User clinton curry - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-23T13:52:00Zhttp://mathoverflow.net/feeds/user/1743http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/111097/textbook-for-undergraduate-course-in-geometry/111109#111109Answer by Clinton Curry for Textbook for undergraduate course in geometryClinton Curry2012-11-01T04:39:19Z2012-11-01T04:39:19Z<p>I like <em>Euclidean and Non-Euclidean Geometries: Development and History</em> by Marvin J. Greenberg. I will warn you: it is certainly an axiomatic treatment. However, I really enjoyed the way that the book develops it. For example, the distinction between the axioms of a geometry and theorems you can prove about them, versus the models of geometry and their various properties, is clearly drawn. I dare say that, despite how advanced your undergraduates feel, they will learn a lot about the axiomatic method from this book. I recommend that you give it a look; even if it is not the primary textbook for the course, you can use it as a convenient source of motivation, problems, examples, and history. (There is a <em>lot</em> of history in this book, and many exercises.)</p>
http://mathoverflow.net/questions/19999/finding-all-roots-of-a-polynomial/33606#33606Answer by Clinton Curry for Finding all roots of a polynomialClinton Curry2010-07-28T03:14:40Z2010-07-28T03:14:40Z<p>This can be done. Check <a href="http://www.math.cornell.edu/~hubbard/NewtonInventiones.pdf" rel="nofollow">this article</a> by Hubbard, Schleicher, and Sutherland, entitled "How to find all roots of complex polynomials by Newton's method".</p>
http://mathoverflow.net/questions/4722/boundary-of-planar-region/5269#5269Answer by Clinton Curry for Boundary of planar region Clinton Curry2009-11-12T22:37:39Z2009-11-12T22:37:39Z<p>Local connectivity of the boundary provides much of the topological structure that such a domain would have. If you specify the following two conditions, you have that the boundary of a domain $U$ is a finite union of simple closed curves, I think.</p>
<ol>
<li>$\partial U$ is locally connected.</li>
<li>$\overline U$ has finitely many complementary components.</li>
</ol>
<p>The thrust is this: if $V$ is a complementary component of $\overline U$, then $\partial V$ is the common boundary of two simply connected domains ($V$ and the component of the complement of $\overline V$ containing $U$). A locally connected plane continuum which is the common boundary of two connected open sets is always a circle.</p>
http://mathoverflow.net/questions/4722/boundary-of-planar-region/5269#5269Comment by Clinton CurryClinton Curry2009-11-12T22:41:40Z2009-11-12T22:41:40ZOh, yes -- It must also be that $\partial U$ is the boundary of the complement of $\overline U$. Otherwise, you could have something silly like $U$ being the complement of an arc. Incidentally, I think that one could then omit the second condition if one replaces local connectivity by local contractibility, as Greg suggested.