bio | website | cornellmath.wordpress.com |
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location | Ithaca, NY | |
age | 30 | |
visits | member for | 3 years, 11 months |
seen | yesterday | |
stats | profile views | 860 |
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.
Aug 25 |
answered | Analytic functions with algebraic Taylor coefficients at some point. |
Aug 10 |
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High dimensional beta integral (a typo in Stein's book “singular integrals”)
You're welcome. I can totally understand the confusion about the domain of integration since Stein mentions the beta integral which only integrates on [0,1] and makes a typo that $x=1$! Then again, one often learns more from correcting mistakes in books than from simply reading. There is something to be said for getting one's hands dirty and doing some hard work. |
Aug 9 |
answered | High dimensional beta integral (a typo in Stein's book “singular integrals”) |
Aug 5 |
awarded | Nice Answer |
Jul 9 |
comment |
Long time behavior of the heat equation on R
If μ is absolutely continuous with respect to Lebesgue measure, then the density should be dominated by some polynomial. Certainly in that case, the integral will grow at most like some power of t: in particular I believe it should grow like $t^{d/2}$, where d is the degree of the polynomial bound for the density function. |
Jul 9 |
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Long time behavior of the heat equation on R
I don't think it needs to decrease. For instance if $\mu$ is simply Lebesgue measure, then $u(x,t)\equiv 1$, no? |
Jul 8 |
answered | Carleson's Theorem (on the Adeles and other exotic groups) |
Jun 16 |
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What is convolution intuitively?
Ignoring the extraneous first $x$, this formula is only correct if the functions are integrable. Convolution makes sense if both functions are square-integrable but not integrable, in which case this equation is incorrect (since Fubini does not apply). But in any case, what intuition do you glean from knowing, for a fixed $f$, that $T_f(g)=f*g$ has the property you stated? The operator $S_f(g)=g(x)\int_{-\infty}^{\infty}f(t)dt$ has the same property but is just a constant times the identity... |
Jun 14 |
answered | What is convolution intuitively? |
Jun 2 |
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A Generalization of Hadamard?
Obviously Mariano was mostly joking, but I think the grammar in the title works fine; one frequently uses 'of' in the sense of 'owing to', e.g. the lemma of Schwarz (I think my professor used this form mostly because it was easier to say, frankly, but there is also a historical trend for that particular lemma). We often leave the originator's name in theorems even if they were later generalized, as was indicated above with the Cauchy-Schwarz inequality (sorry Bunyakovsky). Plus isn't it a bit poetic that Hadamard might exist through mathematics in some generalized way even after his death? |
May 31 |
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A quick and elementary question from Hubbard's Teichmuller Theory : Volume I
The Cantor-Lebesgue function has derivative 0 almost everywhere but is continuous and strictly increasing. The issue is that for a function on $R$ to have distributional derivatives it must be absolutely continuous; simply having a well-defined almost everywhere derivative is not enough. |
May 31 |
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A quick and elementary question from Hubbard's Teichmuller Theory : Volume I
Suppose you are on $R^1$ and $l$ is just the point $\{0\}$. Then the function $f$ which is 0 when $x<0$ and 1 when $x\ge 0$ has distributional derivative 0 away from 0. So $f'$ is defined almost everywhere and $f'$ is the distributional derivative for any function compactly supported in $R^1-\{0\}$. But if $\phi$ is any $C_c^\infty$ supported function which is 1 in a neighborhood of $0$, then $-\int f\phi'=\phi(0)=1$ rather than 0 if $f'$ were actually the distributional derivative on all of $R$. You can come up with clever ones so that $f$ is continuous: see the Cantor-Lebesgue function. |
May 31 |
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A quick and elementary question from Hubbard's Teichmuller Theory : Volume I
Yep, got a surprising little more mileage out of that book. The OP and I should thank you for asking for the book back, otherwise the book wouldn't have been right next to me when I saw this question and this answer would never have happened. |
May 31 |
answered | A quick and elementary question from Hubbard's Teichmuller Theory : Volume I |
May 31 |
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Why is mechanical differentiation so hard to get right?
Well, it doesn't seem to be an issue with the derivative at all. Wolframalpha seems to believe that the given expression is $x^2/2$ when $x>0$, for instance. |
May 30 |
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Why is differentiating mechanics and integration art?
@Ryan: yeah, good point. As I was writing the previous comment, I realized it wasn't so tough to just get the right leading coefficient; I usually prove it just using the product rule ($x^n=x\times...\times x) and do the first few examples to avoid explicitly teaching induction because a lot of students really don't get the finite sum stuff. Although, I have done some stuff with the Riemann sums they seem to like if you phrase it correctly and avoid anything resembling induction :D |
May 30 |
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Why is differentiating mechanics and integration art?
@Deane A bit tongue in cheek, but: division of real numbers is harder than multiplication because it's secretly still multiplication, just with the additional step of inverting. Multiplication by $x$ is a continuous function, but multiplication by $x^{-1}$ is no longer continuous near zero. But if you look at, say, the circle group, multiplication and division are clearly equally difficult! |
May 30 |
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Why is differentiating mechanics and integration art?
Ryan --- he's saying that computing the derivative is simpler because it only depends on local information at a point. Integration depends on information about every point on a fixed interval, simultaneously. Thus differentiation is fundamentally simpler than integration, simply from the definitions. Doesn't it seem a lot easier to organize the cutting down of a forest one tree at a time than having to cut them all down simultaneously? |
May 30 |
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Why is differentiating mechanics and integration art?
That should be $n^{318}$. How shameful! |
May 30 |
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Why is differentiating mechanics and integration art?
@Ryan: ok, I see the distinction you're making. There is an argument one must make to be able to do constant step sizes (the function need be continuous). And I agree that in this case it comes down to computing those sums. I never remember the formulas for those sums aside from the first couple. Do you remember them for $n^318$, for example, or have any way to easily figure out what it is in such a case? :). I believe that business is rather involved (although all you really need is the coefficient of the highest power). |