bio | website | vmm.math.uci.edu |
---|---|---|
location | Univ. of California at Irvine | |
age | 83 | |
visits | member for | 4 years |
seen | Jul 21 at 0:13 | |
stats | profile views | 6,521 |
I'm a Professor at UC Irvine, but spent most of my career at Brandeis. My research areas: Differential Topology, Transformation Groups, and Global Analysis. Recently I developed a math visualization program, 3D-XplorMath, freely available at http://3D-XplorMath.org) and a companion website, the Virtual Math Museum at http://VirtualMathMuseum.org. Last year I co-authored a differential equations text with my son Bob, most of which is downloadable from http://ode-math.com .
Jun 25 |
awarded | dg.differential-geometry |
Jun 25 |
awarded | Revival |
Jun 25 |
awarded | Pundit |
Jun 14 |
answered | Calculus book in the spirit of the 18th century |
Jun 9 |
answered | Mathematics for ebook readers |
May 31 |
awarded | Popular Question |
May 30 |
comment |
A Canonical Form Theorem for $n$-forms?
@Francesco YES ! I don't think you are missing anything---that is essentially the same proof as mine. I'll have to get Kobayashi's book from the library and have a look. Many thanks for this clue. |
May 30 |
comment |
A Canonical Form Theorem for $n$-forms?
@Mariano. Right, that is the way I think of it, and it was Darboux Theorem that made me think it might be true. |
May 30 |
asked | A Canonical Form Theorem for $n$-forms? |
May 24 |
awarded | Notable Question |
May 24 |
awarded | Nice Answer |
May 22 |
comment |
Why don't more mathematicians improve Wikipedia articles?
@Mark M YES! That is a very good approximation of what I was asking for. Thanks for telling me about it. It is now one of my "pinned" tabs and I will try to help with the project. |
May 22 |
answered | Why don't more mathematicians improve Wikipedia articles? |
May 17 |
comment |
Are there proofs that you feel you did not “understand” for a long time?
How about the proof that Einstein is said to have (re)discovered as a teen-ager? It is clear that the areas of similar right triangles is proportional to the squares of corresponding sides---in particular to the squares of their hypotenuses. Dropping a perpendicular from the vertex of the right triangle onto the hypotenuse c divides it into two similar triangles with hypotenuses a and b, hence: k*c^2 = k*a^2 + k*b^2 so c^2 = a^2 + b^2 |
May 10 |
awarded | Nice Answer |
May 2 |
answered | A simple and good reference about solitons |
May 1 |
answered | Does a *topological* manifold have an exhaustion by compact submanifolds with boundary? |
Apr 27 |
awarded | Nice Answer |
Apr 26 |
awarded | Good Answer |
Apr 20 |
awarded | Necromancer |