@Prithubiswasleftmse the theorem holds pretty generally

It works in any normed space for Frechet derivatives and in any locally convex linear space for Gateaux derivatives, I think

@Jakobian I apologize I don't know anything about your generalization of the chain rule.

Probably because I am studying basic calculus.

What is the range of $T: C[0, 1] \to C[0, 1], \quad f(t) \mapsto \int_0^1 f(\tau) d\tau$ ? Are constant functions the only thing missing ?

@TedShifrin How is the domain irrelevant here? If g:ℝ→ℝ and f:ℝ[>0]→ℝ and image(g)=ℝ , then how can we contruct f ∘ g to talk about the derivative of f ∘ g at point a∈ℝ?

mehdi: constant functions are the only things in the range of that map (unless the right hand side of the \mapsto is intended to be something else)

my bad, you are right

prithu that is more a question about what the domain of f circ g is. it is not really about the chain rule.

* $T: C[0, 1] \to C[0, 1], \quad f(t) \mapsto g(t) = \int_0^t f(\tau) d\tau$

i think most of yesterday's discussion goes through. for the chain rule, you'd generally want f to be differentiable at (and in particular, defined at) g(a), and g to be differentiable at a.

mehdi: C[0,1] has stuff that isn't even differentiable, and differentiable things whose derivatives are not continuous. think about the FTC here.

@leslietownes thank you for the hint!

@LeakyNun @AidenChow: Not exactly. You cannot do the division if y−2·y^2 = 0. You're lucky that this particular differential equation does not have a solution that intersects { y = 0 } or { y = 1/2 }, otherwise you would be in trouble.

@user21820 so how would u do this if it does? how would u separate the variables without dividing them?

@AidenChow When you have a curve described by dy/dx = f(y)·g(x), you do get 1/f(y)·dy/dx = g(x) at every point where f(y) ≠ 0, so you can solve the equation for any part of the curve on which f(y) ≠ 0. You would then have to piece together such parts to find the whole solution.

See here for an example of that sort:

@leslietownes Yes. I agree. But I just wanted to clarify that we might not be able to construct f ∘ g if we don't know the domains and ranges of the functions.

Unfortunately, almost nobody ever teaches the rigorous approach, so in high-school and even undergraduate exercises it is very common to see only differential equations where the wrong approach yields the right answer.

@leslietownes ^ This is for leslie's comment.

sometimes there is a big theorem lurking in the background that rules out nasty examples like the one you post, so i would hesitate to say the methods are always wrong. but the books that have such stuff tend not to explain the gap in the symbol-pushing or how to fill it.

or why it won't be there for the examples that are given to the student.

that's a weird thing about beginning DE books.

@leslietownes Of course, but correct pedagogy must explain how to do things right. I've literally seen tons of ODE exercises where they intentionally avoid the problem.

At a higher level, sure, we can teach high-powered theorems such as for analytic/meromorphic functions, for which it doesn't matter to do such dubious division. But can a student that fails to understand "division by zero is ill-defined" successfully understand these higher theorems?

i don't know. i think it's perfectly fine for pedagogy to say "we are putting you in a walled garden where these methods will work. please be aware that the world outside is not this garden, but we will not be exploring it in this class." that isn't so much explaining how to do things right, as explaining why you won't be explaining how to do things right.

but i agree that something should be in a book somewhere. the agenda shouldn't be totally hidden.

@leslietownes Yes that's another approach, though not my preferred one. I don't teach wrong things at all.

@leslietownes (Unrelated to ODEs) More funny functions here:

These are well-known, called Blancmange pudding curve and Cantor staircase function.

The graphs were actually plotted, using Graph. I can provide the formulae if anyone is interested.

i must say, you missed a golden opportunity to label the x-axis, in blue: "Bounded continuous function with zero derivative everywhere"

HAHAHA!

what does the federal trade commission have to do with integration?

@copper.hat None, because there is only disintegration.

and datintegration.

copper, when you gather together infinitely many f(x) delta-x's in one place it presents enormous antitrust concerns

both vertical and horizontal integration potentially raise this issue

ahh, fuzzy non standard oligarchies

prithu: sure. although the chain rule would still hold (under the usual hypotheses) if C and D and g were such that the domain of f circ g were strictly smaller than the domain of g, at least as long as the set was open

what the, i have to consider the domains of the functions for chain rule???

i don't know why someone keeps asking questions about the chain rule. look in a textbook, see how they phrase it. it might not be 'optimal' in the sense that the chain rule holds in contexts that aren't expressly assumed. that's fine. nobody wants a nasty, Most General Chain Rule.

unless you do.

i propose a change of subject to antitrust concerns raised by the riemann integral

@AidenChow ¯_(ツ)_/¯

you would need to set some upper and lower bounds on the discussion...

it's pretty common for calculus books to state results for functions defined on intervals, or finite unions of intervals, whose compositions are defined on sets of the same type. that's where you confront the full generality of proving the theorems of interest to most people even if it's not exhaustive. single variable analysis books are often are a little more general, but still stick a lot to intervals. i'm OK with this

i taught out of a book once that expressly assumed all functions were defined on intervals and piecewise monotone (i.e. monotone on each element of a finite partition of the domain)

stuff like that makes the change of variable theorem a little easier to motivate and has the odd advantage of making a lot of 'intuition' arguments closer to generally true

pretty bad from an analysis point of view but maybe fine for the folks who just have to take the one class or two and then major in biology

@leslietownes I can't sleep without knowing what assumptions [even if it is far from general] a particular book is making.

a well written analysis book, of which there are many, will not keep you up at night

although it might if you wonder if the hypotheses in the book's theorems are the "best possible"

sounds like a constitutional preamble

maybe consumers need to be protected from the riemann integral

@user21820 could you share the formula for the staircase function please ?

@MehdiSlimani y = f(x) where f(x) = if(x<0 or x>1,f(x-floor(x))+floor(x),x<1/4,f(4x)/2,x>3/4,1/2+f(4x-3)/2,1/2).

@user21820 thank you

@MehdiSlimani I might as well give you the pudding curve as well.

@user21820 is this a correct way to read the function definition ? : the function is defined by parts. the first 3 conditions lead to a recursion, with the last expression being the "else" part. the 2nd, the 3rd and the "else" define what's happening between 0 and 1

y = g(x) where g(x) = if(x<0 or x>1,g(x-floor(x)),abs(x-1/2)<err,1/2,x<1/2,x+g(2x)/2,1-x+g(2x-1)/2) and err = 2^-64.

Just paste the definitions of f,g,err into the "Custom functions/constants" in Graph, and then you can plot them immediately.

@MehdiSlimani No. The syntax is described in the help file under "if".

is this software runnable on ubuntu ? i am trying to run the file Setup.exe downloaded from this page padowan.dk/download . i get "exec format error"

@MehdiSlimani It's a windows software. I do not know how to run windows software on linux, but I think there are emulators and I expect them to work because this is clearly a very simple program.

there is a software called "Wine" for running Windows applications on linux. ill give it a try

If it doesn't work, or if you want to plot it using another software, then the syntax is like this: if(C[1],v[1],C[2],v[2],...,C[k],v[k],v[k+1]) tests C[1..k] in that order, and if it finds C[i] = true then it returns v[i] and stops testing, and if it finds C[1..k] all false then it returns v[k+1].

@user21820 ok, thanks a lot! allowing overlapping conditions is neat

But if you can get Graph to work, it's really fast and plots well and has some convenient features like symbolic differentiation and numeric integration. And most importantly it's free haha..

@user21820 the program runs fine with Wine. I am trying to insert a "standard function". in the function equation line, which starts with f(x)=, I past the if statement. i get the error: ' unknown variable, function or constant "f" '

Under "Function" on the menu.

You need to define f,g,err there. Then you simply plot the graph y = f(x).

Note that you can arbitrarily redefine stuff there, and once you apply it will update all the graphs. So you can use that feature to experiment with changing some 'constant'.

no worries, thanks to you. neat software!

Great! Have fun playing around! =)

@MehdiSlimani And thanks for confirming that it works. Now I can tell other people that they can run Graph on linux using Wine.

=)

@user21820 **thanks a lot! neat software, glad i could help in some way

You're welcome!

Why some professors ask to make problem in take-home exam?

@love_sodam I'm sorry... what?

@XanderHenderson In my point-set topology class, professor announced that there will be a problem that ask to make a problem.

@love_sodam I'm not sure I understand... are you saying that your instructor is putting a problem on the exam which asks you to write a question?

If so, this is not uncommon.

@XanderHenderson Yes. Make a problem and that problem will be graded by peer-grading.

In principle, asking students to write (and answer) a questions is a way of getting at how deeply they understand the material, and serves as a kind of anti-cheating check, as student written questions should be more-or-less unique (making it harder to collaborate with other students in the class).

Different professors have different philosophies about student collaboration.

It is questionable how many people will create a truly original problem. I guess many of them will make some small variation of known problems and submit it as a solution.

Of course this is my naive idea. It's not from experience

Let $G$ be any group and let $\Bbb{C}[G]$ denote its complex group ring/algebra, which I am viewing as all the functions from $G \to \Bbb{C}$ with finite support. My question is, what does it mean for a function $f \in \Bbb{C}[G]$ to be of positive type?

@love_sodam I doubt that anyone is looking for a "truly original" problem.

The hope is that each student will turn in a problem that is distinct enough to (a) demonstrate the student's level of knowledge (turn in "What is 1+1?", get no points), and (b) check against collaboration in the class or simply copying problems out of the text (if two students turn in the same problem, or a student turns in a problem identical to one in the text, a red flag goes up).

@XanderHenderson Well that's my first impression of such problem. I thought I need to provide truly original problem so that I get full credits.

@love_sodam Hard to say.

Personally, I don't much like giving those kinds of problems, as I think they require to much subjectivity to grade (in that you have to assess how "good" the problem is), and they are harder to objectively mark.

On the other hand, I have started including an oral component to my final examinations, and many of the same criticisms can be made, so... meh.

And you said that they are peer graded, and other students tend to be much harsher than I am, so... double meh.

@love_sodam what problem did you provide? :)

@Koro I don't have any problem yet. But there was an announcement that such problem will be on the exam.

Class covered quotient space so there could be some interesting problems

@love_sodam: Do you have any ideas on this? math.stackexchange.com/questions/4428323/…

I want to know how to simplify $(2,x,y)/(x,y)$.

@Koro I would think it has a composition of maps and first iso. But comment is already given in the post

the comment indeed answers my question and I understood what the comment says. But what about doing that using the third isomorphism theorem?

Does third isomorphism theorem help?

That is, if $R$ is a CRU (commutative ring with unity) and let $I$ be an ideal of $R$ and let $J$ such that $I\subset J$ is an ideal of $R$. Then $R/I$ is isomorphic to $(R/I)/(J/I)$ is isomorphic to $R/J$.

@love_sodam yes, I think that simplifies the calculations a lot if executed rightly.

The answer posted to my question uses the third isomorphism theorem in the first equivalence, I think.

But I don't understand how :(

How exactly? $(\Bbb Z[x,y]/(2,x,y))/((x,y)/(2,x,y))\simeq \Bbb Z[x,y]/(x,y)\simeq\Bbb Z$.

Well comment only uses first iso

$(Z[x,y]/(2))/((2,x,y)/(2))\simeq Z[x,y]/(2,x,y)$

The numerator on LHS is isomorphic to $Z_2[x,y]$

Oh my expression is nonsense. $(x,y)\subset (2,x,y)$.

Forgot lots of ring theory

Anyway you can consider the map $\Bbb Z[x,y]\to(\Bbb Z/2)[x,y]\to(\Bbb Z/2)[x,y]/(x,y)$ and conclude its kernel is $(2,x,y)$.

Hi all! I have a brief question about category theory, specifically concerning identifying if a particular functor is faithful and/or full.

In proving that a functor $F:A\to B$ is not faithful, does it suffice to find a single pair of morphisms between two objects $f:A_1 \to A_2$ and $f':A_3 \to A_4$ which the functor maps to the same morphism in $Ff=Ff'=g: B_1 \to B_2$ in $B$?

I'm specifically goofing around with the fraction field functor $K:\mathbf{Int}\to\mathbf{Field}$ which maps the category of integral domains with injective homomorphisms between them to the category of fields with injective homomorphisms between them. My claim is that this functor is not faithful, because for every integral domain $R\in\mathbf{Int}$, there exists a "scaling" homomorphism $\Lambda:R\to R'$ via $r\mapsto \lambda r$ for non-zero, non-unit $\lambda\in R$.

This scaling homomorphism $\Lambda$ has the same image morphism in $\mathbf{Field}$ as the identity morphism $\text{id}_R : R\to R$, because they both map to the identity morphism $\text{id}_K$ in $\mathbf{Field}$ where $\text{K}:R\mapsto K$

Is this counterexample (pair of non-parallel morphisms in $\mathbf{Int}$ mapping to the same morphism in $\mathbf{Field}$) enough to prove unfaithfulness of the fraction field functor $\text{K}$?

@dsillman2000 your $\Lambda$ is not a morphism in that category

it doesn't preserve multiplication

also it doesn't map to the identity morphism, it is still multiplication by $\lambda$

(look at what it sends $1$ to)

@LeakyNun Totally right. Thank you for catching that

Just for solution verification purposes, can anybody vouch for the faithfulness/unfaithfulness of the fraction field functor? ($\text{K}:\mathbf{Int}\to\mathbf{Field}$)

@dsillman2000 it is faithful

@LeakyNun What church or temple does it attend?

@XanderHenderson the church of grothendieck

@LeakyNun Oh, dear.

That'll never do.

@love_sodam I think I'll do it via defining explicit homomorphism instead of using the third isomorphism theorem (which imo is useless anyways as I don't think it's used anywhere except for its own statement).

@Koro Third isomorphism theorem is essentially first isomorphism theorem. I wonder if people remember third isomorphism theorem.

How about just $R/(I+J) \cong \bar R/\bar J$, where bar denotes modding out by $I$?

@love_sodam there is 4th isomorphism theorem also, which is actually useful :).

In studying the ideals of the quotient R/I.

@Koro Yeah 4th iso. I usually call it correspondence theorem which is extremely useful.

or lattice theorem :).

But is there any use of the second isomorphism theorem?

Anyways, they all come from the first isomorphism theorem :).

such a bland name

Return of the Isomorphism Theorem

Isomorphism Theorem 2: Judgment Day

2 Isomorphism 2 Theorem

Isomorphism Theorem II: The Wrath of Khan

Isomorphism Theorem: The Squeakquel

I've never learned any more than one isomorphism theorem ... that I can recall.

Maybe when I was Munchkin's age I knew more.

some books trot out a first, second, third, fourth, i think i've even seen a fifth. because you've got to name them something, i guess. it may give the impression that these names are canonical.

"fundamental theorem of calculus" has a milder version of the same problem, although it's approximately only two things, with minor static around the hypotheses

Yes, I don’t deny the proliferation.

i like un-named theorems. nobody ever talks about Theorem 3.5, but it's a good one.

isomorphism theorem 2 electric boogaloo

i thought about that one, but 'boogaloo' has an unfortunate second meaning now in the US.

nevertheless, it defines the genre of funny sequel names.

@leslietownes i didn't know about that; and i think we shouldn't let them claim the word

i agree, although i'm particularly fond of 'squeakquel'

I hadn’t been aware of that new meaning. Leslie, have you done anti-Wordle yet?

I'm looking for a result which I'm fairly sure exists but I haven't managed to find it on google/approach0/mathscinet. Any tips?

@Derivative only advice i can give is put quotes "" around key terms and try synonyms in search queries

look for papers in the area related to adjacent issues. if it seems like it ought to be an old result, maybe earlier papers in that area.

Why am I receiving so many downvotes on this answer??? math.stackexchange.com/questions/4428673/…

tough to say, when people don't leave comments. it could be one or more of (1) people felt the question in its current state did not reflect enough effort on the behalf of the asker to deserve an answer, (2) you don't really use the first homomorphism theorem to define that map or show it's an isomorphism.

but again, when people don't leave comments, one can only guess.

it seems to me like fairly rapid downvoting activity for a recent answer, but i don't know what is normal for this kind of thing.

@dsillman2000 I have no idea, but the word "clearly" and the phrase "because we all know" set my teeth on edge.

a common response in the situation (1) is just to vote to close the question (without necessarily downvoting people who have already answered)

@leslietownes Yeah this was the main thing, I posted it and them almost immediately got a lot of downvotes, it freaked me out. I guess you're right I didn't prove it satisfied the homomorphism criterion

If you define an isomorphism, then you certainly do not need to apply the Fundamental Homomorphism Theorem. That seems to show a lack of understanding.

But you do need to prove it's a homomorphism onto with trivial kernel.

hah, i missed that. the given argument doesn't explain why f respects the group operation (although it certainly does).

still, allow me to silently vote 'how rude!' for downvoting without comment.

:)

@leslietownes Meh. While I am not one of the downvoters, I get really tired of the idea that downvoting without commenting is "rude".

hence my smiley face. it certainly can be rude, although it isn't always rude.

I almost never downvote without first commenting and trying to get the OP to make improvements. But after a few hours of silence, I will sometimes add the downvote. I've been voting to close pure homework statements with zero effort without making comments; after hundreds of comments, I'm worn out.

Then there was this person who allegedly didn't want to type her work because it was "too long," but she expected us to do all the typing instead.

I did make several comments out of pique.

i think some people are inconsistent in downvoting answers to PSQs. there are a number of fairly high rep users who routinely answer PSQs and i never see downvotes. although their answers are also generally substantively fine.

i vote and comment fairly infrequently, so there is a sense in which all voting and commenting behavior seems weird to me.

I don't downvote answers to PSQs, but I do vote to close the questions and/or downvote the questions.

Has Munchkin cooked those agridolce leeks yet?

no. we need to find a grocery store that has good leeks. whole foods and ralphs have been terrible lately.

i wouldn't mind if it was just for stock, or something, but if it's the center of the dish.

I got some at the farmers market last time, but I usually get them at Sprouts.

Yeah, sometimes they are very ratty.

ooh, i live near a sprouts. it's off our radar because we need to cross under 405 to get there, which psychologically feels like a long drive into los angeles. but it's actually very close.

Going under a freeway seems like driving 60 miles? Huh.