Area of circle using incribed polygons

[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 lim_S-->∞(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.

Some help, please?

 

[Jamin2112]

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

 

[tiny-tim]

I was trying to find the area of a circle the ancient way.
…
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 + ...].

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. ​

 

[Jamin2112]

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. ​

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.

 

[Jamin2112]

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

 

[Redbelly98]

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

 

[tiny-tim]

Booooom! Derived that completely by myself.

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 lim_x->0sinx = x And that's essentially the same as the following method …

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.

Assumption? That's the definition of π.

How else would you define π? ​

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

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

Nooo …… that would be (0+1)! / 0

Try again! ​

 

[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.

 

[tiny-tim]

Yup!

Now how about 0! ?

 

[Jamin2112]

Yup!

Now how about 0! ?

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!

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

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

 

[Jamin2112]

Next I'll try to do find the formula for the n^th 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 (x_n x_n-1)_T= A * (1 0)^T, I think.

 

[tiny-tim]

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

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

Let me explain again.
…
----> (n+1)! = (n+1)*n!

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

That's better!