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12:04 AM
Again . . .
in Helpful Commentary, 6 hours ago, by Shaun
Can I get some feedback on this question, please?
in Helpful Commentary, 1 min ago, by Shaun
It just got its third downvote. Any advice?
@EE18 There is approximately $\frac{4^{n-1}}{n^{3/2}\sqrt{\pi}}$ ways of putting brackets on a product of $n$ elements
it increases exponentially, and any other way of doing so would be "wild" as I meant it
wild having a mild interpretation of "not one of the standard ways to put the brackets to interpret $a^n$"
as you said, a can of worms
@Balarka whether we traverse the suspension upwards or downwards, however, always makes a difference of sign $-1$
so it can't cancel out the alternating sign behavior we've observed so far...
1 hour later…
1:13 AM
This is sort of a "soft" question, but I've been finding in the last couple weeks that a lot of .edu websites don't open for me, they just load and load until a server timeout happens
has anyone had this before? i thought it was maybe due to adblock but not so
it opens for me
opens for me, both on PC and mobile
ya it's an issue on my end because many other .edu/professor websites won't open. cleared cache, etc. etc.
though it takes a bit to load
nothing on google and it's only these professor sites which don't open so I don't know
1:18 AM
ee18, that server at uga has been gone forever
you can find pete’s current webpage by googling
i guess it must reroute because it works for jakobian and thorgott?
and this was the first hit on google
for pete clark that is
I had to update all my links with new links
ya it's something on my end and not sure what. will have to ask the IT folks i guess, but just so weird that it's only .edu sites
maybe they've seen me clicking around too much
1:24 AM
it isn't serving https, which depending on your settings / system administrator's settings, may prevent you from accessing it
god forbid we disclose the contents of our http requests for pete clark's materials on the open web
@EE18 maybe it is because connection isn't secure?
1:45 AM
what do you mean Jakobian?
your browser or system settings may prevent you from communicating with web servers that serve http and not https
you could use web.archive.org (which does serve https) to browse recent copies of pete clark's website
or anyone else's for that matter, i don't mean to single him out
why would you want to browse someone elses website? :(
you can also just buy pete clark's collected works on the dark web
1:50 AM
oh, like his drafts and stuff?
yeah, everything
hmm just check that too leslie and no dice
anyway, will figure it out. figured i'd ask here given the connections to .edu type sites lol
if this were happening a lot in a work or school environment i would contact an IT person about it
my ISP wisely or unwisely allows me to shoot alpha.math.uga.edu right into my veins
ya gonna do that once the weekend is over. the latency is not great going that route but seems no other option
2:50 AM
Is this formula important for anything/recognizable to anyone?
Debating whether it's worth it to take the time to solve
It looks as if the hint is trying to lead you to a telescoping series
3:10 AM
i guess this isn't a common formula which i should expect to come up again though?
3:24 AM
@EE18 its a modification of Pascal's formula: $\binom{l}{j}+\binom{l}{j-1} = \binom{l+1}{j}$
3:41 AM
EE18 what might come up again (if only because it is a near infinite source of homework problems) is being asked to 'simplify' sums, perhaps involving binomial coefficients but maybe more generally involving other recursively defined sets of numbers, perhaps using induction
no one of these identities has any more particular use than any other
4:05 AM
@Balarka The issues are two-fold. First, the orientation issue you and Ted pointed out. Second, the issue that the basepoints don't match up. The solution I present is $D^{k+1}/S^k\rightarrow S^{k+1},\,tx\mapsto-(\cos(\pi t),\sin(\pi t)x)$ and $\Sigma S^k\rightarrow S^{k+1},\,[x,t]\mapsto(\sin(\pi t)x,-\cos(\pi t))$, as well as $\Sigma S^k\rightarrow D^{k+1}/S^k,\,[x,t]\mapsto tx$.
These all do the right things on basepoints and when you compare the composition, you see they differ by $k$ transpositions and $k$ reflections, hence have the same degree. The pictures in low dimensions all give
3 hours later…
6:42 AM
If a question said round 1028 to three significant figures, then the answer would be 1030

Would I be correct in thinking that somebody would be wrong to say that 1030 is 3 significant figures. 1030 is 4 significant figures. (obviously the answer has to be 1030 though 'cos that's the precision required).

So that makes me think, is it correct to even say round 1028 to three significant figures?

Is there a more correct way to phrase that question?

If we say round a number to 5dp, it means the result will be to 5dp.
a lot of treatments of "significant figures" do not do a very good job of indicating how to indicate whether trailing zeros are significant or not. some use weird color schemes or markings on the last significant digit. this is not something that folks have any consensus about, so in an exam setting, do whatever your instructor or book says to do
it is not actually that common outside of the classroom to rely upon notational rules to figure these things out, it is more common to expressly state whatever you are doing
well i'm not in an exam situation.. i'm just wondering what the precise language would be
"significant figures" is a made up classroom thing, don't expect people to treat them uniformly in real life
isn't it used in science?
the use of scientific notation allows one at least to distinguish notationally between 1.03 x 10^3 and 1.030 x 10^3, but whether that is a statement of intended 'significance' or not is in the author's head
6:49 AM
well i'm not thinking here really about scientfiic notation cases and zeros after a decimal point, but zeros before a decimal point.
unless it is expressly written out (as it might be in a good paper)
Suppose we say that trailing zeros are significant before a decimal point.. So how would they say to round 1028 so that it becomes 1030?
7:01 AM
@SnoopyKid I saw a similar question here reddit.com/r/GCSE/comments/18otk9l/percentage_problem In your case the fraction is 50/1 so if you want a percentage in there you've got to multiply that by 100%. 100%=1. So yeah 5000%. Similarly if you have a fraction 5/10 and you want that with a percentage then you wouldn't say 5/10=0.5 So 0.5%. And you wouldn't even say (5/10)*100=50% 'cos 50!=50%. You'd say (5/10)=(5/10)*100% = 50%.
a tick attached itself to me today at tilden
7:13 AM
copper: is it detoxing now
@leslietownes Ha ha …
3 hours later…
9:47 AM
Can it be proven without defining a bilinear form?
that is, using only chain rule.
$(f\circ \gamma)''(t)= D(f\circ \gamma)_t' (1)$.

$(f\circ \gamma)'(t)= D(f\circ \gamma)t (1)= Df{\gamma (t)} (\gamma'(t))$

how to go from here?
3 hours later…
12:30 PM
@Thorgott Of course it can all be fixed. The orientation-preserving diffeomorphism group of the sphere acts transitively on all points, so preserving basepoints cannot possibly be an issue.
Part of it is our insistence on writing "neat equations", but that is not mathematics but aesthetics.
hi balarka
12:53 PM
hi @Sahaj
1:28 PM
Is there a mathematical process that converts diffusion to rotation?
1:48 PM
Rotational diffusion is the rotational movement which acts upon any object such as particles, molecules, atoms when present in a fluid, by random changes in their orientations. Whilst the directions and intensities of these changes are statistically random, they do not arise randomly and are instead the result of interactions between particles. One example occurs in colloids, where relatively large insoluble particles are suspended in a greater amount of fluid. The changes in orientation occur from collisions between the particle and the many molecules forming the fluid surrounding the particle...
oh. got it.
1 hour later…
3:16 PM
@BalarkaSen yeah, it's annoying though
I see 1028 to 3sf as 1.03 * 10^3 makes 3sf clear and clear how it's 3sf. (and it seems from wikipedia that one could make it 4sf with 1.030 * 10^3)
it almost makes me wanna use $e_n$ as basepoint of $S^n$, but that's not a great solution either
@leslietownes according to wikipedia, en.wikipedia.org/wiki/Scientific_notation "advantage of scientific notation is that the number of significant figures is unambiguous."
my context btw was figuring out whether the map $\pi_k(X)\stackrel{\partial}{\leftarrow}\pi_{k+1}(C_-X,X)\rightarrow\pi_{k+1}(\Sigma X,C_+X)=\pi_{k+1}(\Sigma X)$ in the Freudenthal suspension theorem is actually the suspension or not
though I can also just not care about sign and say it's $\pm\Sigma$ by Yoneda, which is probably the best way
3:32 PM
A: Why does this definition capture the notion of anticommutativity intuitively?

Aritro ShomeThe concept of anticommutativity in algebraic structures provides a generalized notion of commutativity, emphasizing the reversal in result or sign upon interchanging operands. In contrast to commutativity, where the order of operands does not affect the result (i.e., $ a \cdot b = b \cdot a $), ...

Does this sound like chatgpt to you too?
the answer that is
@barlop As @leslietownes said earlier, the notion of "significant figures" is mostly a classroom tool. I would try to stop getting hung up on on what the "right" answer is, and focus on what the answer is supposed to communicate.
In the sciences (not mathematics), you make measurements, and those measurements contain unavoidable errors (you cannot measure anything with infinite precision---every measurement tool has a maximum resolution). The point of "significant figures" is to communicate the level of precision your tools can provide you with.
3:48 PM
@EE18 it does, it's not a great answer in any case
also that definition is bizarre
Ya I asked it a while back and forgot about it but then bountied it cause i never got an answer
i wasn't sure how that definition was supposed to intuitively capture anticommutativity
4:02 PM
Hey everyone I am having a little trouble getting in touch with a mod about a question I need removed. If you can help me I’d be greatly appreciative
what do you mean you need a question removed
I bountied a question that was part of an exam (unknowingly) from a friend. This is cheating and I’d like it removed as I do not want to be associated with it
I will award the answerer for their hard work but I do not feel comfortable keeping this up as the exam period is still happening
@Thorgott Lol
4:22 PM
there's an entire section in May's Concise where he discusses how to identify these spaces when he does cellular homology, but he chooses the stereographic $D^{k+1}/S^k\cong S^{k+1}$ that maps the boundary sphere to the north pole
i guess we should just always leave the basepoint of $S^n$ unspecified with an implicit appeal to the transitive action of $SO(n+1)$ lol
@MathIsLife12 in that case, flag the question for cheating
@Thorgott Ok I have done that
I'm struggling with some weird slope arithmetic similar to your sign issue
have you tried appealing to the Yoneda lemma
I have a basis $\mu, \lambda$ ("meridian", "longitude") of $T^2$ where $\mu$ is slope $0$, $\lambda$ is slope $\infty$ according to the conventions I am working with. I have a pair of curves $C, C'$ of slope $1/n, 1/(n+1)$ (either this, or both have negative sign, I haven't quite figured this out yet - I am inclined to say both should be -ve).
I want to change basis so that $C, C'$ have slope $0, -1$. And then I want to the slope of $\lambda$ under new basis.
So many matrix multiplications...
so this is a calculation in $H_1(T^2)$?
4:29 PM
I guess if $C$ has slope $1/n$ then $C = \lambda + n \mu$. (If $n = 0$, then $C = \lambda$ which indeed has slope $1/0 = \infty$)
but $(1,n)$ is not a basis element for positive $n$
or you work rationally?
Hm? $(1, n)$, $(1, n+1)$ is an integral basis of $\Bbb Z^2$
$1 \cdot (n+1) - n \cdot 1 = 1$
sorry, brain fried lol
$(1, n)$ is obtained from $(1, 0)$ by Dehn twisting $n$ times along $(0, 1)$. Likewise, $(1, n+1)$ is obtained from $(1, 1)$ by Dehn twisting $n$ times along $(0, 1)$.
So Dehn un-twist to get $(0, 1)$ and $(1, 1)$ back
let $v=(1,n)$, $w=(1,n+1)$, then the basis you want is $-w,v$
and $\lambda=(1,0)=(n+1)v-nw$, no?
so has slope $n/(n+1)$
4:35 PM
$\lambda = (1, 0)$ gets sent to $(1, 0) - n \cdot (0, 1) = (1, -n)$ under Dehn untwisting $n$ times though
That seems to not match with what you wrote
I get slope $-1/n$ (admittedly, this is different from what I was getting last time I tried this calculation lol)
@Thorgott No, I want a basis $v, w$ such that $(1, n) = v$ and $(1, n+1) = v - w$.
That's why $C, C'$ has slope $0, -1$ with respect to new basis
In your case that's $0, \infty$ or something.
$(1,n)=v$ would give slope $\infty$
Right. If you set $v = (1, n)$, $w = (1, n+1)$ like you did, then $C, C'$ has slope $0$ and $\infty$ wrt basis $\{v, w\}$.
Which is not the basis I want. I want slope $0, -1$.
yeah, you're right
but I think your requirements also don't specify a unique basis
cause $0$ divided by anything yields slope $0$
so you have an entire shear of such bases
Ah, that must be why I was getting a different answer last time.
Wait, but I think the condition of it being an integral basis forces something
Like $(1, 0)$ and $(0, 2)$ has slope $0$ and $\infty$ (resp.) wrt the standard honest to god basis of $\Bbb Z^2$. But they certainly don't span all of $\Bbb Z^2$.
A pair of slopes should uniquely determine a basis.
At least my answer geometrically checks out with the picture in my head so I am not going to bother with framing hell anymore
4:51 PM
there's conditions on the slopes, it's weird
for $r,s$ to be slopes coming from a basis, you need that $1/(r-s)$ is an integer
yeah true but once you fix those conditions, theres a unique basis it corresponds to, right?
The condition is simply that if $r = p/q, s = p'/q'$ is a reduced fraction representation of the slopes, then $pq' - p'q = \pm 1$
Where $0 = 0/1$ and $\infty = 1/0$ are by definition the reduced fraction representations
sounds right, perhaps up to sign
if you switch both signs, the basis changes, but the slopes don't
also $-\infty$ is a slope distinct from $\infty$
Hm yeah
Same for $-0$
5:07 PM
the point should be that if $\begin{pmatrix}v_1&w_1\\v_2&w_2\end{pmatrix}$ is in $\mathrm{GL}_2(\mathbb{Z})$, then $v_1$ and $v_2$ resp. $w_1$ and $w_2$ are coprime, so they can be recovered from $v_1/v_2$ resp. $w_1/w_2$ (the slopes) up to the $\mathbb{Z}/2\mathbb{Z}$ action that inverts signs
1 hour later…
6:28 PM
OK I may get lampooned for this question but it's been bothering me
How do I show that $(x^2 - x)(x^2 +x) = x^4 - x^2$ where $x$ is a natural number and where I imagine I'm a person who has no knowledge of the integers or anything else so that $x^2 - x$ is just an entity/notation for the unique difference between $x^2$ and $x$, not $x^2 + (-x)$
Is my only recourse induction?
I tried a direct attack but was unsuccessful
6:42 PM
@EE18 $(a-b)(a+b) = a^2-b^2$
I agree, but am trying to show it carefully within $\Bbb N$. Going to try to induct on $a$ for $a \geq b$ (which is when $a-b$ is well-defined anyway)
@XanderHenderson sorry for the ping, but is that you’re always online here? Or maybe it just so happens that I always come here when you’re online…
OK that was hideous
i have a newfound appreciation for the negative integers. whoever invented them was wicked smaht
7:04 PM
Lucky Sahaj
1 hour later…
8:17 PM
@EE18 Probably invented by the first debt collector or mortgage company.
@EE18 Surely even the most pedantic person will admit that when $x\in\Bbb N$, $x\ge 2$, we have $x^2>x$, so $x^2-x>0$.
I want a notation for "large positive integer". $M$ (manifold), $N$ (tubular neighborhood), $K$ (knot) are all taken.
$n$ is taken, I am using it to denote a positive integer somewhere else.
@Thorgott Erm, works, but that sounds too close to a real constant.
$\lfloor C\rfloor$
I'll go with $C$
8:25 PM
That intuitively feels worse than C
M, but with a different font.
B, for"large bound".
$\Xi$, 'cause it's funny looking.
8:29 PM
ẞ, for "Schranke"
use the egyptian hieroglyph thats the side view of a guy holding his hands in the 'egyptian' way
How about something kanji?
9:18 PM
@TedShifrin Perhaps the pedantry is too far then since for that last step I would agree that it's true but only because of Theorem 5.3 in my book which says that $x^2 > x \implies [(\exists ! d \in \mathbb{N}^\times) x + d = x^2]$ and that $d$ we call $x^2 - x$ (and $d \in \mathbb{N}^\times \implies d\neq 0 \implies d > 0$ by the minimality of $0 \in \Bbb N$)
But of course I would never say that to you because I'm supposed to be moving onto the analysis ;)
9:32 PM
Do you know closure under multiplication?
Then $x(x-1)$, as a product of natural numbers, is a natural number. I think That is only reasonable. I’ve pretty much lost interest in this crap and will no longer engage.
Yup I do, but it's all good, I solved this one after the hint from Jakobian. I will definitely aim to only ask analysis questions herein
9:51 PM
@EE18: I don't think it is necessarily pedantic to ask questions about how to formally prove statements from elementary arithmetic. However, what constitutes a proof at this level depends quite sensitively on the foundation you are using. For instance, a formal proof from $\mathsf{PA}$ might be acceptable in one context, but not in another.
on the off chance that this is self study, EE18, and to follow on joe's remark, this is partly why some folks (perhaps ted, certainly myself) feel that elementary arithmetic not a very good 'sandbox' for practicing proof writing in general. you pretty quickly reach a point where what's a proof and what isn't, what the content of any proof generally looks like, etc., is "too much" a function of the details of your foundational setup and unrecognizable to someone who hasn't made those choices
i.e. past a point, you aren't so much moving up the scale of "can this be made more rigorous" as you are learning what proofs look like in a constructed language that might as well be Esperanto or Klingon
no offense to any klingons
(full offense to esperantists, your hippie world is never going to work, get over it)
That's all well taken to both of you. Yes, this is self-study (I didn't and don't study math in school), and noted regarding what we take as foundational (to your point Joe, in the book I was referencing a certain version of PA is taken as bedrock truth)
I had never heard of esperantists, had to look that up
klingon on the other hand...my klingon is passable
@EE18: The symbol $-$ is not in the language of $\mathsf{PA}$, and hence the statement you are trying to prove is not well-formed ;)
maybe we'd have world peace if children just learned a certain version of peano arithmetic before they learned any human languages
I agree! But (and I think this is OK) my book defines $a - b$ as the unique $d$ s.t. $b + d = a$ (exists iff $a \geq b$)
and this, unless I am completely misunderstanding what we mean by PA (Peano axioms?), is OK isn't it?
10:07 PM
Sure, my comment was meant to be tongue in cheek. I believe that introducing new notation in formal languages is justified by a metatheorem called "expansion by definitions", but I don't know the details.
Yes, $\mathsf{PA}$ refers to the Peano axioms.
No worries, it was taken as such :)
"expansion by definitions" is what the snake gave eve in the garden of eden
Your set from which you take literary references includes at least the Bible and Star Trek, leslie. Putting aside the extent to which every element of that set is true, they're certainly both good from a poetic perspective
my daughter just knocked a picture frame that was hanging above a bookcase off the wall. i'd love to know how she did it, but i was upstairs at the time (i only heard the frame hit the floor), and she, very wisely, isn't talking
@EE18: Does your version of the peano axioms say something about mathematical induction?
10:09 PM
i can probably fix it with some glue
She does have the right to remain silent after all
It does, yes. @Joe It gives induction in the usual way (any subset of $\omega$ which is inductive is itself $\omega$)
Ah, so there is actually a subtlety here which I have ignored so far. The term $\mathsf{PA}$ refers to the first-order Peano axioms, which formulates induction differently to how you're doing it.
The version of the Peano axioms you're talking about is second-order.
Interesting, I will have to look this up :)
The distinction is a little technical, but roughly speaking, in first-order $\mathsf{PA}$, we don't ever talk about sets of natural numbers.
We just talk about natural numbers.
So "induction" cannot be phrased in the way you do it.
10:25 PM
hey yall
To be learned I suppose! Thanks very much for the tip Joe
11:00 PM
@leslie Ask Olivia.

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