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How to prevent addition from showing up in command window

1 Ansicht (letzte 30 Tage)
Kyle Donk
Kyle Donk am 17 Jan. 2020
Bearbeitet: stozaki am 18 Jan. 2020
I ran some code and am having trouble with the results of the code come out like this:
995
996
997
998
999
1000
How do I just get 1000 instead of all of the numbers? I have suppressed literally everything in the code.
  3 Kommentare
Rik
Rik am 18 Jan. 2020
Without a meaningful sample of your code, we can't give a meaningful solution.
Stephen23
Stephen23 am 18 Jan. 2020
Kyle Donk's "Answer" moved here:
N=10;
error=1;
while error>10^-4
N=N+1;
total=0;
for n=1:N;
y=1/n^2;
total=total+y;
end
error=((pi^2)/6)-total;
disp(N)
end

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Akzeptierte Antwort

stozaki
stozaki am 18 Jan. 2020
Bearbeitet: stozaki am 18 Jan. 2020
If you get only 1000, you can try following script.
N=10;
err=1;
while err>10^-3
N=N+1;
total=0;
for n=1:N
y=1/n^2;
total=total+y;
end
err=((pi^2)/6)-total;
end
disp(N);
Is the intent of the question I understood fit?
 
Regards,
stozaki
  2 Kommentare
Image Analyst
Image Analyst am 18 Jan. 2020
Built in function names, like error and sum, should not be used as variable names.
Also not sure why you're printing out the iteration number in
fprintf('>> error value %d is over 10^3\n',N);
instead of the final error.
stozaki
stozaki am 18 Jan. 2020
It was just a descriptive print statement. I'm sorry. I also prioritized leaving the questioner's code as much as possible.

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Weitere Antworten (1)

Image Analyst
Image Analyst am 18 Jan. 2020
Try this:
numberOfTerms = 10;
maxIterations = 1000000; % To prevent infinite loops.
loopCounter = 1;
theError = 1;
while theError > 1e-4 && loopCounter < maxIterations
total = 0;
for n = 1 : numberOfTerms
y = 1 / n^2;
total = total+y;
end
theError = ((pi^2)/6) - total;
% Optional: print out the total and error at each iteration.
fprintf('After %d iterations, and using %d terms, the total is %f, and the error is %.9f.\n', ...
loopCounter, numberOfTerms, total, theError);
numberOfTerms = numberOfTerms+1;
loopCounter = loopCounter + 1;
end
% Print out the final results.
fprintf('The final error after the loop exited is %.9f.\n', theError);
You'll see
After 9989 iterations, and using 9998 terms, the total is 1.644834, and the error is 0.000100015.
After 9990 iterations, and using 9999 terms, the total is 1.644834, and the error is 0.000100005.
After 9991 iterations, and using 10000 terms, the total is 1.644834, and the error is 0.000099995.
The final error after the loop exited is 0.000099995.

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