Approximating Pi using a series

Question

Chewystein 2017-9-9

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/355872-approximating-pi-using-a-series

评论： Jose Marques 2017-9-9

So I have no idea where to start. The problem is approximating pi using (-1^(n+1))*((x^(2*n-1))/(2*n-1)) and the first 10 values in the series. Here is what I have:

function y=f(x)
  y=(-1^(n+1))*((x^(2*n-1))/(2*n-1));
end
x=0;
z=10;
N=1000000;
sum=0;
deltax=(z-x)/N;
for i=1:N;
  xeval1=x+(i-1)*deltax;
  xeval2=x+(i)*deltax;
  sum=sum+0.5*(f(xeval1)+f(xeval2))*deltax;
end
fprintf('The approximate vaule of pi %15.10f.\n',sum)

I also have to do this using Machin's formula with the first 10 values in a power series

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Answer 1

Jose Marques 2017-9-9

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https://ww2.mathworks.cn/matlabcentral/answers/355872-approximating-pi-using-a-series#answer_280919

在 MATLAB Online 中打开

PI=0;
N=10;    %qty of elements in the serie
v=zeros(N+1,1);
for n=0:N
    PI=PI+4*(((-1)^n)/(2*n+1));
    v(n+1)=PI;
    PI
end

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Jose Marques 2017-9-9

在 MATLAB Online 中打开

function  [y] = f(x,n)
  y=(-1^(n+1))*((x^(2*n-1))/(2*n-1));
end

I forgot the function

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Answer 2

John D'Errico 2017-9-9

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/355872-approximating-pi-using-a-series#answer_280920

编辑：John D'Errico 2017-9-9

在 MATLAB Online 中打开

Ok. So you made an effort. Not terrible, but there are some problems.

NEVER use a variable named sum. If you do, then the function sum is now useless.

Next, it looks like your function f(x) tries to return the nth term of the atan function.

http://mathworld.wolfram.com/MachinsFormula.html
https://www.craig-wood.com/nick/articles/pi-machin/

It will not work correctly, because you never pass in the value of n. Instead, try this:

machterm = @(x,n) (-1).^(n+1).*((x.^(2*n-1))./(2*n-1));

Note that I was careful to put parens around -1 there. There is a difference between -1^4 and (-1)^4.

So Machin's formula tells us that

4*atan(1/5) - atan(1/239)

is pi/4. Lets first see if the formula computes accurate values of the arctan.

atan(1/5)
ans =
         0.197395559849881
sum(machterm(1/5,1:10))
ans =
         0.197395559849881
machterm(1/5,1:10)
ans =
  Columns 1 through 6
                       0.2      -0.00266666666666667                   6.4e-05     -1.82857142857143e-06      5.68888888888889e-08     -1.86181818181818e-09
  Columns 7 through 10
      6.30153846153847e-11     -2.18453333333334e-12      7.71011764705883e-14     -2.75941052631579e-15

As you can see, machterm is set up so that we can generate all the terms at once.

4*sum(machterm(1/5,1:10)) - sum(machterm(1/239,1:10))
ans =
         0.785398163397448
pi/4
ans =
         0.785398163397448

As you can see, it does a pretty good job of approximating pi/4.

I suppose, given that this is homework, you are not allowed to use the vectorized solution I give here. But nothing stops you from using a loop.

machterm = @(x,n) (-1).^(n+1).*((x.^(2*n-1))./(2*n-1));
at1 = 0;
at2 = 0;
for n = 1:10
  at1 = at1 + machterm(1/5,n);
  at2 = at2 + machterm(1/239,n);
end
piapprox = (4*at1) - at2;  % yielding pi/4
piapprox = 4*piapprox
piapprox =
        3.14159265358979
pi
ans =
        3.14159265358979

As you can see, it did well, as we would expect.

How might you build a direct power series approximation for pi? I would note that atan(1)=pi/4. This is the classic, basic series solution. We can test that by:

atan(1)*4
ans =
        3.14159265358979

So, now just use

4*sum(machterm(1,1:10))
ans =
           3.0418396189294
>> 4*sum(machterm(1,1:100))
ans =
          3.13159290355855
>> 4*sum(machterm(1,1:1000))
ans =
          3.14059265383979
>> 4*sum(machterm(1,1:10000))
ans =
          3.14149265359003
>> 4*sum(machterm(1,1:100000))
ans =
          3.14158265358979
>> 4*sum(machterm(1,1:1000000))
ans =
          3.14159165358978
>> 4*sum(machterm(1,1:10000000))
ans =
          3.14159255358974

As you can see, even taking 10 million terms in the simple power series for atan(1), we get a pretty darn poor approximation for pi. But that series converges quite slowly, so what can you expect?