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 Mean Diagonals (Posted on 2018-11-23)
Let Mn be the arithmetic mean of the lengths of the diagonals of regular n-gon with a circumradius of π(pi).

Compute limit of Mn as n tends to infinity.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 soln by integration Comment 4 of 4 |
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
 Posted by Steven Lord on 2018-11-25 12:17:45

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