@Thorgott Witten has done work on this, relating geometric Langlands to Khovanov cohomology and Gauge theory. Witten is definitely widely read among the most mathematically-inclined theoretical physicists

GAP users: If you have something like this: <algebra of dimension 5 over GF(2)> can you enumerate its ideals with a command? The interwebs says "Ideals()" does this, but I only get an error message when I try it...

for some reason, i find the notation $A^B$ for the functions $B \to A$ to be disturbing.

I have seen this, maybe for clarity write it as $A^(B)$ is better

It just seems backwards :-)

Think about $A^2$.

Ordered pairs is the set of maps $\{1,2\}\to A$.

To me, $A^2$ is $A \times A$ :-)

Yes, so?

Points of that are still in 1-1 correspondence with maps as I said.

i;m slower than usual tonight. can't even blame the wine...

Blame your hip.

i cycled the 3 bridges today (excepting the SF->Treasure Island ferry bit) with a friend.

bit tired at the moment, but it was fun

And all the roads from

one to the next?

I walked 5.5 miles. That was enough for my hips!

albany, richmond-san rafael bridge, sausalito, gg bridge, sf, ferry to treasure island, bay bridge back to albany.

excellent! i can barely walk 3 miles at the moment.

standing is uncomfortable.

the scenery is awesome for the most part.

as you know

ate a packed lunch standing in the rain in sausalito watching some tall ship pass by!

How do you avoid the freeways?

cross at marin/buchanon, some complicated route near point richmond, similar machinations near san rafael, there is a bike path from treasure island to the east bay now.

we were very lucky to catch a small, essentially empty (because of rain) ferry to treasure island from the ferry building in sf.

@SineoftheTime It is claimed that the Convolution Theorem can be used to justify a given step. It would be nice to show how that is done.

@copper.hat Just put $1$ in place of $|A|$ or $|B|$. Mapping to a set of one element gives one function, whereas mapping from a set of one element gives $|A|$ functions.

@TedShifrin Staying at home? ;-p

x.com/RockstarGames/status/1730587560726892883?s=20

@robjohn sorry, what step are you referring to?

@SineoftheTime Sorry, I was looking at the other answer.

Why do we know that the solution $\phi(t;(x_0,y_0))$ starting at $(x_0,y_0)$ stays on the level set $L^{-1}(0)$? All we know is that $\dot L(\phi(0;(x_0,y_0)))=\dot L(x_0,y_0)=0$, but I don't see what prohibits the possibility that this derivative is non-zero at a later time

I've found an indirect argument using results about trapping regions

So I consider my question solved (however, if someone has a direct argument, I'm still interested)

Can't uderstand why this answer has 62 downvotes. It's elegant

@SineoftheTime Not everyone agrees with you, obviously.

Personally, I find it a little condescending, but it is clearly attempting to be humorous, so I have chalked it up to my having a different sense of humor, and have ignored it.

@TedShifrin I try to go out hiking (often in the Petrified Forest) frequently. My knees are good for about 6 miles---8 if I push it a bit. It turns out that Olympic-style fencing is hell on the knees. :/

What is the meaning of the 20k+ tag in the cured room? @XanderHenderson

@SoumikMukherjee I don't know? Why are you pinging me here?

You used the tag so I pinged you. And I thought I should not ping in the cured room as that room has some specific purposes.

@Jakobian who removed our messages?

@copper.hat can I buy you a coffee?

@TedShifrin do you do Yoga or exercise?

@SoumikMukherjee you are asking about CURED. I don't know what you are talking about.

So the issue of integration by substitution is to integrate a function of the form $(f\circ \Phi) \Phi'$ knowing $\Phi$ is some differentiable function

We then want it to hold $\int_a^b (f\circ \Phi)\Phi' = \int_{\Phi(a)}^{\Phi(b)} f$

@Jako do you like ice-cream?

@SineoftheTime how you saw 62 downvotes??

@LuckyChouhan I'm the ceo of this site

@SineoftheTime Wow, can I also be the part of StackExchange group?

yes

Nice, do you guys like ice-cream?

sure

I also ;-) My goal is to be moderator of this site. How can I be @Sine ?

Just kidding haha, being a moderator is tough task to do 😅

@LuckyChouhan eating less ice cream is the way

At Xander: my question was the same that Jakobian asked in the CURED room, I get the answer, thank you.

@SineoftheTime Oh I don't eat ice-cream daily. I just ate 'American dry fruits' :-) what is your favorite ice-cream

Whenever I go to Ice-Cream Parlor, I get confused 😥😭😭

If $f$ has a primitive $F$, so that $F$ is differentiable and $F' = f$, then of course $(F\circ \Phi)' = f\circ\Phi \cdot \Phi'$ and integration by substitution holds. This is simple application of chain rule.

Thus we get a simple integration by parts theorem: let $f$ have anti-derivative, $\Phi$ be differentiable, then above equality of integrals holds.

Next are some generalizations

First objection is that, why do we need $f$ to have a primitive? After all, fundamental theorem of calculus still holds if $f$ only has a $c$-primitive $F$, that is a continuous function $F$ such that $F'(x) = f(x)$ for all $x$ outside of a countable set.

That objection carries on to the function $\Phi$ as well

All we need to do is to estimate the amount of points on which $F\circ \Phi$ is not differentiable that we can't use chain rule on

If $F$ lacks differentiability on a set $C$ and $\Phi$ lacks differentiability on a set $D$, then $F\circ \Phi$ lacks differentiability on a set $C\cup \Phi^{-1}(D)$. All we need to do now is make latter set countable.

This will be countable if $f$ has a $c$-primitive, and either $\Phi$ is differentiable (in which case $D = \emptyset$), or $\Phi$ is differentiable except for a countable set, and $\Phi$ has countable fibers

So we obtain the refined version of substitution theorem using chain rule: If $f$ has a $c$-primitive, and ($\Phi$ is differentiable or ($\Phi$ is continuous and differentiable everywhere except for a countable set and $\Phi$ has countable fibers)), then the integration by substitution theorem holds: $$\int_a^b f\circ \Phi\cdot \Phi' = \int_{\Phi(a)}^{\Phi(b)} f$$

But a lot of functions that are integrable don't come from a $c$-primitive, so we'd like an integration by substitution theorem which wouldn't restrict our choices of the function $f$

If $\Phi$ is continuous, injective and has derivative everywhere except for a countable set, then $f$ is (absolutely) integrable iff $f\circ \Phi\cdot \Phi'$ is (absolutely) integrable, and the above formula holds then

Thus by assuming that $\Phi$ is injective, that is monotone, we are able to weaken the above hypothesis about $f$

Neither theorem is stronger than the other

@LuckyChouhan if someone finds a message inappropriate, then they can exercise their option of flagging it for moderator attention. If the message is indeed inappropriate, it will be deleted.

Huh @Koro This is going to be my last message in this group. I am going...... I know you won't miss me but........ Bye

@TedShifrin What do you think about those substitution theorems? Do you know any other generalization generalization of the substitution theorem?

@LuckyChouhan This happens on mse outside chat too. Say, in comment section of a post if one user says something derogatory to another user, then their messages also get deleted.

I gave a possible reply to your earlier comment asking about removal of messages.

we have a course which has 13 references !!

@Koro What class is that? (Note: I'm not shocked by that---my Anthro Theory course typically required around 750 pages of reading per week; I skimmed a lot that semester).

It's Discrete mathematics.

@Koro Given the way that "discrete mathematics" is taught at a lot of places, I don't find that surprising. It is often desirable to bring in ideas from different sources in this kind of "survey-lite" class.

@XanderHenderson Anthropology?

@Jakobian Yes.

Oh that sounds awesome

@Jakobian It was way too much reading.

750 pages sounds like lots of digging in the subject

I watch about anthropology sometimes here:

youtube.com/@TREYtheExplainer

@Koro something specific in the subject?

discrete mathematics sounds like it could be anything

it's endless.

that's a lot, graph theory, discrete geometry

sounds like a fun course

:)

@Jakobian That looks like a lot of classics and pseudo-science to me. Not really my cuppa.

Boba's fine, but I prefer joe.

Boba is fine but what is joe?

never heard of it

Sense 2.

Ohh I thought Joe tea

😁

No.

@XanderHenderson The only pseudo-science is cryptozoology but thats in the format "of course this isn't true, and here's possible explanation, but if it were true..."

@Jakobian I get that---it just doesn't interest me.

So I take n boxed out of which m are empty. So we have n-m boxes left to be filled. Each of these n-m boxes has to be filled. So the generating function is $(e^x-1)^{n-m}$, right?

so $r$ is the variable here?

I interchanged the roles of m and n-m in my attempt above. m are non-empty so we should have $(e^x-1)^m$.

@Jakobian it's fixed.

I don't understand then, which one is the variable

ohh, I see what you mean. I'm sorry I am new to this. Yes, r is the variable here.

By which I mean that the coefficients in exponential gen. functions are indexed with r.

how common is it for a function to simultaneously satisfy a linear and nonlinear differential equation?

@JohnZimmerman Can you give a non-trivial example?

@XanderHenderson no, but I am inclined to think it's quite rare even though I don't have evidence to support that claim

$y(x)=\sin(x)$ solves the nonlinear differential equation $y'(x)^2 = \cos^2(x)$ as well as linear equation $y''(x)=-y'(x)$. Any function $f(x)$ satisfies the linear differential equation $y'(x)=f'(x)$ and the nonlinear equation $y'(x)^2 = f'(x)^2$.

maybe it's not rare

@robjohn I had some momentary (hopefully) brain fuzz. Still don't like $C^{B^A}$ as a way of representing composition :-)

@LuckyChouhan No thank you. Not entirely appropriate.

Just a reminder: I'm trying to get the following undeleted.

I've asked for help in CURED. It seems one person has voted to undelete from there.

I can't vote to undelete, since I voted to undelete the last time the bot got rid of the question.

I plan to set a bounty on the question to get feedback on the substance of my proof attempt, not on the notation used.

@Shaun You have brought this up multiple times in multiple fora. Enough. You have either gotten the support you desire at this point, or you have not. Please stop spamming the site with this request.

It's important that I get help with this question. I need it.

@Shaun It is important to you. This does not make it important or valuable to anyone else.

The generating function is $(1+x+...+x^{25})^{4}$ and we want coefficient of x^{25} because we have 25 votes.

next part is what confuses me- Suppose each student who is a candidate votes for herself or himself. Now what is the generating function and the required coefficient?

Would it be wrong to reask the question, detailing why the change in notation is not helpful, @XanderHenderson?

Shall I ask a meta about requesting for more specific help?

This confuses me because I have two different answers: 1) I look at coefficient of x^{25} in $(x+x^2+...+x^{25})^4$. 2) I look at the coefficient of $x^{21} in $(1+x+x^2+...+x^{21})^4$.

Those are the same thing

how? :(

computing them gave the same value?

I haven't computed them yet.

I am asking based on how the expressions look.

Because the polynomials differ by a factor of $x^4$

ahh, it sure looks like that but I think it's not true: 1) is $x^4(1+x+...+x^{\color{red}{24}})^4$

Oh, it’s a bit more than that. I didn’t notice the polynomials inside the parentheses are not just off by a factor of $x$

Does someone see why the annulus $\mathscr A$ is positively inveriant? (maybe that any solution that starts in $\mathscr A$ will stay in $\mathscr A$ as time increases)

had it been-"suppose each candidate gets atleast one vote,..." then 1) above would be correct, I think.

but the linked part says more by specifying who gives that vote.

otherwise also I think 1) and 2) should be same but apparently they aren't.

hm, I might see it

However, the coefficient of $x^{21}$ in $\left(1+x+x^2+\dots+x^{21}\right)^4$ is the same as that in $\left(1+x+x^2+\dots+x^{24}\right)^4$

I'll explain how I got 2): Four students F_i, 1<=i<=4 get $f_i$ votes. Clearly, $\sum f_i=25$, $f_i\ge 1$. Setting $e_i= f_i-1$, I get: $\sum e_i=21, e_i\ge 0$. So we want coefficient of $x^{21}$ in $(1+x+...+x^{21})^4$.

@robjohn thank you so much and I'm sorry for not verifying it sooner.

But I still don't understand how the following: I think (1)=(2) answer the following: "suppose each candidate gets atleast one vote,..." then

How do I answer- "each candidate gets vote from him/her self."?

hmm, I think in that case I should then subtract 24 from (1).

to account for choice that the remaining students make up that 1 vote.

this seems 2b overcou9g

@Koro None of the terms $x^{22}+x^{23}+x^{24}$ are used because they have too high an order.

@Jakobian The generalization that interests me is integration of differential forms/densities over orientable/non-orientable manifolds.

@robjohn yes, indeed.

@Ted: how are you this weekend? I've had some pretty bad lower back pain since Wednesday. Trying some Voltaren to see if that helps (the symptoms, at least).

At PT on Thursday, they tried some heat and electrostim.

It was still bad that evening.

Ack! the extended Cantina Scene from Return of the Jedi is on. Why ruin that scene?

@robjohn Yikes. Well, I have permanent neck/back pain but try to do exercises mostly daily and see PT, chiro, and massage regularly. My surgeon says it's not bad enough yet. ... Did you do something to tweak your lower back? I never found that the electrostim helped me, but I do regularly sleep on a heating pad for a few hours.

There is nothing that I can remember that I did to cause this, but Christmas lights might have had something to do with it.

It sort of developed over Wednesday.

May have had some pain on Tuesday as well.

That's the trouble. We rarely are aware of an awkward movement that caused it. For me, schlepping groceries (as I just finished doing, like every Sunday morning) doesn't help and I'm always sure to put on a brace/strap to support my lower back.

@robjohn In summary, getting older sucks ....

My pain comes and goes throughout the day, so it is hard to tell if something is actually helping. I just hope one of the pain ebbings becomes permanent.

Did your PT person give you some exercises to do regularly? I find exercises/stretches are essential for me.

@TedShifrin My PT is for my pelvic core, as a follow up from my surgery last January, the heat and electrostim was an after session thing to help. My exercises are for the pelvic core. Some may help both, but they are not targeted for the lower back.

Ah, yes, indeed. I do exercises for the core, too.

quick question why is that true (the highlight portion).

should it be $2 < k + 1$ ?

$k=1$

oh yes HAHAHHA.

I got confused on the induction step that k is greater than or equal to 4. Then I was like is that same k in the highlight portion because that will never be equal to 2.

I mean technically the base case starts at 4, so it isn't wrong to say $2<k$ I guess

but $k=1$ as a base case is a convention

I've had the terrible realization that my general discrete factoring problem: 1) is actually better expressed $z = 2^{\frac a b} c$, and 2) is equivalent to solving $x = n + m$ for real $x$, rational $n$, and irrational $m$. In other words, what I need is not an integer function, nor fractional part function, but an "extract rational part" function for the logarithm of $z$ (when the logarithm is a natural base).

Definitely more confident about the first and third; less so about the second.

@oscarmetalbreak But they’re already assuming $k\ge 4$. The statement is not incorrect; it’s not best possible.

Ah, you already said that.

Yep and since it did little harm so I just let it go

induction always been a issue for me.

I knew I had the same feeling but it was surprising reliable and sometimes the easiest way to come up with a proof

@EM4 And you've probably never done an "interesting" proof by induction, either.

@Koro the answers would be $\binom{28}{25}, \binom{24}{21}$.

@Ted: the Voltaren may have helped, at least temporarily. We’ll see what happens when I take the dogs to the park in an hour.

@copper.hat are those supposed to be the coefficients for $x^{25}$ and $x^{21}$, or is this something else?

@robjohn is it a muscular or skeletal issue?

@robjohn the coefficients of $x^{25}, x^{21}$ resp. usual joe disclaimers re correctness

number of ways of choosing $k_1+\cdots+k_4 = n$ for $n=25,21$.

@copper.hat I think it’s muscular. So far, I haven’t had any skeletal issues.

@robjohn may not be helpful to you, but i used to get disabling back pain (muscular) and i found that doing sit ups regularly made a huge difference.

@copper.hat I think the coefficients are the same. Look at the disconnected conversation I had with Koro around noon (PST)

It could be the generating functions don’t represent the right things. I haven’t looked into that.

@robjohn would it not be that in the second case that it is 21 votes for 4 places?

i remember reading some notes by Qiaochu Yuan (one of them) and being amazed at how generating functions could be used. however, following someone and applying them to different situations are not the same thing, so it was fun to read, but didn't really add to my tool kit.

I’m just now looking back at the actual question…

Okay. Your numbers are correct.

except i didn't use generating functions :-(

Stars and bars is what I used

basically what i did. except i only heard of the term stars & bars about a decade ago.

sounded more like a drinking spot for actors

The generating function seems to allow some people to vote for more than one candidate, unless I’m looking at it wrong.

@copper.hat sounds more military to me

something from a top gun movie (i watched the original 5+ times :-)).

Hmm… the generating function gives the correct answer

i'm not sure what the value of the generating function approach is really

Ah, yes, I was thinking about it wrong.

Both give the correct answer

The coefficient of $x^{25}$ in $\left(1+x+x^2+\dots+x^{25}\right)^4$ counts how many ways $25$ voters can distribute their votes among $4$ candidates.

Can you deduce that BV functions are differentiable almost everywhere from Lebesgue differentiation theorem (about Lebesgue intrgrable functions)?

it is a cute visual, but i still need to figure out the number of ways i can get $k_1+...+k_4 = 25$ which brings me back to the stars & bars

@Jakobian do you mean for measurable BV functions, or is measurability what you are asking about?

A BV function can be non-measurable?

bit slow today. difference of monotonics

it is locally $L^1$ so why would the differentiation theorem not apply?