- #1
mjordan2nd
- 177
- 1
If you don't know what trollpi is, you can see it here:
http://qntm.org/files/trollpi/piequals4.png [1]
It seems to me that the flaw in this problem is that while the shape may converge to a circle, the way you measure the perimeter is ds=dx+dy=1+(dy/dx) not ds=sqrt(1+(dy/dx)^2 )dx as usual. So I tried using this new definition of the arclength in the first quadrant with r=0.5 and integrating expecting to get 1. I put the following into mathematica:
Integrate[1 - x/Sqrt[0.25 - x^2], {x, 0, 0.5}]
This did not give me 1. Can someone explain why?
http://qntm.org/files/trollpi/piequals4.png [1]
It seems to me that the flaw in this problem is that while the shape may converge to a circle, the way you measure the perimeter is ds=dx+dy=1+(dy/dx) not ds=sqrt(1+(dy/dx)^2 )dx as usual. So I tried using this new definition of the arclength in the first quadrant with r=0.5 and integrating expecting to get 1. I put the following into mathematica:
Integrate[1 - x/Sqrt[0.25 - x^2], {x, 0, 0.5}]
This did not give me 1. Can someone explain why?
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