Compute the infinite sum of pi/4 up to 5 correct decimals.

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Snirisa  Gödel
Snirisa Gödel le 19 Sep 2013
I want to approximate pi/4 by using the infinite sum pi/4=((-1)^(n))/(2*n+1) from 0 to infinity. Here is my MATLAB code:
if true
tol = 10^(-5) %Up to five correct decimals
d=1; %Arbitrarily chosen, only condition is that d>tol at the start
sn=0; %Initial starting value
n=0;
while d>tol
sn = ((-1)^(n))/(2*n+1); This is the approximation for pi/4.
d=abs(pi/4-sn); Value that decides whether the loop continues.
n=sn; I'm not sure but I want n to increase by one each
time to prevent
circular calculations.
end
Why do I get Inf+ Nani?
  2 commentaires
dpb
dpb le 19 Sep 2013
Why do I get Inf+ Nani?
Because of
n=sn;
Try each step directly at the command line and see what happens...
You need to set a value for the total of the summation of each term initially to zero and then accumulate the terms into that variable -- sn is ok for the name but you don't accumulate a sum of the terms in n, you replaced in with the individual term going forward.
To get the subsequent terms you need to increment n, that is correct --
n=n+1;
where the n=sn; line is instead; the implementation of accumulating the summation is left for you to consider how to do that a little more...
Snirisa  Gödel
Snirisa Gödel le 19 Sep 2013
Thank you dpb, I appreciate you're answer.

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Réponse acceptée

Azzi Abdelmalek
Azzi Abdelmalek le 19 Sep 2013
Modifié(e) : Azzi Abdelmalek le 19 Sep 2013
tol = 10^(-5)
d=1;
sn=0; %
n=0;
while d>tol
sn = sn+(-1)^n/(2*n+1);
n=n+1;
d=abs(pi/4-sn);
end
disp(sn)
  2 commentaires
Jan
Jan le 19 Sep 2013
Modifié(e) : Jan le 19 Sep 2013
This is obviously a homework question. Solving it does not allow the OP to find the solution by his own.
Azzi, please do not post solutions of homework questions, because this reduces the reputation and efficiency of the forum. As a teacher on a university you should think of your colleagues, who do not want to get solutions created by others than their students.
Snirisa  Gödel
Snirisa Gödel le 19 Sep 2013
Thank you.

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