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3:00 PM
@DanielFischer I think we can prove that using contrapositive, too, right?
 
That's what I did, actually.
 
@DanielFischer, Ok!
Will you please post set question's answer in main, please? That is so good! I will accept it.@DanielFischer
 
Yes! Got it.
 
3:28 PM
OMG I got 80 points and now my answer is migrated to math educators, WTFWTFWTFWTF
 
@JasperLoy Which answer?
 
Top 3% this week? LHFs pay well
 
11
Q: How to teach someone that $-3>-4$?

rijul guptaI am trying to teach a teenage person math, but he doesn't seem to be able to grasp the concept of negative numbers and 0. Again and again he finds -4 greater than -3 because he has spent several years seeing 4 greater than 3. Similarly he has never experienced adding 0 to stuff and stuff to 0 ...

 
@JasperLoy I will just use the definition of an inequality.
 
Hi @Jas
 
3:31 PM
@lookofdisapproval you seem familiar.
 
Hi @Parthe
 
@ಠ_ಠ Hi boyfriend, lol
 
I can change my username tomorrow
 
:15812004 I have added an answer to your question. Are you now satisfied with all the questions in your comment on CV?
 
@ಠ_ಠ Make it Bart Parker
 
3:33 PM
Hello there
 
I don't want to make it my real name
 
@JasperLoy: what happened to your avatar? You are not monotone
 
@robjohn Ah, I like the default one very much!
 
@JasperLoy: well, I guess blue and white is technically monochromatic, but you know what I mean.
 
@robjohn Yes I know all your secrets, lol
 
3:35 PM
@ಠ_ಠ Whatever you make it, please make it something typeable.
 
I'm thinking of a cool username
 
Make it ( ͡° ͜ʖ ͡°).
 
@Hippalectryon: I've added some of the explanation I gave in chat to my answer to your question. I hope it helps.
 
Guys I will teach you a way to change your username in less than 30 days. Wanna listen?
 
@JasperLoy I think I know it.
 
3:37 PM
@JasperLoy bug a mod until they change it for you?
 
@robjohn thank you
 
@robjohn Nope. I shall expound now...
 
Change it on another site and copy to all accounts.
 
Create a new account on SE (a site you have not joined). Change your username there, then copy profile to all accounts. Done!
@ParthKohli Missing an important point
 
@JasperLoy Yeah, a site that I haven't joined...
 
3:39 PM
Finally, since you have not posted anything on the new site, click delete to delete that account, lol.
 
@JasperLoy that only works if you don't mind changing your name on all sites.
 
@robjohn Oh yeah.
 
3:52 PM
@ParthKohli Where is Sawarnik?
 
@JasperLoy He's gone for good.
Just kidding: he's on vacation.
 
@ParthKohli I will pray for his soul lol
 
Vacation to where
Fiji or something
 
No, just inside India.
 
He has gone to see the Buddha, lol.
 
3:54 PM
Is India a fun tourist attraction?
 
No.
 
You will get food poisoning there.
 
One of my classmates said he went to visit his family in India and when he woke up and went outside, all the tires on his grandma's car was missing
 
This is not a joke.
 
@JasperLoy Can confirm.
 
3:55 PM
So bring my own food?
 
Eat at more high class places
 
@ಠ_ಠ No, just eat at good places.
 
How do you identify the good ones
 
@ಠ_ಠ Your hotel would be one.
 
How do you identify a good hotel
 
3:57 PM
@ಠ_ಠ They're rated through stars.
 
And the stars are not corrupted by politicians and corporations?
 
The hotel should be rated four or more.
@ಠ_ಠ Thankfully, India is not as bad as you think.
You can trust the star rating at least.
 
Ok
I don't trust anywhere, especially not my own country
(America)
 
You'd be able to tell for yourself. If it's a five-star hotel, it'd be sensational.
 
@ಠ_ಠ Nobody trusts their own. They know it too well.
 
4:00 PM
5-star sounds a bit out of my price range anyway
So what is the scale relative to? Like is a 1-star rating just an average hotel, or is it really bad like a motel with bugs everywhere
 
@DanielFischer Hi. Can I ask you an FA quickie?
 
@ಠ_ಠ I don't even think we've got one-star hotels. But there are places called "guest houses" that are more like homes. They're not expensive and the food is a lot like homemade food, which again, is clean.
You just don't want to eat at cheaper restaurants and on the street.
 
@MattN. I think you can.
 
@DanielFischer W00T!
@DanielFischer In this answer, what's the theorem giving the linear combination of linear functionals $\alpha_i$?
 
4:10 PM
@MattN. I'm not sure what the question is. If $\{e_1,\dotsc,e_n\}$ is a basis of a vector space $E$, then every $x\in E$ can be written in a unique way as a linear combination $x = \sum\limits_{i=1}^n \alpha_i(x)\cdot e_i$ of the basis vectors (by definition of basis). It's an easy verification that the coefficient functionals are linear, I don't think there's a (named) theorem for that.
 
@G.T.R T'as vu le message que j'avais laissé sur la base extraite et la base incomplète ?
 
Oui
 
En plus cette preuve très rapide fonctione aussi en dimension infinie à Zorn près ;)
 
@DanielFischer I don't get it. Is it that obvious that given any $x$ then for any $\lambda \in \mathbb C$ there exists a linear functional $\alpha$ with $\alpha (x) = \lambda$?
@DanielFischer Duh.
 
@MattN. No. You need $x\neq 0$ to get any $\lambda\neq 0$. But that's not the point here. By definition of a basis, we have $n$ coefficient functionals $\alpha_i \colon E\to \mathbb{C}$ such that $x = \sum_{i=1}^n \alpha_i(x)\cdot e_i$ for every $x\in E$. Then observe that these functionals are linear.
 
4:16 PM
@DanielFischer I'm sorry for bothering you the functionals are linear.
So much duh.
Then $$ \alpha (x) = \alpha (\sum x_i e_i ) = \sum x_i \alpha (e_i)$$
Hm. No this doesn't work.
 
@Hippalectryon honnêtement, pas la peine de se prendre la tête avec la démonstration de la théorie de la dimension qui ne servent strictement à rien par la suite
 
@G.T.R Oui mais bon comme on peut l'avoir à une ligne près :)
 
@DanielFischer Okay, I don't get it after all. I don't understand how the definition of a basis implies the existence of coefficient functionals. I only see that this is true in an inner product space where the coefficient functionals are the inner product with the basis vector $e_i$.
In fact I have never heard of coefficient functional before.
Maybe they are defined to be the maps that map the vector to the $i$-th coefficient.
 
@MattN. The definition of basis says that for every $x$, there is a unique $n$-tuple of coefficients. That gives you maps, of which you don't a priori know that they are linear. Then you prove linearity.
 
@DanielFischer That's exactly what I meant when I wrote my last comment : ) Ok. Thanks a lot!
 
4:28 PM
Does anyone have a good set of problems for reviewing the whole calculus?
Especially, Calculus III
 
@MrWho You don't want to use a textbook?
 
@MrWho Nobody knows what Calculus 1,2,3,etc mean
 
@JasperLoy I have used my textbook and it has been finished!In addition, some problems in my textbook don't have anything to write home about.Furthermore, I'm looking for a organized problem set in multivariable calculus to help me review stuff I've learned if I face lack of time due to the fact of focusing on the other areas!
 
@MrWho What I mean is, different places will use different books on calculus and will cover different topics and use different notation, so your question is vague =)
 
+1
 
4:35 PM
@JasperLoy So you're telling me people know different types of calculus and knowledge has been become private and people develop different notations, what a funny remark!
 
lol
 
LOL
 
@MrWho Our school uses Apostol's Calculus series for our entire calculus sequence and I think it has some interesting problems
 
@JasperLoy Calculus is calculus, problem set is a problem set, nothing differs when you're doing a multiple integral and mathematical perspective do NOT change.
 
Erm...
 
4:38 PM
What math do you encompass exactly when using the term CalcIII?
 
@G.T.R If you read carefully I corrected what I said and mentioned multivariable calculus!
 
Well, multivariable calculus can be treated to different depths.
For example, multiple integral can be Riemann or Lebesgue
There may be use of differential forms or not
 
Exactly, and what are you looking for? Computation of integrals, gradients, etc... Or some proof-problems?
 
That's more like measure theory categorized under analysis, isn't it? Seems like from context @MrWho is only considering introductory calculus
 
@JasperLoy Riemann stuff.
 
4:41 PM
Or a typical/average intro calculus
 
Well, I don't know many good calculus books, but the books by Apostol and Spivak with calculus in their titles are good
 
@ಠ_ಠ Yeah, introductory calculus! thank you, what a relief!
@JasperLoy Did you learn calculus from your teacher?
 
@MrWho I learn it mostly myself by reading books. My lecturers were mostly lousy...
 
@JasperLoy Well, there is no good calculus book but what I'm asking for is independent set of problems for reviewing stuff time to time.
 
You may find his notes useful!
 
4:44 PM
@JasperLoy I've read the whole stuff there.In addition, I've solved all his problem sets regarding to calculus III !
@JasperLoy Still I need more!
 
@MrWho Ah, you are very hardworking!
@MrWho Why don't you practise the past exam papers in your university?
 
Here is a cute integral given on a math contest $$\int (2 x^{10}+3x^5) (x^5+3)^{1/5} \ dx$$
The answer itself is not that important, but the way to the answer is important ... :-)
 
@JasperLoy Well I haven't gone to the university yet!
 
@MrWho Oh, then you should stop practising and study more advanced math topics then!
 
If $f$ has continuous derivative on $\langle a,b)$ what does it say about $f$'s differentiability and continuity?
 
4:49 PM
@JasperLoy I'm thinking of it, however, I don't want to become stagnant after these tons of practicing that I've made.Shortly, I want to still practice and learn new things from basics.
 
Does it mean $f$ is continuous on $\langle a, b)$ and differentiable on $(a,b)$?
 
@MrWho So what book did you learn calculus from?
@mirgee It means it is differentiable and the derivative is continuous!
 
@Chris'ssis Looks too hard
 
@JasperLoy Thomas calculus.
 
@JasperLoy To be clear, the continuity of derivative doesn't say anything about continuity of the original function?
 
4:52 PM
@ಠ_ಠ In what sense you mean it is hard?
 
@mirgee The existence of the derivative implies the continuity.
 
@mirgee If a function is differentiable at a point it is continuous at that point
 
So what I've written above is correct? @DanielFischer @JasperLoy
 
Nah I think it amounts to a brutal u sub @chris'ssis
 
@Chris'ssis Hard relative to the extent of my abilities
 
4:55 PM
@mirgee Yes, and the derivative is continuous as well
 
@G.T.R @ಠ_ಠ if you do the proper thing, all becomes very easy.
 
sup
?
 
If anybody lurking has thoughts: I am wondering how to argue $L|K$ separable iff $L|M$ & $M|K$ separable for any $L|M|K$. If $L|K$ is finite then it fits into a tower of simple extensions and by counting embeddings we obtain the lemma that $L|K$ separable iff $\#\hom_K(L,\overline{K})=[L:K]$. This lemma doesn't apply to infinite extensions though. My definition of separability is $L|K$ separable if $L\otimes_K\overline{K}$ reduced.
 
And one more thing: for per parts for Riemann integral, we require the existence of both $\int_a^b f'g$ and $\int_a^b fg'$, whereas for improper integrals, the theorem says that if one exists, also does the other... Why?
 
@chris'ssis hint ?
 
5:07 PM
@G.T.R Think of the way the primitive should look like.
 
I don't really know
 
rephrased: if you know one of their primitives, you know the other's primitive. how?
what formula do those two integrals appear in?
 
I am going to try to get 200 rep on each of the 4 sites I am on now, lol
 
@seaturtles I am really confused about this :D
$fg$ is primitive to $f'g+fg'$, but only if both integrals exist
So how does existence of one imply the other
 
@mirgee answer my question: if you know one of their primitives, you know the other's primitive. how?
if you know $\int f'g$ how do you write $\int fg'$ in terms of it?
 
5:21 PM
$\int fg' = [fg] - \int f'g$
 
Well done
 
@JasperLoy Shut up :D
 
I have forgotten most of calculus actually.
 
Isn't that just the product rule?
 
@seaturtles I do get that, but why do I need both for Riemann and only one for improper?
 
5:23 PM
equivalently, by-parts
 
@seaturtles you seem to be kinder than anon
 
I see you have a nym today
 
what is a nym?
In combinatorial game theory, the Sprague–Grundy theorem states that every impartial game under the normal play convention is equivalent to a nimber. The Grundy value or nim-value of an impartial game is then defined as the unique nimber that the game is equivalent to. In the case of a game whose positions (or summands of positions) are indexed by the natural numbers (for example the possible heap sizes in nim-like games), the sequence of nimbers for successive heap sizes is called the nim-sequence of the game. The theorem was discovered independently by R. P. Sprague (1935) and P...
you mean a nim?
 
weren't you user something with a 14 in it yesterday?
 
oh yeah, this is cooler
do you like cowboy bebop?
 
5:32 PM
yeah. think I like samurai champloo more.
 
what's your favorite anime other than gits?
 
my favorites are gits, champloo, death note, monster.
 
@JasperLoy there are LHF waiting for you
 
@G.T.R Thanks, lol
I have gotten 200 on Math and Math Educators, now I will focus on getting 200 for Eng and Eng Learners
 
5:54 PM
@G.T.R wait... why does he get them all?
@JasperLoy yeah, stop taking all my questions :-)
 
@robjohn Well, actually those that you answer are a little too hard for me
 
@JasperLoy I answer questions all over the board. Some are hard, some not so much.
@JasperLoy I've answered a lot of hard questions recently... it's time for some easier ones :-)
 
Hi, anyone may help on my calculus question, it seems the answer by other one doesn't match the wolfram alpha gave. math.stackexchange.com/questions/815157/…
 
@robjohn Good good. You can now aim for 200k!
 
@Victor what's wrong with the answer you have?
 
5:59 PM
now i think the answer is okay, thanks @robjohn
 
I have a question - anyone up for galois theory? @seaturtles, perhaps?
 
@DanielFischer what's the convergence definition for multi-indeterminate power series?
 
@EnjoysMath What is a "multi-indeterminate power series"? A power series in several variables?
 
yes
 
@EnjoysMath IIRC one sums over the monomials of degree $\leq n$ and take the limit of these.
 
6:10 PM
All I need is for 2-var
 
@BalarkaSen what's up?
@MikeMiller are you familiar with how to argue separability (of field extensions) is transitive, without assuming the extensions are finite?
 
ok, so the partial sum sequence converges, I see!
 
@seaturtles I'm not. my knowledge of field theory stops when the extensions aren't finite.
if you're familiar I'd be interested in hearing the argument.
 
I don't know the argument, everywhere I look assumes the extensions are finite, but a few places mention that separability can be defined for infinite extensions too but don't extend the facts for finite extensions to infinite extensions.
 
@EnjoysMath Not quite. You have no natural ordering on the indices, so one usually takes summability, which in this setting is most likely the same as absolute convergence.
 
6:15 PM
oops, I'm a weenie. i meant to stick an absolute value into what I said.
@seaturtles to clarify, we're still talking algebraic extensions, right? I don't know a notion of separability otherwise.
 
@MikeMiller nope. $L|K$ separable means $L\otimes_K\overline{K}$ is reduced.
(has no nonzero nilpotent elements)
 
yeah, nevermind. it's trivial for algebraic extensions, finite or not.
 
@seaturtles Consider the transcendental galois extension $\Bbb Q(x_1, x_2, \cdots, x_5)/\Bbb Q(e_1, e_2, \cdots, e_5)$, where $e_i$ are the elementary symmetric polynomials of $x_i$. take another similar extension $\Bbb Q(y_1, y_2, \cdots, y_5)/\Bbb Q(e'_1, e'_2, \cdots, e'_5)$ and let the galois groups be isomorphic. Is it possible to show that the pair $\{x_i, y_i\}$ is $\Bbb Q(e_1, e'_1, \cdots, e_5, e'_5)$-algebraically dependent?
 
@MikeMiller hmm, the argument I know uses the fact that $\#\hom_K(L,\overline{K})=[L:K]$ iff $L|K$ separable in order to prove transitivity, but this doesn't extend to infinite extensions.
@BalarkaSen when you speak of the pair $\{x_i,y_i\}$, the index $i$ is fixed right?
 
@seaturtles IIRC separability for algebraic extensions is equivalent to every element having separable minimal polynomial. are those only equivalent for finite extensions?
 
6:23 PM
Well, I don't know, I can just let the be arbitrary, @seaturtles
Take $\{x_i, y_j\}.$
 
I mean, you're talking about e.g. {x1,y2}, not {x1,x2,x3,x4,x5,,y1,...} or whatnot?
 
@seaturtles The former.
 
@MikeMiller yes those are equivalent, even without assuming algebraic. can we use that to prove transitivity of separability?
 
Could someone take a look at Ryan's answer here: math.stackexchange.com/questions/153472/…
 
6:25 PM
actually I'm not sure if they're equivalent with algebraic, hmm.
 
@seaturtles 'equivalent with algebraic'?
 
see my message right above it
 
ah, i thought you were referring to me.
 
@seaturtles I'm fairly certain one can get an argument out of that. but only if that's actually an equivalence.
 
All my 4 accounts rep are now at multiple of 5, I can sleep now...
 
6:28 PM
I don't get how he got that result for the repeating integral at the end. From the first integral he gets $-\sqrt{1-r^2}$, if he places in the required r value, he gets... ah. wait..
 
I haven't thought about fields in about two years, though, @seaturtles, so it might be in your best interest to ignore me.
 
I studied galois theory ten years ago, lol
 
@seaturtles It's a kind of a converse to what Tschrinhausen transforms does.
 
Got it, nevermind.
 
I guess summer months are taking their toll
is it just me, or has the site slowed down a lot?
 
6:33 PM
@seaturtles In fact, I'd guess it's even true in general with base field $\Bbb Q$ replaced through some algebraic number field. Do you have any idea how to approach this one?
 
Ooh my dearie goodness. Minus 417.
@MikeMiller I didn't notice anything.
 
@BalarkaSen perhaps consider $f(x_1,y_1)=0$ in light of an appropriate power basis. haven't thought it through.
 
@JasperLoy If you think about all the things that happened before you came into existence and all the things that will happen after you disappear then this is not really all that important.
 
@MattN. Yes, I know, I was just joking a bit, lol
 
@MattN. Maybe it's just the tags I frequent, or the times of day I check in. (I really only look at main at night.)
 
6:39 PM
@seaturtles That seems like an interesting thought.
 
There is some thing I saw on the interwebs and it went something like this: There is a bit of truth in every joke and a bit of fuck you in every 'whatever'. : )
 
@BalarkaSen hello.
 
@MattN. Perhaps there is a bit of truth in every lie and a bit of lie in every truth, hmm
 
@ParthKohli ahoy.
 
@JasperLoy Maybe.
 
6:40 PM
@MattN. If someone responds to a joke with "lol", that means your joke sucked.
 
And a bit of alcohol in my glass.
 
@BalarkaSen When exactly does a chat freeze? One week?
 
@MattN. who the hell was removed?
@ParthKohli I don't know. ask @robjohn.
 
@MikeMiller Thanks for the translation, lol : )
 
Or it means they're Jasper.
 
6:43 PM
Actually, when I see lol I always immediately think of a teenage girl.
 
What is a Pac-Man contour?
 
Like a 14-year-old.
 
Haha, maybe I am a 14 yr girl, lol
 
@G.T.R An apple pie minus one piece.
 
@G.T.R Take an annulus and cut a path out of them.
It's less known by that name though. We usually call it keyhole contour.
 
This pic looks familiar lol
 
@JasperLoy Of course, it's a keyhole.
 
@BalarkaSen Dadadaaddaaa Pacman!
 
@MattN. Ouch.
 
6:52 PM
There's a multivariate power series with the same zeros in $\Bbb{Z}$ as the twin primes.
and $2$ probably
 
@EnjoysMath Ah? So you claim to have proved TPC?
 
No
I proved the above statemnt
 
Then I don't understand what you have found.
"same zeros in $\Bbb Z$ as twin primes"?
 
see link
Should have said "first primes in twin prime pairs"
 
@DanielFischer Not so bad : )
 
6:55 PM
@EnjoysMath Obviously. I was under the impression that you found all twin prime pairs.
 
I sort of did it abstractly, I said they're the zeros of this other object. It's now ur job to compute the zeros of that object, lol
 
It's a bad way to lure people from different interests to see there question with false claims of relation with open problems.
But it's quite clever =D
 
I didn't false claim any thing
 
@EnjoysMath At least ambiguous.
 
I want people to look because I think it's interesting
 
6:57 PM
"same zeros as twin primes" evidently makes one think that you have some function that has zeros at twin primes.
 
If only people asked more Futile Attempts questions.
 
@MattN. I just read the email exchange from the awesome guy you linked to in your profile. Niiiice.
 

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