T.R | Title | User | Personal Name | Date | Lines |
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484.1 | | CLT::GILBERT | Juggler of Noterdom | Tue May 13 1986 19:52 | 1 |
| This is a well-known problem with no simple (closed form) solution.
|
484.2 | Oh yes there is! | ZEPPO::DAY | | Wed May 14 1986 09:49 | 2 |
| Re .-1 It does have a closed form solution. - Bob Day
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484.3 | use integration | SIERRA::OSMAN | and silos to fill before I feep, and silos to fill before I feep | Thu May 15 1986 14:52 | 11 |
| I don't recall my integral claculus, but it seems to me that finding
the area of a region formed by the overlap of two circles was a
closed form expression of the equations of the two circles.
One circle's radius is given as 1, and the other we can call R.
Write down the expression in terms of R and set it equal to
one half of pi (the area of circle A), and solve for R.
Does this help ?
/Eric
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484.4 | solution to .0 and also to 496.0 | THEBUS::KOSTAS | Kostas G. Gavrielidis <o.o> | Tue Jun 03 1986 10:53 | 51 |
| The difficulty in solving this problem arises from the fact that
it can be solved only by successive approximations. Values should
not be tried at random, but one should be quided by systematic
interpolation. This is a good example of the use of the Regula
Falsa.
Half the area of the unit circle is pi/2 = 1.5708
Let G be the corresponding amount of area formed by the overlap
of the two circles.
if R = 1.15 then the area G = 1.5513 which is 0.0195 too small
if R = 1.16 then the area G = 1.5740 which is 0.0032 too large
after more interpolation
we get R = 1.1588 and G = 1.5709
Let see
R = 1.1588
sin x/2 = .5794
x/2 = 35.41 degrees
y = 54.59 degrees
x = 70.82 degrees
log R = 0.0640
log R^2 = 0.1280
log rad y = 9.9790-10
----------
0.1070
1.2793
rad x = 1.2217
+ 143
------
2.5153
sin x = .9444
------
G = 1.5709
Note: Some of the computations are not included in here also
no pictures were provided.
Enjoy,
Kostas G.
|
484.5 | Solution... | ZEPPO::DAY | | Wed Jun 11 1986 10:57 | 11 |
|
The answer is given by the equation:
sin(theta) - theta * cos(theta) - (pi/2) = 0
Theta is the angle between the center of circle B and the points
where circle B intersects circle A. Solve the equation for the
value of theta that is between 0 and PI (the solution that makes
physical sense). Then the radius = 2 * cos(theta/2).
|
484.6 | History? | STAR::BRANDENBERG | Civilization is the progress toward a society of privacy. | Wed Jun 11 1986 16:21 | 6 |
|
I'm curious where the question comes from. I know it from a math
competition in High School ( I don't remember if I solved it ).
Monty
|
484.7 | re .6 | ZEPPO::DAY | | Wed Jun 11 1986 17:04 | 8 |
|
I heard it about 10 years ago from a friend. He said that the
only solution that exists is a series solution. The equation
given in .5 doesn't seem to be well known... I'd also like to
know more about the history and solutions to this problem.
|
484.8 | | CLT::GILBERT | Juggler of Noterdom | Wed Jun 11 1986 19:28 | 5 |
| re .2,.5
The problem is that there is no closed form solution for theta.
That is, no equation: theta = some `simple' expression.
|
484.9 | Here are the numbers... | MODEL::YARBROUGH | | Thu Jun 12 1986 16:11 | 2 |
| Re .5: the principle value of the solution is theta = 1.9056957293...,
and r = 1.1587284730... - in case anyone is interested.
|
484.10 | meaning of 'closed form' | ISWISS::DAY | | Fri Jul 25 1986 12:15 | 13 |
| re .1, .8
I checked in a number of mathematical dictionaries and encyclopedias
to try to find a definition of 'closed form solution', and it did not
appear in any of the references I checked. So I guess it's one of those
terms that, to some degree, can mean what one means it to mean.
My understanding is that a closed form solution is one that does not
require a series to express it.
Be that as it may, 'explicit' is the term that is usually applied to a
solution in which the independent variable is isolated on the left
side of an equation, and the independent variables are on the right.
|
484.11 | that reminds me of an anecdote | MODEL::YARBROUGH | | Fri Jul 25 1986 14:21 | 4 |
| I knew of a Math prof. in college with a name that sounded like
"Trijinski" but had several more letters in it. Once, when a Graduate
student asked him to spell his name, he replied, " You can't spell
'Trijinsky' in closed form; you must expand it in a Fourier series."
|