Conference rusure::math

>(1) Prove there is a non-zero limit to the sequence r(n) of radii of the
>circles.
>
>(2) Find that limit. (I can't do this one)

A little geometry reveals that the sequence of r's has the recursion formula
	r   = r *cos(Pi/(n+1))
	 n+1   n
and the cos term tends rapidly toward 1 as n grows. (No, that's not a
proof.) Starting at r = 1, n = 3, r appears to converge fairly slowly to
.1149477172, whatever that is. 

Lynn 

	(1) Prove there is a non-zero limit to the sequence r(n) of radii of the
circles.

	The sequence is product(n=3..infinity) cos(pi/n).

	If I have sequence {x_n} of non-negative numbers, none = 1, such that 
sum(n=1..infinity) x_n exists, then product(n=1..infinity) (1-x_n) exists and
is non-zero.

	To see this pick N such that sum(n=N..infinity) x_n < �.
	Now p(N') = product(n=N..N') (1-x_n) >= 1 - sum(n=N..N') x_n > �.
	So p(N') is a monotonically decreasing function of N', and is bounded
below, by �, so there is a non-zero limit.   So product(n=1..infinity) (1-x_n)
exists and since no x_n from 1 to N-1 takes the value 1, the limit is non-zero.

	In the case here, x_n = 1 - cos(pi/n).   Now x�/2 < cos x < 1, so 
0 < sum(n=3..infinity) x_n < sum(n=3..infinity) (pi/n)�/2 
= (pi�/2).(pi�/6 - 5/4).   So we have satisfied the conditions, and know that

	product(n=3..infinity) cos(pi/n) = P > 0.

	(2) Find that limit.

	Lynn's numerical estimate is nice to know.   I've tried mucking about
with trig identities, but the fact that the n is in the denominator and not
the numerator has prevented me from finding anything interesting.   No
progress.

	(3) What about circumscription instead of inscription?

	Basically, assuming that one starts the inscription and circumscriptions
with the unit circle, the answer is 1/p.

    I doubt if this is any help, but you could get rid of PI in the
    expression by using the product expansion for COS and get

	prod (k >= 3) cos(pi/k)

	prod(k >= 3, n >= 1) (1-(2/((2*n-1)*k))^2)

    Perhaps rearranging this could give a clue as to its closed form.

    By the way, the estimate in .1 seems to be in some error - a closer
    value is

	0.11494204485329620070104015746959876...

    transcendental no doubt :-)

    - Jim

>    By the way, the estimate in .1 seems to be in some error - a closer
>    value is
>
>	0.11494204485329620070104015746959876...

Without knowing how many repetitions of the recursion you used (or did
you use some other method?) I wouldn't trust the accuracy of this result
regardless of the precision to which it is stated. I used 100,000 to get my 
estimate in .1 - how many did you use? Do the last two iterations agree to
all the digits shown above?

Lynn 

Here's what I got from my program.

                    M-1
       M          Product cos(Pi/k)
                   k = 3

    100000  1.1494771718417489374826257088622E-0001
    200000  1.1494488097665505341598383433579E-0001
    300000  1.1494393559285103379173237009919E-0001
    400000  1.1494346290445569614185054107113E-0001
    500000  1.1494317929254062042907630268673E-0001
    600000  1.1494299021839812091782317595503E-0001
    700000  1.1494285516566819665052428361184E-0001
    800000  1.1494275387624598975875102186782E-0001
    900000  1.1494267509565848729562132774335E-0001
   1000000  1.1494261207123524010141823594241E-0001

To get better accuracy, the program actually calculates:

              M-1
      sqrt[ Product (1 - sin�(Pi/k)) ]
             k = 3

And within the loop, the calculation is as follows, for better accuracy.

      Y'  <-  Y  -  Y * sin�(Pi/k)

After M iterations, the result may be off by about M/2 least-significant bits.
The actual error is probably much less than this.

    Re .3

    It isn't difficult to accelerate the convergence of a sequence
    like this.  The ratio of the differences between successive doublings
    of the number of iterations approaches 2:1, as can be seen from the data
    in .5

	p0(2k)-p0(k)
	------------- --> 2
	p0(4k)-p0(2k)

    So you may assume this is true and extrapolate to get a new sequence
    p1.  But why stop here?  The ratio of differnces of the p1's approaches
    4:1, so you can recursively repeat the pattern to extrapolate to the limit
    quite quickly (within the limits of roundoff error and so on.)

    This is in need of analytic justification, but in many cases, like
    doing a hairy integral or finite element mesh refinement numerical
    evidence is all you may have to go on (or have time to get).

    - Jim

    Here is a more rigorous way to approximate the product in .1 for the
    skeptics here.

    Integrate the Taylor series for tan(z) term by term to get a series
    for log(cos(pi/k)) about infinity:
			
				           (2PI)^2n*B[2n]
	ln(cos(pi/k)) = SUM  (-)^n*(2^2n-1)-------------- k^(-2n)
			n > 0		      2n*(2n)!

	B[2n] = Bernoulli numbers: 1/6, -1/30, 1/42, -1/30, etc.

    Now taking the product is the same as summing the series.  Convergence
    isn't any quicker, but you can always integrate the series *again*
    and use this integral (which can be computed by our rapidly convergent
    series) as an approximation to the sum itself.

    Think of the series as applying the trapezoidal rule to the integral.

    Since the error comitted in applying the trapezoidal rule is expressable
    in terms of an asymptotic expansion in the odd derivatives of the
    integrand at the endpoints this can be turned around to make an
    excellent approximation to a slowly convergent series:

	f(a)/2 + f(a+1) + ... + f(b-1) + f(b)/2 = Trapezoidal rule ~=

				 B[2k]     (2k-1)     (2k-1)
	INTEGRAL  f(x)dx + SUM   -----(f(b)      -f(a)      )
	a < x < b	   k > 1 (2k)!

	B[2k] = Bernoulli numbers (again!)

    Since this approximation improves going into the tail of the series
    it's usual to sum the first few terms by hand and add in the correction
    for the tail.

    This can give some pretty amazing results - taking the product for
    3 <= k < 10 directly (a mere 7 terms) and applying a first derivative
    correction to the tail gives 0.1149418 already
    (should be 0.1149420448532962...)

    - Jim