Mathematics - 2023-07-02

robjohn

00:02

It makes me itch

Xander Henderson

It makes me want to flea.

robjohn

Today hit 103° here

Ted Shifrin

Ugh.

Xander Henderson

It is currently 87°F where I am.

93°F at home.

robjohn

😝

Xander Henderson

00:03

It got down to 65°F up in the mountains, around 11am.

That was real nice. I had the windows rolled down and everything.

robjohn

Even in Mammoth it got to 81° today

Xander Henderson

Yay!

When my sister was living just outside of Death Valley, she would often housesit for friends in Mammoth during the summer. Much less hot there. :D

冥王 Hades

I wanna hit someone with a Kusanagi and trap them in a Sake bottle

I pulled a dangerous but amazing cobra maneuver under the Golden Gate Bridge in an F-22 Raptor. Does anyone wanna see?

robjohn

00:25

@XanderHenderson yeah, on our way to Mammoth, I jokingly ask my wife if she wants to take the short (100 mile) detour to Death Valley. She always seems to decline.

@冥王Hades that sounds like sim fun

Ted Shifrin

00:38

@robjohn Maybe you should instead ask if she wants to be stranded in Death Valley.

copper.hat

always found it amusing that the highest & lowest points in California are pretty close.

Ted Shifrin

something about instability of the elevation function, @copper

copper.hat

:-)

shintuku

01:35

my god

everything is a module

and there's some definite statement out there about the fact that nice modules are isomorphic to nice products of cyclic groups

copper.hat

01:53

category theory gone wild

shintuku

everything is cyclic

reality is a beyblade

one potato two potato

02:08

CAT

robjohn

02:46

@copper.hat The highest and lowest points in Florida are also pretty close, but in elevation.

345 foot elevation variation

leslie townes

03:04

over and underpass, respectively

robjohn

03:25

almost. I can imagine tall buildings being taller than 345 feet in Tampa or Tallahassee.

Yep, the tallest building in Tampa is 579 feet.

copper.hat

i am sure i am repeating myself here. during an intership in the Netherlands a long long time ago, i went cycling to visit some place. the instructions were to keep cycling until i saw a big hill and then so something. a long time later no hill was in sight, so i stopped and asked and had to backtrack a few miles to find out that the big hill was a marginal bump on the ground. its all relative, i suppose.

it was a very interesting project and perfectly suited to me. using a 6502 to sample transformer currents, do some fourier analysis to extract the first and second harmonic and compare.

the idea was to make a software short detector. if a short occurs the current is very large and you trip the relay. except that when a transformer starts, there is something called inrush current which is also large but normal. however the inrush results in a very non sinusoidal current and one can avoid a false positive trip by looking at the relative harmonic size.

i learned about continued fractions and many other interesting things. at some point my local advisor came and said that part of the internship was to see the Netherlands so i should go home earlier.

i should have taken his advice

Ted Shifrin

03:45

It sounded more like an order than advice?

@leslie Did one commune with the canards?

lone student

@robjohn Hello, good morning (in my location), If the comment section is inappropriate, please ping me as I should continue here.

robjohn

04:00

Yes, this is a better forum for this discussion.

leslie townes

we went on a short walk and did not see any ducks.

lone student

04:11

@robjohn Thank you. Please, see the last comment, because I am going to delete the comment section after your first comment .

lone student

04:55

Ahh, that was a not correct

Sonozaki

I would like to formalize with logic the equivalence between equations when dividing. Assume $f:A\to C$ and $g:B \to C$, hence an equation of the form $f(x)=g(x)$ is defined in $A\cap B$. Is it correct to affirm that $\forall x \in A\cap B\left(f(x)=g(x)\right) \iff \forall x\in A\cap B,\left(\left(\frac{f(x)}{g(x)}=1\right) \wedge (x \in A\cap B\setminus \{x \in A \cap B \ | \ g(x)=0\})\right)$?

lone student

@robjohn It should be $$\begin{align}P(z):=\underbrace{\left(z^2+z+1\right)}_{\Delta_z<0}\left(z\color{red}{+}\frac {1}{z^2+z+1}\right)^2+\frac{\overbrace{3z^2+3z+2}^{\Delta_z<0}}{\underbrace{\color{black}{z}^2+{z}+1}_{\Delta_z<0}}\end{align}$$

shintuku

@Sonozaki why do you have f(x)/g(x) = 1

Sonozaki

@shintuku I divided by $g(x)$ in $f(x)=g(x)$.

shintuku

so A,B are sets of say real numbers?

Sonozaki

05:02

Yes, sorry I didn't specify it.

shintuku

so neither A nor B contains 0?

what if f(x) = 0 and g(x) = 0

Sonozaki

Well, they could. For instance, if $g(x)=x-1$ then $B$ can contain $0$. You're right, they can be both $0$ and so $f(x)=g(x)$ is trivially true and we can't divide. Sorry if I was not completely precise. I was not considering the trivial cases.

shintuku

but f(x) = g(x) with 0 = 0 is not trivial: there are a ton of functions R -> R that cross the x axis at the same time

and there are a ton of functions that are equal at the same time, and then have f(x) - g(x) = 0

Sonozaki

I meant trivial in this context, that is, equivalence between equations as set of solutions. If $f(x)=g(x)$ is such that $0=0$ for the case $f(x)=0$ and $g(x)=0$, then the equation is always true in $A \cap B$ and so the set of solutions of $f(x)=g(x)$ is the set $A\cap B$. Or am I wrong?

I am not searching for the functions $f$ and $g$, I am searching for the subsets of $\mathbb{R}$ such that $f(x)=g(x)$ is true. Sorry if this was not clear.

shintuku

@Sonozaki why would the equation always be true in $A\cap B$?

editing

f:R -> R, g: R -> R, f(x) = x - 10 , g(x) = 10 - x. they cross at x = 10 where they are both zero

yet $A\cap B = A = B$ and $f(x) = g(x)$ only at $x = 10$

Sonozaki

05:15

But you said "what if $f(x)=0$ and $g(x)=0$", I assumed that you meant $f(x)=0$ and $g(x)=0$ for each $x$.

shintuku

no, i meant any such $x$

so yeah, everywhere is one of those examples

but not the only example

consider also $f(x) = \sin x$ and $g(x) = 0$

or $f(x) = \sin (x+\pi/2)$, $g(x) = \cos x$

etc.

$f(x) = \cos x$, $g(x) = \pi/2 - x$

Sonozaki

Ok, so I didn't understand what you meant. I will try to give more context to be clearer: imagine we have to solve $\sqrt{x-2}=(x-2)\sqrt{x}$ in $[2,+\infty)$. Imagine that we don't see the root $x=2$ and we want to divide by $\sqrt{x-2}$ both sides and still have an equivalence. I would write $[\sqrt{x-2}=(x-2)\sqrt{x}] \iff \left[\left(\frac{\sqrt{x-2}}{\sqrt{x-2}}=\frac{x-2}{\sqrt{x-2}}\sqrt{x}\right) \wedge (x \ne 2)\right]$.

Is this a correct logical interpretation of the fact "two equations are equivalent if we divide both sides for a nonzero function defined in the set where the equation makes sense"?

Where with "set where the equation makes sense" I mean the intersection of the set of definitions of all the functions involved in the equation.

shintuku

05:34

the statement is not true as you've written it, because $x$ can be 2 on the left side of $\iff$

Sonozaki

You're right, thanks. I should add "and $x \ne 2$" on the left side too.

shintuku

yeah but i see what you're trying to formalize

@Sonozaki by equivalent you mean preserves roots right?

i.e. is that the only significant property you want to keep?

Sonozaki

Exactly!

Sorry for the delirium, but it should be something like this.

Oh my god, it renders it horribly. I try to write it in more lines.

shintuku

first you need to start thinking in terms of solution sets instead of equations

Sonozaki

$\left[(x \ge 2) \wedge (\sqrt{x-2}=(x-2)\sqrt{x}\right] \iff \left[\left((x>2)\vee(x=2)\right) \wedge (\sqrt{x-2}=(x-2)\sqrt{x})\right]$
$\iff \left[\left((x>2)\wedge (1=\sqrt{x-2}\sqrt{x}\right))\vee\left((x=2)\wedge (0=0)\right)\right]$
$\iff \left[\left((x>2)\wedge \left((x=1-\sqrt{2})\vee(x=1+\sqrt{2})\right)\right)\vee(x=2)\right]$
$\iff [(x=1+\sqrt{2})\vee(x=2)]$

@shintuku Yes, indeed that's what I mean when I write $\sqrt{x-2}=(x-2)\sqrt{x}$: I mean $\{x \ge 2 \ | \ \sqrt{x-2}=(x-2)\sqrt{x} \ \text{is true}\}$.

shintuku

05:48

ok, so you understand then why $\{x \in \mathbb R: \sqrt{x-2} = (x-2)\sqrt{x}\} \neq \{x \in \mathbb R:\frac{\sqrt{x-2}}{\sqrt{x-2}}=\frac{x-2}{\sqrt{x-2}}\sqrt{x}\}$

division kills roots in solution sets

Sonozaki

Yes: $2$ belongs to the left hand set, while $2$ does not belong to the right hand set. And that was my mistake when you replied "is not true as you've written in, because $x$ can be $2$ on the left side of $\iff$"

shintuku

@Sonozaki $0=0$ is non-active statement in your second line

Sonozaki

What you mean with non-active? I canceled it because it is always true, and since the conjunction is true when both the statements are true we have that $[(x=2)\wedge(0=0)] \iff [x=2]$. You meant this?

shintuku

oh nevermind, i understand why you put it there

@Sonozaki yeah that works

Sonozaki

06:06

Thank you!

shintuku

np

lone student

@robjohn What I mean, in this answer, completing the square is "terminological" incorrect or "procedural" incorrect or "identity-like" incorrect ?Can you add further explanation, if possible...?

user1176589

08:20

Hello

leslie townes

09:30

hm, any californians feel an earthquake?

mm, small one off the coast of LA. big enough to feel it, though.

PM 2Ring

09:43

Here's a new bluegrass song about a Californian railway line, from Molly Tuttle & Golden Highway - San Joaquin

Molly Tuttle & Golden Highway - San Joaquin (Live)

►

PM 2Ring

09:58

Here's a more sedate song from them (which I posted here last year), about the high cost of living in the Bay area. San Francisco Blues

leslie townes

haha

user 726941

:D

PM 2Ring

10:19

Molly now lives in Nashville, but grew up in Palo Alto. A couple of years ago, she was visiting her parents, and sadly learned that some of the restaurants they used to dine at when she was a kid have gone out of business.

sunny

10:42

What is the "relative absolute-value topology"?

The word relative confuses me. Is it simply the topology induced by the absolute value metric in $\mathbb{R}$?

Thorgott

11:02

yeah, that's it

the wording is kinda odd

Thomas Finley

I tried solving this by two different methods.

Strangely, I am getting 2 different answers

One one hand, I am getting, y=\sqrt{\frac{2x}{xc-1}}-\frac{y}{cx-1}$

And on the other hand, I am getting, $y^2=2cx-c^2$

I don't think they are the same provided c is just an arbitrary constant and $p=y'$

What d' you guys think?

sunny

11:47

Definition: A function $f: X\to Y$ is continuous on $X$ if $f^{-1}(V)$ is open in $X$ for every open set $V\subset Y$.

> If $Y$ is discrete, then $f:X\to Y$ is continuous if and only if $f^{-1}(\{y\})$ is open in $X$ for each $y\in Y$.

How does this follow from the definition? Why only consider the sets $V=\{y\}$?

Alessandro Codenotti

Because preimages commute with unions

sunny

Ok, I think that clarifies it, thank you.

Thomas Finley

12:11

Nvm, ignore it, I was making a silly error

冥王 Hades

Kinda funny how no one here knows my real name

I’m Brandon. Or as I like to call myself, Brandoff

george

13:12

I would appreciate some help on this Post:math.stackexchange.com/questions/4728916/…, so far haven't got satisfactory answer :(

Mats Granvik

14:01

Is there any point in finding the divisors of $n$ as a solution to a knapsack problem?

Jaakko Seppälä

What should one do if he wants to study mathematics but an exercise in a book is too hard and one gets no help from math.stackexchange.com

Jakobian

14:37

sometimes it happens even to the experts

I remember how Ronnie Brown was talking about his experience with Spanier's book, and how he still can't solve one of the exercises there

just leave it

what can you do, how much time you want to spend on it only you can decide

george

15:44

There is problem in math education often things are not well explained to students, also in response to Jaakko above. I have similar problem above, there was solution in book which itself was unclear, see my link

In above comment

Prithu Biswas

If the product of two right cosets is also a right coset, then:
(H·a)·(H·b) = H·a·b
Then:
a·H ⊆ H·a·H ⊆ H·a (1)
H·a ⊆ a·i(a)·H·a ⊆ a·H·i(a).a [∵ (1)] ⊆ a·H (2)

From (1) and (2)
a·H = H·a

Will this work?

PNDas

16:14

Any hints to show that: the set of left invertible maps on Hilbert space $V$ form an open set in $\mathcal{B}(V)$?

$T$ is left invertible iff $T^*T$ is invertible. So I trying like $||S-T||<\delta$ and wanted to show $||S^*S||<1$.

And I also know that $||T^*T||=||T||^2.$

PNDas

16:45

nvm. got it.

Sha Vuklia

17:20

Let $n\in\mathbb N$, and write $n = p_1^{e_1}\cdots p_r^{e_r}$ for its unique prime factorization. Let $f,g\in\mathbb Z[x]$. I want to understand why we have an isomorphism:
$$(\mathbb Z/n\mathbb Z)[x]/(f,g) \cong \prod_{i =1}^r (\mathbb Z/p^{e_i}_i
\mathbb Z[x]/(f,g))
$$
By the CRT and using the UP of polynomial rings and direct products, we obtain an isomorphism
$$
\mathbb Z/n\mathbb Z[x]\cong\prod (\mathbb Z/p_i^{e_i}\mathbb Z[x])
$$
We also know that the ideal $(f,g)$ is mapped into the ideal $\prod (f,g)$ of $\prod (\mathbb Z/p_i^{e_i}\mathbb Z[x])$. How to show this is onto?

Thorgott

finite product is same as finite direct sum, which commutes with tensor

Sha Vuklia

"the same"; not categorically

but as sets, I suppose then?

actually... not even as sets, right

direct sum in CRing is an annoying construction, no?

like free product of groups

wait, I'm conflating things

direct sum is the tensor product

ok, I was thinking in the general case of not necessarily commutative rings

oh ok, I think I interpreted your "direct sum" as "coproduct"

I was trying to distinguish the notions categorically, and in the process I threw the notion of "direct sum" away

stupid, but thanks!

I don't think I'll ever write $R\oplus S$ anymore, that's hella confusing, because it's not a coproduct. Also the operation $(a,b)\cdot (c,d)=(ac,bd)$ begs for the direct sum to be called a direct product. I understand that in abelian categories, the two notions coincide, but CRing is not even abelian (or additive)

sunny

17:41

@Thorgott here and in other questions too they are using the same phrasing; "relative topology" (for what I believe to be the topology induced by the absolute-value metric). I'm curious where the 'relative' comes from...

Thorgott

@Sha I was thinking of modules, not of rings

for rings, coproducts and products are very different indeed

but if you have a ring hom that's a module iso, it's also a ring iso

@sunny it's a synonym for the subspace topology

Sha Vuklia

@Thorgott yea I will still write $\oplus$ in the case of modules

@Thorgott fair

Jakobian

17:56

@PrithuBiswas that's not right. You don't know that (Ha)(Hb) = Hab, just that (Ha)(Hb) = Hc for some c

That's what it means for product of right cosets to be a right coset

If (Ha)(Hb) = Hab, this is a stronger assumption

Prithu Biswas

18:18

@Jakobian We know that (Ha)(Hb) is a right coset.

Let (Ha)(Hb) = P

Now a ∈ Ha and b ∈ Hb, so ab ∈ (Ha)(Hb) = P

But, we also know that ab ∈ Hab

And since right cosets are disjoint,

P = Hab

*Right cosets are either equal or disjoint.

Am I doing it something wrong?

Jakobian

18:53

No that argument is correct

Okey with this your whole argument is correct

Ted Shifrin

19:42

@george I don't follow it, either. There is some slick way of avoiding thinking about it recursively, but I don't see it yet. Have you tried examples? The formula seems to be right for the examples I've tried.

Ted Shifrin

20:11

@george So if you work out examples, it then becomes clear how to prove the formula inductively — I still will think about where their "explanation" comes from. Let $f(n,m)$ be the number of hours you can light the room with $n$ candles, given that we make one new candle after $m$ are burned. Then $f(n,m) = f(n-m+1,m)+m$ (can you see that?). Now check that with $f(k,m)=k+[(k-1)/(m-1)]$, we have ...

oneofvalts

Do anyone know how does cofactor expansion is called in French?

lone student

@TedShifrin Hello.

oneofvalts

Wikipedia article does not have any coreespondent page.

Ted Shifrin

by complete induction $f(n,m) = f(n-m+1,m)+m = (n-m+1)+[(n-m)/(m-1)]+m = n+1+\left[\frac{(n-(m-1)-1}{m-1}\right] = n+1+\left[\frac{n-1}{m-1}\right]-1 = n+[(n-1)/(m-1)]$, as desired.

lone student

@TedShifrin Would you like to leave a useful comment here I don't know, how can I eloborate .

Ted Shifrin

20:18

Hi, lone. I am trying to think through george's question.

lone student

@TedShifrin Thank you. Not problem, I would gladly wait for you until you have a few seconds :-) .

Ted Shifrin

@lone This sounds like a very old book that is abusing notation and the reader hasn't figured out what they mean.

lone student

@TedShifrin Interesting...Are the comments I left correct in this sense?

Ted Shifrin

I wrote a comment.

lone student

@TedShifrin I see, many thanks I appreciated .

george

20:28

@TedShifrin didn't try examples I was merely a bit angry that author just gave that brief explanation as solution. And i was surprised why i could not see how the solution directly flowed from his explanation

I mean initially i was surprised

Ted Shifrin

I think they're focusing on what gets you to the stage where you're burning the absolutely last candle. But I don't see an immediate way to justify the $m-1$. But it seems to make sense when you do the inductive step I wrote above.

george

@TedShifrin that was my problem I wanted to justify usage of n-1 and m-1 ... i will look at your solution.

Ted Shifrin

@lone You have analyzed one case correctly. But you can also have $y=-x$, $z=0$.

george

@TedShifrin "Now check that with $f(k,m)=k+[(k-1)/(m-1)]$" -- where did this come from?

Ted Shifrin

I'm proving the formula holds by induction. So I'm going to assume it holds for $f(k,m)$ with $k<n$.

george

20:43

@TedShifrin yeah i thought so my math got rusty, need to lookup complete induction. Anyway feel free to write up an answer too, maybe will be useful for someone else in future

Ted Shifrin

I would like someone to give a nice conceptual explanation. I didn't take the time to read the first answer; maybe that person really did so.

george

@TedShifrin nah the first one didnt illuminate me either :)

Ted Shifrin

@lone I wrote an answer.

Maybe you read it after the last candle had burned out, @george.

leslie townes

people who are deep into combinatorics often have a sense of 'explanation' that does not correlate too highly with any intuitive one. e.g. a formula itself might count, or a formula plus a proof that it works might count, even if there is no explanation of where the formula came from. sometimes there is theory that semi justifies this way of thinking (e.g. WZ theory) but not always.

Ted Shifrin

Combinatorial arguments can indeed be very clever, and I am not combinatorially clever. Most of those arguments come from looking at the problem a different way :) That might be the case here.

@george What book is this, anyhow?

I admit, though, having given a valid proof, I would like a nice way to understand the formula.

leslie townes

20:55

because i am a symbol pusher, i definitely count a formula plus a proof that it works as an explanation. maybe not the best explanation, but definitely an explanation. even an explanation that might stop me from looking for a better explanation.

feels like there "ought to" be one here, though.

Ted Shifrin

We want to think of additional candles as coming from dividing $n-1$ (all but the last candle left burning) into groups which will give additional candles. But each of those groups will in turn lead to further candles, etc. I think that's why there's an $m-1$ rather than $m$, but I don't see it yet.

Let's just focus on that. I have $k$ books. I stack them into groups of $m$. That's $[k/m]$ groups, hence $(k-m[k/m]) + [k/m]$ books in round 2. Now I group those into $m$, etc. There's got to be a nice way to do this.

george

@TedShifrin sadly i dont think it is in english

laterna.ge/?cat=98&show=18935

Author is coach of programming olympiad team

In my country

Jakobian

I don't recognize those letters

Ted Shifrin

So this is a text on competition mathematics, I assume?

I'm assuming that in an earlier problem some argument like this was explained more carefully.

Jakobian

ah, Georgian

george

21:02

@TedShifrin no abou programming puzzles but due to the nature of solution i thought posting here

lone student

@TedShifrin Thank you for your feedback. I thought in such way : Since $x=-y\iff y=-x$ and since if $ \frac{y}{x-z}=\frac{y+x}{z}=\frac{x}{y}$, then $y,z≠0$ otherwise the equality would not hold, therefore I thought that, I need only $x+y≠0$ for applying the Fraction Arithmetic . Therefore, I tried to show that, $x+y=0$ can not be possible, which contradicts with the problem statement. Therefore, the Fraction Arithmetic can be applied .

Ted Shifrin

@lone The fraction arithmetic presupposes the fractions all make sense in the first place. The OP is reading a book that allows them NOT to make sense. That's why we move to this system of quadratic equations instead.

george

@TedShifrin i can find authors email in case you are interested

Ted Shifrin

No, @george. Now that I've rephrased the question, it seems like this should be a "well-known" argument ... even if it's not well-known to me.

Jakobian

it's written in Georgian and you are George. Isn't that funny?

Ted Shifrin

21:04

Where are our clever combinatorial thinkers? I can certainly email one of my friends who's a professional combinatorialist :)

leslie townes

i really don't understand the degree to which people bring material from really old books to MSE. i understand why people don't always use the most recent books, or the books regarded as "classics" in some parts of the USA. but copyright considerations (for example) wouldn't explain why people dig into this really old stuff.

and if their instructors are working from templates developed decades ago and nobody has bothered to change, that's... a lot of decades of not changing.

lone student

@TedShifrin Yes, indeed .

David Raveh

@leslietownes how old is "really old"?

Jakobian

around three old

george

@Jakobian yes that's a coincidence or how it is called :)

Ted Shifrin

21:06

Well, sadly, leslie, as the OP said, even in "modern-day" calculus classes (NOT MINE) and books, people do write crap like this with parametric equations.

leslie townes

david: i have no specific range in mind but someone was recently asking about a differential equations text published in the 1890s.

Ted Shifrin

This sort of stuff comes up naturally in projective geometry.

@leslie And he keeps asking about it.

george

@TedShifrin "Well known argument" now you mean the books solution is clear to you?

Jakobian

I think 1950+ is still acceptable though

Ted Shifrin

No. But it will be to people who think about this kind of stuff. I'm not done thinking yet.

leslie townes

21:08

absolutely. 1950+ is new. :)

Ted Shifrin

@leslietownes But I'm 1950+ and you call me a triple dinosaur.

george

@TedShifrin I see at least I will not be suffering by thinking I was too dumb that I didn't realize how from this books explanation the solution followed

I threw out the book actuallyu

because of that :)

Ted Shifrin

LOL ... it's actually a cool question/problem.

leslie townes

i just don't understand how self study would lead one into the 19th century, unless it was self study on the history of the subject, and not trying to learn it in the first instance. and i refuse to believe that people are intentionally teaching out of these books for any good reason.

Jakobian

I think stuff like paracompactness was only formalized in 1960 to 1970

george

21:09

@TedShifrin Problem is cool but I was dissapointed by explanation

Ted Shifrin

No, paracompactness is older than that.

george

That's not how you write explanations for people who you want to teach programming via puzzles

Ted Shifrin

Well, the first thing you should do, @george, particularly if you're that sort of person, is work out examples, and working the examples leads you to the formula in a reasonable way. But I agree they haven't explained it.

leslie townes

@TedShifrin no, you're a dinosaur because you used to hang out with archimedes.

george

@TedShifrin How did you see the formula from examples, didn't you just assume "k+[(k-1)/(m-1)]" in the induction and you assumed that because the books explanation said so, wasn't that the case?

Ted Shifrin

21:11

The sentence "the number of burned-out candles reduces by $M-1$" is the key. I found that awkward, but if we understand that the solution makes sense.

David Raveh

I have recently been self-studying group theory, and it has inspired me how deeply fundamental the material is. The only other math subject that I find as profound (still an undergrad) is linear algebra. Any other subjects in math that are comparable in this regard? (Topology? Will take it next semester.)

george

@TedShifrin Does N-1 make sense too?

Ted Shifrin

No, I worked out three or four examples and saw that the extra term had to be what was there.

Jakobian

I've checked and it was introduced in 1944 as a covering property by Dieudonne. As a covering property, seems to imply it was known earlier, just not in the standard form

Ted Shifrin

Yes, $N-1$ makes sense. You have to have one candle left burning while you're converting the others.

1944 is a lot before 1970, @Jakobian. It was in all the standard manifold treatments by the 50s ...

george

21:13

@TedShifrin I thought by -1 he was referring that in the FINAL moment you have one unused burnt candle left

by N-1

lone student

@TedShifrin I voted your answer. Thank you for your remarks .

Ted Shifrin

But think about getting to what you think is the last candle (before you've swapped for more).

george

@TedShifrin Not sure what you mean

@TedShifrin Also one thing out of curiosity, in this response "Maybe you read it after the last candle had burned out, @george.
" it was humor when I said about illumination right ? :) Just asking out of curiosity if I "got" it no big deal

Jakobian

yeah, I was mistaken. “In the late 1940s and early 1950s, general
topology had a strong upsurge brought on by the definition of paracompactness”

Ted Shifrin

You are down to your last candle burning. So you have $N-1$ to convert and get more to burn. It's the repeated process that leads to dividing $N-1$ by $M-1$ instead of by $M$.

leslie townes

21:15

@DavidRaveh topology certainly comes up a lot, although often more as a language for talking about stuff than, like, as a bag of specific tools or theorems that you use. i guess group theory is also like this, it is way more common to encounter groups and use their basic theory/notions than it is to use deep results specific to group theory.

Ted Shifrin

@george Yes, of course.

@leslie I suspect there's some sort of formal power series trick here.

george

@TedShifrin why N-1 is used in nominator, that part I still can't follow :( sorry.

Ted Shifrin

I've explained it.

There are $N-1$ candles already burned.

Are you sure you translated the sentence I quoted correctly? "reduces by $M-1$"?

george

@TedShifrin When are N-1 already burned? after which step?

Jakobian

thankfully I'm not studying history of mathematics, that could of been embarassing otherwise

Ted Shifrin

21:18

When we think we're burning the last candle (the $N$th) ... before we trade for new ones.

george

Let's go by example. N = 4 M =2
1) Burn 4 candles (4 hours)
2) Now we have 4 waxes
3) Make 2 New candles
4) Burn those two candles (2 more hours)
5) We have one more new candles due to step 4
6) Burn that one too (1 more hour)
7) We have that final unused and burnt candle

Ted Shifrin

No. You can't do that.

The room will be dark.

You can only burn 3 and keep the last one burning to light the room. Now group your 3 ... only one new candle and an old one.

george

@TedShifrin in which my step is the error?

Ted Shifrin

So you get 3 -> 1 new candle plus 1 left over. When the last burning candle is burning out, you light the 1 new candle. You then have the left over candle and the one you just burned. Those two trade to one last new candle.

Your step 1 is wrong.

george

@TedShifrin Sorry it is getting late here and I need to go to bed soon :( (due to work) It will be best I think if you write up an answer which conceptually explains and justifies use of N-1 and M-1

@TedShifrin Do you think you can do that?

Ted Shifrin

21:27

No.

But you need to reread this discussion tomorrow and understand your error.

george

@TedShifrin Ok I will try. But why is there dark in my room? First step 4 hours it is lit right? When the fourth hour ends I can already use new candle obtained from burning first two candles. Then I can use candle obtained by burning 3rd and 4th candles. And then I can create final new candle and immediately burn it

@TedShifrin Anyway if you don't feel like explaining anymore it is OK too, no big deal

Ted Shifrin

Oh, the first answer really does explain it in a valid way. And now I understand that awkward sentence. Every time you burn $M$, you get a new one, so you have a net loss of $M-1$ candles. So we need to know how many groups of $M-1$ there are in $N-1$.

george

@TedShifrin " are in $N-1$." Why in N-1?

That's my problem currently :)

Ted Shifrin

As I have said three times, you must always have (at least) one candle available to be burning.

I will think about trying to explain it more clearly. Good night.

george

@TedShifrin Hmm.. Ok I hope tomorrow when I come back to this chat it will be more clear for me. if you write an answer it will be even better potentially for others. Either way, thanks and good luck!

Ted Shifrin

21:39

I think the first answer is actually pretty good.

Anyhow, I'm doing other things now.

george

@TedShifrin My main problem remains the N-1 term, somehow I couldn't follow your points where you explained it that is why I hope tomorrow I can see something there

but I am not sure

i mean see something in chat tomorrow

And weirdly this chat shows my rep from another stack exchange site, maybe it is a bug.. Anyway thanks and good night.

Xander Henderson

22:56

@george Chat shows your total XP across all sites.

For example, I have about 26k XP on Math, 7k on Math Ed, 2k on Academia, 1.4 K on seasoned advice, and about 1.5 across other SE sites.

So I have around 38k XP total (give or take).

This is the number which should be shown in chat.

Indeed, chat shows that I have 37.8k XP.

This is as intended.

I'm bored. Someone do a little dance and entertain me. :/

Sine of the Time

23:12

@XanderHenderson 💃💃💃💃💃💃

Xander Henderson

@SineoftheTime Awesome. Thanks. I am entertained.

Ted Shifrin

Count on me not to be entertaining in the least!

Xander Henderson

@TedShifrin Oh, I'd already counted you out. Dinosaurs should not have to dance. :P

Ted Shifrin

We don't need more earthquakes!

Xander Henderson

@TedShifrin Earthquakes are fun. I miss living in a tectonically active area.

Ted Shifrin

23:21

Leslie said there was one early morning outside of LA.

Xander Henderson

Though, technically, I do live in an area which is volcanically active.

Though there haven't been any eruptions in about 600 years.

Er... sorry, 950 years.

Ted Shifrin

Almost as long as since I roamed the earth.

Xander Henderson

Heh.

leslie townes

yeah, there was a little bump this morning. not big enough to wake anybody up, clearly.

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$Mathematics$ Mathematics