User jesse madnick - MathOverflow

Answer by Jesse Madnick for Fundamental motivation for several complex variables

2012-11-28T05:04:31Z

If those are your reasons for caring about complex analysis, then I'm not sure to what extent I can convince you to care about SCV.

But here is why I like SCV: I like multivariable calculus, and I like complex numbers. I've always wondered what would happen if you put the two together. And apparently pretty wild things happen (compared with either the single-variable or real-variable case).

I'm also really interested in complex geometry. The local theory of complex manifolds involves SCV.

One final comment: Your reason (1) for caring about holomorphic functions is in fact my reason for caring about harmonic functions (for which I would otherwise desire motivation).

Undergraduate Differential Geometry Texts

2009-12-05T04:35:39Z

Can anyone suggest any basic undergraduate differential geometry texts on the same level as Manfredo do Carmo's Differential Geometry of Curves and Surfaces other than that particular one?

(I know a similar question was asked earlier, but most of the responses were geared towards Riemannian geometry, or some other text which defined the concept of "smooth manifold" very early on. I am looking for something even more basic than that.)

Answer by Jesse Madnick for Jokes in the sense of Littlewood: examples?

2010-09-16T22:30:53Z

The chain rule "joke" reminded me of a similar notation joke: Radon-Nikodym derivatives.

If $\mu$, $\nu$, $\lambda$ are $\sigma$-finite measures with $\nu \ll \mu \ll \lambda$, and $f \geq 0$ is measurable, then:

$$\int f\ d\nu = \int f \left[\frac{d\nu}{d\mu}\right]\ d\mu$$

and

$$\left[\frac{d\nu}{d\lambda}\right] = \left[\frac{d\nu}{d\mu}\right]\left[\frac{d\mu}{d\lambda}\right]$$

Answer by Jesse Madnick for Jokes in the sense of Littlewood: examples?

2010-09-16T21:28:08Z

The Cayley-Hamilton Theorem:

If $A$ is a square matrix with characteristic polynomial $p(\lambda) = \det(A-\lambda I)$, then $p(A) = 0$.

Because you know, you "just plug in."

Answer by Jesse Madnick for What's your favorite equation, formula, identity or inequality?

2010-07-15T05:00:55Z

The Gauss Formula from Riemannian geometry:

$\overline{\nabla}_XY = \nabla_XY + \text{II}(X,Y)$

It may just be a decomposition into tangential and normal parts, but I find it very aesthetically pleasing. (It's also not completely immediate that the tangential part of the ambient connection should actually be the intrinsic connection.)

Answer by Jesse Madnick for Theorems with unexpected conclusions

2010-07-15T04:44:36Z

If $f\colon [a,b] \to \mathbb{R}$ is increasing, then $f$ is differentiable almost everywhere [w.r.t. Lebesgue measure].

(We can further conclude that $f'$ is measurable and $\int_a^b f'(x)\ dx \leq f(b) - f(a)$, but it's the first part that struck me when I learned it.)

And sure it makes sense, but knowing how real analysis often is, one might think that there must be some increasing function that fails to be differentiable on a set of positive measure.

Answer by Jesse Madnick for How should one present curl and divergence in an undergraduate multivariable calculus class?

2010-07-14T00:08:49Z

This is perhaps a crude (and certainly non-rigorous) explanation, but it's always how I've thought of motivating it.

Let $F = (F_1, F_2, F_3)$ denote a vector field in $\mathbb{R}^3$, and write $\text{curl}\ F = (G_1, G_2, G_3)$. We would like a situation where $G_1$ describes the "instantaneous" rotation of $F$ about the $x$-axis, $G_2$ the rotation about the $y$-axis, and $G_3$ the rotation about the $z$-axis.

So let's think of vector fields which do just that. Three simple (linear!) ones which come to mind are $$H_1(x,y,z) = (0, -z, y)$$ $$H_2(x,y,z) = (z, 0, -x)$$ $$H_3(x,y,z) = (-y, x, 0)$$ So in order to measure how much $F$ rotates about, say, the $z$-axis, it makes sense to look at something that compares how similar $F$ is to $H_3$. The dot product $F(x,y,z) \cdot H_3(x,y,z)$ seems reasonable, which is precisely $-yF_1(x,y,z) + xF_2(x,y,z).$

This suggests that defining $$G_1(x,y,z) \approx -zF_2(x,y,z) + yF_3(x,y,z)$$ $$G_2(x,y,z) \approx zF_1(x,y,z) - xF_3(x,y,z)$$ $$G_3(x,y,z) \approx -yF_1(x,y,z) + xF_2(x,y,z)$$ might give something close to what we want. But this is a very crude way to measure "instantaneous" rotation -- in fact, one might say it's a sort of linear approximation. Thus, we are led to replacing the linear terms with their corresponding derivations: $$G_1(x,y,z) = -\frac{\partial}{\partial z}F_2 + \frac{\partial}{\partial y}F_3$$ $$G_2(x,y,z) = \frac{\partial}{\partial z}F_1 - \frac{\partial}{\partial x}F_3$$ $$G_3(x,y,z) = -\frac{\partial}{\partial y}F_1 + \frac{\partial}{\partial x}F_2,$$ which is precisely the curl.

This heuristic also works with divergence, but instead consider $(H_1, H_2, H_3) = (x,y,z)$.

Convergence of Fourier Series of $L^1$ Functions

2010-06-16T19:08:02Z

I recently learned of the result by Carleson and Hunt (1968) which states that if $f \in L^p$ for $p > 1$, then the Fourier series of $f$ converges to $f$ pointwise-a.e. Also, Wikipedia informs me that if $f \in L^p$ for $1 < p < \infty$, then the Fourier series of $f$ converges to $f$ in $L^p$. Either of these results implies that if $f \in L^p$ for $1 < p < \infty$, then the Fourier series of $f$ converges to $f$ in measure.

My first question is about the $p = 1$ case. That is:

If $f \in L^1$, will the Fourier series of $f$ converge to $f$ in measure?

I also recently learned that there exist functions $f \in L^1$ whose Fourier series diverge (pointwise) everywhere. Moreover, such a Fourier series may converge (Galstyan 1985) or diverge (Kolmogorov?) in the $L^1$ metric.

My second question is similar:

Do there exist functions $f \in L^1$ whose Fourier series converge pointwise a.e., yet diverge in the $L^1$ metric?

(Notes: Here, I mean the Fourier series with respect to the standard trigonometric system. I am also referring only to the Lebesgue measure on [0,1]. Of course, if anyone knows any more general results, that would be great, too.)

Answer by Jesse Madnick for What are your favorite instructional counterexamples?

2010-04-05T05:59:01Z

I've always been fond of the popcorn function (aka Thomae's Function), which is given by $f\colon \mathbb{R} \to \mathbb{R}$ via

$f(x) = \begin{cases} \frac{1}{n} & \mbox{if } x = \frac{m}{n} \in \mathbb{Q} \\ 0 & \mbox{if } x \notin \mathbb{Q}. \end{cases}$

This function has a couple of amusing properties.

(1) It is upper semicontinuous on $\mathbb{R}$, yet has a dense set of discontinuities (every one of which is removable) (namely $\mathbb{Q})$.

(2) Since it is bounded and has a set of measure zero as its set of discontinuities, it is Riemann integrable. So if we consider $g(x) = \int_0^x f(t)\ dt$, we see that $g \equiv 0$, so that $g'(x) \not \hskip 2pt = f(x)$ on a dense set.

References: http://en.wikipedia.org/wiki/Thomae%27s_function and of course "Counterexamples in Analysis" (Sec 2.15-2.17)

Comment by Jesse Madnick

2013-02-24T23:29:36Z

And here I was thinking the standard proof was just the Integral Test for series convergence.

Comment by Jesse Madnick

2013-02-23T07:40:17Z

This question isn't appropriate for MathOverflow, which is (largely) for research-level questions. You'll have better luck if you ask your question here: math.stackexchange.com

Comment by Jesse Madnick

2013-01-12T09:48:18Z

It's interesting. Pretty much everyone I know would agree that it's not a good thing when elementary school teachers tell their students that they didn't like math, either. And yet I can think of more than a handful of those same people (grad students and professors) who would have no problem telling their calculus classes that they don't like calculus, or that it's not "real" math somehow...

Comment by Jesse Madnick

2013-01-06T07:43:09Z

This website is for research-level questions. You might have more luck if you ask your question on math.stackexchange.com instead.

Comment by Jesse Madnick

2012-12-12T11:55:45Z

Funny, I would've said that when doing geometry, you either fall into algebra or analysis. In fact, it seems to me that the more "rigid" your geometry is, the more likely you are to fall into one of the sides. By contrast, "softer" geometries can rely on topology.

Comment by Jesse Madnick

2012-11-28T23:04:09Z

That's fair. In truth, I'm still rather ignorant about harmonic functions and their properties. I don't mean to suggest that reason (1) is the only reason one would care about them, but simply that it's a primary reason that I do (again, given my ignorance).

Comment by Jesse Madnick

2012-10-11T01:16:31Z

"Maybe the bred desire that you mention is (or is related to) the emphasis on 'theory building'. Yes, theory building is great, but I personally see it through the lens of problem solving." Okay, but what if one is far more interested in theory building than problem solving, or if theory-building IS the lens by which one views things?

Comment by Jesse Madnick

2011-06-29T08:54:44Z

"What makes some of you think the 'math' people learned in school is 'not real'?" Just for that one line I would upvote your comment ten times if I could. For me it is extremely refreshing to hear a professional mathematician question that line. Thank you for that.

Comment by Jesse Madnick

2011-04-05T10:08:54Z

To be pedantic, I think technically it's $\text{curl} = \sharp \circ \ast \circ d \circ \flat$, but no matter.

Comment by Jesse Madnick

2010-10-28T07:34:15Z

I have seen $A \Subset B$ mean that the (topological) closure of $A$ is contained in $B$, but I'm sure there are plenty of other uses as well. Could you perhaps provide some context?

Comment by Jesse Madnick

2010-10-05T05:37:40Z

+1 for the first sentence. Memorization is certainly not a substitute for understanding -- and really, who would ever claim that it is? -- but it's sometimes just so useful! Especially in areas like real analysis where theorems can have multiple hypotheses and omitting even one of them can render the theorem false (think Dini's Theorem), sometimes it just saves time and stress to sit down one day and memorize it. Also, the "modes of convergence" diagram is wonderful.

Comment by Jesse Madnick

2010-09-21T04:31:36Z

Well, I never said I didn't like do Carmo (although I must admit, it's not among my favorites) -- it's simply that at the time of this posting, I wasn't really aware of any alternatives. Basically, I just wanted to explore other treatments of the subject. When learning a subject, I almost always use at least two texts simultaneously.

Comment by Jesse Madnick

2010-09-02T06:23:37Z

Just to expand on Jonas' comment that "You don't need anything near the strength of Picard's theorem," note that $a$ is an essential singularity of $f(z)$ if and only if the image of any punctured disc centered at $a$ is dense. In fact, this is often taken as the definition of essential singularity.

Comment by Jesse Madnick

2010-08-29T21:06:06Z

Or -- not research-level per se -- but rather: questions that would be of interest to mathematicians. In particular, homework questions are discouraged.

Comment by Jesse Madnick

2010-08-29T21:04:11Z

Please see the FAQ for the types of questions that should/shouldn't be asked on this site. This site is primarily for research-level questions. A site that is perhaps more suited for your question is math.stackexchange.com