Area of circle using incribed polygons

1. May 27, 2010

Jamin2112

I was trying to find the area of a circle the ancient way. For example, here is the area of an octagon inscribed in a circle.

You formula is the same regardless of how many sides your figure has: (S/2)r2sin(2π/S).

And so, the area of a circle must be limS-->∞(S/2)r2sin(2π/S).

I can expand that out using the Taylor series.

(S/2)r2 [(2π/S) - (2π/S)3/3! + (2π/3)5/5! - (2π/3)7/7! + ... - .... + ...... ]

I can factor a (2π/S) out of the sum.

r2 ([STRIKE]S[/STRIKE]/[STRIKE]2[/STRIKE])([STRIKE]2[/STRIKE]π/[STRIKE]S[/STRIKE]) [ 1 - (2π/S)2/3! + (2π/3)5/4! - (2π/3)6/7! + ... - .... + ...... ].

As you can see, the sum now looks sort of like the infinite series for cosine. Let me pull out a [1 + 1/3 + 1/5 + 1/7 + ....]

r2π cos(2π/S) [1 + 1/3 + 1/5 + 1/7 + .......].

cos(2π/S)--->1 as s--->∞.

This is where I'm stuck. I won't get πr2 unless [1 + 1/3 + 1/5 + 1/7 + .......] is equal to 1, which it obviously isn't.

Last edited by a moderator: May 27, 2010
2. May 27, 2010

Jamin2112

Re: Constructive criticism for you folks

Correction: In the figure I drew S is supposed to be 8, not 5.

3. May 27, 2010

tiny-tim

Hi Jamin2112!

You can't do that!!

that's something like Aa + Bb + Cc + … = (A + B + C + … )(a + b + c + …)

Anyway, "the ancient way" didn't include Taylor series, or series approximations for π.

You need to use the ancient definition of π, which is the limit of the perimeter of that polygon (divided by r) …

go for a relation between the area and the perimeter.

Last edited by a moderator: May 27, 2010
4. May 27, 2010

Jamin2112

Re: Constructive criticism for you folks

You mean where you cut the the circle up into an infinite amount of wedges and line the wedges up next to each other, one half facing the other half? That's sort of lame because you have to start with the assumption the the perimeter of a circle is 2πr.

I refurnished my method.

The area of any S-sided polygon inscribed inside a circle is (S/2)r2sin(2π/S).

This can be expanded using the infinite series sin(x)=∑(-1)nx2n+1/(2n+1)!.

Area= (S/2)r2[(2π/S) - (2π/S)3/3! + (2π/S)5/5! - (2π/S)7/7! + ... - ... + ......]

= πr2 [ 1 - (2π/S)2/3! + (2π/S)4/4! - (2π/S)6/6! + ... - ... + .......].

Area of circle with radius r = limS-->∞ πr2 [ 1 - (2π/S)2/3! + (2π/S)4/4! - (2π/S)6/6! + ... - ... + .......] = πr2.

Booooom! Derived that completely by myself.

Last edited by a moderator: May 27, 2010
5. May 27, 2010

Jamin2112

Re: Constructive criticism for you folks

And I figured out why 0! = 1.

n! = n*(n-1)*(n-2)*(.........)*2*1.

------>

5! = 5*4*3*2*1 = 5*4!
4! = 4*3*2*1 = 4*3!
........

-----> n! = (n+1)! / n

-----> 0! = (0+1)! / 1 = 1! / 1 = 1

6. May 27, 2010

Redbelly98

Staff Emeritus
Moderator's note: posts moved from thread in Forum Feedback. Edited to remove off-topic comments.

7. May 28, 2010

tiny-tim

Well, no … you had the help of Taylor and others! (I don't think he was Greek! ) …

but anyway your proof didn't need the Taylor series, all you've used is that limx->0sinx = x And that's essentially the same as the following method …
Assumption? That's the definition of π.

How else would you define π?
Nooo …… that would be (0+1)! / 0

Try again!

8. May 28, 2010

Jamin2112

Back to what I was saying about the perimeter thing. Here's a drawing of a circle cut into 8 identical wedges.

It sort of looks like a parallelogram with one a side of length r and another side of length πr, though that second side isn't flat---so we can't just do length * width. As you increase the number of wedges, however, the side flattens out. A figure of this type with an infinite amount of infinitely thin wedges will have area πr*r=πr2.

9. May 28, 2010

tiny-tim

Yup!

10. May 28, 2010

Jamin2112

Let me explain again.

8! = 8*7*6*5*4*3*2*1 = 8*7!
7! = 7*6*5*4*3*2*1 = 7*6!

.
.
.
.
.
.
.
.

----> (n+1)! = (n+1)*n!

n = 0 ----> 1! = 1 * 0! ----> 0! = 1! / 1 = 1.

11. May 28, 2010

Jamin2112

Next I'll try to do find the formula for the nth number of the Fibonacci sequence. I remember learning it in two different classes. In MATH 300 we learned some way that gave us a crazy formula involving √5. In AMATH 301 we learned a way involving matrix multiplication; something like (xn xn-1)T= A * (1 0)T, I think.

12. May 29, 2010

tiny-tim

That's better!