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 Math Help > Sequences and Series > Series > Gauss series for pi

 > > > > > > > > > > > > > > > > > > > > > > > > On 12/31/01 2:53:13 AM, Bruce Chiarelli wrote: If you are looking for algorithms to calculate pi I know a few: here is the gauss algorithm (Which I find easiest). Set these values a=1 b=1/sqrt(2) t=1/4 x=1 Next do these iterations a number of times that is greater than the base 2 logarithm of the number of digits to compute for example: If you want to calculate 1 million digits, iterate this 20 times because the base two logarithm of 1 million is 19.932. Here are the values you need to calculate: y=a a=(a+b)/2 b=sqrt(b*y) t=t-x(y-a)� x=2*x Once this has been repeated sufficiently, substitute the values of a, b , and t into this: pi=(a+b)�/(4*t)

WOW!

That's the fastest-converging series I've ever seen. It only took 3 iterations (not counting your initial values) to equal the precision of Excel, which is 15 digits.

Here's the spreadsheet I used to verify this:

 A B C D E 1 a b t x pi=(a+b)�/(4*t) 2 1 =1/sqrt(2) =1/4 1 =(A2+B2)^2/(4*C2) 3 =(A2+B2)/2 =sqrt(B2*A2) =C2-D2*(A2-A3)^2 =2*D2 =(A3+B3)^2/(4*C3) 4 5

Just copy the cells in row 3 to the rows below it, and here are the values you get:

 A B C D E 1 a b t x pi=(a+b)�/(4*t) 2 1 0.707107 0.25 1 2.91421356237309 3 0.853553 0.840896 0.228553 2 3.14057925052217 4 0.847225 0.847201 0.228473 4 3.14159264621354 5 0.847213 0.847213 0.228473 8 3.14159265358979

The value of the cell E5, is

(((((((1+(1/sqrt(2)))/2)+ sqrt((1/sqrt(2))*1))/2)+ sqrt(sqrt((1/sqrt(2))*1)*((1+(1/sqrt(2)))/2)))/2)+ sqrt(sqrt(sqrt((1/sqrt(2))*1)*((1+(1/sqrt(2)))/2))*((((1+(1/sqrt(2)))/2)+ sqrt((1/sqrt(2))*1))/2)))²/(4*((((1/4)-1*(1-((1+(1/sqrt(2)))/2))²)-(2*1)*(((1+(1/sqrt(2)))/2)-((((1+(1/sqrt(2)))/2)+ sqrt((1/sqrt(2))*1))/2))²)-(2*(2*1))*(((((1+(1/sqrt(2)))/2)+ sqrt((1/sqrt(2))*1))/2)-((((((1+(1/sqrt(2)))/2)+ sqrt((1/sqrt(2))*1))/2)+ sqrt(sqrt((1/sqrt(2))*1)*((1+(1/sqrt(2)))/2)))/2))²))

This method of approximating pi is also called the Gauss-Legendre approximation.

You may wonder how I came up with that silly expression.  I used Excel to create a text string, rather than carry out the computation.  That is, instead of =sqrt(B2*A2), I wrote ="sqrt(("&B2&")*("&A2&"))", etc.  Then, in cell E5, I got that delightful expression you see above.  Then I found (and fixed) places where 1 is a factor, such as 1* or *1, and removed extraneous parentheses to get the (relatively) shorter expression you see here.

((((1+1/sqrt(2))/2+ sqrt(1/sqrt(2)))/2+ sqrt(sqrt(1/sqrt(2))*((1+1/sqrt(2))/2)))/2+ sqrt(sqrt(sqrt(1/sqrt(2))*((1+1/sqrt(2))/2))*(((1+1/sqrt(2))/2+ sqrt(1/sqrt(2)))/2)))�/(2*2*(((1/(2*2)-(1-(1+1/sqrt(2))/2)�)-2*((1+1/sqrt(2))/2-((1+1/sqrt(2))/2+ sqrt(1/sqrt(2)))/2)�)-(2*2)*(((1+1/sqrt(2))/2+ sqrt(1/sqrt(2)))/2-(((1+1/sqrt(2))/2+ sqrt(1/sqrt(2)))/2+ sqrt(sqrt(1/sqrt(2))*((1+1/sqrt(2))/2)))/2)�))

You may have noticed that the number 4, which appeared only twice in the original expression, is now replaced by 2*2, leaving an expression with just ones, twos, arithmetic operators and sqrt!  I dare you to copy this and paste it into Excel just to see whether it gives you pi!

What fun!

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Proof that π is irrational

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