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Post by I wuv M4( Satar Jaèoèdoæ) on Apr 22, 2009 14:15:45 GMT -5
I discovered on my own a forula to prove pi. It is a forula for perimeter were n is the # sides p is perimter and radius is 1 P=2n tan((360/2n) as n aproaches infinatey that forula aproaches 2pi
Veers would be proud!
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Post by The Dark Master on Apr 22, 2009 14:18:44 GMT -5
Exactly how much spare time do you have on your hands?
I think its spelt formula, not forula.
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Post by I wuv M4( Satar Jaèoèdoæ) on Apr 22, 2009 14:20:55 GMT -5
Shut up.
I had to do stupid state test and I finished early. I actualy started thinking about it last saturday while I was at states for SO.
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Post by ~Memzak~ on Apr 22, 2009 16:37:03 GMT -5
WTF...
no wait I didn't mean that...
WTF!?!?!?
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Post by General Veers on Apr 22, 2009 17:29:40 GMT -5
Ah, someone understands the concept of a limit and used it to find the value of pi. You didn't quite "prove" pi, but you realized that an infinity-gon is indeed a circle...
lim(x -> infinity)2n*tan(360/2n)
lim(x -> infinity)2n*tan(180/n)
lim(x -> infinity)2n*sin(180/n)/cos(180/n)
lim(a -> 0)(2/a)*sin(180a)/cos(180a) lim(x -> infinity)1/x = 0
lim(a -> 0)(2sin(180a))/(a*cos(180a))
lim(a -> 0)((1/90)cos(180a))/(-(a/90)sin(180a)+cos(180a))
(1/90) / (0*0 + 1)
1/90
Odd, I didn't get 2pi...maybe I did something wrong? You are right, an infinity-gon would be a circle with a "perimeter" (technically a circumference) of 2pi.
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Post by I wuv M4( Satar Jaèoèdoæ) on Apr 22, 2009 17:42:57 GMT -5
It was actually quite easy to make the formula.
An interior angle of a regular is 360/number of sides.
Divide one of the triangles created by the midsegments with the angle bisector of the interoir angle and you get a right triangle were theta=360/2n
Then use tan of theta to get half the lenghth of any side. Multiply that by twice the number of sides and you get the perimeter.
I used a graphing cal. to prove my formula.
I put the ('s wrong the first time I entered it in though.
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Post by Qwerty on Apr 22, 2009 18:53:43 GMT -5
Okay, I'll let Veers and Ganon have fun here. Don't make a mess!
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Post by dbsndust on Apr 22, 2009 21:34:08 GMT -5
I discovered on my own a forula to prove pi. It is a forula for perimeter were n is the # sides p is perimter and radius is 1 P=2n tan((360/2n) as n aproaches infinatey that forula aproaches 2pi Veers would be proud! That's on wikipedia for one. Also, a circle has an infinite number of sides, so your computations would be impossible to be 100% accurate. |
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Post by General Veers on Apr 23, 2009 15:07:05 GMT -5
...unless the computations are done an infinite amount of times, but that is impossible for anything that has a finite lifespan...
Again, the formula does not actually prove the value of pi, but shows that an infinity-gon is a circle.
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Post by I wuv M4( Satar Jaèoèdoæ) on Apr 23, 2009 15:20:48 GMT -5
It's on wikipedia!
I knew this was too easy.
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Post by General Veers on Apr 23, 2009 15:23:02 GMT -5
Don't worry, I'm still glad you discovered that relationship...not everyone is curious enough to do anything like that...
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Post by The Dark Master on Apr 23, 2009 16:07:57 GMT -5
Or clever.
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Post by I wuv M4( Satar Jaèoèdoæ) on Apr 23, 2009 16:20:48 GMT -5
Well now I'll try to find to forlua for perimeter if the angle bisectors =2.
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Post by Qwerty on Apr 23, 2009 17:58:04 GMT -5
Okay, you do that.
Now, in the meantime, what do we do in this thread? Wait for you to do that?
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Post by dbsndust on Apr 23, 2009 19:07:54 GMT -5
I suggest computing the volume of a shape that appears on all sides like this _______
/ _ _ \
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|_| |_| |_| Lol, my first ASCII art ever is a crappy mcdonalds logo. Just compute it |
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Post by GGoodie on Apr 23, 2009 19:27:57 GMT -5
What grade is everyone in here? I'm a pretty smart guy but i don't ever think about proving formulas. Though i did try and write a proof for the pythagorean theorem with a trapezoid but that was simple, this isn't something you would normally think about, unless you just got out of math class.
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Post by dbsndust on Apr 23, 2009 21:01:13 GMT -5
no.... my ascii art got messed up brb
anyway im in 9th, tho technically im in 9.25 since im in honors bio & math (math is grade up)
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Post by Qwerty on Apr 23, 2009 23:20:34 GMT -5
you be in algebra II trig accelerated?
Nice, fairly high class. I remember that class...
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Post by I wuv M4( Satar Jaèoèdoæ) on Apr 25, 2009 8:35:32 GMT -5
I am in algerabra II, but my grade is far bellow 10th or 11th.
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Post by General Veers on Apr 25, 2009 9:44:10 GMT -5
I'll just say that I am high enough to be able to tell you that the nth derivative of xn is n! (a.k.a. n factorial)...
...and that any function can be rewritten as an infinitely large polynomial. For instance, ex can be written as 1 + x + x2/2 + x3/6 + x4/24 + x5/120 + ... + xn/n! + ... = sum(xn/n!, x, 0, infinity).
...and I could tell you that the region bounded by sinx and -sinx over the interval [0, pi] would be 4.
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