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These two 201 digit numbers are consecutive
answers in the growing-length formula.
3.
14159 26734 30964 26755 13656 11706 16386 67510 63702 84833
56094 43930 89724 14575 05354 47102 60328 56627 25521 28292
94903 82273 63223 83191 37611 33955 40994 23619 23652 19248
55797 86846 76158 10560 61830 61990 25525 22492 19054 02768
3.
14159 26337 48622 60304 59811 49487 75608 75755 96691 53788
60971 03699 81724 37713 82696 92833 50008 48582 61281 65507
76333 33504 73869 42065 65077 84733 53155 55050 64416 08331
65610 56850 94003 24274 80086 34031 29953 15619 91927 79003
The average of these gives 8 additional decimals correct.
These digits are 53589793 near the beginning of pi.
Pi is 3.14159265358979323...
I am using the 4 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 formula.
It is maybe around 20 million terms here.
I am using 222 digits, 22 before the decimal, and 200 after the decimal.
The 22 before the decimal is because of the denominator getting
bigger every term, so I needed space for that.
It is running under lcc C on a pentium.
The last 8 digits or so of each 200 digit number is bogus now, due to
rounding errors for the last 20 million or so iterations.
Sorry I didn't print the exact iteration number.
See the 02768 at the end of the first long number?
Well the digit where the 8 is became bogus after about the first
10 iterations of the formula. I should write routines to keep
track of max-errors. That is a future project. I have to think about
dividing and multiplying errors so all the routines have error tracking
someday.
Last edited by John E. Franklin (2008-03-09 03:57:14)
igloo myrtilles fourmis
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What is the growing-length formula?
I'm confused.
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From shortest to longest:
pi = 4
pi = 4 - 4/3
pi = 4 - 4/3 + 4/5
pi = 4 - 4/3 + 4/5 - 4/7
pi = 4 - 4/3 + 4/5 - 4/7 + 4/9
pi = 4 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11
pi = 4 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + 4/13
This is a simple formula, but computes
very slowly as I thought it would.
igloo myrtilles fourmis
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Here's a faster formula I'll try out someday.
Last edited by John E. Franklin (2008-03-09 04:04:51)
igloo myrtilles fourmis
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There is an even slower method, but fun.
Create a random point in a square (ie 2 random numbers (x,y)) and figure out if they would lie within a circle that fits in the square. Take the ratio of within/total and it should be pi/4.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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Ooh, that reminds me of another one.
Get a matchstick (or some other long, thin object) and draw parallel lines on a piece of paper, such that the distance between lines is the same as the length of your matchstick.
Then toss the matchstick on the piece of paper and note the number of times it lands on a line and the number of times you throw. (Throws count as void if the matchstick doesn't completely land on the paper, since the surface with parallel lines on it should theoretically be infinite)
The ratio of lines crossed to matchsticks thrown should approximate 2/π.
One of my teachers made the whole class do that in the 10 minutes at the end of a lesson, and when we pooled all our results together we got π ≈ 3.143.
Why did the vector cross the road?
It wanted to be normal.
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On mathsy's one, I quick-checked your 2/π number
by adding up the sine of the odd degrees from 1 to 89,
which totals 28.64934425 on my calculator.
Then I divided that by 45 for the average.
And you are right! 0.636652094 is that average,
which is very close to 2/π or .636619772
(This calculation I thought up based on random
"height"s of rotated toothpicks.)
igloo myrtilles fourmis
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