2:36 AM
I have neat algorithms, the type Numberphile might do videos on I suppose. Curious whether there's any sort of value in these? Often when I've shared in the past the usual "what is this good for'" arises. It can be offputting to hear when it's unlikely that anyone has developed the same methods prior
zac: depends on what you mean by 'value,' i guess? most algorithms, even very clever ones, do not have a wide audience, just because even people who work with them all day can't usually be bothered to care about other people's algorithms, without some very good motivating reason to do so.
it's not, like, a killer electric guitar riff, where random people might click in and be like 'oh, what's this.' you won't even get the click
if i were designing math content with an exclusive goal of getting as many clicks as possible, i would try to work in the realm of 'algorithms for doing things that people might remember from elementary school, that are different in some way from what they might remember, and can be made to look 'easier' than what they remember, at least on judiciously chosen examples at the length of a short video."
i sometimes see stuff go semi-viral on social media in that realm, because people can relate to it (even if it is mathematical hogwash)
and even then, that level of virality is not what you'd get from putting the same effort into a prank on tiktok
Haha nice music reference leslie! My hopes I guess are either that I might be seen as a prize student and get better opportunities than I can currently afford, or that others might see what can be explored in the ideas throughout the meantime. These algos while neat r far from simple enough to gain steam I reckon. Is this a fine space to share one?
Is there a shorthand for the extension field $\mathbb{Q}((-1)^{1/2}, (-1)^{1/4}, (-1)^{1/8}, \cdots)$?
0

How can I solve this integral? $$\iint\limits_{0\le x\le y\le1}\!\sqrt{1+x^2-y^2}\,{\rm d}x\,{\rm d}y$$ I know from Wolfram Alpha that the answer is $\frac13\ln2+\frac16\approx0.39772$. However, doing this explicitly seems like a nightmare. Is there any nice way of doing this?

akiva: i have to ask, how does the title relate to the post? :)
2:48 AM
Oh I forgot to edit that out
lol one minute
@ZacU. i don't think anyone would object too highly, although they also might not care very much. :) anything that is likely to run more than a few lines, or resembles 'code' more than math, is probably best stored somewhere else (e.g. a pastebin like site) and only linked to in the chat, so people who aren't interested can just ignore one link and don't have to scroll past a ton of text
akiva: haha one commenter also all over the fisher and the fish
@MikeBattaglia I originally had phrased it in terms of a probability question: a fisher is somewhere on the edge of a square-shaped lake, and a fish is somewhere inside of it. What is the probability that the fisher is closer to the center than the fish is? Using the above integral, the answer is $\frac{1-\ln2}3$. While drafting the question I figured it was cleaner just to discuss the integral on its own. — Akiva Weinberger 22 secs ago
Nothing too crazy and formatting isnt great. But maybe someone out there could use either the methodology or engage further with whatever direction may develop
@leslietownes Solving the probability puzzle, with or without the help of this integral, is left as an exercise for the motivated reader ^_^

3 hours later…
6:23 AM
@XanderHenderson damn. i can understand

1 hour later…
7:32 AM
Basic question. Can we speak of the pdf of the positive/negative part of a random variable of arbitrary sign? Suppose $X$ has pdf $f(x)$. In other words, can we say the pdf of $X^+$ is $f(x)$ for $x\geq0$ and that of $X^-$ is $f(-x)$ for $x>0$?
7:50 AM
psie the question is not phrased very clearly. the "positive part" of a random variable X (by which i understand something like max(X,0)) is pretty clearly a random variable. you can ask if it has a pdf and if so what it is. you separately seem to be asking that if X has pdf f(x) then ... i don't know what. i'm not sure that it does make sense to say that the pdf of a random variable "is" a function g(x) "for [a limited interval of x values]."
if you mean, can you recover the probability that a random variable is in some collection of intervals by integrating that function over those intervals, that's one thing. but that's not usually what one means when one says that one function "is" a pdf of a random variable
ok, I believe my assertion that the pdf of $X^+$ is $f(x)$ for $x\geq 0$ is missing a normalization factor.
you should also maybe clarify what you mean by saying that in the first place
i don't mean to cast doubt on whether you can do this, as much as i mean to suggest that you should write out what you are trying to say instead of "the pdf of X^+ is f(x) for x >= 0"
@leslietownes but $X^-$ and $X^+$ are nonnegative random variables, continuous if $X$ is (since $\max$ and $\min$ are continuous if the arguments are), so I don't see the problem in asking what the pdf is for $x\geq0$. Doesn't it sound sound? :)
8:06 AM
i guess i mean, if you're asking yourself this question in the first place, you may not understand why X^+ has a pdf
so maybe you should investigate that from first principles as opposed to 'oh its a continuous function of a continuous thing'
i.e. consider what the pdf, if it exists, must be for all values, not just nonnegative values
identify the pdf in full
ok
your use of the term "normalization factor" suggests that you've intuited that in general, it will not be the case that X^+ has a pdf given by, like, literally having, f(x) if x >= 0 and 0 otherwise. probably because you sense that this function won't be a pdf in general
but the term "normalization factor" doesn't, at least to my ear, adequately capture what you need to do to fix this
for any t > 0 and any Y one has Pr(Y < t) = Pr(Y <= 0) + Pr(0 <= Y < t)
for any t < 0 and any X one has Pr(max(X,0) < t) = 0
two relevant facts that maybe help organize the thinking a little here
ok @leslietownes, I'm trying to formulate why I wrote the question in the first place, let's see if I can write something that makes sense
to tie it back to my first remarks, when X has pdf f(x), it's certainly true that if 0 < a < b then one has Pr(a <= max(X,0) <= b) = Pr(a <= X <= b) [because the two events inside Pr( ) are the same thing], and the latter is the integral from a to b of f(x) dx, i.e. you can compute the probability that max(X,0) will land in an interval of positive reals by integrating f(x) over that interval
but that's not quite the same thing as identifying any pdf
8:27 AM
@leslietownes Maybe it seems a bit unrelated, but this is the reason I was asking my question in the first place. If $X$ is a random variable with density $f$ over $\mathbb R$, and if it has a moment generating $M_X$, then $$M_X(-t)=\mathcal{L}\{f(x)\}(t)+\mathcal{L}\{f(-x)\}(-t),$$where $\mathcal{L}\{f(x)\}(t)$ is the one-sided Laplace transform.
I'm trying to understand the two terms on the right of the displayed equation. The Laplace transform of a nonnegative random variable is $\mathbb E[e^{-tX}]$. I am trying to identify which random variable corresponds to $\mathcal{L}\{f(x)\}(t)$ and which corresponds to $\mathcal{L}\{f(-x)\}(-t)$.
8:39 AM
Maybe there is no such correspondence...
9:06 AM
@psie indeed, there doesn't have to be
yeah, there doesn't have to be one, true, unless the pdf is somehow symmetric around $0$.
If you do the actual calculation, $\max(X, 0)$ is whatever, the same as $X$ on $(0, \infty)$, and $\max(X, 0) = 0$ has probability $P(X\leq 0)$ so if this is positive, there is no pdf
so $X^+$ and $X^-$ just don't have a pdf when $X$ does
now if $P(X < 0) > 0$ and $P(X> 0) > 0$ then you can consider random variables $Y$ and $Z$ with pdf's $\frac{f(x)1_{(0, \infty)}}{P(X > 0)}$ and $\frac{f(x)1_{(-\infty, 0)}}{P(X<0)}$
and then $M_X = M_Y+M_Z$
in the way you want
ah yeah, thanks :)
sorry not $M_X = M_Y+M_Z$
$M_X = P(X > 0)\mathcal{L}_Y + P(X < 0)\mathcal{L}_Z$
where $\mathcal{L}_Y$ is the one-sided Laplace transform or whatever
I will double down on what Leslie said and ask - why?
you can't boil it down to equalities of non-negative random variables this way
$x+y = z+t$ doesn't imply $x = z$ and $y = t$
@Jakobian because now we can claim the uniqueness of $M_X$ via the uniqueness of Laplace transforms (and not go the route via characteristic functions, although you could go that route too if you find it easier to work with Fourier transforms). We know (or I know at least) that the Laplace transform for $Y$ and $Z$ are unique, so hence $M_X$ is unique.
9:21 AM
@psie you know its unique for non-negative ones, see my other point
4 mins ago, by Jakobian
$x+y = z+t$ doesn't imply $x = z$ and $y = t$
if you actually go and write this argument you will see what I mean

5 hours later…
2:00 PM
Choose four numbers uniformly from $[0,1]$ and label them $a,b,c,d$ from least to greatest so that $a<b<c<d$. What is the probability that $a^2+d^2>b^2+c^2$?
3:00 PM
Hey all. Can anyone help me bound the integral $$I = \oint_{\Gamma} \frac{\pi}{\sin (\pi z)(z^2+1)} \mathrm dz$$ where $\Gamma$ is the contour of a big circle centred around origin with radius $R$, to show that $I\to 0$ as $R\to \infty$? I saw this being used in an MSE answer where the detail was left for the reader.. I’ve not been able to get ML inequality to work because of the $\operatorname{cosec}$..
3:10 PM
@AkivaWeinberger Let $X_1, X_2, X_3, X_4$ be the order statics in an increasing order. \begin{align*} P(X_1^2+X_4^2 > X_2^2+X_3^2) &= \int_{x_1^2+x_4^2 > x_2^2+x_3^2} 4! dx_1dx_2dx_3dx_4 \\ &= 4!\times \text{volume of }\{(x_1, x_2, x_3, x_4)\in [0, 1] : x_1^2+x_4^2 > x_2^2+x_3^2\}\end{align*}
\begin{align*} P(X_1^2+X_4^2 > X_2^2+X_3^2) &= \int_{x_1^2+x_4^2 > x_2^2+x_3^2, x_1 < x_2 < x_3 < x_4} 4! dx_1dx_2dx_3dx_4 \\ &= 4!\times \text{volume of }\{(x_1, x_2, x_3, x_4)\in [0, 1] : x_1^2+x_4^2 > x_2^2+x_3^2, x_1 < x_2 < x_3 < x_4\}\end{align*}
this is because the joint distribution of $(X_1, X_2, X_3, X_4)$ is given by $f(x_1, x_2, x_3, x_4) = 4!$ for $0\leq x_1 < x_2 < x_3 < x_4 \leq 1$ and $0$ otherwise
so its the volume of this region, whatever that is, multipled by 16. You may have success in calculating the exact value by first integrating over, say $x_2$ and $x_3$, using polar coordinates, and then repeating
@Sahaj $|z^2+1|\geq |z|^2-1 = R^2-1$ so that's one term bounded
$$|\sin(\pi z)| = \frac{1}{2}\left|e^{iz}-e^{-iz}\right| = \frac{1}{2}e^{\text{Re}(z)}| e^{2iz}-1|\geq \frac{1}{2}e^{\text{Re}(z)}(e^{-2\text{Re}(z)}-1)$$
$$|\sin(\pi z)| = \frac{1}{2}\left|e^{i\pi z}-e^{-i \pi z}\right| = \frac{1}{2}e^{\pi\text{Re}(z)}| e^{2\pi iz}-1|\geq \frac{1}{2}e^{\pi \text{Re}(z)}(e^{-2\pi\text{Re}(z)}-1)$$
actually scrap this idea, I think we should just calculate $\int_\Gamma \frac{1}{\sin(\pi z)} dz$ from residue theorem
I'll stop pretending like I know what to do, @robjohn maybe you could help Sahaj with this problem in complex analysis?

1 hour later…
4:37 PM
@Jakobian I should say I know the answer.
It's ln(2) exactly
The real question is if there's a nice derivation.
@Jakobian that’s right, but I’m not sure to work with the rest you’ve given..
I’ll appreciate if someone could help! Thanks.

1 hour later…
5:58 PM
0

My computer science college professor starts class at nine o’clock in the morning, so people would normally wake up by eight o’clock. However, I feel that a twenty-year-old should not wake up as late as eight, but they should always wake up as early as four o’clock am to make sure they start thei...

6:38 PM
@Sahaj Note that $\pi\csc(\pi z)$ has a residue of $(-1)^n$ at $n\in\mathbb{Z}$. If $\Gamma$ passes through the real axis at $\mathbb{Z}+\frac12$, then as the radius of $\Gamma$ gets bigger, the max of $\pi\csc(\pi z)$ on $\Gamma$ is $1$.
I'll try to compute the integral in a bit.

2 hours later…
8:40 PM
Does anyone know what the dual of the Hilbert space $H^\infty(\mathbb R) = \bigcap_{k}H^k(\mathbb R)$ is? I can't seem to find it in the literature...
@robjohn Thanks for the response; Growing the contour on the real line as $n +\frac12$ with integer $n$ does the job. With $R\to \infty$, $I\to 0$. With the use of residue theorem and using it to find $I$ one can calculate the sum $$\sum_{r=1}^{\infty} \frac{(-1)^r}{r^2+1}$$ which is what motivated me to take up this topic in the first place.
As a younger student, I find it quite amazing how complex analysis can be used to find this sum. It always amazes me how distant various fields of mathematics appear, but just many times beautifully connect with each other.
anak: if you can't find it in the literature, why do you need to consider it at all? :) all joking aside, (1) i don't recognize any of that notation in the hilbert space context, so more context would be needed, (2) all hilbert spaces (however realized) are isometrically isomorphic to their duals, so i'm not sure i'd understand the question even if i ignored that i didn't understand the notation
@leslietownes Sobolev spaces
okay, i don't recognize "sobolev spaces" in the hilbert space context, so more context would be needed
for some reason anak thinks that while $H^k(\mathbb{R})$ are all Hilbert spaces, with different norms of course, then $H^\infty(\mathbb{R})$ has a natural choice of a norm as well
8:53 PM
oh yikes this is some ongoing thing
yeah anak if this has to do with what sobolev spaces are you are going to have to let go of the idea of everything being a hilbert space
and do whatever the opposite of letting go is, to the idea of a certain amount of arbitrary choice sometimes being involved in how different resources define the seminorms of interest to them
e.g. sometimes the topology is more important than the seminorms used to induce/define it, and it is not always going to be as simple as "this notation means this norm"
Oh, is $H^\infty$ not a Hilbert space?
this goes along with a general thing that i sometimes harp about, which is, even when notation for function spaces as sets of functions is well established (which is way less often than people think it is), the norms or seminorms that you put on top of those sets of functions are rarely conveyed by the notational choice itself
and it often helps matters to put those things front and center in the conversation and not regard them as somehow just 'part' of the notational backdrop
anak: this is maybe a good homework exercise. how has H^infty been defined for you? is it defined such that you can expressly compute the norms of specific elements, or otherwise test whether the parallelogram law is satisfied in particular cases?
I just assumed it was based on the notation, you are right about that. Terrible choice of notation, since the H in H^k is supposed to stand for Hilbert...
do you know abstract theory about topological vector spaces or normed vector spaces that might let you distinguish hilbert spaces from other spaces on some basis other than calculations?
i'm not sure that the H in H^k is supposed to stand for hilbert
I am pretty sure it is.
9:00 PM
who's using it?
en.wikipedia.org/wiki/Sobolev_space#The_case_p_=_2 seems to suggest it, though by no means definitive.
oh god, wikipedia
okay, well good for them
It's also the notation I have seen used in any functional analysis course, and I can't really come up an alternative.
Also it was expressed here: mathoverflow.net/questions/90527/…
well, sometimes people use H^p to refer to hardy spaces which are differently defined and generally not hilbert spaces
just generally, if everyone could erase from their minds that function space notation is at all standardized, even if in a lot of contexts it can feel like it is
it would be a better world
Well notation is supposed to be suggestive and helpful for communicating ideas.
9:04 PM
communicating also helps
that MO page by the way mentions the use in the context of hardy spaces
minimalism of this type is something I don't like, explaining every bit of unclear information is what people should be striving for
and has a comment suggesting that at least one person says that the H maybe came from the cyrillic for nikolski
all of which is just, it helps to provide the definition of whatever the H^k are alongside any question about what some H^p is or isn't
@Jakobian What is unclear for one person isn't unclear for another. It's about balancing context. The great thing is that just like leslie did, one can ask for clarification.
In any case, I am looking at this from the Lie group side of things. I have a Lie algebra which has a (semi-direct) factor of $H^\infty(\mathbb R^n)$ (defined as the intersection of the $H^k$), and I was just looking at the dual algebra but it would not require a Hilbert structure anyway..
even when things have names like "the sobolev space" or "the schwartz space" like, there is enough variation from author to author in definitions that i'd still not be insulted if someone 'reminded' me what they meant by that
if someone says the lebesgue space L^p[0,1], okay i probably wouldn't ask them what they meant by that, unless they were asking something that really seemed like it might depend on how they were constructing that space
even then i guess i might ask a lot, because it bears upon how much measure theory a person might know or be able to use
anak: separate from any context, the simplest description of the dual of a concrete space is often just that, "the dual of [the concrete space]"
with "better" descriptions often coming only at the cost of work and theorems
9:12 PM
Awhile back I was reading something which specified the elements of the dual as densities, but I am having a hard time seeing it.
But I am not the most savvy with analysis (you might be able to tell), so I defaulted to seeing if someone had already written something on it.
yeah, that can also be kind of a rabbit hole, in terms of how useful that other description is, i could see a google search turning up a lot of junk
"oh the dual is this space of signed 'weight-like currents' on a space that isn't quite the space you started with but is this completion of it with respect to ... but other than that it's basically the same thing and of course this analogy only holds when everything is locally compact hausdorff and it usually isn't" ... OK, great
i'll add that to my bag of tools i'll be using every day
The mappings $H^{k+1}\to H^k$ given by $f\mapsto f$ are weak contractions, so I think what we are doing is we are taking a colimit $H^\infty$ in the category of Banach spaces and operators of norm $\leq 1$
and this is probably what anak means by $H^\infty$
apparently, this exists by Semadeni-Zidenberg theorem, although I'm not quite sure what the norm on such space is
Another paper on the same topic I was looking at conflated $H^\infty$ with $C^\infty$, but I would have thought it would be smaller.
9:27 PM
oh, the explicit form is $\{f : \sup_k \|f\|_{H^k} < \infty\}$, of this colimit
but that's not exactly the intersection but something smaller
the question of "what type of topological vector space is $H^\infty$" prevails
oh I guess we are talking about it as a Frechet space?
Yes it would be Frechet if you want a manifold.
I believe, then $H^\infty$ will be Frechet but not Banach, and so its dual won't even be Frechet, in some sense
14

Let $\mathcal{F}$ be a Fréchet space (locally convex, Hausdorff, metrizable, with a family of seminorms $\{\|~\|_n\}$). I've read that the dual $\mathcal{F}^*$ is never a Fréchet space, unless $\mathcal{F}$ is actually a Banach space. I'd like to know in which ways the dual can fail to be Fréche...

10:22 PM
@Sahaj Great! I’m glad it helped.
10:34 PM
@Sebastiano averi bisogno di una mano per fare un disegno con Tikz. Se riesci a darmi una mano mi faresti un favore.
Il disegno è come questo solo che le linee verticali le vorrei tratteggiate, sto avendo problemi a fare disegni di funzioni con discontinuità di tipo salto
10:54 PM
1

Reduce a string up to idempotency Objective Given a string consisting of printable ASCII characters (!0x21 ― ~0x7E), treat it as an element in the free idempotent monoid (as defined below) over the printable ASCII characters, and output the same element represented in the fewest characters as pos...

Does this uniqueness proof look good?
11:15 PM
@DannyuNDos I don't get what you're proving
okay I get what you're proving
> Since every member of the class x belongs has α as the first character, and likewise for y and β, it follows that α=β.
not sure if this convinces me
11:32 PM
Play with idempotency, and you'll see.
quick question: if $a\in \Bbb R$, then $\delta(ax)=\frac1{|a|^n}\delta(x)$ right?
I consider delta in $\Bbb R^n$
11:53 PM
@DannyuNDos I do see. But its not about me, is it?