No result for while loop Statement

Matlab remains busy running my script forever, what's wrong with my script?

 n=1;
while abs(pi-sum(4./((2*(1:n)-1)*(-1).^(n+1))))>=0.001
      n=n+1;   
       sum(4./((2*(1:n)-1)*(-1).^(n+1)));
end
fprintf('The approximation using Leibniz''s formula falls within 0,001 of pi when it equals to')
fprintf('%.5f with n equls %d\n\n',sum(4./((2*(1:n)-1)*(-1).^(n+1))),n)
    end

Answer 1

[Guillaume]

Well, if you hadn't put a semicolon at the end of the line sum(...); effectively making it useless by suppressing the output, you would have seen that as n increases your sum gets further and further away from π instead of converging towards it.

That would be because you've implemented the formula incorrectly, in particular the (-1).^(n+1) which should be a vector. Manually testing each term of your formula for a few values of n should have shown you that.

Also note that you're recalculating all the terms of the expression each time you increase n, so your algorithm is not going to be particularly fast. I suppose it doesn't matter as it would converge quickly if the formula had been implemented right.

[Zhuoying Lin]

I write (-1).^(n+1) to design a sign for each nth term, so the result should be a number with +/- sign.

[Guillaume]

I repeat, manually testing each term of your formula for a few values of n should have shown you the problem.

Try

n = 4
4./((2*(1:n)-1)
(-1).^(n+1)
4./((2*(1:n)-1)*(-1).^(n+1)

You're multiplying all the elements of the whole vector [1, 1/3, 1/5, ...] by the scalar (-1).^(n+1)

You need to multiply the sequence by another vector [1, -1, 1, -1, ...], not a scalar.

[Zhuoying Lin]

Okay I see what you mean now, yes it works! Thank you!

[Zhuoying Lin]

However, I get n=1000 and the approximation is 3.140592654, this does not fall within 0.001 of pi though

[Guillaume]

Actually, I was wrong about one thing, the series does not converge quickly. According to wikipedia, "Leibniz's formula converges extremely slowly: it exhibits sublinear convergence. Calculating π to 10 correct decimal places using direct summation of the series requires about five billion terms"

[Zhuoying Lin]

so is there any reason why I get this result?

[Guillaume]

You need at least 500 terms (i.e n=1000) to get within ±0.001 of pi.

For n = 1000, the value is indeed ~3.14059265383979, which is ~0.000999999749998981 less than π, so within 0.001 of π.