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| Title: | Mathematics at DEC |
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| Moderator: | RUSURE::EDP |
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| 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 |
1612.0. "Some things that need saying" by CIV009::LYNN (Lynn Yarbrough @WNP DTN 427-5663) Mon May 18 1992 19:01
Two years ago we made a move from Lexington, MA to Washington, DC, and it
was a difficult experience saying goodbye to over a hundred business
associates, church members, neighbors, bridge partners, whatever - and I am
sure we failed to say goodbye to a few people who, for whatever reason, we
simply failed to run into before we left. Now I am retiring from DEC, as of
the end of this month, and faced with having to go through the same process
with another several hundred people whom I know "electronically" as well as
personally. It is not something I enjoy.
The community of mathematicians, salespeople, computer scientists, and
whatever that I now leave behind has been amazingly enriching to me, and I
hope I have made some positive impact on a few of you. I value you all as
friends - even though most of us have never actually met. I hope that there
will in the future be some way that we can continue to communicate about
some of the issues we have dealt with here - I hope to remain active. Yes,
I have a copy of MAPLE on my PC at home!
I am still working on the "Math Notes" book, with Stan Rabinowitz. He has
been very encouraging, although I have not been able to take the time and
energy to go ahead very fast. Unless I take another job quickly, I should
have time now to finish it - I have transcribed most of the notes to my PC,
too.
Special thanks to Topher Cooper, who has agreed to take over the MAPLE
conference. (It will probably be renamed to indicate a broadened scope
including all symbolic math tools.)
My warmest regards and best wishes to you all! Please keep in touch.
Now, as my parting shot, the Yarbrough Memorial Puzzle:
Suppose you have a *large* sheet of quadrille paper (i.e. ruled into 1/4"
squares); what is the smallest circle that can be drawn on that sheet so
that at least 1,000,000 complete 1/4" squares are inside the circle?
Have fun,
Lynn Yarbrough
| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1612.1 | | DESIR::BUCHANAN | | Wed May 20 1992 14:10 | 21 |
| 1612.2 | | TRACE::GILBERT | Ownership Obligates | Thu May 21 1992 16:25 | 13 |
| 1612.3 | | GRANPA::BAYOUNG | Really CHOVAX::YOUNG... | Sun May 24 1992 20:59 | 4 |
| I for one will miss you greatly, Lynn. Any Will you be staying in the
DC area, or moving elsewhere?
-- Barry
|
| 1612.4 | Ooops - forgot to leave an address | CIV009::LYNN | Lynn Yarbrough @WNP DTN 427-5663 | Sun May 24 1992 21:34 | 12 |
| I can be reached, for the indefinite future, at
Lynn Yarbrough
931 Park St. SE
Vienna, VA 22180
(703)938-4019
Of course, if & when we move again later (likely in a few years since DC is
an expensive place to retire) I will inform all I can manage to.
Thanks,
Lynn
|
| 1612.5 | Series Approach | FASDER::MTURNER | Mark Turner * DTN 425-3702 * MEL4 | Fri May 29 1992 21:30 | 83 |
| Assumptions:
. circle centered on "+"
. solution for 1st quadrant is sufficient
. "div" = division with truncation
Consider the radius of length R sweeping clockwise from the vertical. As it
intersects each vertical quadrille line "Q", we'll determine how many squares
are completely covered in the column bordered on the right by Q.
At the intersection with the first line, the angle formed by R and the x-axis
is
1) theta(1) = arccos(0.25 / R)
= arccos(1 / 4R)
and the altitude ("opposite side") of the right triangle is
2) b(1) = sin(theta(1)) * R
How many blocks will be *completely* covered in the first column of squares
(once the radius has swept through the whole quadrant)?
3) blocks(1) = b(1) div 0.25
= int(4 * b(1))
Continuing the clockwise sweep, at the second quadrille line we find:
4) theta(2) = arccos(0.5 / R)
= arccos(1 / 2R)
5) b(2) = sin(theta(2)) * R
6) blocks(2) = b(2) div 0.25
= int(4 * b(2))
So the total number of blocks is the series:
n
----
7) \
/ int(4 * b(n))
----
i=1
But what's "n", i.e. how many vertical lines does R intersect in its sweep (at
least: how many which have at least one completely covered block to their
left)? The answer is simply:
8) blocks(1)
which we can visualize by pivotting the diagram around the 45-degree line.
So the series is
blocks(1)
----
9) \
/ int(4 * b(n))
----
i=1
where blocks(1) = int(4 * (sin(arccos(1/4R)) * R))
or
blocks(1)
----
10) \
/ int(4 * (sin(arccos(i/4R)) * R))
----
i=1
So... if this is right, then all we have to do is find the smallest R making
(10) equal to 250,000 (going back to the assumption that one quadrant
suffices). It doesn't look easy, and I wonder if:
a. I've overlooked an obvious simplification of the trig expressions,
b. there's some quick iterative method that will find the R for us.
Mark
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| 1612.6 | | TRACE::GILBERT | Ownership Obligates | Sat May 30 1992 03:26 | 44
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