bio | website | cornellmath.wordpress.com |
---|---|---|
location | Ithaca, NY | |
age | 31 | |
visits | member for | 5 years, 2 months |
seen | Apr 22 at 2:27 | |
stats | profile views | 948 |
I was a graduate student at Cornell University studying harmonic analysis with Camil Muscalu. Now I'm a postdoc at Washington University in St. Louis.
Jun 14 |
answered | Why do we care about L^p spaces besides p = 1, p = 2, and p = infinity? |
Jun 12 |
comment |
Weighted Hardy Inequality for bounded domains
At least in the case where the function is zero at L, I suppose you've tried the change of variable x to v/(v+1), so that one can attempt to use the inequality in your link (after some algebra) and then do some more algebra to simplify things a bit? It seems like that would work to at least give you some control over the function u(x)-u(L). Would this be at all helpful for your purposes? Why do you need such an inequality? |
Jun 7 |
comment |
What are your experiences of handouts in mathematics lectures?
Are you not free to disallow the use of laptops in class? |
Jun 7 |
comment |
What are your experiences of handouts in mathematics lectures?
I completely agree with these sentiments, Terry. Such a system is extremely rewarding to students who work hard. Students all have varying levels of raw ability; the only thing you can do as an instructor is make hard work as valuable to them as possible. |
Jun 7 |
comment |
Real analysis has no applications?
Greg, I completely agree with you regarding BC calculus. I took BC calculus in high school, and it was a wonderful experience. We did avoid some of the more rigorous details --- I think delta-epsilon proofs of continuity were completely absent, for instance --- but I think I still came out of that course with a great understanding of analysis. I clearly understood the delicate interplay between approximation and error terms. Once deltas and epsilons came, I needed only understand the framework; the tools were already there. |
May 19 |
comment |
Why is Lebesgue integration taught using positive and negative parts of functions?
Fatou's lemma is pretty useful. |
May 16 |
awarded | Commentator |
May 16 |
comment |
Differentiable structures on R^3
The inverse of $x^3$ is not smooth, but the term diffeomorphism as it pertains to differentiable structures does not require this: a bijective map $f:\mathbb{R}\rightarrow\mathbb{R}$ is a diffeomorphism of $(\mathbb{R},A)$ and $(\mathbbf{R},B)$ if $x\in A$ iff $x(f)\in B$. Here A and B are maximal atlases. |
May 15 |
comment |
Beginning a sentence with a mathematical symbol
Greg, do you have any particular examples worth sharing, or are you simply finding your prose has become a monotonous glob of ideas punctuated by thus and therefore? |
May 15 |
answered | Beginning a sentence with a mathematical symbol |
May 15 |
comment |
If ErdÅ‘s is published as Erdös in a paper, which do I cite?
The definitive answer to "are there good math jokes?" |
May 13 |
answered | Would Euler's proofs get published in a modern math Journal, especially considering his treatment of the Infinite? |
May 11 |
comment |
Evaluation of the following Series
Gerald, Max wrote the incorrect general term, hence my error. $\displaystyle{\sum_{n=1}^\infty\frac{1}{((2n-1)2n)^2}=\sum_{n=1}^\infty\frac{1}{(2n-1)^2}-2\sum_{n=1}^\infty\frac{1}{((2n-1)2n)^2}+\sum_{n=1}^\infty\frac{1}{(2n)^2}}$. The first and second sums add to give $\displaystyle{\sum_{n=1}^\infty\frac{1}{n^2}}$, which then gives Michael Greenblatt's result. |
May 11 |
comment |
Evaluation of the following Series
Ah, I see, you have written the general term down incorrectly, Max. It is not $1/(n(n+1))^2$ . It should be $1/(2n(2n-1))^2$. |
May 11 |
comment |
Evaluation of the following Series
The second sum which you have written does not sum to $\log 2$. It is an alternating series which sums to 1. |
May 11 |
comment |
Evaluation of the following Series
haha I had just written this up a second before the site indicated you posted your solution... here is what I wrote: Write $\frac{1}{n^2(n+1)^2}=\left(\frac{1}{n}-\frac{1}{n+1}\right)^2=\frac{1}{n^2}-\frac{2}{n(n+1)}+\frac{1}{(n+1)^2}$. Summing the first term gives $\pi^2/6$, the last term gives $-1+\pi^2/6$, and the middle term is an alternating seriesgives $-2$, so that the whole sum is $\frac{\pi^2}{3}-3$. |
May 5 |
answered | What are trig classes like within a universe that's “noticeably” hyperbolic? |
May 4 |
comment |
How did Gauss discover the invariant density for the Gauss map?
You're very welcome! |
May 3 |
answered | How did Gauss discover the invariant density for the Gauss map? |
May 1 |
awarded | Supporter |