3 ms·

I think he is not asking what he wants to know. He wants to know how we know why Archimedes' solution does converge to the right length. Couldn't there be a pro

• caf 16y ago
  It's because the tangents to Archimedes' polygons also approach the tangents to the circle. (This is the "velocity vector" convergence condition mentioned by cousin_it above).
  • jules 16y ago
    That doesn't explain why. Why is it that if tangents approach the circle (whatever that means precisely) that the length of the approximation approaches the length of the circle?
    • caf 16y ago
      Loosely speaking, the total path length is defined as the sum of a bunch of infintesimal segments that have both position and direction. For the sum (integral) of those infintesimals to approach the sum of the similar infintisemals around the actual circle, those of the approximation path have to approach those of the circle in the limit, and of course this means in both position and direction.
• sesqu 16y ago
  Here's a thought: Archimedes's approximation approaches the target. That is, the perimeter of each subsequent approximation is closer to that of the circle than all that came before. The side of a regular polygon is √(sin²(τ/n)+(1−cos(τ/n))²)=√(2(1−cos(τ/n))), while the corresponding arc is τ/n. d/dn τ/n−√(2(1−cos(τ/n))) = −(τ/n²+sin(τ/n)/√(2(1−cos(τ/n)))) < 0, so the difference is strictly decreasing (note that sin(τ/n)>0 when n>2). So there's your lower bound for the difference. You'd still have to prove the limit is at 0, though.