- #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.
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?
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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