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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1086.0. "one for you analysis weenies..." by HERON::BUCHANAN (Andrew @vbo DTN 828-5805) Fri May 26 1989 10:28

	Draw an equilateral triangle
		Within the triangle, inscribe a circle
	Within the circle, inscribe a square
		Within the square, inscribe a circle
	Within the circle, inscribe a regular pentagon
		Within the pentagon, inscribe a circle...


	(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)

	(3) What about circumscription instead of inscription? (It's just
occured to me.)

Andrew
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1086.1An approach...NIZIAK::YARBROUGHI PREFER PIFri May 26 1989 13:1613
>(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 
1086.2HERON::BUCHANANAndrew @vbo DTN 828-5805Mon Jun 05 1989 11:2932
1086.3MECAD::ROTHIf you plant ice you'll harvest windMon Jun 05 1989 17:3517
    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
1086.4Precise enough, but how accurate?NIZIAK::YARBROUGHI PREFER PIMon Jun 12 1989 14:1012
>    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 
1086.5many more iterations are needed4GL::GILBERTOwnership ObligatesMon Jun 12 1989 21:1929
1086.6extrapolation to the limitMYVAX::ROTHIf you plant ice you'll harvest windWed Jun 14 1989 15:5021
    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
1086.7Yooler-Maclurin summationMYVAX::ROTHIf you plant ice you'll harvest windWed Jun 14 1989 15:5642
    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
1086.8You mean it's *NOT* 10/87?POOL::HALLYBThe Smart Money was on GoliathWed Jun 14 1989 16:491