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5:19 AM
Hi! So I am learning multivariable calculus, and I have a doubt. In one of my assignments, I was asked to prove that if a multivariable function had continuous first partial derivatives on a domain $D$, then that function is continuous as well.
Now, when I tried working out the problem, I noticed that continuity of first partial derivatives wasn't a necessary condition. With only the condition that the partial derivatives are defined everywhere on the domain $D$, I could prove that function was continuous on the domain $D$.
So my doubt is whether only the existence of first partial derivatives everywhere in the domain $D$ guarantee the continuity of the the function? (I think yes)
 
I do not believe you have the proof you say.
 
I would also love if there's any counterexample which disproves what I think to be true.
 
yes.
 
@TedShifrin Could you please provide a counterexample?
 
Do you know how the proof should go assuming continuous partials?
 
5:25 AM
@TedShifrin Yeah. My TA used the Increment Theorem to prove the continuity of the function.
 
I don’t know what that is. But continuous partials implies differentiable implies continuous.
 
And then applied the limit to show that $f(x+h, y+k)\to f(x,y)$ as $h\to 0$and $k\to 0$.
 
Do you know examples of functions continuous everywhere except the origin?
 
@TedShifrin It's just like writing the difference of function value in terms of its first partial. Sort of taylor series.
 
It's using differentiability, yeah.
 
5:30 AM
@TedShifrin $$f(x,y) =\begin{cases} 1 \qquad \text{if }x\neq 0 \text{ or } y\neq 0\\ 0 \qquad \text{if } x=0\text{ and } y=0\end{cases}$$
 
More interesting examples?
 
@TedShifrin Continuous partials do imply differentiability of the original function. But I only want to prove the continuity of the original function, so I guess even the existence of partials everywhere would do the work.
 
No, it will not. I told you. You only get continuity along lines parallel to the axes
 
@TedShifrin $$f(x,y) =\begin{cases} \displaystyle \frac{xy}{x^2+y^2} \qquad \text{if }x\neq 0 \text{ or } y\neq 0\\ 0 \qquad \text{if } x=0\text{ and } y=0\end{cases}$$
 
There you go. That function fails your conjecture.
 
5:33 AM
@TedShifrin But if I get continuity along all the lines parallel to both the axes, won't that imply overall continuity?
 
Try this example!
Continuity and differentiability are quite subtle in multivariable.
 
@TedShifrin And that it is quite hard to wrap my mind around since I am used to the single variable simplicity.
:)
 
There are even functions with all directional derivatives $0$ at the origin that fail to be continuous!
You can see a few of my YouTube videos on these matters for discussion/examples.
 
@TedShifrin I have your playlist saved in my watch for later section :) I guess this is the time.
 
Do you see why your example is a counterexample?
Partials exist everywhere!
 
5:39 AM
@TedShifrin Yeah, because the derivatives along the axes exist, but the function is discontinuous at the origin when approached along $x = y$ and $x = -y$ direction.
 
Yes. As I said, you can make more fascinating functions with all directional derivatives $0$. That stunned me at first.
 
@TedShifrin I vaguely remember seeing a function whose limit existed along all polynomial paths, but didn't exist along all exponential paths.
 
That's a different phenomenon.
 
@TedShifrin Oh, ok.
 
 
2 hours later…
7:28 AM
good night!
 
 
1 hour later…
8:41 AM
Hello!!
When it is asked to calculate sin3 , is it meant in rad or deg?
 
9:36 AM
Star graph $S_k$ connected to a path graph $P_r$ on one of its leafs, does it have a name?
Nvm, I guessed right. It's a Dandelion.
 
 
1 hour later…
10:43 AM
For finite group G
is there always exists some polynomial f such that Gal(f) = G?
over F
 
@love_sodam This is known as the inverse Galois problem and it's open in the general case.
 
Well, I the full statement is 'For finite group G, there exists a polynomial f over F such that $Gal(f)\cong G$' You mean this is an open question?
@KReiser
 
Any sources on presentations of the Rubik's Cube group?
 
11:24 AM
@love_sodam Perhaps you should try reading the link in my previous message. There are some details there. Good day.
 
12:07 PM
Damn, I always forget how non-restrictive the conditions for exchanging integration and differentiation are
My profs are doing that without even mentioning it and now I know why
 
12:24 PM
@TedShifrin 73,904,195
2
 
in Explorers of the Deep, 1 hour ago, by Hrishabh Nayal
Can $a^2+b^2=c^2+d^2=e^2$ have integer solution when $a\leq b$ and $c\leq d$ and $(a,b)\neq(c,d)$?
I thought this should already be on the main site but unfortunately I couldn't find it.
Can someone help me?
Something like finding two different Pythagorean triplets for same square
 
@love_sodam What is F?
If you don't specify F it's an easy exercise. It's open for F = Q
 
5^2 + 0^2 = 3^2 + 4^2
 
3-4-5 for the win
 
It's also not hard to show it is true for F = C(t) but I don't know how to do it by pure algebra.
 
12:39 PM
@user2103480 ah nice, I forgot to add non zero,
I did fear something like this would happen though
Infact let's change integer to natural
 
you're not allowed to change the question to make it pathological
 
12:56 PM
@user6232128 well that is what I had in my mind initially I just couldn't write it down properly :( sorry
If it helps, think of it as an extension to the original problem
 
@user6232128 she/he actually is allowed to do that
 
Why is that 2d guys only see 1d shapes while we who are 3d guys see 3d things? Where am I wrong?
I guess we're imagining 3d
we can do that from experience
but we see 2d actually
 
1:19 PM
@BalarkaSen F is an arbitrary field. How it's easy exercise?
 
1:39 PM
@aderchox a 2d guy can't lift himself up off the page.
 
@aderchox a single eyeball works by projecting 3D objects onto a 2D screen. We get a 3D effect by having two eyeballs, so we implicitly measure a difference between two 2D images.
This is called "stereoscopy".
A 2D person can do this, too, actually.
 
If you were a 1d guy, you could only see one way at a time.
and a 0d guy is blind
 
Greetings
 
Hi
 
2:03 PM
How can I compute the $[E:\Bbb F_9]$ where $E$ is a splitting field of x^{16}-1 over \Bbb F_9
What I done so far is that x^16-1 = (x^8-1)(x^8+1) and F_9 is a splitting field of x^{9}-x is x^8-1 splits in F_9. And x^8+1 = (x^4+2x^2+2)(x^4+x^2+2)
 
Can a semi-Riemannian manifold be 1-dimensional?
 
@geocalc33 If your definition of semi-Riemannian includes Riemannian as a special case, then yes.
I think they might technically tthink a signature of just -1 is still semi-Riemannian.
I'd say it should be, just for ease of definition. :P
And according to wikipedia this is the definition used.
 
@BalarkaSen Why would you want to do anything by pure algebra though
 
@anakhro yes, okay. So then I can take products of these manifolds $\zeta^{1,1}\times \zeta^{1,0} \times \zeta^{1,0}$
where the last two manifolds are 1 dimensional
 
Sure, I don't know what zeta is though
 
2:11 PM
up to this point I've only been able to compute the product and derive the metric for $\zeta^{1,1} \times \zeta^{1,1}$
it's a manifold
 
Just any manifold?
What is your definition for a product metric? Is it not just the sum?
 
it's the sum of the metrics
and these are all 0 curvature manifolds
but I'm confused on how to derive the metric for $\zeta^{1,0}$
 
Shouldn't you already know what the metric is?
 
well for
$\zeta^{1,1}$ it's $ds^2=\frac{dxdy}{xy}$
so would $\zeta^{1,0}$ simply be $ds^2=\frac{dx}{x}$?
I'm not sure if that's a valid bilinear form
 
2:43 PM
It doesn't even take in two vectors.
It's a one form.
 
@TedShifrin Whats the name of your YouTube channel, sir?
On an unrelated note, I have a linear algebra problem (again)
For a noninvertibe nxn matrix A, we need to prove that there exists a non zero matrix B, s.t AB=0
 
I like starting with small dimensional examples for questions like that.
so try n=two
 
I am having a hard time trying to incorporate the fact that A is non-invertible
 
Well note that it only woks if it is non-invertible. If A were invertible then what do you get if you apply its inverse to both sides of AB=0?
 
thats easy, we pre-multiply by inv(A) on both sides and conclude that B has to be zero
this was the 1st part of the problem
 
2:55 PM
Great.
So try a two by two matrix where the first row is ones and the second row is zeros
Can you find a B for this A?
And try a few more non-invertible A's after. Just to get a feel for it.
 
oh btw, I missed that B has to be nxn too, lthough I think you already knew this was going to be the case..
 
Yes.
 
it boils down to a system of linear equations which always seem to have solutions
 
What kind of linear system is it?
 
3:01 PM
Although the hard part for me is to generalise the matrix A: How do I even write a matrix while incorporating the info that its singular....
@anakhro homegenous?
 
Can anyone comment on the accuracy of the circled evaluation?
And also how I might be able to incorperate x shift in the function
I am fitting voigt profiles to spectral data in the lab using python. The circled red option is much easier to code given scipy.special.wofz function
 
@anakhro Wow that's really interesting! I'd almost forgotten I had two eyeballs!(seriously:)) Thanks for telling me the word stereoscopy.
 
@satan29 When you solved for the matrix B for that special A that I gave, what did you get?
@aderchox It's easy to forget. ;)
 
@anakhro Yes :))
 
3:21 PM
@anakhro -i got:
-j -k
j k
 
Great, so did you observe anything about this matrix?
If not, maybe try another matrix, like the matrix of all ones.
 
@anakhro it will be the same matrix
 
Oops, that is right. Try to come up with some more examples which are not the same. :P
 
well wait: for a 2x2 matrix, the only way that a determinant is zero is that one row is
the scalar multiple of the other
 
Yes.
 
3:29 PM
so I can generalise A:
x y
kx ky
 
Sure, why not. :P
 
in which case the corresponding B will be
-cy -dy
cx dx
 
Do you notice anything?
 
B also has zero determinant?
 
That goes without saying, given your other part.
In particular, can you say anything about the form of B as compared to A? Keep in mind, the goal is that given A, you can construct B.
 
3:39 PM
okay so I first multiply A with:
0 1
1 0
and then , I apply its transpose
 
Maybe we can consider a different angle to jog your memory about noninvertible matrices.
What are some equivalent definitions for invertible matrices?
 
If B is inv(A) then AB=BA=I
inv(A)= 1/(det(A))*adj(A)
 
What else?
 
its unique....?
I cant think of anything else
 
The question yesterday you were dealing with a special space related to A...what was it?
 
3:45 PM
null space?
 
Do you know how invertible matrices relate to their nullspaces?
 
absolutely no :p
 
Really?
What is the null space?
You said "unique" above...unique what?
 
set of x st Ax=0
 
So if A is invertible...notice anything cute?
 
3:49 PM
x has to be the zero vector !!!!
 
Are you unsure?
 
no..
 
Why the question mark?
 
edited.
 
Heh :P
So how does null A relate to A being invertible?
 
3:54 PM
invertibility implies that the null space has onlyy one element: the zero vector
 
jo, i am new here!
 
@satan29 indeed, if it is not invertible, it has something non-zero in its null space.
@AndreasSchuldei hi
Look at your examples of 2x2 matrices, @satan29, do you see anything from the null space?
 
X is of the form ( p, (-x/y)p )
not anything else :(
 
4:10 PM
@satan29 keep looking at it. Can you break your B into pieces?
 
-cy -dy
cx dx
this B ?
 
Yes.
My hint was to use the null space somehow
The null space is compose of what kind of elements?
 
the matrix B is :
-y *
a1 a3
a2 a4
where (a1,a2) and (a3,a4) belong to the null space
 
Nice!
Great observation. :P
Does this help you?
 
@anakhro took a freaking decade, but thanks :)
say x1 is (a1,a2) and x2 is (a3,a4)
then Multiplying AB, we get:
 
4:19 PM
It's fine. To be honest it's not the easiest observation to make without becoming desperae and plugging columns and rows into A
Or if you are familiar with block multiplication, AB = [Ax1 Ax2] in this case.
And that might have been more key.
 
the first colum contains the elements of Ax1
and the second one ........Ax2
so we start out by writing the zero vector as [Ax1 Ax2 Ax3.....Axn]
and then we decompose it into A[x1 x2...xn]
and there you go, we have a non zero B= [x1 x2 ...xn] .
QED?
 
Hmmm, I am not sure I understand your proof.
By non-invertibility, you have that the nullspace is not only zero.
So you have some non-zero element x.
So now can you make B from x?
 
something like
[x px qx rx....] ?
 
anyone interested in (simple?) differential equations for a practical puropose? math.stackexchange.com/questions/3923616/… i am looking for help
 
What are p,q,r? scalars?
 
4:27 PM
yes
 
Can you calculate AB and see if it is zero?
 
yes, I think so
 
Then you have a B
Note that p,q,r,... can all be zero if you wanted. Or 1.
Or whatever. There are many B's. :)
 
yeee
The idea was simple.....You have a zero matrix, on the RHS, so you needed to invoke the null space somehow
 
@AlessandroCodenotti pure algebra is better than pure analysis
 
user486313
4:38 PM
@user2103480 agreed
 
user486313
at least, proving algebra statements comes more naturally to me.
 
user486313
i can prove algebra theorems before reading the proofs in the book, and it turns out I am often correct ... when I go to compare
 
Many things are better than pure analysis
 
user486313
it seems that all the action is in Analysis though, however
 
user486313
Analysis and PDEs ...
 
user486313
4:40 PM
theory and numerics ...
 
user486313
I'd feel useless being strong in algebra ...
 
Jacobson 2:18-22
18 But someone will say, “You do algebra, and I do analysis.” Show me algebra without analysis, and I will show you my algebra by way of analysis. 19 You believe that there is one Field. You do well. Even the mathematicians believe—and tremble! 20 But do you want to know, O foolish graduate student, that algebra without analysis is dead? 21 Was not Cauchy our father justified by analysis when he offered geometry his son on the altar? 22 Do you see that algebra was working together with analysis, and by analysis, algebra was made perfect?
 
user486313
@anakhro may be true for functional analysis ...
 
user486313
nice quote, btw ...
 
user486313
nobody in my department spends any time on algebra. everyone's doing Analysis and PDEs.
 
jay
4:53 PM
Hi all anyone know a link which proves that a transport plan between measures $\mu_1$ and $\mu_2$ has to be the product measure if either $\mu_1$ or $\mu_2$ is a dirac?
 
5:07 PM
can someone help me with FDM with central difference scheme derivative BVP?
in case of a non linear f in y' = f(, , , ) we need to linearize it. How do we do that?
the jacobian is evaluated at x+ 1/2 . How can I find the Jacobian at x + 1/2 when it itself is non linear?
 
5:49 PM
Analysis and PDEs are what pays, with modern physics and comp sci
 
can you help @Astyx?
 
6:14 PM
@anakhro in addition, I find it dull just to study one area with a super straight focus, without looking at different facets of mathematics
For many areas, one has to get acquainted with many different topics anyways
 
@user2103480 all the more reason to familiarize yourself with a variety of fields from analysis, algebra, combinatorics, geometry, and foundations.
 
Yup. Although personally, I've had enough of that, and I'm looking forward to reaching research-level topics
 
Just be careful not to back yourself into the "my field is the only important/interesting field" corner.
 
except if it's number theory because that is genuinely the only important and interesting field
 
isnt that just applied topos theory
 
6:24 PM
it's applied counting
 
@Thorgott isn't that just applied graph theory
it goes full circle
@anakhro uhh are there actually that many researchers who do that?
 
"A graph is, roughly speaking, a "category without a composition law"."
 
Some might have strong opinions about certain topics but apart from that
@Thorgott the only thing I can see is "a category is a graph with composition laws"
 
@user2103480 many researchers do that. As well, it's a common trait of zealous grad students.
 
ZFC set theory is just a theory of well-founded trees smh
 
6:33 PM
a category is a graph whose path space is equipped with a partial monoid structure
ok, i believe thats not entirely not correct, but who cares
 
@Thorgott nobody. it's about category theory.
 
yeah right, there are many category theoretic statements to whose end you could append "but who cares"
wtf
write
 
@EdwardEvans lmao
british 💯
 
yih roight
 
hi
 
6:35 PM
wagwan
 
it's funny that exactly the nation who "invented" the language is the worst at translating phonetics to symbols
 
@EdwardEvans rogue one? wagon?
 
whats going on
in patois
 
Wagwan is appropriated from the jamaican pronunciation of "what is going on"
oh nice
 
lol ok
 
6:37 PM
I used to use it ironically but now it's part of my standard vocabulary
just like "init"
 
I've never seen as many people having problems with their/they're as in UK people's fb posts
 
there just stupid
 
lmao exactly that thanks
 
I forgot there instead of their
 
6:38 PM
it's all of them tbf
 
the holy triangle
 
in the midlands people say "you am"
 
trinity whatevs
 
Dreieinigkeit
 
matrix neo girl
also a perfectly valid synonym for trinity
 
6:39 PM
rofl
 
I'm sure that movie failed the bechdel test
 
whenever I see the Matrix
and Morpheus goes
"Welcome... to the world... of the real"
I cringe a little bit
or the desert of the real
 
Do you regularly see the Matrix?
 
We all see the Matrix regularly, @Astyx
The Matrix is everywhere. It is all around us. Even now, in this very room. You can see it when you look out your window or when you turn on your television.
 
A Matrix view is said to be regular when ...
 
6:41 PM
rofl
A matrix is said to be neopotent when...
 
@EdwardEvans you really used that opportunity well
 
I know man, fairly proud of it
 
Tough crowd though
 
they just don't know what humour is
 
how could they...... you're writing in british again goddammit
why is it humour and not tumour huh
 
6:51 PM
how cold they*
 
Whenever I see it spelled humour, I think of it in the bolily fluids meaning. Not sure why
 
@TobiasKildetoft maybe you read to many texts about the middle ages
 
Actually, I think it is due to an old strip from irregularwebcomic
 
7:29 PM
@EdwardEvans that's quite the statement coming from a brit
 
@Alessandro take that back
 
@user2103480 it is tumour.
 
@anakhro I honestly didn't know that lol
shite
 
"Elabouration" is one that doesn't exist, afaik.
 
7:39 PM
@anakhro btw, I wish I was still zealous
 
Why don't Americans use "continuos" though
 
they just say not-breaky-uppy
 
@AlessandroCodenotti Because that is one too few syllables
 
@AlessandroCodenotti I choked on my beverage
 
That is a word. Plural of a musical term.
 
7:41 PM
Alessandro should have known, being italian and all.
 
You can't add an s at the end of an Italian noun and call it a plural! That's illegal
 
continuoismos
 
Spaghettis
 
Spaghetti is already plural
 
spaghettoni
idk sorry
 
7:52 PM
spaghets
 
We don't need two Codenottiiii. One is more than enoughs.
 
8:26 PM
Anybody got a clue how that equilibrium position is defined?
I'm one of two people actually taking the class and we both have no clue lmao
The other one actually switched from physics b.sc. to math m.sc. and said "thermodynamics was the lecture that broke me" hahaha
in part b)
 
8:42 PM
Next week we'll hopefully do fourier-transforms instead of this all-on physics stuff
 
8:52 PM
Fourier transforms huh
 
they're probably as mysterious to you as to me
probably one of the most beautiful parts of analysis
as long as one doesn't do this cantor type convergence BS, that's probably very hairy
 
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