Pi

[The Thing]

I've just saw an article online about finding a recursive expression for pi through a formula of centripetal force. It doesn't tell you how though! Anyone's ever heard of this?

[doG]

I've just saw an article online about finding a recursive expression for pi through a formula of centripetal force. It doesn't tell you how though! Anyone's ever heard of this?

Did you look here?

[The Thing]

That's where I got it from lol. I was doing random searches on google when I found the article (the science fair summary).

Never mind though, I'm not looking for it anymore.

Incidentally, what ways have been devised to calculate pi besides the 4arctan(1/5)-arctan(1/239) one?

[BigMoosie]

There are several, however they all involve infinite sums, continued fractions or other transcendal numbers etc, here are some examples:

 

[math]\frac{\pi ^2}{6} = \sum_{n=1}^\infty \frac{1}{n^2}[/math]

 

[math]e^{i\pi}+1=0[/math]

[The Thing]

Wow. I believe I've just come up with a new formula for pi. I don't think it's been done before, but just in case, I'll show you and you guys can comment on it:

A circle is essentially a polygon with an infinite number of sides, and so pi would equal the perimeter of the polygon / 2r. If we set the r to 1, then pi = Perimeter/2.

The perimeter of an equilaterial polygon is side length * number of sides. How do we find the side length? Well, we drop two lines from the center of the polygon so that the two lines form an isosceles triangle with the length of one side.

To find the topmost angle of this isosceles triangle, we divide 360 degrees by the number of sides, to get the topmost angle of this triangle. (360/x), if x is the number of sides.

To find the two base angles of the isosceles, subtract the topmost angle from 180 and divide by two, so we have (180-(360/x))/2, which can be simplified to (90-(360/2x))

Now we can use either the sine law or cosine law. Using the sine law, we find the length of 1 side of this polygon is: (remember the radius is 1, thus the two sides are 1)

1/sin(90-(360/2x))=s/sin(360/x), thus s=(sin(360/x))/(sin(90-(360/2x)))

Using the cosine law, we can see that s=sqrt(1^2+1^2-2*1*1*cos(360/x)), simplifying: sqrt(2-2cos(360/x)).

Now we multiply the number of sides (x) by the length of the sides and divide by the diameter of 2:

(xsin(360/x))/(2sin(90-(360/2x))), and (x*sqrt(2-2cos(360/x)))/2

So we can account for when x approaches infinity by using a limit of these functions:

lim x-->infinity ((xsin(360/x))/(2sin(90-(360/2x))))=pi

lim x-->infinity ((x*sqrt(2-2cos(360/x)))/2)=pi

We can also reduce the bottom of the first formula, so get a third limit:

lim x-->infinity ((xsin(360/x))/2)=pi

Well, here's a method (well, 3) for calculating pi using simple trig!

[NeonBlack]

I didn't read your entire method, but it is pretty clever!

I think Archimedes did essentially the same thing like 2500 years ago to approximate pi.

[BigMoosie]

Convert them to radians and you will see why your methods are invalid.

[The Thing]

Err...they're supposed to work in degrees...

Why are they invalid?

[JustStuit]

It should therefore work in both. Does it work either way?

[doG]

Wow. I believe I've just come up with a new formula for pi.

Sorry but it's not new. Google pi polygon for 337,000 hits, including this one with an interactive form that lets you inscribe and circumscribe a circle with increasing numbers of sides as you watch the values approach pi.

[cosine]

Wow. I believe I've just come up with a new formula for pi. I don't think it's been done before' date=' but just in case, I'll show you and you guys can comment on it:

A circle is essentially a polygon with an infinite number of sides, and so pi would equal the perimeter of the polygon / 2r. If we set the r to 1, then pi = Perimeter/2.

The perimeter of an equilaterial polygon is side length * number of sides. How do we find the side length? Well, we drop two lines from the center of the polygon so that the two lines form an isosceles triangle with the length of one side.

To find the topmost angle of this isosceles triangle, we divide 360 degrees by the number of sides, to get the topmost angle of this triangle. (360/x), if x is the number of sides.

To find the two base angles of the isosceles, subtract the topmost angle from 180 and divide by two, so we have (180-(360/x))/2, which can be simplified to (90-(360/2x))

Now we can use either the sine law or cosine law. Using the sine law, we find the length of 1 side of this polygon is: (remember the radius is 1, thus the two sides are 1)

1/sin(90-(360/2x))=s/sin(360/x), thus s=(sin(360/x))/(sin(90-(360/2x)))

Using the cosine law, we can see that s=sqrt(1^2+1^2-2*1*1*cos(360/x)), simplifying: sqrt(2-2cos(360/x)).

Now we multiply the number of sides (x) by the length of the sides and divide by the diameter of 2:

(xsin(360/x))/(2sin(90-(360/2x))), and (x*sqrt(2-2cos(360/x)))/2

So we can account for when x approaches infinity by using a limit of these functions:

lim x-->infinity ((xsin(360/x))/(2sin(90-(360/2x))))=pi

lim x-->infinity ((x*sqrt(2-2cos(360/x)))/2)=pi

We can also reduce the bottom of the first formula, so get a third limit:

lim x-->infinity ((xsin(360/x))/2)=pi

Well, here's a method (well, 3) for calculating pi using simple trig![/quote']

Haha I did the same thing a few years ago... even got it copyrighted (go ahead, laugh at me!)

 

No sorry its just using algebra to express archimedes method of calculating pi

[The Thing]

How did you get it copyrighted? What happened then?

Did I do the exact same thing as you did?

[BigMoosie]

This is why your methods are invalid:

 

To solve a trig value such as sine you use radians, your pocket calculator will take it in degress then convert to radians using pi then solve using:

 

[math]\sin(x) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}[/math]

 

For small values of sine you can approximate with a straight line, for example:

 

[math]\lim_{n \rightarrow 0 } sin(n) = n[/math]

 

Your sine in your function is limiting the same way so you can effectively remove the sine and convert to radians and you get:

 

[math]\pi = \lim_{n \rightarrow \infty } \frac{n \sin(\tfrac{360}{n}) } {2}= \lim_{n \rightarrow \infty } \frac{n \tfrac{2\pi}{n} } {2} = \frac{2\pi}{2} = \pi[/math]

 

That being said your origional function when in radians can be any value, not just pi, so would you say this is an identity for e: ?

 

[math]e = \lim_{n \rightarrow \infty } \frac{n \sin(\tfrac{e}{n}) } {2}[/math]

 

How did you get it copyrighted?

 

To get something copyrighted you smack a © <NAME> <YEAR> or something similar on your paper.

[NeonBlack]

To get something copyrighted you smack a © <NAME> <YEAR> or something similar on your paper.

 

If I remember right, I think there was a U.S. Law passed in like 2001 that says anything published on the internet is automatically copyrighted unless you explicitly state that it may be used by some person for some purpose.

[cosine]

How did you get it copyrighted? What happened then?

Did I do the exact same thing as you did?

I don't remember the formula exactly, but it definitely was the same concepts you used, so we probably came up with the same result. What happened then? Well the Library of Congress sends you a certificate with your name the title you gave the work and some other details including a reference number of the work. So essentially a copy of it is sitting somewhere deep in some archive of the Lirbary of Congress.