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.

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1. x = 32:2:75 2. x = [2 5 1 6] a = x + 16 b = x(1:2:end) + 3 c = sqrt(x)orc = x.^(0.5) d = x.^2ord = x.*x 3. x = [3 2 6 8]', y = [4 1 3 5]' a = y + sum(x) b = x.^y c = y./x 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)ord = 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)orc = 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 afor-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 = 4xfor any input! Relational operations are subject to precedence and ordering just as arithmetical operations. Thelogicalphrasea<x<bshould be rendered as the compound logical expression((a < x) & (x < b))in MATLAB. 8. See the discussion under IF-block exercise 4, above.

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