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! 
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 
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 
You must be studying to become an engineer or something similar, right? Ouch.

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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:
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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. 
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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