1st a note: the method employed by myself and broll are the same (his k is my i+1). They differed in that I averaged only unique lengths and threw in the radius: n pi, while broll used a unit circle and successfully evaluated the result for the unit circle to be 4/pi in the limit. (I had hoped for an identity for the series limit.... I bet there is one with a corresponding proof).
2nd note: I quit with the series because I figured Charlie had the better idea: in the limit the average becomes a polar coordinate integral. Indeed, here it is:
Put circle of radius r centered at the origin and place a point A at r and theta=0. Draw a diagonal D from A to any point B in the first quadrant.
By connecting B to the origin, you have made an isosceles triangle whose base is D. Bisecting the base yields D(theta) = 2 r sin(theta/2)
I will call theta "t" from now on.
Now, average D from 0 to pi. (From pi to 2 pi the result is symmetrically the same.)
<D> = Int{0 to pi} (2 r sin (t/2) dt) / Int{0 to pi} dt
Use sin(t/2) = sqrt[(1-cos t)/2]
<D> = sqrt (2) (r/pi) Int{t=0 to pi} sqrt (1-cos t) dt
mult by sqrt(1+cos t) / sqrt(1+cos t)
<D> = sqrt (2) (r/pi) Int{t=0 to pi} sin t / sqrt (1 + cos t) dt
With U = 1+ cos t, dU = - 1 sin t dt, and the limits on U are
(1 + cos 0 ) to (1 + cos pi ) = 2 to 0. We change the sign of the integral and flip the limits:
<D> = sqrt(2) (r/pi) Int{0 to 2} U^(- 1/2) dU =
<D> = sqrt(2) (r/pi) 2 U^(1/2)|0 to 2
<D> = sqrt(2) (r/pi) 2 sqrt(2)= 4 r /pi (broll's result)
Putting in r = n pi
<D> = 4 n
Edited on November 29, 2018, 2:21 am
soln by integration
