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 May 19th, 2018, 07:48 AM #11 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,594 Thanks: 1038 I emailed this (about 15 years ago) to someone on the Internet who wanted opinions on the stupidity of these mathematicians who keep calculating PI forever and ever. (at bottom is a little true anecdote between me and a wise older friend, Mason Johnson). .................................................. ............................................... Hi. Well, they remind me of the guy who spent years sticking together over a million matches to come up with a miniature 'London Bridge': got queer looks, some respect for his tenacity...but only 5 bucks at the corner pawnshop.... This stupidity seems to have started in England (where else!) in the early 1900's. A British mathematician William Shanks worked it out to 707 decimal places doing calculations by hand...over 20 years. POOR BILL: an error in the 528th place was discovered in 1945...wonder if they changed the inscription on his tombstone. In Aug/89,YASUMASA KANEDA carried PI to 536,870,000 places, filling 110,000 pages of computer paper taking 67 hours and 13 minutes on Japan's fastest supercomputer. ...gee, I would have bet my 20 against your 10 that some Chineese fellow would have done this...doesn't the rumor say they're the best...like, every respectable computer store has one constantly working on a computer keyboard, all visible from the front store window. Did a search on PI the other day: they're proudly announcing the crashing of the 2 billion mark...WOW-WEE. MY SOLUTION ============ Purpose: make Pi=3.14 even. First, let's use a circle of diameter 12 inches. Next, let's use 3.14159 as 'the going' PI. Then: 3.14159 x 12 = 37.699 inches. 3.14000 x 12 = 37.680 inches. Difference: .019 of an inch (about thickness of 4 sheets of paper). So, after drawing your 1-foot-diameter-circle, all you need to do is make 4 little nicks of .00475 inch (.019/4)(thickness of one page) at each 90 degree and no one will ever know.... WHO CAN I SUGGEST THIS TO IN THE PI KINGDOM?? DENIS AND MASON (a keep-it-simple wise elderly friend) =============================================== DENIS: Hey Mason, just decided to take this amazing course. MASON: Ya? What's it called? DENIS: The Sylva Method of Mind Control. MASON: Hmmm...what will you learn there, Denis? DENIS: Amazing things, Mason; I'll be able to control my own mind and make it do all sorts of things. MASON: Well...can you give me a couple of examples? DENIS: Sure. At the demonstration, they told me I'd be able to figure out what my dog is thinking; also, when I get a letter in the mail, I'll hold it at both ends with my fingertips, concentrate, and I'll be able to tell what is in the letter...imagine that! MASON: Gee. Tell me, Denis, how much is this course costing you? DENIS: Only 325 bucks, Mason. MASON: Well, why don't you buy yourself a plastic letter opener instead: only 79 cents at K-Mart.... NOTE: Fortunately, refunds were allowed before the 1st official lesson!! Thanks from JeffM1
 May 19th, 2018, 10:01 AM #12 Senior Member     Joined: Sep 2015 From: USA Posts: 2,463 Thanks: 1340 \begin{align*} &\lim \limits_{n \to \infty} \dfrac n 2 \sin\left(\dfrac {2\pi}{n}\right) = \\ \\ &\lim \limits_{n \to \infty} \pi \dfrac{\sin\left(\dfrac {2\pi}{n}\right)}{\dfrac{2\pi}{n}} = \\ \\ &\pi \lim \limits_{n \to \infty} \dfrac{\sin\left(\dfrac {2\pi}{n}\right)}{\dfrac{2\pi}{n}} =\pi \end{align*} Where the limit evaluation comes from the well known fact that $\lim \limits_{x\to \infty} \dfrac{\sin(x)}{x} = 1$ So yes. This magic number is indeed a sort of $\pi$ and reaches it in the limit Thanks from Denis and JeffM1
May 20th, 2018, 08:13 AM   #13
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 Originally Posted by Forummur You really gave him an excellent answer using that formula involving the radius and sin(2pi/n) which I wasn't aware of. Also, in general as you probably know there is a simpler formula to determine the area of any regular polygon if you know its perimeter (p) or the length of each side and the length of an apothem (a) where an apothem is the perpendicular length from the center of the polygon to each side. An apothem is the same as the radius of a circle inscribed in the polygon whereas a radius of a polygon is the same as the radius of a circle circumscribed around the polygon. Anyway the apothem-perimeter formula is Area = (1/2)ap.
Simpler perhaps but I somehow prefer romsek's analytic/trigonometric approach.
It feels easier to recreate without consulting a textbook especially when you've forgotten more than half of what you've learned from plane geometry. You just sketch the unit circle and voila; everything seems clear.

 May 29th, 2018, 08:22 AM #14 Member   Joined: May 2018 From: Idaho, USA Posts: 42 Thanks: 4 Hello, I’m sorry I have not been a long time, I had some really rough things happen. I was looking at everybody’s equations they recommended. I did a little experiment, and I tried your equations. I also did an equation that I told you about earlier that I made up, which is 2.37765r^2. After doing all your equations, and mine, I realize that we are all right. All these equations, even mine, get the same answer. If there was a mixup, I apologize. Thanks for helping me out with the formulas. Jared
June 6th, 2019, 07:39 AM   #15
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Quote:
 Originally Posted by Forummur You really gave him an excellent answer using that formula involving the radius and sin(2pi/n) which I wasn't aware of. Also, in general as you probably know there is a simpler formula to determine the area of any regular polygon if you know its perimeter (p) or the length of each side and the length of an apothem (a) where an apothem is the perpendicular length from the center of the polygon to each side. An apothem is the same as the radius of a circle inscribed in the polygon whereas a radius of a polygon is the same as the radius of a circle circumscribed around the polygon. Anyway the apothem-perimeter formula is Area = (1/2)ap.
If a regular polygon with n sides is inscribed in a circle with a radius of r, the sine function is used to find the length of a side. Since each Theta is equal to 360/n, B=360°/(2n)=180°/n and r*sin(180°/n) is one half of a side. Therefore, one side is 2*r*sin(180°/n). The perimeter p is then = n*2*r*sin(180°/n). The apothem (the altitude of each resulting isosceles triangle), on the other hand is given by r*cos(180°/n). The area of each isosceles triangle then is 1/2*b*h=1/2*[r*cos(180°/n)]*[2*r*sin(180°/n)]=r^2*cos(180°/n)*sin(180°/n).
There are n such triangles, so
A= n*r^2*cos(180°/n)*sin(180°/n).
From another angle, as Forummur pointed out,
A = 1/2*a*p = 1/2*[r*cos(180°/n)]*[n*2*r*sin(180°/n)] = n*r^2*cos(180°/n)*sin(180°/n) = n*r^2*1/2*sin(2*180/n) as pointed out by romsek since sin(2*x)=2*sinx*cosx → 1/2*sin(2*x)=sinx*cosx as v8archie reminded us a few weeks ago at
A Demonstration of an application of a half-angle identity
Also, if the apothem a is the one that's given, then
A = na^2*tan(180°/n).
In this particular case, a = 3*cos(180°/5), n = 5
If the length of the side b is instead given, then
A = 1/4*n*b^2*cot(180°/n)
In this case, b = 2*3*sin(180°/5), n = 5

 June 13th, 2019, 07:25 AM #16 Member   Joined: May 2018 From: Idaho, USA Posts: 42 Thanks: 4 Wow, I forgot about this thread! @jonah, thanks for sharing that. When I first started this thread, I was excited to share what I learned and I wanted others opinions on it. Since then, I have learned many more things from the people here. In fact, my Therapist told me to do something that takes a lot of focus to practice controlling my ADHD. I chose math and chess to do this. Thanks for sharing that. I will be posting something else I found later on today. Jared
June 13th, 2019, 08:35 AM   #17
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Quote:
 Originally Posted by romsek Where the limit evaluation comes from the well known fact that $\lim \limits_{x\to \infty} \dfrac{\sin(x)}{x} = 1$
$$\lim \limits_{x\to \infty} -\dfrac{1}{x} \le \lim \limits_{x\to \infty} \dfrac{\sin(x)}{x} \le \lim \limits_{x\to \infty} \dfrac{1}{x}$$
Since the outer two limits are zero, I conclude that your well known fact not a fact whether it's well known or not.

Edit: Oh God! I'm correcting a year-old post.

June 13th, 2019, 12:20 PM   #18
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Quote:
 Originally Posted by romsek $\pi \lim \limits_{n \to \infty} \dfrac{\sin\left(\dfrac {2\pi}{n}\right)}{\dfrac{2\pi}{n}} =\pi$
my bad (for a change), limit goes to 0, not infinity

let $x = \dfrac {2\pi}{n}$

$\pi \lim \limits_{x\to 0}\dfrac{\sin(x)}{x} = \pi$

(really you could have just noted the typo, the math was pretty clearly correct)

 June 13th, 2019, 12:25 PM #19 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,594 Thanks: 1038 Yer doin' better since yer cat started to smile.....!!! Thanks from Greens
June 13th, 2019, 02:56 PM   #20
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 Originally Posted by v8archie Edit: Oh God! I'm correcting a year-old post.
I completely missed that when I decided to have some brandy a few days ago and checked out what my last post was. Very embarrassing.

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