bio | website | education.msu.edu/search/… |
---|---|---|
location | Ann Arbor, MI | |
age | ||
visits | member for | 1 year, 10 months |
seen | Aug 5 at 1:30 | |
stats | profile views | 123 |
I am an Assistant Professor at Michigan State University, with a joint appointment in the Dept. of Teacher Education (in the College of Education) and the Program in Mathematics Education (in the College of Natural Sciences). My Ph.D (2009) is from the University of Michigan, and was a joint degree in Mathematics and Education awarded by the Mathematics Department and the School of Education.
My dissertation focused on the capacity of the secondary Geometry course to faithfully represent authentic mathematical values and practices, and the extent to which school mathematics can cultivate a mathematical sensibility in students. Current research interests include an analysis of the secondary "Algebra 2" course, and an investigation into the mathematics education practices of home-educated students.
Mar 31 |
awarded | Quorum |
Mar 22 |
awarded | Yearling |
Mar 21 |
awarded | Nice Question |
Mar 21 |
awarded | Supporter |
Mar 21 |
accepted | Rings for which no polynomial induces the zero function |
Mar 21 |
comment |
Rings for which no polynomial induces the zero function
Let $R$ be the ring $\mathbb{Z}/p \mathbb{Z}$ with infinitely many elements $u_i$ adjoined, each satisfying the relation $u_i^p -u_i = 0$. Then the polynomial $p=x^p-x$ satisfies $\hat{p}=0$. |
Mar 21 |
comment |
Rings for which no polynomial induces the zero function
P Vanchinathan, yes, that was one of my examples. More precisely if $R$ is a ring that has my desired property, than $R \times R$ also has the desired property, even though it is not a domain. So that shows that "infinite domain" is not necessary for this property. |
Mar 21 |
awarded | Editor |
Mar 21 |
revised |
Rings for which no polynomial induces the zero function
added 10 characters in body |
Mar 21 |
asked | Rings for which no polynomial induces the zero function |
Oct 21 |
comment |
Lost soul: loneliness in pursing math. Advice needed.
When I was in my 1st year of grad school I got married; in my 3rd year we had our first child. I missed a lot of classes, and a lot of deadlines, and was worried that he consequences might be dire. But my professors all -- all, without exception -- told me not to worry: "Some things are more important than math", said one, and that advice has kept me (reasonably) sane and balanced. I hope it helps you too. |
Oct 15 |
comment |
A weird function related to the denominators of rational squares
Only in the most elementary case: \sigma (a) = 2 iff $a=n^2 + n$ for some $n$. |
Oct 12 |
awarded | Scholar |
Oct 12 |
comment |
A weird function related to the denominators of rational squares
Thanks -- those citations will be very helpful, I think. I am a total newbie when it comes to the OEIS; is there a way to search for papers / pages that discuss or reference a specific OEIS sequence? |
Oct 12 |
accepted | A weird function related to the denominators of rational squares |
Oct 12 |
comment |
A weird function related to the denominators of rational squares
Plus I've got a list of 20 special cases that have the form: If $a=n^2 + $ [something], then $\sigma (a) = $ [some specific linear polynomial in n with simple coefficiets]. But I can't find a way to tie those 20 special cases together into a general statement. |
Oct 12 |
comment |
A weird function related to the denominators of rational squares
(4) The lower bound is harder to state, but: If $a$ is not a perfect square, write $a=n^2 + b$ and $a=m^2 - c$ where $n^2$ and $m^2$ are the closest squares below and above $a$, respectively. Then: $\sigma (a) \geq \underline{ max \left \{ \frac{n+\sqrt{a+1}}{b+1}, \frac{m+\sqrt{a}}{c} \right \} } + 1$. |
Oct 12 |
comment |
A weird function related to the denominators of rational squares
Maybe I should give some more detail on what I have already found so that people don't waste their time re-inventing my wheel. (1) For all $a$, $\sigma (a) \leq \overline{\frac{8a\sqrt{a}}{4a-1}}$, where the overbar denotes the ceiling function. (2) When $a=n^2$ the upper bound above is sharp, and simplified to $\sigma (n^2) = 2n+1$. (3) When $a=n^2 + n$ we have $\sigma(n^2 + n) = 2$. |
Oct 12 |
awarded | Student |
Oct 12 |
asked | A weird function related to the denominators of rational squares |