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Old 03-14-2009, 12:44 PM   #1
Patrick Donnelly
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Default Happy Pi Day

Who could let 3/14 pass by without wishing everyone a happy pi day?

This is a somewhat special day for me in particular, since back in eighth grade, I calculated some 1083 digits of pi with an algorithm I developed (repeatedly making use of the Pythagorean theorem to inscribe a regular polygon with 4*2^n sides inside a circle) and, with the help of my dad, wrote into a simply computer program. I can still recite the number to 25 decimal digits, 3.1415926535897932384626433.

Happy pi day, everyone!
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Old 03-14-2009, 09:16 PM   #2
Matthew Bacorn
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I miss the days of bringing pie to math class on pi day

thats pretty cool about the algorithm; i was on the math team in eighth grade. We sat around, ate pizza, listened to They Might Be Giants, and did math.

Not your kind of math though. Normal 8th grader math
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Old 03-15-2009, 10:53 AM   #3
Craig Loizides
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That reminds me of a math professor I had who could do over 1000 digits of pi from memory. One day in class he wrote out the first few digits on the blackboard and when he ran out of room he just kept going writing in chalk on the walls. He did a few laps around the room continuing to talk about whatever we were doing the whole time.

http://en.wikipedia.org/wiki/John_Horton_Conway
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Old 03-15-2009, 08:28 PM   #4
Derek Maffett
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You must be studying to become an engineer or something similar, right? Ouch.
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Old 03-16-2009, 06:27 PM   #5
Patrick Donnelly
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Quote:
Originally Posted by Matthew Bacorn View Post
Not your kind of math though. Normal 8th grader math
I haven't thought of anything that spectacular since the eighth grade. It was really kind of luck that I found it anyway.

d = 1
r = 1/2
a[0] = r
b[0] = r
c[0] = sqrt(a[0]^2 + b[0]^2)

for i, as i goes to infinity
a[i+1] = c[i] / 2
b[i+1] = r - sqrt(r^2 - a[i+1]^2)
c[i+1] = sqrt(a[i+1]^2 + b[i+1]^2)
end

pi approx. = c[i]*4*2^i

Basically, it gives you the length of one side of a square for the first set of statements. When you go through the loop, it gives you one side of an octogon, then one side of a 16 sided figure, then one of a 32 sided figure, then one of a 64 sided figure, then one of a 128... etc. Multiplying that number by 4, 8, 16, 32, 64, 128,... etc. respectively will give you more and more accurate approximations of the circle's circumference. In this case, that's equal to pi, since d = 1.

Quote:
Originally Posted by Craig Loizides View Post
That reminds me of a math professor I had ...

http://en.wikipedia.org/wiki/John_Horton_Conway
Lucky.

Quote:
Originally Posted by Derek Maffett View Post
You must be studying to become an engineer or something similar, right? Ouch.
Physicist. Really, it's the easiest thing for me. That's just how I think.

For example, the other night, I was looking for a particular house, number 318, though I couldn't read the numbers on any of the houses because of the darkness, and I wasn't sure if I had passed it or not. So, I got out of the car, walked near to one house. Saw 308. Checked the next house down the street (in the direction of my travel) and saw it was 306. Therefore, dy/dx on that side of the street in my direction was equal to -2, a constant since house numbers are linear. So, 318 = -2x + 308, x = -5, and the house I wanted was 5 places in the opposite direction. I turned around, and counted the houses, 1, 2, 3, 4, 5. Number 318, bingo.


It's easy if you have the mind for it.
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Old 03-16-2009, 07:47 PM   #6
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Well, I can certainly respect that. It just seems deadly dull to me (which I suppose is why I don't have the mind for it).
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