I have come out with my own equation(I have no idea whether it is new) for pie : π = √2/2 x n x √((1-cos(dΘ)), where n is the number of triangles in the circle, and dΘ is the angle of the triangle that is -->0. Ok, now, say, I put n = 1440, so dΘ will be 360/1440 = 0.25, and put it into the equation, i will get π= 3.141590118…; if I put n=2880, so dΘ will be 0.125, putting it into the equation, I will get π=3.141591603.., if I put n=5760, so dΘ will be 0.0625, putting it into the equation, i will get π= 3.141592923..., we know π= 3.141592654..,but i can never really get the n and dΘ to give that answer. Anyone can come out with some good idea??
Dear Sir,
- You want to calculate following limit:
lim (n -> infinity) (sqrt(2)/2*n*sqrt(1-cos((2*PI)/n))
- put k = 1/n
lim (k -> 0) (sqrt(2)/2 * sqrt(1-cos(2*PI*k)/k
- To calculate the limit, we replace the cos by the first 2 terms of its Maclorin series
cos(x) = 1- x*x/2 + x**4/4! - .. + ..
lim(k -> 0) (sqrt(2)/2 * sqrt(1 - (1 - 2*PI*2*PI*k*k/2)) / k
lim(k -> 0) (sqrt(2)/2 * sqrt((2*PI*k)**2/2)/k
lim(k->0) (sqrt(2)/2 * 2*PI *k / sqrt(2) / k
= PI
-> So, it is indeed true. The limit goes to PI. Note that this is not a practical way to calculate PI, however.
-> You can calculate the error by also including the third term in the Maclorin series. The third term is the major source of the error
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