T.R | Title | User | Personal Name | Date | Lines |
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1086.1 | An approach... | NIZIAK::YARBROUGH | I PREFER PI | Fri May 26 1989 13:16 | 13 |
| >(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
|
1086.2 | | HERON::BUCHANAN | Andrew @vbo DTN 828-5805 | Mon Jun 05 1989 11:29 | 32 |
1086.3 | | MECAD::ROTH | If you plant ice you'll harvest wind | Mon Jun 05 1989 17:35 | 17 |
| 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
|
1086.4 | Precise enough, but how accurate? | NIZIAK::YARBROUGH | I PREFER PI | Mon Jun 12 1989 14:10 | 12 |
| > 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
|
1086.5 | many more iterations are needed | 4GL::GILBERT | Ownership Obligates | Mon Jun 12 1989 21:19 | 29 |
1086.6 | extrapolation to the limit | MYVAX::ROTH | If you plant ice you'll harvest wind | Wed Jun 14 1989 15:50 | 21 |
| 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
|
1086.7 | Yooler-Maclurin summation | MYVAX::ROTH | If you plant ice you'll harvest wind | Wed Jun 14 1989 15:56 | 42 |
| 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
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1086.8 | You mean it's *NOT* 10/87? | POOL::HALLYB | The Smart Money was on Goliath | Wed Jun 14 1989 16:49 | 1 |
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