if you use a similar parametrization there and choose different parameter pairs that map to the same point in R^3, you might get the same normal or the one of opposite sign, depending on which points you pick
someone probably has a nice mathematica or whatever workbook about this
in fact if you compute the normal vector to the strip you get $\mathbf{N}(u,v)$ and then you compute the limits $u \to 0^+, \to 2\pi^-$ of $N(u,0)$ you get different results
I'll go to bed now my eyes are shutting down :p, thanks @leslietownes for the help as always :)
@SineoftheTime $\sin(\pi\sqrt{1+n^2})\sim (-1)^n \frac{\pi}{2n}$ is correct, but the steps, who knows
I wouldn't expect the author to be able to justify them in any way, so yeah, looks like hand-waving
the sort that leads you to an incorrect answer but is correct in this case
also, alternating test does need you for your sequence to be positive and decreasing, so definitely marty cohen's way is the correct way to go about this
just stating an asymptotic isn't enough to decide about convergence
Hmm... Let $K$ be a field with discrete valuation $\nu$ and suppose $\mathcal{O}$ be the associated valuation ring. Let $\nu\in K^\ast$ be a uniformizer and $L\subset K^2$ be a lattice meaning there is a basis $e_1,e_2\in K$ of $K^2$ that $L = \mathcal{O}e_1\oplus\mathcal{O}e_2$. Suppose $L'$ be another lattice of $K^2$ that $L'\subset L$ and $L/L'\simeq\mathcal{O}/\mathcal{O}\pi$. Then $L\pi\subset L'$?
Maybe sounds complicated but for example $L' = \mathcal{O}e_1\pi\oplus\mathcal{O}e_2$. $L/L'\simeq\mathcal{O}/\mathcal{O}\pi$ means it's ''maximal'' sublattice in some sense.
So if $L' = \mathcal{O}e_1\pi\oplus\mathcal{O}e_2$, then $L\pi\subset L'$ is obvious because $L\pi = \mathcal{O}e_1\pi\oplus\mathcal{O}e_2\pi$. But the problem is when $L'$ is not of that form.
There is some evidence $L\pi\subset L'$ is true but no clear reason. It should follow from something straightforward but I can't see.
"The notion of pointwise convergence of a sequence of functions on [a,b] cannot be described by any metric" does this mean that there exists no metric on function space so that convergence in metric if and only if pointwise convergence?
@onepotatotwopotato I suppose this is an $\mathcal{O}$-linear isomorphism and $\pi$ annihilates $\mathcal{O}/\mathcal{O}\pi$, so $\pi$ annihilates $L/L^{\prime}$, but that's equivalent to $L\pi\subseteq L^{\prime}$
also, we did talk about why it's measurable, didn't we
@nickbros123 real-valued I presume?
essentially this should be the same as saying that the sequential coreflection of $\mathbb{R}^{[a, b]}$ is not metrizable
sequences alone are sometimes not enough to generate any topology at all
if you find a subset $A$ such that iterating sequential closure twice, say $A_1 = \text{sqcl}(A)$ and $A_2 = \text{sqcl}(A_1)$ and $A_1\neq A_2$, then that already says this can't be given by any metric
even if the topology of pointwise convergence is not metrizable, there is still hope that the convergence of sequences can be written in terms of a metric
So uh, howdy people. I am trying to defined a weird version of a derivative in a general setting for reasons, mostly for use in functional analysis later. Would any of you have a very general definition of a derivative lying around, one with minimal assumptions or minimal enough to serve? Something like what a metric is to the concept of distance but for rate of change?
@Thorgott if it's an analysis textbook that talks only about metric spaces then that's even more of a proof that topology of pointwise convergence is not what they meant
in this case I don't think they really had topology in mind, maybe measure theory
how convergence a.e. can't be defined using topological space, say
@Thorgott while the definitions from the standpoint of someone who knows those things it may appear so, the dumbed down version is really just "we use sequences"
you don't need to know about what sequential coreflection is to phrase this question, just like nick didn't
I don't see how that's not closer to what the audience knows
@Thorgott Just a set and the space of its functions but I am not particularly opposed to using vector spaces (its just I want to avoid them where I can because my main goal was to develop such operators and see if they induce natural diffeologies (for those who know what that is)).
And by space of functions on $X$ a set, I mean $F(X,\mathbb{R})$. The space of real output having functions which is an incredible set by itself
It can be given all forms of cool stuff like pre-hilbertian structure and norms and such but I wanted to see if I could avoid those for now and just work off of operators in the form of $d:X\timesF(X,R)\rightarrow F(X,R)\timesE$ where E is some error set or overflow or extended function set of some kind where I can fit functions I don't want to view as "differentiable".
I believe similar ideas underpin discussions of continuous extensions as operators on function spaces and limits in the usual sense as operators to be applied to functions.
And they are not compatible to my understanding. Which is a bummer.
Like for example can this not also be seen as a generalization of a derivative?
Consider X and Y metric spaces equipped with $d_x$ and $d_y$
consider a function f from X to Y. Let us look at the "derivative with respect to metrics" of f as the following "function" df from X to R. taking the value $\lim_{x\rightarrow y}\frac{d_y(f(x),f(y))}{d_x(x,y)}$
I say "function" since it might not be defined at all points.
But its definitely local. And I know for a fact it exists since the concept of a limit extends very directly to metric spaces
@Thorgott I think if you take typewriter sequence $1_{[0, 1]}, 1_{[0, 1/2]}, 1_{[1/2, 1]}, ...$ then every subsequence has a convergent subsequence
One just has to show that there exists $x\in [0, 1]$ such that we can find as small intervals as possible in the subsequence containing $x$
or if we're advanced enough perhaps an argument that the sequence of intervals has a convergent subsequence in the Hausdorff distance because it's compact metrizable or something
ah I think the thing I said above is true but not the one I said above it
there should be some sequence of intervals $I_n$ which get smaller in diameter and converge to a singleton, but not necessarily this singleton contains all of them, so it can converge to the zero function
now in metrizable space, if every subsequence has a convergent subsequence, then the sequence is convergent
@Jakobian would be better to replace it with some elementary argument, but I have hard time thinking about it
@Jakobian oh no this doesn't actually work, because this would be equivalent to convergence if every subsequence had a subsequence convergent to the same $f$
why this example works for convergence a.e. is because then they converge a.e. to $0$ since $1_{\{x\}} = 0$ a.e., but here it doesn't
oh okay, replace closed intervals by open intervals
that thing about approximation by 1/(z-a)'s caused me to look to see if anyone had improved on a result i was interested in 15 years ago (they hadn't)
(the only papers i can find citing the papers i was looking at use them as like a string citation in front of "here's how we work on a generalized, essentially different, and much easier version of this problem")
every day just comes and goes, life is one long overdose
@Logarithmnepnep The issue with such a definition is that the limit has absolutely no reason to exist. Even if you're just doing analysis on $\mathbb{R}^n$ and $f$ is a nice (i.e. differentiable) function, you get different rates of change in different directions when $n>1$ and have to assemble them into a linear map.
That's why you would usually do stuff like fix a parametrized path going through the point $x$ and then only look at how $f$ changes along that path, but this is also not a particularly well-behaved notion unless your metric space has enough nice paths and they assemble into something remotely resembling a tangent space.
@leslietownes I see. I was thinking about what math is the other day. We're trying to grasp the complexity of things in simple observations, patterns. And it strucks me that this will never be complete, and it's a lot like listing cases. Even though it's supposed to make things less tedious, math itself is tedious
@Jakobian a lot of results in uniform approximation are like this, tons of very clever stuff that only adapts to whatever problem motivated it and only really works once
and yeah its a matter of opinion whether that makes it cooler or less cool
and in nice settings, the directions are parametrized by a linear space akin to a tangent space, and then for nice functions, you may expect the total result to be a linear map on that tangent space
the impression I get, basically, is that math is ugly, there is nothing beautiful about it, what appeals to the human brain is really just the patterns one can get or come up with
Seeing here the differential of a function as the function of "an infinitesimal rate of change".
Because the directional derivative is a measure of sloping in a graph of some function aka a surface in $\mathbb{R}^n$ with respect to a canonical direction.
which makes me think for example... why is there no stiljes derivative
like the rate of change of a function with respect to another
there is technically the notion of a differentiation on a manifold which kind of functions that way (when considering differentiation along smooth curves and such)
@Jakobian I feel it often seems like someone made an observation about something that seemed really "nice" and then we stuck with it.
often.
Like seriously, these insights into things may not even be the only valid ones one can follow to get interesting math, but we follow in the footsteps of our forefathers and there is definitely a sense of I guess "à la mode" or "classic" in math. Like we used certain assumptions and built so much and taught it for so long that it has become like a shackle on our civilization. They often are solid assumptions and get the job done (we wouldn't be using them otherwise) but it makes one think.
to flip that a little, a lot of common assumptions are commonly assumed not because they shackle our perceptions to some false reality, but because the things we want to study do have those properties, and the math of "why" they should have those properties from some set of first principles is usually more complicated than just assuming them, particularly in a classroom environment
e.g. i don't think you lose much by studying the set of linear solutions to the functional equation f(x+y) = f(x) + f(y) (assumed to hold identically for all x, y for a function f from R to R) because a million things that you might assume about f other than linearity will end up implying that such an f is linear
and in a lot of contexts, diving into the general cauchy problem in the abstract is more of a mind-shackling theoretical exercise than just assuming that there is no cauchy problem
or like "why does my ODE/PDE book assume so many things are smooth" probably because for lots of ODE/PDE everything not assumed to be smooth will end up being smooth for complicated reasons that a beginning ODE/PDE student can't understand
@Logarithmnepnep yeah, but I think those were really more dictated per need, and what we really abstractify goes further down the line, those are things that those people don't study and are of no interest to them. And it wouldn't change if we made different choices along the way
usually the further away you get from commonly studied things, the less interesting things get, you may have to do enormous amounts of work just to find examples of what you're talking about, let alone get other people to care about them
jakobian: because i build bridges for a living, and making this assumption means i don't have to check as many calculations, and adopting these mental blinders lets me quote a lower rate for the project, steal jobs from my competitors, and cuts down on lead time once i get going
i'm pouring concrete under that assumption right now
Hello there. I'm having trouble with a Lagrange multiplier. Let me simplify the problem (of course it's not a calculus problem but I will express it as one). Let's say I want to extremize $f(x,y)=2x^2-xy$ with the constrant $x^2+y^2=1$. So, I have to extremize: $$2x^2-xy-\lambda(x^2+y^2)$$ but if I write it in a symmetric fashion (using $2x^2=x^2+x^2=x^2-y^2+1$, the constraint) $$x^2-y^2+1-xy-\lambda(x^2+y^2)$$ I get a... different $\lambda$. So I wonder if this is normal and if using the constraint beforehand can cause some sort of trouble
Just to be clear, again, it's not a calculus homework. The underlying context is superconductivity, so I tried to make it as less physical as possible
the values of lambda that you get can definitely depend on how you characterize the constraint. the solutions to the underlying optimization problem shouldn't change
i am not rendering latex right now in chat so i'll use the notation of en.wikipedia.org/wiki/Lagrange_multiplier as an example (the "single constraint") section. note, i am not generally recommending wikipedia's treatment or offering it as a model of how to handle it
but there, the constraint is expressed as "g(x,y) = 0." what matters for purposes of the optimization problem is that the solutions to this equation identify what the constraint is. what doesn't matter for the problem is what "g" you use. for example, x - y = 0 has the same set of solutions as y - x = 0, which has the same set of solutions as (y - x)^2 = 0, which has the same set of solutions as 240520987 (y - x)^10 = 0
if you find yourself using x-y vs. y - x vs. (y - x)^2 vs. any of those other choices as your g(x,y), you might find that the lagrabge multipliers setup leads to different algebra and different values of lambda
I'm working a counterexample to the Fubini theorem, namely $$f(x,y)=2e^{-2xy}-e^{-xy},$$with both measures involved being Lebesgue measure on $(x,y)\in(0,\infty)\times (0,1]$. It's claimed that $f(x,y)\geq c$ for some constant $c>0$ if $x\leq 1/(8y)$ and hence $$\int_{(0,\infty)\times(0,1]}|f(x,y)|\,\mathrm{d}x\mathrm{d}y=\infty.\tag1$$What's confusing me is that when I compute $(1)$ without the absolute value, I get $0$.
If $f(x,y)\geq c$ for some constant $c>0$ if $x\leq 1/(8y)$, shouldn't the integral diverge also without absolute values?
@psie the way I'd reason would be to split up the limits of integration from $0\leq x\leq 1/(8y)$ and $1/(8y)\leq x<\infty$ and use the bound $f(x,y)\geq c$ on the integral with limits $0\leq x\leq 1/(8y)$.
@VladimirLysikov Okay, I read about this in question on the site. In my case, I'm using the same characterization of the constraint in the lagrangian. The difference is that I (my book) used the constraint equation inside the original equation, other than imposing the constraint, so it's a little more complicated than scaling the constraint by a factor
Of course it can be more complicated In the example above you basically replace the objective by a linear combination of the original objective and the constraint It can be even more complicated
emphasis on "you," how could it turn out that you are wrong? what's wrong? who says?
if it's just some computer marking that wrong as an answer, couldn't the computer be wrong? don't we have enough information to determine this on our own?
if it's just some computer marking that wrong as an answer, couldn't the computer be wrong? don't we have enough information to determine this on our own?
Again, is the observation that $\ln(0)$ is undefined actually relevant to this problem? It certainly explains how a computer might mess things up (if it attempted to solve the DE using the same steps that you did)...
But is it possible to demonstrate that $y=cx$ solves the DE in any other way?
@XanderHenderson So the reason for this not to work is the $\ln (0)$? right? Now "is it possible to demonstrate that $y=cx$ solves the DE in any other way?" what is the answer to this ?
You have noted that if you solve the DE by separation of variables, then one step of that solution is $\log(x) + C = \log(y)$, and that $\log(y)$ is undefined when $y=0$. This could certainly cause one to believe that they IVP has no solution, which could explain why your answer was marked as wrong.
But that could be an error.
@pie What does it mean for $y$ to be a solution to the IVP?
$0 = y(0) = c\cdot 0 = 0$, right? So it further appears that for any choice of $c$, it is true that $y=cx$ (a) solves $xy' = y$ and (b) $y(0) = 0$. Doesn't this mean that $y = cx$ solves the IVP?
@leslietownes the computer but I want to understand why, I am now so confused, so is it because of the $\ln(0)$? is the answer no solutions ? or what I am so confused rn
i think you're going too far down the rabbit hole of "why could the wrong computer have been led to its wrong conclusion." keep in mind that the computer is perhaps not an actual computer but some human who has input questions and an answer key
nothing about the problem or list of solutions directly presents the issue of what ln(0) is or might be, that is more a question i would ask if i had access to a human being who asserted that there weren't infinitely many solutions to the DE
@XanderHenderson btw the professor made the answers and the questions.. so he made the answer, he also changed his whats-app status to "no solution" because a lot of students has contacted him and he won't answer anyone...
If you are a student in a class, you should have some official means of communication with the instructor of that class. At my institution, that is the email system.
@XanderHenderson well we don't use email to contact at all, tbh my college is so chaotic and there is no order that what allow me to do whatever I want...
@pie Again, you must have some official means of communication. It must exist. I said that at my institution it is email. It might be different where you are.
In any event, you seem to have convinced yourself that your solution is correct, so who cares if the answer was marked incorrectly? Who care why it was marked incorrectly? If you genuinely cannot reach the person who set the problem for you, stop perseverating on it, and move the f*ck on.
@Thorgott also very poor education, the instructor, students etc don't care and it is just so chaotic there is a famous joke here "The college is like a play: the professors pretend they are teaching, the teaching assistants pretend they are working, and the students pretend they are understanding."
i mean their department should have an official page listing all the personnel and each teacher would list there the training courses, their syllabus, maybe some material, schedules and the like... together with a section on their research activity and maybe some page with jokes...
it is virtually the same in any institution around the globe.
@pie well, at first I wanted to say "now you know, so you should try to not categorize everyone as either he or she", but you seem to be just excusing yourself. Either way, now you know.
I feel bad for yoy guys that you had to read my poor english
@Jakobian In arabic if we talk about people from many genders (like some men and women) we use "he" that is why I use it all the time because I translate litrally from arabic to english
i think no one got offended here it was more an advice for the future. yes if you use they, the plural, is gender neutral in english and more acceptable because no assumptions are needed about the gender
> "Consider $(\mathbb R^d,\mathcal B(\mathbb R^d),\lambda)$, $\lambda$ being Lebesgue measure. Let $f$ and $g$ be two real measurable functions on $\mathbb R^d$. The convolution $f\ast g(x)=\int_{\mathbb R^d}f(x-y)g(y)\,\mathrm{d}y$ is well defined provided $\int_{\mathbb R^d}|f(x-y)g(y)|\,\mathrm{d}y<\infty$. In that case, since Lebesgue measure on $\mathbb R^d$ is invariant under translations and under the symmetry $y\to -y$, $g\ast f(x)$ is also well defined."
I struggle with putting the pieces together. I know the abstract change of variables formula for $\nu$ being the pushforward of $\mu$ under $\varphi:E\to F$. Then for any $h:\mathbb R\to\mathbb R$, $$\int_E h(\varphi(x))\,\mu(\mathrm{d}x)=\int_F h(y)\,\nu(\mathrm{d}y).$$
I figured that $h(y)=f(x-y)g(y)$ and $\varphi(x)=x-y$. Since Lebesgue measure is invariant under translations, $\nu$ is just Lebesgue measure again. So I get indeed $f\ast g(x)=g\ast f(x)$. But what do they mean by invariance under the symmetry $y\to -y$?
Consider a right-handed orthonormal reference system in the space $S$. Two planes are given: $\pi_h : h x - h y - (2h + 1)z + 2 = 0$ $\pi' : 2x - y + 3z + 1 = 0$
Set $h$ to the value obtained in the previous point. Denote by $r$ the line of intersection of the two planes. Determine a line $s$ skew to $r$.
i found $h = -1$
but I can't find a short way to do this point, any advice?
@pie while this usage is capable of offense (particularly if someone has been asked not to use it, and then continues to do so, or if it is used in some way to draw attention to the potential gendered use of 'guys'), in my experience [native speaker of english in USA] it generally would not give offense when used to describe a group of people of mixed or unknown gender
even most stuff with potentially gendered pronouns is like that in my view - i.e., fine unless someone objects to it or you're deliberately trying to gender something
older english usage follows arabic in using "male" pronouns to describe individuals or groups of mixed or unspecified or unknown gender and generally this isn't very objectionable either, unless someone is doing it with an intent to annoy peoplee
Which as you may know is a gendered language. The way to go for mixed gender groups is to use masculine words. Nonetheless, there is no 1:1 correspondence between grammatical gender and real gender in any case. Grammatical gender is only determined by usage and rules
in an international forum i err toward accepting basically any forms of address that aren't intentionally offensive, particularly as some languages basically require you to gender things or to apply formal address in a way that sounds hostile in translation
$r: \begin{cases} x = -4t - 3 \\ y = -5t - 5 \\ z = t \end{cases} \quad t \in \mathbb{R}$ I think I managed to find the line $r$ of intersection of the two planes
Set $h$ to the value obtained in the previous point. Denote by $r$ the line of intersection of the two planes. Determine a line $s$ skew to $r$.
Determine a like $s$ skew to $r$ so : they are not coplanar, that is, they do not lie in the same plane. They do not intersect, that is, they have no points in common.
@HerrFeinmann same as Polish. Here situation is different, we don't have something like "they", and everyone has to be referred as either "he" or "she" in my language (unless, say, they want to be referred as "it")
My idea is that as long as grammar dictates some rule, there is no reason to object. As I said grammatical and actual gender are not mapped 1 to 1. When, instead there is no grammar, I would consider also the speaker's intentions. If someone is actively trying to harass another user is one thing, while if someone lightheartedly says something is another can of worms
@Jakobian Italian has two genders. Even if we have a word for it, which in Latin would correspond to neuter gender (some languages such as German still have it, I don't know about Polish, though), Italian does not have it. This entails that even "it" stuff in Italian will be gendered
We have some kind of influence in English too. For example using "she" with ships :D
psie: both kinds of invariance are examples of invariance under affine transformations, but, when you say you don't see "the" connection, what connection is being asserted?
leslie: well, that the convolution is commutative. For me it's only a matter of translation invariance of Lebesgue measure. I don't see why we need invariance of y -> -y.
@Jakobian Depends of where you live. I'm California, "guys" had been neuter for a generation or two. "Dude" had been neuter since the 90s, and even "bro" seems to be neuter more.
psie: as with many of your questions it would help to spell out the full context. what is being asserted, what is assumed, what is the gap between them
psie: there's perhaps something more general here about whether an assumed invariance under "group stuff" means invariance under stuff for the binary operation of the group only, or also under the inversion operation of the group, but when you say "translation invariance" that starts to sound like the former
> "Consider $(\mathbb R^d,\mathcal B(\mathbb R^d),\lambda)$, $\lambda$ being Lebesgue measure. Let $f$ and $g$ be two real measurable functions on $\mathbb R^d$. The convolution $f\ast g(x)=\int_{\mathbb R^d}f(x-y)g(y)\,\mathrm{d}y$ is well defined provided $\int_{\mathbb R^d}|f(x-y)g(y)|\,\mathrm{d}y<\infty$. In that case, since Lebesgue measure on $\mathbb R^d$ is invariant under translations and under the symmetry $y\to -y$, $g\ast f(x)$ is also well defined."
What I don't understand is the very last sentence.
In particular, the relevance of the invariance under the symmetry $y\to -y$.
But maybe I'm misreading something.
This is the very beginning of the section by the way.
okay, so natural things to do here would be "what integral does g star f ask me to contemplate" and "what do translation or symmetry invariance have to do with that"
have you done that
int g(x-y) f(y) dy would, under a substitution that turns the variable of integration into a translate of itself, like t = y - x, and under an assumed translation invariance, turn into something like int g(-t) f(t+x) dt, and under an assumed reflection invariance, turn into something like int f(x - s) g(s) ds which looks like something
or some such thing
you should be fiddling around with stuff like that
and not just "i don't see what is going on with this invariance." the key is to particularize the problem
well, I tried to apply "the abstract change of variables formula" above to get the equality g star f=f star g, but in my attempt above (if one scrolls up), I only get that we are pushing forward Lebesgue measure by varphi(x)=x-y, which is a translation. I don't see where reflection enters into the picture.
another path that is always open to you is, not taking every line of your textbook as an invitation to spiral into technical details but as an additional axiom until you get to something more concrete that feels wrong
would it completely derail your self study if an author said slightly more than was absolutely necessary to establish what they were using
what goes wrong if the author doesn't need the full force of their assumptions? or is it just, you want to understand at every stage where you are losing generality
no analysis book is based on minimal assumptions, and the assumptions you see in a book with "analysis" with a title are not secret signals that everything you're assuming are necessary
you can relax for the ride and go where the book is taking you (if the assumptions they make do, in fact, justify what they are saying follows from those assumptions), or you can reject the premise of analysis and go somewhere else
if your reaction to "let f solve this equation and then blabhlbah" is "but what if f doesn't solve this equation," then studying a book that assumes those sorts of things is just a waste of time
to put it another way, what is the author assuming X set of hypotheses preventing you from doing
if it's just "an abstract investigation into those hypotheses" then maybe you don't need to go there?
or is it "well actually in this application i'll only have some of those hypotheses and not all of them for X reason"
then you have context that sort of establishes the stakes of why you care, beyond just "maybe this isn't logically necessary"
with stuff like fourier series, studied since the late 1700s and maybe earlier, tons of known sufficient conditions for things, and known necessary conditions for things, but its pretty rare that they meet in the middle with an 'if and only if'
and if that didn't prevent 200 years of other people from studying them, but does prevent you from studying them, it's maybe what gen z calls "a you problem"
