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Asked by Snirisa Gödel
on 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?

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Answer by Azzi Abdelmalek
on 19 Sep 2013

Edited by Azzi Abdelmalek
on 19 Sep 2013

Accepted answer

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)

Jan Simon
on 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.

## 2 Comments

## dpb (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/87672#comment_169878

Why do I get Inf+ Nani?Because of

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 (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/87672#comment_169894

Thank you dpb, I appreciate you're answer.