*(t - 1913.25));
P = 197273000./term;
plot(t,P)
xlabel('year'), ylabel('population')
P2020 = 197273000/(1 + exp(-0.0313*(2020 - 1913.25)))
2. A = [ 2 4 1 ; 6 7 2 ; 3 5 9]
x1 = A(1,:)
y = A(end-1:end,:)
c = sum(A)
d = sum(A,2) or d = sum(A')'
N = size(A,1), e = std(A)/sqrt(N)
5. A = [2 7 9 7 ; 3 1 5 6 ; 8 1 2 5]
B = A(:,2:2:end)
C = A(1:2:end,:)
c = reshape(A,4,3) or c = A' (they are different but are both 4x3)
d = 1./A , rats(d)
e = sqrt(A)
6. randn('seed',123456789)
F = randn(5,10);
N = size(F,1)
avg = mean(F)
s = std(F)
tscore = (avg - 0)./(s/sqrt(N))
None were different at 90% LOC (all < 2.132).
4. x = [3 15 9 12 -1 0 -12 9 6 1]
a = x, idxa = x > 0, a(idxa) = 0
b = x, idxb = ~rem(x,3), b(idxb) = 3
c = x, idxc = ~rem(x,2), c(idxc) = 5*c(idxc)
5. x = 1:35;
y = zeros(size(x));
idx1 = x < 6;
idx2 = (x >= 6) & (x < 20);
idx3 = (x >= 20) & (x <= 35);
y(idx1) = 2;
y(idx2) = x(idx2) - 4;
y(idx3) = 36 - x(idx3);
disp([x(:) idx1(:) idx2(:) idx3(:) y(:)])
plot(x,y,'o')
Loop constructs: The answers here provide one version of the solutions. Alternatives are possible and encouraged, especially where time and efficiency of the code is important.
1. x = [1 8 3 9 0 1]
a. total = 0;
for j = 1:length(x)
total = total + x(j);
end
b. runningTotal = zeros(size(x));
runningTotal(1) = x(1);
for j = 2:length(x)
runningTotal(j) = runningTotal(j-1) + x(j);
end
c. s = zeros(size(x));
for j = 1:length(x)
s(j) = sin(x(j));
end
2. A = rand(4,7);
[M,N] = size(A);
for j = 1:M
for k = 1:N
if A(j,k) < 0.2
A(j,k) = 0;
else
A(j,k) = 1;
end
end
end
3. x = [4 1 6], y = [6 2 7]
N = length(x);
for j = 1:N
c(j) = x(j)*y(j);
for k = 1:N
a(j,k) = x(j)*y(k);
b(j,k) = x(j)/y(k);
d(j,k) = x(j)/(2 + x(j) + y(k));
e(j,k) = 1/min(x(j),y(k));
end
end
c = sum(c); % or use 1.a. loop
4. These code snippets do the job but their repeated use is much
more interesting. An example is given for the first exercise.
a. total = 0; % initialize current sum (the test variable)
count = 0; % initialize the counter (output of the program)
while total < 20 % loop until 20 is exceeded
count = count + 1; % another loop repeat => another number added
x = rand(1,1);
total = total + x; % modify the test variable!
end
disp(['It took ',int2str(count),' numbers this time.'])
------------------------------------------------------------
To do this many times, place the above code in a for-loop.
Some simple (though perhaps subtle) changes are needed wth respect
to retaining the counts for each repeat. Also, the summary has
been changed from a single text message to a histogram.
Nrep = 1000; % collect 1000 repeats of the above code
count = zeros(Nrep,1);
for j = 1:Nrep
total = 0; % reset the test variable each repeat!!!
while total < 20
count(j) = count(j) + 1; % use a vector to capture each result
total = total + rand(1,1);
end
end
hist(count,min(count):max(count))
xlabel('Number of random numbers from U(0,1) added to make 20')
ylabel('Count')
title(['Histogram of results for ',int2str(Nrep),' repeats'])
------------------------------------------------------------
b. count = 0;
while 1 % infinite loop use
count = count + 1;
x = rand(1,1); % get a number
if (x < 0.85) & (x > 0.8) % check the value
break % bail out if in selected range
end
end
disp(['It took ',int2str(count),' numbers this time.'])
c. count = 0;
avg = 0; % test variable
while abs(avg - 0.5) > 0.01
count = count + 1;
% The following line is one way to update the average.
% (count-1)*avg is the sum of the first count-1 numbers
% and rand just adds another number. Dividing by count
% then gives the new average.
avg = ((count-1)*avg + rand(1,1))/count; % modify the test var.
% There are other ways to do this and you are encouraged
% to come up with them
end
disp(['It took ',int2str(count),' numbers this time.'])
5. while 1 % use of an infinite loop
TinF = input('Temperature in F: '); % get input
if isempty(TinF) % how to get out
break
end
TinC = 5*(TinF - 32)/9; % conversion
disp(' ')
disp([' ==> Temperature in C = ',num2str(TinC)])
disp(' ')
end
1. n = 7 gives m = 8
n = 9 gives m = -1
n = -10 gives m = -11
2. z = 1 gives w = 2
z = 9 gives w = 0
z = 60 gives w = sqrt(60)
z = 200 gives w = 200
3. T = 50 gives h = 0
T = 15 gives h = 31
T = 0 gives h = 1
4. x = -1 gives y = -4
x = 5 gives y = 20
x = 30 gives y = 120
x = 100 gives y = 400
In this last exercise, y = 4x for any input! Relational operations
are subject to precedence and ordering just as arithmetical
operations. The logical phrase a < x < b should be rendered as
the compound logical expression ((a < x) & (x < b)) in MATLAB.
8. See the discussion under IF-block exercise 4, above.