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12:24 AM
I don't want to be an ass, but it's almost as if the authors of this stats paper are just throwing around technical terms willy-nilly.
 
never heard of anyone doing that, ever. certainly not in stats
 
Here is a sentence from page 8: "We first described a simple search over a finite search group $\mathcal G$, and then a more complicated algorithm for when $G$ is a compact connected group." So naturally you'd expect to see the word "connected" used in proximity to the description of some algorithm, right? Wrong. The word "connected" appears in only one other place in the paper: the abstract.
Either I'm confused or the authors are. I don't see any alternative
 
its conceivable that they aren't close enough to math to realize that some people use those words with a reasonably standardized, non-impressionistic meaning
 
Oh well. As we already discussed, these things don't necessarily undermine the main content of the paper
 
or they know they're using some of those hypotheses but maybe not sure which ones so they're spamming all of them
yeah, applications oriented stuff is often reliable for what it focuses on but maybe not past that point
 
1:21 AM
Presumably a compact, connected topological or Lie group.
 
2:04 AM
I'm researching this now, but if $(X, \mathcal F, \mu)$ is a measure space, and $f, g$ are measurable functions into the metric space $(S, d)$ (equipped with the Borel $\sigma$-field with respect to the topology from the metric), then do I need $S$ to be separable in order for $d$ to be $\mathcal F$-$\mathcal B(\mathbb R)$ measurable? Does that sound right?
 
2:51 AM
I think the (somewhat superficial) explanation is that I need $\mathcal B(S \times S) = \mathcal B(S) \otimes \mathcal B(S)$ in order for the $\sigma$-fields to agree and conclude that the metric is a composition of measurable maps, hence measurable, and this equation is ensured when $S$ is separable
 
 
2 hours later…
4:30 AM
@copper LOL
 
4:53 AM
:-)
 
the vision of the jumpsuit can be so strong it is blinding.
 
i just can't wait.
 
5:26 AM
I ain’t got nothing.
 
6:08 AM
so I'm trying to transcribe some function I found on some youtube video (no existing transcription of it) but it seems I'm struggling with part of it: i.imgur.com/V4hlLGz.png
I know the dot symbol is for multiplication, but I'm not sure what to make of the purple number in the image? is this exponentiation?
 
sounds like a good question for whoever made the video
its probably some made up thing unique to them
maybe they explain it somehow in the video?
 
hmm, that's fair. I did think of asking them, but I thought it might be faster to ask here since I'm curious
they don't explain no, at least as far as I'm aware.
also I don't think it's unique to them since I have found some similar notation on math.se but I could be wrong
 
well, it's nonstandard notation, and maybe even at odds with conventional notation, e.g. it looks like the function is just x plus some number, where they've put in a lot of effort to make the number look visually confusing
 
I see hmm
 
with additional context (such as what the video is, what the function is purportedly doing, where you have seen the notation before) it might work as a question on MSE
but by and large, people who put equations in videos and visual art tend to be more interested in how they look than in what they mean or do, and its very easy to visually conjure up stuff that has no meaning or a meaning that is particular to one context but maybe not to a wider community
 
6:18 AM
@leslietownes That's a good advice, thanks. I'm just not sure how to frame the question, maybe something along the line of "struggling with transcribing this function" but that might attract more downvote, and might not be useful
@leslietownes That's true for a lot of math related videos yeah. Which is exactly why I'm trying to transcribe it, so I can verify if it work as it did in the video
 
yeah i dunno how to frame it as to be MSE appropriate. maybe as a question about the origin or meaning of the notation, where it was first used, etc. but without more context, i wouldn't be surprised if someone answered "the origin of this notation is this video you're not telling us anything more about"
 
I would of course provide the video link, but as you mention too, it's hard to frame as a good question :/
 
maybe they are footnotes to explanations that appear in a written document :)
the presence of powers of two and the name "zero threshold" and the fact that it might just be sending x to x plus something suggest maybe it has something to do with testing or breaking a machine implementation of some kind of arithmetic
thats really the only explanation i could think of for why you would somehow distinguish between a computational representation (e.g. "1+2+3") and what that number is (e.g. 6), if somehow having a machine compute 1 + 2 + 3 according to its machine + and its machine 1 and machine 2 and machine 3 was "part of it" somehow
but it also just looks like someone started typing random numbers around powers of two and arithmetic signs until they filled up the frame of a video
 
yeah, this was found through bruteforcing/computer search (according to their video). This is for a weird neural network architecture it seems: youtu.be/Ae9EKCyI1xU?t=2006 (timing to the same function as the image above)
it's not that amazing as it is quite slow compared to existing ones (if not the slowest) but I was just curious as mentioned.
 
6:37 AM
@robjohn yes
 
Does this series $\sum_{n=1}^\infty e^{-n^{\epsilon}}$ converge? for small fixed $\epsilon>0$.
I know it converges but how can I prove?
oh it seems related to the gamma function
 
no matter what epsilon > 0 is, you have n^epsilon > 2 ln(n) for all n sufficiently large. is that enough?
 
6:53 AM
oh that's better
 
7:24 AM
Hi :) Please may I have some feedback on the following?
-1
Q: Does there exist a topos with these $n+2$ truth values?

ShaunThis question is based on the answers to this question. The Question: Let $n\in\Bbb N$. Let $N$ be a set with $n+2$ elements, labelled $0$ to $n$, and the $(n+2)$th element labelled $\infty$. Suppose we have a function $$\begin{align} t:N &\to N,\\ 0 &\mapsto 0,\\ m &\mapsto m-1,\quad\text{(for }...

(I don't get the downvote.)
 
8:01 AM
@user123234 Then consider $\lim\limits_{n\to\infty}\left(1+\frac1n\right)^{nx}=\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n$ by substituting $n\mapsto n/x$
 
 
2 hours later…
9:43 AM
4
A: The prime ideals of a quotient ring

C. FalconFirst, let us recall the following: Proposition. Let $f\colon R\twoheadrightarrow R'$ be a surjective ring homomorphism and let $P$ be a prime ideal of the ring $R$ containing $\text{ker}(f)$, then $f(P)$ is a prime ideal of $R'$. Proof. First, $f(P)$ is a subgroup of $R'$, since $f$ is a group...

This answer had wrong theorem so I decided to edit it. Since it's old, its used as reference for others so I felt obligated to do so
 
makes sense to me. the editing is minimal enough. it is a shame for such an old and probably commonly-read post that the exposition isn't clearer, but there isn't much to be done about that.
 
@leslietownes well... I had to also include that $f(P)$ is proper in the proof
 
yes. i mean, the editing is minimal enough to be consistent with the truth, but if one had a freer hand, one would probably also write the argument more clearly.
 
There was a lot of overlook in the answer. I don't think a lot of ppl read what is being said
 
i know that MSE generally frowns on editing other users' posts, and i think that substantially changing the exposition would probably run afoul of at least some people's interpretation of what is OK. but your edits are very minimal, just to the point of making it not wrong anymore.
 
9:56 AM
Its like there is some invisible wall. This could have been fixed sooner
 
if the written statement has an error more than a typo, then it's worth to edit. But there are some types of people doing pointless meaningless edits like adding tags for example or changing latex code even though it's very readable.
 
yeah, there is even some earlier commentary about the argument, that somehow didn't get to this issue.
that's what made the answer so potentially dangerous, it's not clear where the author is making a mistake, or if they were making a mistake, how to fix it. the exposition was not clear enough to shed light on where essential hypotheses were being used.
i could see an answer like that generating subsequent questions on the site.
one: that's interesting. personally i don't mind typo fixes, or improvements on the latex, at least when it is done to otherwise recently active posts. even the revivification of ancient posts for that purpose strikes me as harmless, maybe even helpful, at least if it's on a question people might return to.
 
I was worried that some student will google it, and take it as truth
 
which, of course, is never how it happens :) it's usually someone doing like 10 different trivial edits on ancient posts for some reputational purpose or something.
 
its even accepted after all
 
10:04 AM
I'm working with a tricky sequence of functions. Consider $$f_k(x)=\frac{x^{2^{k+1}-1}}{(2^2-1)^{2^{k-1}}(2^3-1)^{2^{k-2}}\cdots(2^{k+1}-1)}.$$ The first few terms are $f_1(x)=\frac{x^3}{3}$ and $f_2(x)=\frac{x^7}{3^27}$. I'm supposed to find an $a>0$ such that for $|x|>a$, $f_k(x)$ diverges. A TA has noted the following; $$f_k(x)=\frac{(f_{k-1}(x))^2 x}{(2^{k+1}-1)}\geq \frac{(f_{k-1}(x))^2 x}{2^{k+1}}.$$
The TA has suggested that if $f_k(x)$ is to diverge, then it needs to increase by, say, a factor of $2$. So $$\frac{(f_{k-1}(x))^2 x}{2^{k+1}}\geq 2\cdot f_{k-1}(x).$$ Why would the sequence diverge if it were to increase by a factor of $2$?
 
the thing that's a little funny, is that the author did seem to be careful about their argument, but not in that one particular place where he claims that $f(r_1r_2)\in f(P)$ implies $r_1\in P$ or $r_2\in P$. That $f(P)$ is proper is a simple overlook.
The other funny thing is that justification of both of these requires that $\text{ker}(f)\subseteq P$
so if the author thought more about these steps, this wouldn't happen
I suppose this shows that you don't have to be right to get a good answer on this site. You just have to write it well
People will join the train of voting up your answer blindly
people prefer not to think critically, especially when the question seems to be easy
okay I'm done rambling. This is stupid
@sunny try to treat $f_k$ as partial sums of a power series?
no probably can't
$|f_k(x)| \leq \frac{|x|^{2^{k+1}-1}}{3^{2^{k-1}+...+1}} = \frac{|x|^{2^{k+1}-1}}{3^{2^k-1}}$
 
10:23 AM
I think the reason we want it to increase by a factor of $2$, is because we want to say something like $f_{k+1}(x)\geq 2^k f_1(x)$
 
well the question is clearly about asymptotics of $a_k = (2^2-1)^{2^{k-1}}...(2^{k+1}-1)$
 
yeah, does this grow faster than $x^{2^{k+1}-1}$?
 
So we might try taking $\log(a_k) = 2^{k-1}\log(2^2-1)+2^{k-2}\log(2^3-1)+...+\log(2^{k+1}-1) \leq 2^{k-1}\cdot 2^2 + 2^{k-2}\cdot 2^3 + ... + 2^{k+1} = k2^{k+1} $
@sunny is $a$ supposed to be minimal?
also in above inequality we can go sharper because $\log(x)\leq x-1$
 
@Jakobian no, it just needs to be some $a$
the answer given is $a=\sqrt{8}$, for example (so I think $a$ could be smaller and bigger)
 
$\log(a_k)\leq (k-1)2^{k+1}+2$ so that $a_k\leq e^{(k-1)2^{k+1}+2}$
well I suppose in this exercise we want to bound it from above
below*
 
10:38 AM
@sunny the TA found $a=(42)^{1/4}<\sqrt{8}$ through the last inequality I stated here, by setting $f_{k-1}=f_1$ and solving for $x$
 
 
2 hours later…
1:07 PM
@onepotatotwopotato By the Archimedian property, there is an $N\in\mathbb{Z}$ so that $N\epsilon\gt1$. $$e^{-n^\epsilon}\lt\frac1{1+n^\epsilon+n^{2\epsilon}/2!+\dots+n^{N\epsilon}/{N!}}$$
 
 
1 hour later…
2:12 PM
Is there an error in this proof? It doesn't change the proof, but methinks that the final centred equation should have a $\geq$ and not the strict $>$
 
Ok, this is pretty embarrassing but I can't figure out this elementary school problem;
 
that is, $\deg(s_{(1)} - s) + \deg q \geq \deg q$
 
It is in Turkish but I think it is self explanatory
 
The reason is because the difference $s_{(1)} - s$ could be a constant (nonzero) polynomial (of degree 0)
 
@yasar No, it is not self explanatory. I have no idea what is being asked.
 
2:18 PM
There is supposed to be a pattern in there, question asks what is star minus square
Only hint is wheel is arranged according to a rule
 
What is the answer? $16$?
 
Don't know yet, how did you arrive at that number?
 
It’s absurd. No clue what “the” pattern must be.
 
I agree it is absurd, I thought maybe I was missing something simple
 
Well, I am too.
This kind of thing gives math a bad name.
 
2:25 PM
@yasar One possible pattern is that the product of opposite numbers is $24$, but of course there can be other patterns and other answers
 
Makes sense, my thinking was it was about difference between slices so we could solve for "star minus square" without solving for those numbers individually
Anyways, I am just glad that problem sucks and It wasn't me :)
 
It's almost impossible to tell what pattern the author thought, say I ask you what will be the next number in this sequence, $\pi, \pi, \pi, \pi, \pi, \pi, \pi, \dots$, you may think it will be $\pi$, but the answer is $\sim \pi - 4.62\times 10^{-11}$
 
2:51 PM
@EE18 yeah
 
Thanks @Jakobian :)
 
 
2 hours later…
4:52 PM
anyone know where i might find information on the SDE $dX_t = - cX_t dt+ a(x)\sqrt{2}dW_t$ for a nice function $a(x)$ say bounded, lipschitz, continuously differentiable etc.
particularly its steady state
 
I don’t know stochastic stuff, but what does steady state mean?
 
@TedShifrin this isn’t even math, it’s the kind of stuff you’ll see on IQ tests, I had mine tested and I saw stuff like that on it
The author of the question is asking you to guess what pattern they had in mind
 
5:12 PM
If there is a unique answer….
As I said, I strongly dislike such.
 
@TedShifrin more often than not, there isn’t.
 
5:28 PM
To be fair, coming up with the existence part can often be hard and therefore (perhaps) worthwhile
I am struggling with one aspect of the generalization of formal power series in one to $m$ indeterminates
For reference, here are the two parts of the discussion which I am trying to contrast:
In particular, I am trying to make sense of the discussion in the second picture which begins "Set $X:=(X_1,...,X_m)$..."
Whereas (8.10) can be proved (depsite their putting the $:=$, I am fairly sure it can be proved), it's not clear to me that the corresponding $X_\alpha^\beta$ expression can be proved (and that it's therefore reduced to a definition). Is this understanding correct?
The notation is fairly unclear in that line beginning "Set $X:=(X_1,...,X_m)$...". What is being defined in writing $X:=(X_1,...,X_m)$? Are we defining $X$ as the function for which $p_\alpha = 1$ iff $\alpha = (1,1,...,1)$?
 
6:17 PM
9
A: Normed space $C^2[0,1]$ with norm $\lVert f\rVert:=\max_{t\in[0,1]}\{\lvert f(t)\rvert+\lvert f''(t)\rvert\}$ is Banach space

Rhys SteeleIt is a standard exercise to see that $C^2([0,1])$ is a Banach space for the norm $$\|f\|' = \max_t (|f(t)| + |f'(t)| + |f''(t)|).$$ You can use the same approach as the question linked in the comments once you can work with this norm (just adding an extra derivative). The problem you may have h...

in this answer, why we can consider $y\ge 1/2$? And shouldn't the sign of the last term (the one with the second derivative and the Lagrange remainder) be negative?
 
6:43 PM
No, EE, $X$ is a vector variable, not a function. ... I think it is absurd to be bogging down in this at the beginning of an analysis course. Just move on.
@Sine They're doing two different estimate, one for small $y$, one for large $y$. You're right about the sign, but you'll get the same thing when you do the triangle inequality estimate.
 
@TedShifrin The reason I say it's a function is because (at least for the single indeterminate case), $X$ is a function. Namely that element of $R^\mathbb{N}$ for which $X_n = 1$ iff $x = 1$.
Thus I figured this was some generalization of that, and I had thought that maybe the $m$-indeterminate definition of $X$ included that single-variable definition of $X$ element-wise somehow, but it felt unclear.
 
That is not a helpful way of thinking about it. I really dislike this book. Just get to the analysis and stop bogging down in every crazy technical thing that is irrelevant, anyhow.
 
@TedShifrin is this permissible? when you Taylor expand, shoudn't $\xi$ be in a neighbourhood of the point?
 
I think their "schtick" is to do the analysis in all its abstract glory which I honestly don't mind mostly because I don't have a good background in algebra, so it's cool to get that as a second benefit
 
You're Taylor expanding on $[0,1]$, but he's writing it two different ways depending on the size of $y$.
 
6:53 PM
But that's well taken, I won't get bogged down if I don't follow.
Will leave the details to eventually studying something like Dummit and Foote I guess
 
The level of pedantic formalism makes me dubious about how they will treat actual analysis.
This book makes Rudin seem intuitive.
 
Ha!
I can only report that it does seem to be well thought of by fellow Math SE users, but I won't know until page 129 :)
 
ok. Another question. What if instead of $[0,1]$ there was a generic inteval $[a,b]$? I guess the problem is with $0$, so should I do different cases?
 
@Sine Why are you worrying about this? Are you really not using the standard norm on $C^2$?
Understanding that any two closed intervals are diffeomorphic by a linear map, the interval should be inconsequential.
 
it's an exercise and I've to prove they're equivalent but on a generic interval
 
6:58 PM
So write your own proof on a general interval. Try to bound $|f'|$ in terms of $|f|$ and $|f''|$.
 
yes, the problem is $0$ I guess
It may be an end point, or in the interior of the interval
 
It's more natural to work from the center of the interval and use as the variable the displacement from that center point.
 
ok I'll try, thank you
 
@leslietownes Did you see my comment?
 
@Sine There is essentially this same exercise both in Rudin and in Spivak.
 
7:10 PM
@TedShifrin Is that from getting the book from and putting it back on the shelf?
 
I remember it, actually, from both places. I think I was the one who added it to the third edition of Spivak long ago.
 
. o O ( Ted is finally ignoring me )
 
I didn't understand your quip at all.
I'm also grumpy because I'm having to file an appeal to Medicare because they refused to pay for my non-routine x-ray. Bureaucracy is absurd, and whenever I have to do this sort of thing I wonder how the "common person" deals with it.
 
That sucks. I am dealing with an expensive lab test done at a place where we were supposed to get stuff very inexpensively for UCLA insurance. This is like a surprise charge, that I think we are protected from.
The lab charge is over 3 times the charge of the whole rest of the charges because it was sent to a non-UCLA friendly place.'
 
@TedShifrin I don't have these books, I guess they're more used in the US and the UK
what volume of Rudin? RCA?
 
7:23 PM
No, Baby Rudin. Anyhow, if you write $f(x+h)=f(x)+hf'(x)+h^2f''(\xi)/2$, you can proceed.
@robjohn Well, the cost is not prohibitive, but I'm not doing this as a lark. Cancer was serious and I was told by the surgeon 10 years ago to do this x-ray every year to make sure it has not recurred. shrug
 
And now the insurance decides that the doctor is wrong and the x-ray is not necessary.
 
@robjohn lovely thinking cloud
 
Yes. I think it may be that the code that my PCP's office put in may not make things clear enough, but they knew the situation. ...
 
@Jakobian Thanks. I do think once in a while.
@TedShifrin Yeah, often the proper code is the key./
 
Well, one of my multitudinous phone calls today was to the referrals office of that doctor, who insist they put the same code as they put a year ago, when things went through fine.
 
7:37 PM
sorry I've a last question :( in the answer, when $y<1/2$, why we can divide by |y| ?
 
That's why they did a different approach for $y<1/2$, @Sine. That was the whole point.
You're obviously ignoring my advice to do it a different way that is way more efficient.
 
@TedShifrin oh yes I'm stupid
 
I have no idea why that answer was written so awkwardly. My way is the "clear" way to do it :D
 
ok I'll try to follow your method
 
7:41 PM
Can you give me a hint what to do if i want to sow that there is no matrix $M\in M_{2\times 2}(\Bbb{R})$ s.t. $e^M=\left(\begin{matrix}
\lambda & 0 \\
0 & \lambda^{-1}
\end{matrix}\right)$ where $\lambda<0$ and $\neq -1$

My idea was to do something with the eigenvalues. So if I know $\lambda$ is an eigenvalue of $M$ then thee complx conjugate is also one
but I don't see how to use this
 
What is the trace of $M$?
 
is it the trace of $\log\left(\begin{matrix}
\lambda & 0 \\
0 & \lambda^{-1}
\end{matrix}\right)$
 
Is that helpful?
What if $M$ is diagonal?
 
if $M$ is diagonal then the trace is the sum of the diagonal entries
 
That's true always. But what do you conclude in this case?
 
7:50 PM
@TedShifrin I don't think so
but I don't know anything about the diagonal entries
do I?
AH sorry now I get it, if M is diagonal, the eigenvalues are it's diagonal entries
so the trace is $\operatorname{tr}(M)=\mu+\bar \mu=2\operatorname{Re}(\mu)$ where $\mu$ is an eigenvalue. @TedShifrin
But shouldn't it be $2\mu$ since $M$ need's to have real coeff?
 
8:25 PM
I have a question from a long problem, which I've tried to strip the unnecessary details from.
Let $x(t)$ be an unknown function and consider the following two inequalities $$\begin{cases}\frac{1}{1-t}\le x(t) & \text{if } 0\leq t < 1 \\ x(t)\le\tan\left(t+\frac{\pi}{4}\right) & \text{if } 0\leq t <\frac{\pi}{4}.\end{cases}$$
In the context of the problem in which this question arose, it is claimed that $x(t)$ has a singularity at a value close to $1$, but not equal to $1$. This confuses me, because it is bounded below by $1/(1-t)$ and so I'd expect that it also blows up at $1$. Would you agree?
By singularity I mean that $x(t)\to\infty$ as $t\to c$.
 
sunny i would probably just ignore a comment like that unless it is crucial to something. note that saying it has a 'singularity at a value close to 1 but not equal to 1' is not saying that it only has a singularity at a value close to 1 but not equal to 1, and the second condition requires x(t) to go to -infinity as x approaches pi/4 (a little more than .78) from the right
oh, i misread the conditions. i think they goofed on the piecewise inequalities.
i thought the <= tan inequality also applied on (0,1)
as stated, the conditions do not imply the existence of 0 < c < 1 for which lim_{x to c} |x(t)| is infinity. maybe their remark has to do with some physical application where the theoretical badness at 1 would manifest itself in real-world badness before then.
 
@leslietownes but if $1/(1-t)\leq x(t)$ for all $0\leq t<1$, then surely, $x(t)$ must tend to infinity as $t\to1$, no?
 
yes
from what you've shared, the only thing you can be sure of is that it blows up at 1. there is nothing in the above that requires blow up at any earlier time
so if the book says that you do have blow-up at some earlier time, they're maybe using something that isn't encoded above, or they're speaking informally
 
8:41 PM
@leslietownes correct, however, it could blow up for any $\pi/4\leq t\leq 1$, right? Since $1/(1-t)$ is only a lower bound.
 
hello, can someone tell me if i am asked to find the number of ways 3 balls can be distributed in 2 boxes, can i assume that one of these boxes may be empty one time that is to say all the 3 balls are in 1 box in a possibility?
 
Sure
 
sunny: sure, but nothing requires that. i thought your question was about what, in that information, allowed the author to conclude "singularity at a value close to 1, but not equal to 1," and the answer is "nothing"
 
@leslietownes gotcha!
 
@TedShifrin assuming that was for me, thanks
 
8:43 PM
Yes :)
 
Ted may I ask a last question?
If the center of expansion is $0$, why there's $f'(y)$ and not $f'(0)$?
 
you can even ask another, post-ultimate question
 
is there a way to reduce $(\cos(\theta_1)\cos(\theta_2)+\sin(\theta_1)\sin(\theta_2))$
 
@Obliv $\cos (\theta_1-\theta_2)$
 
@SineoftheTime Who said $0$ is the center?
 
8:54 PM
@SineoftheTime howd u do that
 
@TedShifrin why there's $f(0)$ in the lhs?
 
burn the witch!
 
I thought you were redoing it as I suggested?
 
I want to solve it both ways
but since I started the old method, I want to finish it
but I don't understand why the formula holds
 
Well, I’m done with it.
 
8:57 PM
@TedShifrin I have thought about it, if $M$ is diagonal say $M=\left(\begin{matrix}
a & 0 \\
0 & b
\end{matrix}\right)$ for $a,b\in \Bbb{R}$, then the trace is $a+b$ and we know that $\det(e^M)=e^{\operatorname{tr}(M)}=e^{a+b}$ but then by assumption $\det(e^M)=\lambda\cdot \lambda^{-1}=1$. But this means $e^{a+b}=1$ which is equivalent to say that $a=-b$. But is this then a contradiction?
 
Well, of course there won’t be a diagonal real matrix.
 
why is this clear?
 
@robjohn can you explain me why here there's $f'(y)$ and not $f'(0)$? why the formula holds?
 
Because you can’t take ln of a negative number.
 
Ah I see, so I can say that there is no diagonal real matrix satisfying this property. Thus I need to look at an arbitrary comples matrix which is diagonalisable?
 
9:10 PM
I suggested you think about diagonal to get started. You have to figure out whether diagonal/diagonalizable is needed for the argument.
 
I would say it is since the space of diagonalisable matrices is dense in the space of all matrices. And for the diagonalisable one I need the diagonal case.
 
9:28 PM
does $$\frac{d}{dt}(2\dot{\theta}_1\cos(\theta_1-\theta_2)) = 2(\ddot{\theta}_1\cos(\theta_1-\theta_2)-(\dot{\theta}_1-\dot{\theta}_2)\dot{\theta}_1\sin(\theta_1-\theta_2))$$
I guess specifically $\frac{d}{dt}\cos(\theta_1 - \theta_2) = -(\dot{\theta}_1-\dot{\theta}_2)\sin(\theta_1-\theta_2)$?
is that the correct use of the chain rule
 
Yes.
 
The paper I'm reading briefly mentions the concept of an epsilon-net of a Lie group. So I go read about SO(3) and apparently there are at least six distinct metrics you can put on SO(3). Definitely above my pay grade, for now
 
9:50 PM
I doubt 6. Constant multiples don’t count.
 
well, they might count for what epsilon is
:)
 
it's always possible I misunderstood. here:
 
ack. nobody asked for that :). yes, it sounds like they evaluated six functions, and that some of them are equivalent to each other at least in a slightly broader sense than the one ted had in mind. [metrics that differ by ted's 'constant multiples' would be boundedly equivalent to one another]
hamilton would be happy to learn that people are still claiming efficiency benefits for quaternions in 2023.
 
sorry, just sharing. I don't have a specific question about it
 
So I have a trip. And I have to take the Shinkansen for this, since everyone else is as well. Not too happy about that
 
10:01 PM
Well, so if four are boundedly equivalent, then, I guess only 3 "truly" different metrics, maybe
 
that reminds me of a former officemate of mine, who had a side line in numerical stuff relating to ODE and would sometimes point out weird things in papers he was reading. there's a lot of unusual stuff written at the intersection of theory and applications.
 
I suppose it's rare to really understand both
 
Why does that remind me of Rota
 
that goes to a theory i had about it, actually.
at the cutting edge of many applications, absolutely everything matters, including the choice of constants, because the intimate specifics of how each bit is stored/manipulated can affect performance. but people who work on that are broadly aware that the choices/tweaks that are crucial for them are not of theoretical interest (e.g. because they would not affect asymptotic properties that are used to benchmark families of methods against one another).
similarly at the cutting edge of theory there is often a broad awareness that big leaps in improvements on proved theoretical bounds have no relation to applications because the stuff that people "actually use" is light years faster even if it is not provably that way.
and then there are people in the middle, who kind of incorporate some of caring about specific choices of constants or function implementations (like applications people), but also incorporate some caring about abstract stuff (like, siting their work in an abstract lie group).
and my theory was that a lot of these people get over by convincing more specialized members of their audience that their work is somehow of interest to members the opposite camp. the theory people think "oh, this must matter for applications" and vice versa.
and really nobody is reading that stuff or using it except other people who do the same thing.
anyway, my former officemate could name a bunch of people who would have been guilty of that in ODE-related numerics, 20 years ago.
 
I've never heard that theory before, but it does sound plausible, because my limited experience in publishing papers is that it's desirable to invoke (i.e. cite) many other papers to advance your argument, but those invocations are very often very superficial
So you glance at an abstract and then cite a paper because it looks like it bolsters your own, which sounds like it could be related to the opposite camp stuff you're talking about
 
10:13 PM
yeah, without tarring any particular person with undeserved criticism, it's generally a lot easier to be lazy when you mix inputs from multiple communities or disciplines, than it is to be lazy when you don't.
there's maybe some overlap with citation practice here, although that might also be its own independent thing. some fields just seem to cite more than others, and in a field where it might be normal for a two-page paper to cite 50 papers, i don't know that anybody (other than administrators who love cite counts) really attributes much significance to individual citations.
 
cough Biology
 
i certainly wasn't not thinking about biology when i said that.
 
I have an even worse offender in my mind, but I'll resist
 
it seems to be pretty common in "lab sciences" generally, if only because you can't do "lab science" without armies of people and funding, and one might feel driven to reflect that in armies of citations.
 
whereas math is just chalkboards and frozen pizza
 
10:23 PM
not true, some of the pizza is freshly cooked.
 
 
1 hour later…
11:40 PM
@SineoftheTime They are expanding $f(x)$ around $x=y$ and evaluating at $x=0$.
 

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