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Nested loops to estimate cos(pi/4)

3 vues (au cours des 30 derniers jours)
Mary Jeppson
Mary Jeppson le 9 Déc 2016
Commenté : Mary Jeppson le 10 Déc 2016
I have written this code that is giving me an error:
true = cos(pi/4);
n = 2;
es = .5^(2-n);
x = pi/4;
ea = 1;
est = 1;
k = 1;
while(1)
for ea >= es
est = 1+x^(k*2)/factorial(k*2)*(-1)^k
ea = abs(true-est)/true*100
end
k = k+1;
end
Clearly, I do not understand the relationship of for to while loops. Can anyone help me understand?
Thank you, Mary

Réponse acceptée

Steven Lord
Steven Lord le 9 Déc 2016
  1. You should avoid calling your variable true, as that already has a meaning in MATLAB.
  2. This line of code "for ea >= es" is not valid MATLAB syntax.
  3. Your while loop is an infinite loop; you have not specified a way to break out of it.
Most of your line of code "est = 1+x^(k*2)/factorial(k*2)*(-1)^k" matches the term in the series definition of the cosine function, but the "1+" part doesn't.
Basically you want your code to look something like this pseudocode:
1) initialize variables
2) while (the approximation is not close enough to the known value)
2a) add another term to your approximation
There's a while loop in step 2, but the condition is not 1. There is no for loop in the pseudocode. I can't say much more without simply giving you the code.
In general, if you're still unsure about how while loops work, I think one way you can understand them better is to walk through the first example (calculating factorial(10)) on its documentation page yourself, with you playing the role of MATLAB. Take a piece of paper and a pencil and "execute" each line yourself, recording the values of variables as they change on that paper. [When you reach the end statement, return to the while statement. If the while condition is not satisfied, go to the statement immediately after the end.]
  1 commentaire
Mary Jeppson
Mary Jeppson le 10 Déc 2016
Thank you for your thorough and conceptual answer. I appreciate it!

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Plus de réponses (1)

Image Analyst
Image Analyst le 9 Déc 2016
You don't need both. You need just one of the other. the for loop is most straightforward in this case:
theta = pi;
trueValue = cos(theta);
numberOfTerms = 10; % Whatever.
theSum = 0;
for k = 0 : numberOfTerms-1
thisTerm = (-1)^k * theta ^ (2*k) / factorial(2*k);
theSum(k+2) = theSum(k+1) + thisTerm;
theError(k+1) = theSum(k+2) - trueValue;
end
% Make fancy plots.
subplot(2, 1, 1);
plot(theSum, 'b*-', 'LineWidth', 2);
ylabel('The Taylor Sum', 'FontSize', 20);
xlabel('Number of Terms', 'FontSize', 20);
grid on;
subplot(2, 1, 2);
plot(theError, 'b*-', 'LineWidth', 2);
ylabel('The Error from True Value', 'FontSize', 20);
xlabel('Number of Terms', 'FontSize', 20);
grid on;
% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
% Get rid of tool bar and pulldown menus that are along top of figure.
set(gcf, 'Toolbar', 'none', 'Menu', 'none');
% Give a name to the title bar.
set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off')
  1 commentaire
Mary Jeppson
Mary Jeppson le 10 Déc 2016
Thank you for this! I am looking closely at it and learning from it. Thank you!

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