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21:00
What I'm looking for is a formula that works with the vectors themselves (although you can use their standard coordinates if you wish).
@Tobias, working with that, thanks
If the polynomial is a real polynomial, should the coefficients be real, or should the unknows be real, or both?
The coefficients
Hi chat
Salut, @Astyx. Tu vis toujours! :)
21:04
@Krijn so, a real polynomial can have a complex zero, right?
Oui, la maladie m'a épargnée pour le moment :p
Comment vas-tu ?
@Kirill Yes
Ça va, merci, Astyx. Je suis rentré chez moi.
@Krijn ok
What property does a topological space $X$ need in order for proving separability by finding a countable set $D$, for which every point of $X$ is a limit point of $D$?
21:06
Ah tu étais où ?
(Pourquoi épargnée et non épargné? Ça veut dire que tu es une femme? Je ne savais pas ça!)
Je voyageais ...
Non c'est juste que j'ai tapé la lettre "e" trois fois au lieu de deux x)
there s no 'e' both cases
@Agawa: No. Past participle agrees with the preceding direct object, so there is a difference.
Yes there is
Ted is right
Ted is always right
21:08
LOL ... guère !!
il a epargné , elle a epargné
no 'e'
@Astyx Careful.
(except for $0\in \Bbb N$, but that's something else)
read carefully, @Agawa. He didn't say that construction.
smacks @Astyx
What? Does he think $0 \not\in \Bbb N$?
21:09
elle m'a epargnée
@MikeM: Positive and negative are supposed to be disjoint subsets. Shaddup.
I never said 0 was positive. I just said it's natural.
Since the object of the action is placed before the verb, the past participle agrees with the object
Clearly the positives are $\Bbb Z^+$.
Well, I want natural = positives. Shaddup.
21:10
Sad!
Haha this discussion again :p
@Astyx: Je regrette maintenant ce que je viens de dire :P
@MikeMiller I see Trump is wearing off on you already. Did you just tweet about it?
Hey all
Saw an interesting question on main that I'm having trouble try to answer.
1
Q: Obtaining an integral from a system of differential equations

Irregular UserHow can we get from the system \begin{align*} \lambda_t + u\lambda_x &= -u_x,\\ u_t + uu_x &= -\varphi'(\rho)\lambda_x, \end{align*} where $\lambda = \log \rho$, to $$f(\rho) = \int\varphi'(\rho)\mathop{\mathrm{d}\lambda}?$$ I've tried replacing the $u_t$ and $u$ in the first equation and doing...

Clearly the precalculus tag is totally inappropriate. Remove it!
It's not my question but I can ask the person to remove it. I know them personally :-)
21:12
I removed it ... and the improper integrals tag.
I am pushy about tags when I am aware.
1
Q: How to show that the fabius function is nowhere analytic?

mickConsider the fabius function https://en.m.wikipedia.org/wiki/Fabius_function https://people.math.osu.edu/edgar.2/selfdiff/ How does one show that this function is nowhere analytic ? Probably related , Maybe even a step in the answer : how to evaluate this function for nonreals ? Is it defined...

Got any hints for tackling it, @TedShifrin?
What is the meaning of $f$? What does it have to do with $u$?
I feel like it'd involve a total derivative but can't see where the $\text{d}\lambda$ will come from.
Hey guys, can anyone answer this question?

What property does a topological space XX need in order for proving separability by finding a countable set DD, for which every point of XX is a limit point of DD?
21:14
But $d\lambda = d\rho/\rho$.
Yep, I can see that but where it pops up while working through the simultaneous equations is a bit iffy to me
Are you asking for sufficient conditions for separability @JohnSnow ?
@Kari: I honestly have no idea what $f$ has to do with anything in the question.
I'm guessing $f$ is a shorthand for some function of some variables that the system reduces to
(or possibly necessary conditions)
21:16
Probably a re-arrangement of some kind
I've been playing with the pdes but to no avail
So many of these questions use notations without defining them, it's annoying
@KajHansen No think I don't understand why a topological space $X$ needs to be metrizable in order to prove separability in terms of converging sequences.
@Astyx: D'acc :)
ok astyx , ted ,d'accord +1 a vous deux
@TedShifrin $f$ is to be found, presumably as a function of $\rho$
21:17
LOL @Agawa. I am pretty much a stickler for grammar.
I don't think it does need to be metrizable @JohnSnow
mais quand il s'accord au cod, qui est pas feminin, alors le 'e' est absent
That makes no sense to me, @IrregularUser. So write down that integral and you have $f$. What does it have to do with the problem at hand?
J'ai pas fait 15 ans de français pour rien quand même !
Oui, ça c'était juste une typo
@ted surtout lorsqu'il s'agit du français, le type de l'ecriture sms m'a alteré la langue
21:19
@IrregularUser: Is it supposed to be some sort of conserved quantity in terms of $u$? The question just makes no sense to me.
@KajHansen The Wiki-page on dense sets, need metrizability in order to say that $D$ is dense in $X$ if and only if $X$ is given by $A$ together with all limit points of $A$.
@Agawa: sms?
@IrregularUser: Let me try something else. Is $\varphi(\rho)$ just a random function?
@TedShifrin $f$ is what we need to end up with from the system -- it's similar to showing that $(x^2+3x)(x+2) = ax^3+bx^2+cx+d$
can you link that page? @john
21:21
@TedShifrin All we know about that function is that $\varphi'(\rho) > 0$, but it doesn't seem relevant
I give up, @IrregularUser. This makes no sense to me. Sorry.
Huh. You can definitely talk about dense sets, limit points and all that without a metric
You go! @Kaj :)
Where is the integrating variable in the third line ? Isn't that just $\phi'(\rho)\lambda + \text{cst}$ ?
You should :p
(removed)
21:22
Well, I don't want to be left out of deletion.
ROFL @Kaj
@TedShifrin Thanks for the attempt anyway, I'll go ahead and remove the question
@KajHansen Okay so the section on density in metric spaces, does not even require metric spaces?
You haven't even heard me speak, @Steamy :D
@IrregularUser: Make it understandable rather than removing it.
@john that's because the wiki is speaking about limit of sequences in that paragraph, which are all the limit points in a metrizable (first countable is enough) space, but generally speaking the set of limits points is bigger than the set of limit points of sequences
@TedShifrin It's pretty much word for word from my lecture notes and I've spent more time than it's worth on the problem already
21:24
Go ask your professor and let me know Monday :)
Nope, @JohnSnow. $A \subset X$ is dense if every point in $X$ is either in $A$ or is a limit point of $A$. Topologically, $x$ is a limit point of $A$ if every open neighborhood of $x$ contains points in $A$.
@TedShifrin I think I'll do that
@Kaj: There are of course the confusing issues with limit points versus cluster points. :)
Oh lord
21:26
@AlessandroCodenotti Okay so it is in general correct to prove seperability of $X$ by finding a countable set $D$ and showing that that every point in $X$ can be written as the limit of a sequence in $D$? No problem here then?
I'd have to look up those definitions again @TedShifrin. And accumulation points. I can never keep them straight, and definitions aren't even consistent among authors IIRC
Are there conditions on the dimension of the vector space associated to an algebra ? (hope this question makes sense)
@Astyx No, not unless one puts them there
The question makes sense. There's always a trivial algebra structure, though. So what do you mean?
@IrregularUser what is the problem?
21:31
Still alive, @TedShifrin. I'll be in touch soon.
@Danu Ah.
I mean what you understood and answered. What would a trivial algebra structure look like ? (I am really new to these)
Learning basic representation theory of Lie groups.
@Astyx depends on which adjectives you want on you algebra (associative, unital...)
OK @Danu ... you may already know more than I do ... :)
21:35
Don't think so. I'm just trying to understand roots and stuff in order to be able to read Hirzebruch-Borel
oh, lie algebra stuff ... I never remember that stuff ...
Although I did teach a smidgeon of it centuries ago. But I never use it.
Could some one please answer these 2 questions of mine math.stackexchange.com/q/2087861/333392 and math.stackexchange.com/q/1963640/333392 of mine . Both are of calculus of variations and I guess particularly in the 1st one I am just unable to get the link between 1st order change leading to 2nd order variation. It would be great help . Thank you !
@TedShifrin actually not. The groups also have weights and roots!
@Danu Sure, but those come from the action on the Lie algebra
See — told you I don't know anything.
21:37
So says Tom Dieck, in any case ;)
(and they are "the same")
whew @Tobias :P
Well to me it was defined as such (terminology might vary from my language to english) :
An algebra over a field $\Bbb K$ is a quadruplet $(A, +, \times, \cdot)$ where :
$(A, +, \times)$ is a ring
$(A, +, \cdot)$ is a vector space over $\Bbb K$
$$\forall (u, v)\in A^2, \forall \lambda \in \Bbb K, \lambda \cdot (u\times v) = (\lambda\cdot u)\times v = u\times(\lambda\cdot v)$$
@Shashaank: There's just toooo much writing there. First and second variation are like computing first and second derivatives in regular calculus. When you look at Euler-Lagrange and you ask for $\partial F/\partial y'$ there is no chain rule. You look at $F(x,y,y')$ as a function of three variables and you ask for the partial with respect to the third variable.
@Astyx: Your English is superb, so I don't accept excuses like that. Sickness, sure. :)
@Astyx So that is an associative algebra (but yes, often the word "associative" is left out)
21:40
@Ted Don't stress me !
Awww, pauvre petit(e) :P
Given a vector space you can turn it into an algebra by choosing a basis and declaring that basis to be a set of orthogonal idempotents in the multiplication
Did one of you send me a chess.com friend request in the last hour?
(this defines the multiplication uniquely by extending linearly)
Very presumptuous question, @Kaj. On several grounds.
21:41
Though if the space is not finite dimensional, you will not get a unital algebra
What does unital mean ?
@Astyx "has a $1$"
Doesn't every ring have a 1 ?
Possibly took care of some of the presumptuousness @Ted. The vast majority of "random" requests come from here due to my profile.
Yes, @Kaj, you took care :P
21:42
@Astyx Usually, yes, but not quite as often in the context of algebras
I resent being called "fine," even if I might be a "lad."
@Astyx: Some people want to call $2\Bbb Z$ a ring, and it has no $1$. (I'm not one of them.)
With the requirement of a $1$, it is a bit less obvious how to get a "trivial" algebra structure on an infinite dimensional space
What is the difference between a polynomial and a polynomial function?
21:43
@Kirill: Great question. Consider the polynomial $f(x)=x^p-x$ on $\Bbb Z/p\Bbb Z$.
It is certainly a polynomial. Is it the 0-polynomial?
By "orthogonal idempotents in the multiplication" do you mean something like $e_i \times e_j = \delta_{i, j}e_i$ where $\delta$ is the Kronecker symbol ?
@TedShifrin one moment, I need to google the ring rest class notation
@Astyx right
@Ted Neither am I
@Tobias Right, thank you very much for your time, this makes a lot sense to me !
@Kirill: Do you write $\Bbb Z_p$ or $\Bbb F_p$?
21:46
@TedShifrin depends on $p$
No, it doesn't depend on $p$.
@Astyx In many contexts, people will focus on finite dimensional algebras, but not always
And are infinite dimensional algebras always non unital ?
non-unital?
$\mathbb{Z}/(m)$, $\mathbb{F}_p$ if $p$ prime
21:47
@Ted Uups
OK, @Kirill. Of course, $\Bbb F_p = \Bbb Z/(p)$. At any rate ... that is not the 0-polynomial, right?
@Astyx No, they can be unital, they will just not be of the form I described earlier
@Astyx In fact, I claim that you already have seen at least one infinite dimensional unital algebra
@TedShifrin if it is, it is $0$ for all $x$ from the domain. Then, it depends on the domain.
Let me think
NOOO ... The zero polynomial is the polynomial all of whose coefficients are $0$. @Kirill
But the function is $0$ on all of $\Bbb F_p$. Hence the difference.
21:50
@Astyx it is related to a word said several times by Ted the past few messages
I hope Astyx is ignoring Ted.
Do polynomials form an algebra ?
Oh yeah obviously
@TedShifrin ok, that is not a zero polynomial
21:53
So the polynomial is not 0, whereas the polynomial function is identically 0.
This doesn't happen working over infinite fields, of course.
@TedShifrin I feel myself stupid all the time I learn math
No need to feel stupid. This particular point is something I stressed a dozen times when I taught algebra.
And there is also the space of endomorphism of any infinite vector space right ? Which implies that there exists algebrae of which the dimension is any infinity
LOL @algebrae :P
@TedShifrin ok, I will try to re-read your explanation
21:57
Why not? antennae is a good word. :)
@Kirill: You understand it's not the zero polynomial but for every $a\in\Bbb F_p$ we have $f(a)=0$, so every value is $0$.
I like to use latin plural when I can. For instance pizzae, which is totally inappropriate
@Astyx That does not imply any cardinality quite, as it for example misses out on countable
LOL ... comme tu es bien amusant, même si tu es malade, @Astyx :P
(and whether it gets all others probably gets into CH territory. Probably the existence of an algebra structure of any cardinality needs AC or something like that)
@Tobias That's why polynomials exist
21:59
@TedShifrin but isn't the derivative of q a function of q like I have written in the 2nd question. If q' is a function of q ( I believe it is and the answers there agree) then we should be applying the chain rule
Well comparing infinities needs AC anyway, does it not ?
@Astyx Ahh, right, that is the only reason we have polynomials :)
Obviously
@Shashaank: You can argue if you want, but I'm telling you the correct way to interpret Euler-Lagrange. You think of $y$ and $y'$ as independent variables.
If you go through the derivation, you'll understand. Maybe do a concrete example.
@Ted Je pense n'être quasiment plus malade de toutes façons :)
22:01
OK, tant mieux :)
1
Q: How to show that the fabius function is nowhere analytic?

mickConsider the fabius function https://en.m.wikipedia.org/wiki/Fabius_function https://people.math.osu.edu/edgar.2/selfdiff/ How does one show that this function is nowhere analytic ? Probably related , Maybe even a step in the answer : how to evaluate this function for nonreals ? Is it defined...

@mick: Repeated spamming of the chatroom is really not cool.
@TedShifrin confused about the "$\mathbb{F}_p = \mathbb{Z}/(p)$ at any rate". I was tought that $\mathbb{Z}/(p), p \in \mathbb{N}$ is a quotient ring, the "rest class" ring. And it is a field, if $p$ is a prime number.
Sorry , was not sure it appeared. Because i got error messages @TedShifrin
22:03
@Kirill: residue class ring (in English) ... yes. But $\Bbb F_p$ is the symbol for the (unique, up to isomorphism) field with $p$ elements. That's what $\Bbb Z/(p)$ is.
But what is also not cool is the closure of Some of my questions :/. For instance this is a good question but put on hold. math.stackexchange.com/questions/2075374/…
@TedShifrin
@ all
I'm not interested in battles over what's on hold and what's not.
I just get annoyed at people coming in expecting to order people to look at their questions on main.
We're here for conversations here.
And procrastinating
Wow
Loooots of procrastinating (wasn't there some paper I was supposed to finish... woops)
22:06
Oh yeah, and especially for procrastinating. Where's Semiclassic?
@TedShifrin No, no I was not arguing. I just wasn't able to understand why do we treat y' as independent of x when I think it is not. There might be something wrong in my argument . I was just asking where I was going wrong . Because I can't understand why we treat y' as independent of x
@TedShifrin but still the $\mathbb{F}_4$ doesn't exist, whereas $\mathbb{Z}/(4)$ exists, or did I get it wrong?
You treat $x$, $y$, and $y'$ as three independent variables, @Shashaank. If $F(x,y,y')=(y')^2$, then $\partial F/\partial x = 0$, $\partial F/\partial y = 0$, and $\partial F/\partial y' = 2y'$.
@Shashaank he was referring to me
@Kirill: $\Bbb F_4$ does exist, but it has nothing to do with $\Bbb Z/(4)$ (which is not even an integral domain).
22:09
I do not like closed/on hold questions. Imho " no context " = nonsense argument. Like you notice an integral = 0 and they ask for context !? Always similar to that.
@mick yes , I got it
No context =\= bad question !
The correlation is quite high though
Even if the question is interresting, if barely no-one can understand it, there's no point
The "no context" thing is the default message when people close a question because the OP didn't mention any attempt at solving it themselves.
No context is usually used when it's an obvious homework question. I worked a week on one question about an integral that was very hard and I finally figured it out. I asked teh OP where the integral came from and never got an answer, and that upset me, but I still worked on the question because it was interesting to me.
22:11
That is another matter , we all understand integral and Sum
Wrong , it was not homework at all
@TedShifrin I do not understand. Your wrote that $\mathbb{F}_4$ is the symbol for the unique, up to isomorphism field with 4 elements. That is what $\mathbb{Z}/(4)$ is. Now you say that "F" has nothing to do with "Z".
@TedShifrin
It was not homework
NO, @Kirill, $\Bbb Z/(4)$ is not a field. It is not even an integral domain. $\bar 2\cdot \bar 2 = \bar 0$!!
I'm withdrawing from this debate, @mick.
@TedShifrin yes, it is not, because 4 is not a prime number.
But there is a field with $4$ elements.
22:14
Look on the good Side , i do not think you ever closed one of my questions @TedShifrin
@TedShifrin and not 0
Out of interrest people like to have context anyway, and it helps giving answers that the OP can understand.
You can write down a table for it, but it's $\Bbb Z/(2)[x]/(x^2+x+1)$, @Kirill.
Do you understand that, @Kirill, or shall I say more?
But if it is a computer assisted observation , there is No context or theory or homework ... ( within the knowledge of the Op )
Often you should answer this, @mick: Why should someone other than you be interested in it?
22:17
@TedShifrin So it's like we treat x y and y' as independent variables without thinking whether they are dependent or not . Like we forget that y'=dy/dx can be function of x . Is that right ? Have I got it right now? And if say we had some y' and we wanted to take it's partial derivative w.r.t to x (not talking about Euler Lagrange equation apart from it) and y' depended on some other variable (say z) , then would we have applied chain rule there ?
So again i disagree with " off-topic , no context ". I might agree with too General in Some cases.
Totally unrelated to what I asked, @mick.
It was not an answer @TedShifrin
@TedShifrin thank you for the explanations, but cannot understand anything there at the moment...
There is : "I was fiddling with an algorithm I wrote and stumbled upon a result which caught my eye :"
An argument backing what I'm saying is the comments in your question
22:18
@Shashaank: If you're asked to differentiate $y'$ with respect to $x$, then of course you do that. We should probably discuss this with a particular instance in mind.
@Kirill: That's fine. Basically you look at $\Bbb Z/(2)$ along with a new creature $\alpha$ and you consider $0,1,\alpha,\alpha+1$, with the obvious addition. For multiplication, you use $\alpha^2 = \alpha+1$ and work out the multiplication table.
Should I have said "is" instead of "are" here ?
Yes, @Astyx, because it's "an argument."
But that's always a tricky thing in English when you have a singular = a plural.
I try to rephrase when that occurs :P
But that question with respect , is silly : why should you care about my question ?? Why should i care about yours ?? That is nonconstructive ... My SURPRISING integral or limit is just as good as anyones elses. Context or not. @TedShifrin
Well, we disagree. And I probably won't look at many of your questions. Have it your own way.
@mick And most of us do not consider surprising integrals interesting at all or for that matter to be worthwhile questions
22:21
Seriously, I'm done arguing.
That's what I thought ... but I guess you could argue that this is a structure where the object is put first to emphasize on it, then "is" would have been correct (even though that's very far-fetched), am I right ?
hm, ok @Ted, I think I have a formula, but's it's really ugly. The idea is that I take the first $2$ vectors, calculate the orthogonal projection of the second on the first and use that to find the height of the parallelogram they span so to calculate it's area. Now I take the third vector, calculate it's orthogonal projection on the plane spanned by the first 2, use this to find the height of the parallelepipid the 3 vectors span and calculate it's volume, then I take the fourth...
@TedShifrin Ok . So what was confusing me was this. The L(Lagrangian) is a function of x , y and y'. When we differentiate L w.r.t to y' ,aren't we forgetting that L also depends on x.
@Alessandro: You have an algorithm based on Gram-Schmidt. It's ugly, but correct. That's not what I had in mind at all. :)
But when you close , people who are potentially intrested are put away
22:22
@mick Yes, that is part of the purpose of closing a question
it looks like this if I didn't mess up after writing it in a single formula $\prod\limits_{s=1}^k\bigg|\bigg|v_s-\sum\limits_{i=1}^{s-1}\frac{\langle v_s,v_i\rangle}{\langle v_i,v_i\rangle}v_i\bigg|\bigg|$
@Alessandro: Can you figure out what to do with two vectors in $\Bbb R^3$ without using the cross-product?
Nothing stops those people from posting the question again, with context or their attempts or reasons why they are interested in it.
Right. But there's a beautifully simple solution @Alessandro :P
@Shashaank: I already answered that question with the example above.
With $L(x,y,y') = (y')^2$, $\partial L/\partial x = 0$, not $2y'y''$.
You really should write $L(x,y,z) = z^2$.
@TobiasKildetoft what kind of demonic purpose is that ? A math question is a math question. If you do not like it , do not upvote.
22:25
@mick No, that is not how a site like this ends up with anything remotely like quality questions (which is in fact one of the goals)
What are you discussing about ? @Ted @Alessandro
Define Quality formally without requiring prof like skills from the Op @TobiasKildetoft
@mick Have you even bothered to take a look through the FAQ and tour of the site?
@Astyx: Give me the formula for the $k$-dimensional volume of a $k$-dimensional parallelepiped in $\Bbb R^n$ (spanned by $k$ given vectors). Well, don't give me it, since @Alessandro is still hunting.
@TobiasKildetoft yes
22:27
this @Astyx (please don't say anything if you already know the answer)
Don't worry I won't :p
@mick The description of what sort of questions are acceptable is there
I do not agree they are both justified and applied correctly !!
It's $\Bbb R^{k}$ right ? @Ted
No, @Astyx. $k$ vectors in $\Bbb R^n$, $k<n$.
22:30
@mick whether they are justified is completely irrelevant. They are the rules of the site and you have chosen to use the site.
An observation has no context. Not neccessarily at least.
Ok then i disagree with the way they are applied
An observation is math is more than the word seems to suggest, and will always have some context
If one closed a question for no context , then they should prove there IS context !
Ok , then say what is the context of an integral ??
So for $k=2$ and $n=3$ we mean the surface of the parallelogram spanned by the two vectors right ?
@mick the context is whatever made you look at precisely that integral rather than some other one
22:34
@TedShifrin Yes , I got it. I was just telling where I was going wrong ! Thank you
@mick As I said before, the "missing context" also applies to people posting questions without showing any effort at solving it themselves.
Is there anything else I can do, @Shashaank?
@Astyx, @Alessandro: The lesson here is that you guys should stop talking to me :D
But they are stuck , like as in not knowing how to try to solve it.
Haha, I think I have something but I'll keep it for myself for now :)
22:35
Then you mention what you tried; how far you got, etc.
@Ted, by the way, is that normal, not to understand anything in a math study? Do talented people understand everything "from the beginning" the whole time?
Definitely not, @Kirill. Well, perhaps the ones who will be superstars do, but most of us have to work at it.
Just mentioning something like "I tried substitution x = ... but that didn't simplify the integral" or "I feel like I should use partial integration, but I don't know what functions to use"
@TobiasKildetoft so context is : i considered this integral because i play around with logs more than cosines ??
Is already a great help to anyone trying to answer a question.
22:37
@Kirill: You also have to learn how to learn to succeed in mathematics. Most of the students I taught had a big adjustment when they got to my courses in college, because everything was too easy for them before.
@mick No, it should actually be something that makes the integral seem like it was not just randomly chosen. Picking some random combination of functions and slapping on some limits and an integral sign does not make for quality questions.
I didn't mean you, @Kaj :D
But maybe you can give some helpful advice? :)
I'm just lamenting the downright truth of that statement
To ask a question one of the main thing you have to do is to show good will, not only have it. If your question looks like an exercise from a textbook, no one will want to answer it @mick
22:38
@Astyx: au contraire — lots of people here love getting rep by doing people's homework.
It clearly does not look like a textbook !
Then what question are you referring to?
Lots of people should have something better to do then :p
@TedShifrin Thanks ! Yes just one thing. Why do we take the 1st variation and say that a first order variation causes a second order variation in the function from the minimum value. Why do we not do that a 2nd or 3rd order variation and expect a 3rd or 4tH change
That doesn't make sense to me, @Shashaank.
22:40
@SteamyRoot math.stackexchange.com/questions/2075374/… for instance !!!! How to show context ??!!??
It's like regular calculus. If you have a critical point, you have to look at second (or perhaps higher) derivatives to decide if you have a local max/min/saddle, etc.
@mick I don't want to start a war here, but it does not look like you put much effort in it anyways. If you had explained why and how you happened, it would have been a way better question. And despite your denying of it, people care about that
@mick You even supplied a bit of extra context in a comment, despite denying the existence of such.
@mick First of all, see the comment "Explain what makes you believe that it is true.". Your reply was "tested on a computer". What you can do in this situation, is edit your question to contain a short summary of those tests to show why this is likely to be true.
@Astyx how and why it happened ??? An identity discovered by luck ...
22:42
@Shashaank: I guess it's like considering $f(\vec x+t\vec v)$ (say for function of two variables). This is a first-order variation (we're moving on a line), but we can tell something by looking at second-order (or higher) terms in this.
Second, have you actually tried to solve your question yourself? You should always do so before asking it, and if you didn't manage, mention what you tried.
Well that's still something. And as Tobias, Steamy and I already said, the comments are backing our point
@TedShifrin I am at the point when I do not understand anything in the script, although I was reading and understanding it consecutivly. I read excerpts from different books, tried books from Polya to increase the efficience. Today I try to understand the two pages of a script the whole day long. Can you recommend some books, methods, pages about how to learn more effectivly?
But i do NOT KNOW how to attack that problem ! So how do i show effort ???
We already answered that more than thrice @mick
22:44
No you think you did
I was hoping @Kaj might offer you some words of wisdom. He definitely acquired much better skills along the way.
I would say that working things out for yourself as you read is important. If you think you understand a proof, try to write it out yourself. Work out examples. Work through the examples in the book and then do others yourself.
@mick I even phrased word for word something you could have added
Which would have prevented the 4 comments below
@TedShifrin I meant that " If we go away from the minimum in first order then the deviation of the function from its minimum value is only of second order " -Feynman Lectures . Referring to it why do we go away from the minimum in the 1st order only (why not 2nd or 3rd) and expect deviation of the function from its minimum value in 3rd or higher orders
@mick so add in the parts about your computer testing and mention "I don't know where to start on proving this."
Yeah, I don't believe that, @Shashaank. I can get a whole Taylor series in $t$ for $f(\vec x+t\vec v)$. To say $\vec x$ is a critical point just says that $t^1$ term vanishes.
22:47
Asking for a hint on how to start/proceed shows more willingness to put effort into a question, then just saying nothing and seemingly expect other people to do all the work for you.
Is it possible to upload a picture when editing a question ? There is already a picture in it, but is it possible to change it ?
Most times I give a hint, someone infuriates me by coming along and putting up a complete solution. Makes me angry, especially since the OP asked just for a hint.
@pilko: You should be able to remove it and replace it, yes.
@TedShifrin by the way, is learning a polynomial or non-polynomial thing? :)
I have no earthly idea.
I'd say polynomial
22:49
My advice is wanted?
Out of the little intuition I have
It seems advice is wanted, @Kaj. I volunteered you, as you made great leaps in learning how to learn, etc.
@TedShifrin So it's because we look for the critical point where the 1st order term vanishes
Right, @Shashaank.
For all directions $\vec v$ (calculus of variations is infinite-dimensional, but ...).
@Astyx then there must be a way to learn effeciently
22:51
There is, but our brain is silly
@Ted ok, thanks. I wanted to know before doing the work.
Look for a delete button where you uploaded it, @pilko. Don't trust me.
@Kirill, try using as many sources as possible. I get stuck a lot, and if it's bad, I'll try switching to a different text, or Googling what's getting me stuck, or I'll ask on here. Even Youtube videos can be a good resource.
Machines learn way faster than we do, even though we have "better hardware"
(I don't mean to delete the whole post, @pilko.)
22:51
@TedShifrin Thanks a lot
Trying to learn something from just a single text is hard-mode
@Kaj: I would think active versus passive is important, though. Not just reading/watching ?
It always was for me.
Quote : so add in the parts about your computer testing and mention "I don't know where to start on proving this." End quote Why do i have to belitte myself ?? I Said i do not know how to do it ? Why rephrase it as a desperate loser :/
This is humiliating
No it isn't
Yeah, generally doing is going to be essential. I always think I understand something...until I try the exercises @Ted, lol
22:54
Im leaving
Bye
I guess im not made to become a mathematician
Zut, alors, @Hippa !!
Maybe close my account
I haven't seen hippa in forever
22:56
@TedShifrin Salut !
@KajHansen o/
Why so ? @mick
Master of the memes.
Oh, hey there. Didn't see you enter
@Hippa: Tu connais @Astyx?
@TedShifrin Ca ne me dit rien
22:57
Moi non plus, quoique j'ai peut-être vu de vos messages
Hippa, meet Astyx. Astyx, meet Hippa.
@Astyx: Je ne vais pas me plaindre qu'il te manque le subjonctif :D
@Astyx hi :-)
I have no talent , i Cannot ask questions nor give answers , and do not even make Friends. I can only be annoying and useless !
Heya
@mick That's not true and you know it
@mick you may have talent, I don't know. We just told you how to improve the question you asked.
22:59
@Hippa: Je n'ai pas vu le petit, non plus!
I ask stupid questions and people hate me. I Will never succeed in math.
@Ted En effet, de pire en pire .. :p
You know that

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