| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1466.1 | I found such an operator ! | SMAUG::ABBASI | | Sun Jun 30 1991 05:52 | 4 |
| Yes, in the non-Hyber squared Riemann analytic mutli-valued special
exponential algebra (under the condition it is asymtotic close algebra)
just kidding :-)
|
| 1466.2 | wondering ... | HPSRAD::ABIDI | It's a wild world | Mon Jul 01 1991 13:32 | 3 |
|
Is it possible to have a set of points that are all equidistant from
each other ? In 2-D? In 3-D ?
|
| 1466.3 | | GUESS::DERAMO | duly noted | Mon Jul 01 1991 13:44 | 9 |
| The three vertices of an equilateral triangle in R^2, the
four vertexes of a regular tetrahedron in R^3, etc.
Visualize the first two points as the centers of
intersecting circles/spheres/n-spheres, and then add each
new point as the center of another n-sphere of the same
size.
Dan
|
| 1466.4 | Finite sets | CADSYS::COOPER | Topher Cooper | Mon Jul 01 1991 13:45 | 18 |
| RE: .2
In N dimensions you can have a set of N+1 points equidistance from
each other.
0-Dimension: degenerate (but it works)
1-Dimension: the two end-points of a line segement are equidistance
from each other
2-Dimension: the three vertexes of an equilateral triangle are
equidistance from each other
3-Dimension: the four vertexes of a regular tetrahedron are
equidistance from each other
etc.
The shape in N dimensions containing N+1 pairwise equidistant points
is sometimes called the regular N-Simplex.
Topher
|
| 1466.5 | notes collision :-) | GUESS::DERAMO | duly noted | Mon Jul 01 1991 13:46 | 0 |
| 1466.6 | | CLT::TRACE::GILBERT | Ownership Obligates | Mon Jul 01 1991 14:22 | 4 |
| I recall entering an interesting problem concerning the placement of points
in a plane such that all distances between the points are intergral.
Or was it that all distances were either 1, 2, or 3?
|
| 1466.7 | | ELIS::GARSON | V+F = E+2 | Tue Jul 02 1991 06:59 | 1 |
| And for a similar problem see the first problem in 287.0
|
| 1466.8 | | ALLVAX::JROTH | I know he moves along the piers | Tue Jul 02 1991 07:53 | 12 |
| <<< Note 1466.0 by VINO::XIA "In my beginning is my end." >>>
2
> Does there exist an operator A such that A = d/dx ?
It might be possible to define such an operator with Laplace transforms.
Fractional derivatives can be defined that way.
Also, you can sometimes loosely think of linear operators as
generalizations to continuously many dimensions of matrices transforming
vectors - that may give a clue as to how to construct such an operator.
- Jim
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