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Volume 4, Number 4A  -  January 1999 Discussions

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27 January 1999                                 Vol. 4, No. 4A


  This special issue of the Math Forum's weekly newsletter
  highlights interesting conversations taking place during
  January of 1999 on Internet math discussion groups.

  For a full list of these groups with links to topics covered
  and information on how to subscribe, see:


  Replies to individual discussions should be addressed to
  the appropriate group rather than to the newsletter editor.

______________________________ + ______________________________

                     JANUARY SUGGESTIONS:  

CALC-REFORM - a mailing list hosted by e-MATH of the 
American Mathematical Society (AMS) and archived at
- Revisionist Simpson's Rule (10 Jan. 1999)

  "Some recent calculus books approach Simpson's rule by
   looking at the ratio: (E-T)/(E-M) where E is the exact area
   under some curve and T, M are the trapezoidal and midpoint
   estimates respectively. Experimentally one sees that this
   ratio is (usually) very close to -2. Setting it equal to -2
   and solving for E gives: E = (2M+T)/3. Thus, the exact answer
   is very nearly a certain weighted average of the trapezoidal
   and midpoint rules. Writing this out explicitly results in
   Simpson's formula. 

   When I first saw this I thought it was kind of neat; I even 
   had my students 'discover' it using a spreadsheet lab to 
   experiment with the ratio. However, after thinking about it 
   for a while, I decided that, yes, it was cute, but perhaps 
   a bit too cute.... I have decided to go back to the 
   straightforward motivation that Simpson used, and give the 
   error ratio guess *afterward*, as an interesting computer 
   lab: I didn't feel right about having the students 'discover' 
   it. What do you think?" - Mark Bridger
  "Doug Kuhlmann asked about Richardson Extrapolation, so I'll
   sketch the main idea...." - David A. Olson

 The conversation continued, spreading to:

- (E-T)/(E-M) (11 Jan. 1999)
- Elegance is not the issue (11 Jan. 1999)

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- Tori...Torae composed of straight lines (9 Jan. 1999)

  "Anyone know where I can find a picture of a one-holed torus
   defined by the minimum number of straight lines? I guess the
   lines would have to form a minimum of seven planar polygons,
   but the polygons would not have to be convex." - Bob Underwood
  "The example I know of with seven faces has seven (non-convex)
   hexagons (each meeting all the other faces) and 21 edges.
   The dual - with seven vertices - also has 21 edges. The
   standard torus formed from a ring of three triangular prisms
   has few edges - 18 edges and 9 quadrilateral (actual
   trapezoidal and rectangular) faces. Anyone have a smaller
   number of EDGES?" - Walter Whiteley

  "...I see a clear diagrammatic illustration of the Szilassi 
   polyhedron, a 7-edged toroid of seven hexagons (6 non-convex,
   1 convex) on p. 28 of Scientific American, Nov, 1978. It has
   21 edges (as I count them from the diagram)." - Mary Krimmel

  "John Conway has proposed the problem of finding "holyhedra,"
   meaning every face contains a hole. Here are some 
   constructions of mine..." - Antreas Hatzipolakis
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- Hinge Theorem (21 Jan. 1999)

  "If two sides of one triangle are congruent to two sides of 
   another triangle, and the included angle of the first is 
   larger than the included angle of the second, then the
   third side of the first is longer than the third side of 
   the second." - Antreas P. Hatzipolakis

   A conversation that began on January 21 with posts by 
   John Conway and Antreas Hatzipolakis, stating the theorem 
   and its converse, discussing how the name came about, and 
   providing a proof.
______________________________ + ______________________________

SCI.MATH, a discussion group focused on general and advanced 
mathematics that can be read as a Usenet newsgroup or on the Web:
- Cutting equilateral triangle (20 Jan. 1999)
  "How does one prove that an equilateral triangle cannot
   be subdivided into finitely many smaller equilateral 
   triangles, such that no two are congruent? Is it known 
   if there exists a tiling of the plane by pairwise
   noncongruent equilateral triangles?" - David Radcliffe

  "It's not possible. This was originally discussed in an 
   article that I discuss at my site this week."
   - Ed Pegg Jr,

   - David Eppstein

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  We hope you will find these selections useful, and that you
  will browse and participate in the discussion group(s) of
  your choice.  

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