@leslietownes I'm sure she'll recover fully, but her condition reminded me of this old joke:
Q: What do you call an Irish girl with one leg longer than the other? A: Eileen. Q: What do you call a Chinese girl with one leg longer than the other? A: Irene.
Q: What's the Irish girl's surname? A: O'Bleakley.
Here's a girl drummer to inspire your daughter. German prodigy Sina, covering Rosanna. Sina was born in 1999. She was already great a few years ago, she's even better now.
Eileen = "I lean" Irene, when pronounced in a number of Asian languages will sound like "I lean" (as a lot of Asian languages don't distinguish between "r" and "l"---they are produced very similarly in the mouth) And "I lean obliquely" is the last punchline.
shuffles like that sound simple but are really hard to do. one of my friends is a great drummer and cannot do them.
at least he says they're hard to do.
xander is that a little bit of a weight off the mind? i never had much trouble with it but i remember a lot of fellow instructors dealing with drama around plusses and minuses. particularly B plusses and A minuses.
UGA started long before I retired. News bulletin … After blowing covid almost maximally, the GA regents want to improve faculty morale by abolishing tenure. Because, you know, Rethugnicans despise education.
well, if you can't pay them well, you gotta give them something. no tenure makes sense. if they have a health plan they should cut that too and hand out lotto tickets.
i could see a policy like that putting some students at a disadvantage because it's so uncommon. even if you're sent a thing with a transcript that tells you a school's grading policy you probably aren't reading it.
@leslietownes Yep. And that shuffle is one of the hardest to get right. There's a reaction video by a professional drum teacher analysing Sina's version. He's most impressed, and recommends watching that video to any drummer who wants to learn Rosanna. He also says that even if you don't want to play that song, learning it will teach you a bunch of useful stuff.
i never had much drama either. i do remember other TAs getting swamped with requests though. particularly in classes required for pre-meds. often these wounds were self inflicted. e.g. describing the policy on regrading or re-evaluation during class time while passing out exams. just wave that red flag in front of the bull.
PM, i'm guessing there is some rush on this youtube channel. has to be.
@TedShifrin I generally have no difficulty with students vis-a-vis grades. Though I did have one student once who made my life quite difficult. After I turned in grades, I was in the office, on my out the door (looking forward to a 90 minutes commute), and she buttonholed me.
And then a French consul entered the office, and she proceeded to ignore me and speak in French for 10 minutes, while I patiently waited to hear her out.
i had a bit of grading issues but nothing at the level of an international incident.
funniest one was a parent of a student leaving a bunch of pissy voicemails on my office phone about their son's grade. i didn't know because i wasn't in my office during winter break. never had more fun deleting my messages.
cool. my daughter loves blue herons. we see them around here on the weekends. she has yet to develop musical tastes but i'm sure i'll end up hating whatever she likes.
Joan's cousin John Baez posts occasionally on MathOverflow. Her dad Albert co-invented the X-ray microscope, and inspired John to become a mathematical physicist.
@TedShifrin Sure, it's easy to to implement the obvious "greedy" algorithm, but the denominators tend to blow up badly. David Epstein has a lot of info about Egyptian fractions, and various strategies for finding them. https://www.ics.uci.edu/~eppstein/numth/egypt/
very cool i'll read that, thanks! i used the mentioned identity to solve a series i forgot which, but i guessed that similar identities likely could have some applications in that sense
There are also some nice reciprocal identities that come up in creating Machin-like formulae. Traditionally, it was easier to do computations with reciprocals of integers. That's not such an issue in the computer age... unless you want to calculate stuff with thousands or millions of digits.
With a bit of algebra, we can find various interesting relations. A very useful one is $$a(u) = 2a(2u) - a(4u^3+3u)$$ Eg, $$a(5) = 2a(10) - a(515)$$ In 1706, John Machin found $$a(1) = 4a(5) - a(239)$$ This historically important formula has its own [Wikipedia page](https://en.wikipedia.org/wiki/Machin-like_formula). And there's a huge collection of Machin-like relations at the old http://www.machination.eclipse.co.uk/ site
Bet you can't guess who, because they almost never mention their music.
If twin primes is false, then for some $x_0 \in \Bbb{N}$, and for all $x \geq x_0, k=0,...,2x$, we have $(x^2 + k)(x^2 + k - 2)$ is not a unit modulo $x!$.
does anyone else have some of the letters in the "arctan" images generated from tex markup in that wikipedia page render funny? i'm seeing weird behavior in both chrome and edge.
@PM2Ring I like Dr Baez, and think that his is an incredibly engaging lecturer. I enjoy his talks immensely. However, I do think that he tended to hand-hold his students a lot in courses, and lead folk through arguments by the nose a little. It wasn't my style---you feel like you are really groking the material, and then you leave the lecture hall and try to do the work and realize that you didn't really follow anything after all.
So I kind of enjoyed his network theory seminar (which I didn't need to get anything out of), but was a little frustrated with his real analysis course (which seemed to move very slowly).
yes, that's also what i see in firefox. or (interestingly) in chrome, but only sometimes, and only if i right click on the image, copy url of image, and just load the image in chrome.
some of the other a's don't render either. in the "area" markup. goofy.
i think it's possible to be too good an expositor. i know that effect of sitting through a talk where everything feels perfect and well motivated and then it falls to pieces when you think it through later. there should always be a few hiccups or genuinely difficult points.
@XanderHenderson Yes, his writing style is a bit hypnotic, too. He makes stuff sound easy, then you suddenly realise that you didn't actually understand the last 3 paragraphs. :)
i also don't like the 'gee whiz' tone where literally EVERYTHING is given the treatment of, isn't this amazing, and isn't this amazing, and wait until you see what's next. although most people like that. every ted talk is like that.
I quite like the style of Australian sci-fi author Greg Egan (who happens to be friends with Baez). He's a very good expositor, but he mostly lets you decide for yourself whether stuff is "gee whiz" or not. He has some excellent introductory articles on relativity & quantum mechanics on his site.
there was a kind of golden era, after google began including the text within PDFs in its search results, and before many instructors were pushed onto course management software, where there were a lot of searchable solutions to homework and exams grouped conveniently by topic on the web. now it seems that math.SE and similar are more likely homes for that material.
He continued working as a programmer for a few years after his first books were published. His coding style is impeccable.
@leslietownes And now Google makes it really tricky to copy & paste a URL for a PDF. So people post those ridiculous Google search result URL monstrosities.
I post lots of links to Wikipedia articles, so a huge proportion of my Wiki visits are just so I can grab the URL from the address bar.
It's a waste of bandwidth, and it artificially inflates the Wiki page hit count. But it's less fiddly than extracting the URL from the search URL, especially when I'm using my phone.
I would like to demo for you guys a website I am building for mathematicians. It's mainly just the UI, but as you can see, you can also now edit the diagram somewhat. I'm modding q.uiver.app.
But I have a video to demo
It's uploading (2 min video of UI)
I've used mostly good coding practices but today I applied a hack fix to block a button event
Is there a closed form for $$\sum_{i = 0}^n (a^i)(b^{n - i})$$ for $a, b > 0$?
It is very similar to the expansion for a binomial to the $n$th power.
I am trying to solve for the probability that AaBbCcDdEeFfGgHhIiJJ crossed to AaBbCcDdEeFfGgHhIiJJ will have an offspring with a genotype AaBbCcDdEeFfGgHhIiJJ, and the sum I asked shows up.
Well, at least if you multiply by $a/b$ you'll get that all terms in the result except possibly the start and end term $i=0, n$ are also terms in your original sum.
The expression is ${a^{n+1}-b^{n+1} \over a-b}$ for $a \neq b$ and for $a=b$ you can either evaluate directly or let $b \to a$ and note that it is a derivative to get $(n+1) a^n$.
It helps to have a few glasses of a nice red Bordeaux before solving such problems.
My drive from SD to Costa Mesa worried me because my neck was sore after half the drive. So I dunno.
These days I drink mostly California wines (I had a subscription to Francis Ford Coppola for years, but now I have it on hold — they are truly excellent).
I had some nice Coppola wine many decades ago in a place he had in Belize near Guatemala. The wine was lovely but the pizza catastrophic. Not sure how a pizza can be screwed up so badly.
I don't really know. My daughter is back in the UK and I know is having fun. My son is not terribly communicative. I miss them both awfully, my daughter less so because we are in constant communication.
He is a bright & capable fellow, but I do not think he has 'found his groove' yet.
I have negotiated a dinner with him on my birthday in the coming weeks. He likes the burritos from the Taco truck next to the Hotsy-Totsy in Albany on San Pablo Avenue :-).
I life a much different life than I did say a decade ago. Many of the procedures I followed are not longer relevant. But some habits die hard. I always have a passport and a little stash within reach.
Of cash, that is. Never touched anything except alcohol. But I have immersed myself in the latter.
I got a little peeved at someone on MSE today who dinged me for saying that the Hessian of $\log \sum_{k=1}^p e^{x_k}$ was straightforward to compute.
Or rather, straightforward to show positive semi definite.
I did manage to restrain myself, so I don't think another suspension is imminent.
i have lost a bit of my 'edge'. having a toddler takes a lot out of you.
she's turning 3 soon, which means a lot of stuff she does will no longer be precocious. it will just be normal "threenager" stuff.
while hobbling around the house today she started giggling and said "you're the kid who doesn't walk." with reference to herself. must have heard that at school.
anyone have a hint for this? Let $E \subset \mathbb{C}$ be compact, and define $p(z) = \int_{E} \log(|\frac{1}{z - \zeta}|) d \mu(\zeta)$ for $z \in \mathbb{C} \setminus E$, where $\mu$ is a borel measure supported on $E$. Show that $p$ is harmonic
I want to say that if $z_0 \in \mathbb{C} \setminus E$, then for any $\zeta \in E$, in a small ball around $z_0$, there is a holomorphic single valued branch of $\log(\frac{1}{z- \zeta})$ and then use that my integral is the real part of the integral of this single valued branch over $E$, but my issue is that I may need to define different branches for different $\zeta$'s
so ultimately im not sure if my $H(z,\zeta) : B_{\epsilon}(z_0) \times E \rightarrow \mathbb{C}$ will be integrable, where for any $\zeta$, $H(z,\zeta)$ is the single valued branch of $\log(\frac{1}{z - \zeta})$ I am defining in the ball $B_{\epsilon}(z_0)$
of course the real part of my $H(z,\zeta)$ is bounded, but I worry the imaginary part may explode for some $\zeta$.. maybe this worry is unfounded though
well, that is another approach, I am indirectly using that $log(|z-\zeta|)$ is harmonic away from $\zeta$ by encapsulating it into saying that it is the real part of a single valued branch of $log(z - \zeta)$ (away from $\zeta$), another approach as you say would be to justify exchanging $\Delta$ and the integral , maybe the latter is easier
I think ultimately because we are away from $E$, we could use something like DCT to justify swapping $\Delta$ and $\int$.. but i have not made this precise
yeah, I think the DCT approach works fine actually, its not very elegant, but one can show that $\partial_{x} (\log(|z - \zeta|)$, $\partial_{y} ( \log(|z - \zeta|)$ are dominated by constants on $E$ (given $z$ is in some small ball away from $E$), and show the same thing for higher derivatives, which will then justify the swap (just apply DCT)
i.e. we have justified swapping $\partial_{z}$ and then swapping $\partial_{\overline{z}}$ when the integrand is $\partial_{z}(\log(|z - \zeta|))$, so we can swap in a $\Delta$
fair enough, do you think the justification ought to basically boil down to being away from $E$ meaning that $|z - \zeta|$ is bounded away from zero and above?
i wouldn't presume to be an expert on what you'd call it. trivial category is fine with me. i don't know of a standard name. maybe you could use more specific terminology depending on your application.
one usually uses 1 to denote terminal objects, bold font to denote categories, and the category with one object and one morphism is the terminal object in the category of (small) categories
@TedShifrin one think i do find myself thinking about with the gaussian trick. suppose i have homogeneous polynomials $f_1(x),f_2(x)$ such that $\int_{\mathbb{R}^n} e^{-\|x\|^2} f_1(x)f_2(x)\,dx=0$. then homogeneity converts this into $\int_{S^{n-1}} f_1(\sigma)f_2(\sigma)\,d\sigma=0$
so the gaussian trick would seem to allow one to get orthogonal functions on the unit sphere by looking for orthogonal polynomials w/r/t the multivariate Gaussian
subject to the requirement that the polynomials be homogeneous
i would have thought this is the same notion as harmonic polynomials but my computations thus far haven't quite agreed with that
It is unfortunate that most textbooks and teachers do not teach how to use asymptotic analysis. This is an intuitive and powerful technique that can be used even if you want to obtain an ε-δ proof.
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
\def\abs#1{\left|#1\right|}
$
First let's use the technique...
^ I do not understand why this got downvoted almost immediately after I posted it, and I don't see an error. This is despite being a clean and simple solution, in contrast to the other answer (which is so long and ugly that I don't even want to check).
i think i may see how what i'm saying above explicitly relates to spherical harmonics
something like: (complex) harmonic polynomials of degree $m$ are of the form $(x\pm i y)^k P_m^k(z)$
where the $P_m^k(z)$ are even or odd polynomials
i haven't really seen orthogonality of homogeneous polynomials w/r/t multivariate Gaussian though
(if one drops homogeneity, then we can use products of hermite polynomials as a basis. but losing homogeneity loses the point of all this)
@TedShifrin alas, i don't think this will actually help me directly with the higher cases i mentioned. the integration was 'really' a group integral over $SU(2)$
which you could convert to an integral over SO(3) b/c double cover etc
for higher cases i think i can't get away from $SU(n)$
@Semiclassic I've never done an explicit integral over $SU(n)$ in general, although I have used my favorite differential forms integrating over Stiefel manifolds and Grassmannians.
@TedShifrin OK...let me formulate it properly...in a few minutes
Firstly, can I just talk to you about the notion of $|x|$. Now I understand the definitions and the idea that is represents a 'distance' from $0$. But.... and here ..forgive me if this sounds clunky... I want to know what you understand, for example when you see $|x|$ versus $|-x|$
OK... let me do it for you :-) So, given something like $-x$, as you say applying strictly the definition, as you point out, it will remain $-x$ for $\geq 0$ and then as you say $-(-x)$ for $\leq 0$.
OK... that helps a lot. Nailing down those concepts early on is really helpful and easy to gloss over :-) Which leads me to the second ( hopefully easier) notion.
Okay. So, I was able to construct a bounded linear functional on $L^{\infty}[0,1]$ that is not of the form $h \mapsto \int_{[0,1]} gh$ for any $g \in L^1[0,1]$. How does one derive from this that $L^1[0,1]$ is not reflexive as a Banach space, that $L^1[0,1]$ is not isomorphic to its double dual?
@TedShifrin That's what I was doing on my blackboard just a moment ago, but I couldn't see the connection. I thought I need the Riesz representation theorem, but that didn't seem helpful.
If $g<h$ then there is not $M_g\to M_h$ with degree $1$ map where $M_g$ denotes the closed orientable surface of genus $g$. This is one of the exercise problem in Hatcher 3.3.11 and I can prove this using 3.3.10 actually. But it seems there is way to prove this using cup product. How does the proof go basically?
@TedShifrin I guess I also know that $L^1[0,1]''$ is isomorphic to $L^{\infty}[0,1]'$, but now I have a lot of isomorphisms lying around it's getting a bit complicated.
@TedShifrin hmm... I think in my case, UCT say that $H^*(M_g)\simeq Hom(H_*(M_g),\Bbb Z)$ so using naturality of UCT, I can somehow relate $f^*$ and $f_*$.
Basically the degree is defined in terms of homology
