Using a while loop to determine and display the number of terms it takes for a given series to converge within 0.01% of its exact value ((pi^2)/6)?

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Bryan Vaughn
Bryan Vaughn 2017-11-25
评论： Bryan Vaughn 2017-11-25
I'm trying to get a while loop to work to do the above and then display the error and number or iterations ran, but it doesn't seem to be working and I am getting an infinite loop. Any help? My current code is:
ER=100;
s=0;
nn=0;
while (ER >= 0.01)
nn_sum = (1/(nn^2));
s = s + nn_sum; %compute the running sum
ER =(abs(exp((pi^2)/6)-s)/exp((pi^2)/6))*100; %Calculate error
nn = nn + 1; %counter
end
disp(nn-1)
disp(ER)

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采纳的回答

the cyclist
the cyclist 2017-11-25
编辑：the cyclist 2017-11-25
Two observations:
  1. I think you want your initialization of nn to be 1, not 0. Otherwise your some is infinite immediately.
  2. You have an exponentiation in your calculation of ER.
If I fix those two things, I get convergence after about 6,000 iterations.
FYI, I found those by using debug mode.

更多回答（1 个）

Jose Marques
Jose Marques 2017-11-25
In the first iteration, seem that nn == 0 and nn_sum == (1/nn^2) which means a zero division. Maybe you can include a if statement to fix this:
ER=100;
s=0;
nn=0;
while (ER >= 0.01)
if (nn ~= 0)
nn_sum = (1/(nn^2));
else
% do something
end
s = s + nn_sum; %compute the running sum
ER =(abs(exp((pi^2)/6)-s)/exp((pi^2)/6))*100; %Calculate error
nn = nn + 1; %counter
end
disp(nn-1)
disp(ER)

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