Start with a circle of some radius - call it Ro.
Inscribe a triangle inside this circle. Now inscribe a new circle inside the triangle. Inside
the new circle, inscribe a square. Inside the square, inscribe another circle. Inside that
circle, inscribe a pentagon. Etc. Etc.
The question is: does this progression converge to a nonzero radius? Or does it just keep
shrinking to zero?
Here's my stab at the problem:
The radius of the nth circle goes like:
Rn = Ro * product[m=3,n] ( cos(pi/m) )
So as n goes to infinity:
Rinf = Ro * product[m=3,inf] ( cos(pi/m) )
If we take Ro = 1, then the computer shows a
convergence to Rinf = 0.11499878...
To get a cheap approximation to this number, I took
the natural logarithm of both sides of the Rinf
equation:
ln(Rinf) = ln(Ro) + sum[m=3,inf] ( ln(cos(pi/m)) )
Let's be lazy bastards and say that ln(x) = x-1 if
we're really close to x=1. Let's also set Ro=1 to get
rid of that term. Now we have:
ln(Rinf) = sum[m=3,inf] ( cos(pi/m) - 1 )
Let's get even cheaper and say that cos(pi/m) = 1 -
0.5*(pi/m)^2. We're left with:
ln(Rinf) = sum[m=3,inf] ( -0.5*(pi/m)^2 )
That's just a harmonic series of power 2.
sum[m=1,inf] ( m^-2 ) = pi^2 / 6, which is zeta(2).
Three cheers for the Riemann Zeta Function.
Our approximation is now:
ln(Rinf) = -0.5*(pi^2)*( (pi^2)/6 - 1.0 - 0.25)
We have to subtract out the m=1 and m=2 terms from the
sum because we started with a triangle at m=3. That
leaves:
ln(Rinf) = -1.9489...
Rinf = 0.1424...
That's not too far from Rinf = 0.115...
Is there a closed form for sum( ln(cos(pi/m)) )? Anyone?