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Shaun
Shaun
00:04
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?
Jakobian
Jakobian
@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 $a^n$"
EE18
EE18
Touche
as you said, a can of worms
Thorgott
Thorgott
@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…
EE18
EE18
01:13
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
For example: alpha.math.uga.edu/~pete
has anyone had this before? i thought it was maybe due to adblock but not so
Jakobian
Jakobian
it opens for me
Thorgott
Thorgott
opens for me, both on PC and mobile
EE18
EE18
ya it's an issue on my end because many other .edu/professor websites won't open. cleared cache, etc. etc.
Thorgott
Thorgott
though it takes a bit to load
EE18
EE18
nothing on google and it's only these professor sites which don't open so I don't know
Ted Shifrin
Ted Shifrin
01:18
ee18, that server at uga has been gone forever
you can find pete’s current webpage by googling
EE18
EE18
i guess it must reroute because it works for jakobian and thorgott?
and this was the first hit on google
Ted Shifrin
Ted Shifrin
Weird
EE18
EE18
for pete clark that is
Ted Shifrin
Ted Shifrin
I had to update all my links with new links
EE18
EE18
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
leslie townes
leslie townes
01:24
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
Jakobian
Jakobian
@EE18 maybe it is because connection isn't secure?
EE18
EE18
01:45
what do you mean Jakobian?
leslie townes
leslie townes
your browser or system settings may prevent you from communicating with web servers that serve http and not https
Jakobian
Jakobian
yeah
leslie townes
leslie townes
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
Jakobian
Jakobian
why would you want to browse someone elses website? :(
leslie townes
leslie townes
you can also just buy pete clark's collected works on the dark web
Jakobian
Jakobian
01:50
oh, like his drafts and stuff?
leslie townes
leslie townes
yeah, everything
EE18
EE18
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
leslie townes
leslie townes
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
EE18
EE18
ya gonna do that once the weekend is over. the latency is not great going that route but seems no other option
hahahaahaha
EE18
EE18
02:50
user image
Is this formula important for anything/recognizable to anyone?
Debating whether it's worth it to take the time to solve
robjohn
robjohn
It looks as if the hint is trying to lead you to a telescoping series
EE18
EE18
03:10
i guess this isn't a common formula which i should expect to come up again though?
Jakobian
Jakobian
03:24
@EE18 its a modification of Pascal's formula: $\binom{l}{j}+\binom{l}{j-1} = \binom{l+1}{j}$
leslie townes
leslie townes
03:41
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
Thorgott
Thorgott
04:05
@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…
barlop
barlop
06:42
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.
leslie townes
leslie townes
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
barlop
barlop
well i'm not in an exam situation.. i'm just wondering what the precise language would be
leslie townes
leslie townes
"significant figures" is a made up classroom thing, don't expect people to treat them uniformly in real life
barlop
barlop
isn't it used in science?
leslie townes
leslie townes
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
barlop
barlop
06:49
well i'm not thinking here really about scientfiic notation cases and zeros after a decimal point, but zeros before a decimal point.
leslie townes
leslie townes
unless it is expressly written out (as it might be in a good paper)
barlop
barlop
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?
barlop
barlop
07:01
@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%.
copper.hat
copper.hat
a tick attached itself to me today at tilden
leslie townes
leslie townes
07:13
copper: is it detoxing now
Ted Shifrin
Ted Shifrin
@leslietownes Ha ha …
 
3 hours later…
Koro
Koro
09:47
user image
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…
Balarka Sen
Balarka Sen
12:30
@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.
Sahaj
Sahaj
hi balarka
Balarka Sen
Balarka Sen
12:53
hi @Sahaj
John Zimmerman
John Zimmerman
13:28
Is there a mathematical process that converts diffusion to rotation?
John Zimmerman
John Zimmerman
13:48
Rotational diffusion
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…
Thorgott
Thorgott
15:16
@BalarkaSen yeah, it's annoying though
barlop
barlop
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)
Thorgott
Thorgott
it almost makes me wanna use $e_n$ as basepoint of $S^n$, but that's not a great solution either
barlop
barlop
@leslietownes according to wikipedia, en.wikipedia.org/wiki/Scientific_notation "advantage of scientific notation is that the number of significant figures is unambiguous."
Thorgott
Thorgott
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
EE18
EE18
15:32
0
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
Xander Henderson
Xander Henderson
@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.
Thorgott
Thorgott
15:48
@EE18 it does, it's not a great answer in any case
also that definition is bizarre
EE18
EE18
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
MathIsLife12
MathIsLife12
16:02
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
Thorgott
Thorgott
what do you mean you need a question removed
MathIsLife12
MathIsLife12
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
Balarka Sen
Balarka Sen
@Thorgott Lol
yeah
Thorgott
Thorgott
16:22
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
MathIsLife12
MathIsLife12
@Thorgott Ok I have done that
Balarka Sen
Balarka Sen
I'm struggling with some weird slope arithmetic similar to your sign issue
Thorgott
Thorgott
have you tried appealing to the Yoneda lemma
Balarka Sen
Balarka Sen
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...
Thorgott
Thorgott
so this is a calculation in $H_1(T^2)$?
Balarka Sen
Balarka Sen
16:29
I guess if $C$ has slope $1/n$ then $C = \lambda + n \mu$. (If $n = 0$, then $C = \lambda$ which indeed has slope $1/0 = \infty$)
Yeah
Thorgott
Thorgott
but $(1,n)$ is not a basis element for positive $n$
or you work rationally?
Balarka Sen
Balarka Sen
Hm? $(1, n)$, $(1, n+1)$ is an integral basis of $\Bbb Z^2$
$1 \cdot (n+1) - n \cdot 1 = 1$
Thorgott
Thorgott
sorry, brain fried lol
Balarka Sen
Balarka Sen
$(1, n)$ 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
Thorgott
Thorgott
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 $n/(n+1)$
Balarka Sen
Balarka Sen
16:35
$\lambda = (1, 0)$ 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.
Thorgott
Thorgott
$(1,n)=v$ would give slope $\infty$
Balarka Sen
Balarka Sen
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 $0, -1$.
Thorgott
Thorgott
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
Balarka Sen
Balarka Sen
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
Thorgott
Thorgott
16:51
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
Balarka Sen
Balarka Sen
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
Thorgott
Thorgott
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 $\infty$
Balarka Sen
Balarka Sen
Hm yeah
Same for $-0$
Thorgott
Thorgott
17:07
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
Balarka Sen
Balarka Sen
right
 
1 hour later…
EE18
EE18
18:28
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?
user image
I tried a direct attack but was unsuccessful
Jakobian
Jakobian
18:42
@EE18 $(a-b)(a+b) = a^2-b^2$
EE18
EE18
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)
Sahaj
Sahaj
hii
@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…
EE18
EE18
OK that was hideous
i have a newfound appreciation for the negative integers. whoever invented them was wicked smaht
Jakobian
Jakobian
19:04
Lucky Sahaj
Sahaj
Sahaj
?
 
1 hour later…
Ted Shifrin
Ted Shifrin
20:17
@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 $x^2-x>0$.
Balarka Sen
Balarka Sen
I want a notation for "large positive integer". $M$ (manifold), $N$ (tubular neighborhood), $K$ (knot) are all taken.
Thorgott
Thorgott
$C$
Balarka Sen
Balarka Sen
$n$ 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.
Thorgott
Thorgott
$\lfloor C\rfloor$
Balarka Sen
Balarka Sen
Lol
I'll go with $C$
Xander Henderson
Xander Henderson
20:25
L
Balarka Sen
Balarka Sen
That intuitively feels worse than C
Xander Henderson
Xander Henderson
$\Omega$
Balarka Sen
Balarka Sen
...
Xander Henderson
Xander Henderson
M, but with a different font.
D
B, for"large bound".
$\Xi$, 'cause it's funny looking.
Thorgott
Thorgott
Ü
leslie townes
leslie townes
20:29
$\frac{\overline{\Xi}}{\Xi}$
Xander Henderson
Xander Henderson
$\gimel$
Thorgott
Thorgott
ẞ, for "Schranke"
leslie townes
leslie townes
use the egyptian hieroglyph thats the side view of a guy holding his hands in the 'egyptian' way
Xander Henderson
Xander Henderson
How about something kanji?
🌕
EE18
EE18
21:18
@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 ;)
Ted Shifrin
Ted Shifrin
21:32
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.
EE18
EE18
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
Joe
Joe
21:51
@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.
leslie townes
leslie townes
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)
EE18
EE18
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
Joe
Joe
@EE18: The symbol $-$ is not in the language of $\mathsf{PA}$, and hence the statement you are trying to prove is not well-formed ;)
leslie townes
leslie townes
maybe we'd have world peace if children just learned a certain version of peano arithmetic before they learned any human languages
EE18
EE18
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?
Joe
Joe
22:07
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.
EE18
EE18
No worries, it was taken as such :)
leslie townes
leslie townes
"expansion by definitions" is what the snake gave eve in the garden of eden
EE18
EE18
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
leslie townes
leslie townes
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
Joe
Joe
@EE18: Does your version of the peano axioms say something about mathematical induction?
leslie townes
leslie townes
22:09
i can probably fix it with some glue
EE18
EE18
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 $\omega$)
Joe
Joe
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.
EE18
EE18
Interesting, I will have to look this up :)
Joe
Joe
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.
Allie
Allie
22:25
hey yall
EE18
EE18
To be learned I suppose! Thanks very much for the tip Joe
Ted Shifrin
Ted Shifrin
23:00
@leslie Ask Olivia.

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