Area of circle using incribed polygons

In summary: I think you meant 1 * 1! ----> 1! = 1 * 0! ----> 0! = 1! / 1 = 1.)So …n! = (n+1)! / nn = 1 ----> 2! = 2 * 1! ----> 1! = 2! / 2 = 1n = 2 ----> 3! = 3 * 2! ----> 2! = 3! / 3 = 2n = 3 ----> 4! = 4 * 3! ----> 3! = 4! / 4 =
  • #1
Jamin2112
986
12
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.

screen-capture.png


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.

Some help, please?
 
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  • #2


Correction: In the figure I drew S is supposed to be 8, not 5.
 
  • #3
Jamin2112 said:
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! :smile:

You can't do that!

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

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

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. :smile:
 
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  • #4


tiny-tim said:
Hi Jamin2112! :smile:

You can't do that!

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

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

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. :smile:
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.
 
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  • #5


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
Moderator's note: posts moved from thread in Forum Feedback. Edited to remove off-topic comments.
 
  • #7
Jamin2112 said:
Booooom! Derived that completely by myself.

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

but anyway your proof didn't need the Taylor series, all you've used is that limx->0sinx = x :wink: 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? :smile:
Jamin2112 said:
-----> n! = (n+1)! / n

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

Nooo …… that would be (0+1)! / 0 :redface:

Try again! :smile:
 
  • #8
Back to what I was saying about the perimeter thing. Here's a drawing of a circle cut into 8 identical wedges.

screen-capture-2-1.png


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
Yup! :biggrin:

Now how about 0! ? :wink:
 
  • #10
tiny-tim said:
Yup! :biggrin:

Now how about 0! ? :wink:

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
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
Jamin2112 said:
-----> n! = (n+1)! / n

-----> 0! = (0+1)! / 1 = 1! / 1 = 1
Jamin2112 said:
Let me explain again.

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

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

That's better! :biggrin:
 

1. What is the formula for finding the area of a circle using inscribed polygons?

The formula for finding the area of a circle using inscribed polygons is A = (r^2 * sin(360/n)) * n, where r is the radius of the circle and n is the number of sides of the inscribed polygon.

2. How does using inscribed polygons help in finding the area of a circle?

Using inscribed polygons helps in finding the area of a circle by breaking down the circle into smaller and more manageable shapes, making it easier to calculate the area. It also provides a more accurate approximation of the circle's area.

3. Can the number of sides of the inscribed polygon affect the accuracy of the area calculation?

Yes, the more sides the inscribed polygon has, the closer the calculated area will be to the actual area of the circle. As the number of sides increases, the inscribed polygon more closely resembles a circle, resulting in a more accurate calculation.

4. Are there any limitations to using inscribed polygons to find the area of a circle?

One limitation of using inscribed polygons is that it can only provide an approximation of the circle's area, and not the exact value. Additionally, the accuracy of the calculation depends on the number of sides of the inscribed polygon, so a larger number of sides may be needed for more precise results.

5. How is the area of a circle using inscribed polygons beneficial in real-life applications?

The concept of finding the area of a circle using inscribed polygons is used in various fields such as engineering, architecture, and physics. It allows for more accurate calculations of curved surfaces and can also be used in computer graphics to create smooth and realistic shapes.

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