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2:00 PM
@Secret but what is the proof that the diagonal of the sqare WILL pass through the center of the circle?
 
@DHMO do you know first principle?
 
@Ramanujan yes
 
@DHMO try using that :D
 
@Ramanujan that's how I proved it
 
But I never did it like that :P
 
2:02 PM
then how do you prove it?
 
My teacher showed me pattern from n=1 to 2,3,4,…
And by observation …
 
How did your teacher prove the cases when n=1 and 2,3,4,...?
 
For n=1,2,3… he took first principle
 
Well, it would be a nice exercise to extend the result to real numbers.
 
To extend the result you need to understand how $x^\alpha$ is defined for $\alpha\notin\mathbb Q$. But when you know that you can also see what the derivative is
 
in a nutshell, what is a catchphrase to remember Bolzano Weierstrass Theorem?
 
@MartianCactus is it 1:2π?
 
every bounded sequence has a convergent subsequence?
 
no
pi : 2
 
@DHMO how will you do for a^n - b^n ?
 
2:06 PM
@Ramanujan That's the point; you can't do it when n is not an integer.
 
First you need a definition of $x^n$ for non-integral $n$ :P
 
@Ramanujan Can you prove that the derivative of x^(1/2) is (1/2)x^(-1/2)?
 
If you have a good definition you can do it.
 
does someone know a link to a question/document about the proof that A^T*A is semi positif definit?
 
2:11 PM
@DHMO yes,key is : removing radicals from numerator by doing rationalization
 
8
Q: Is $AA^T$ a positive-definite symmetric matrix?

user72870Let $A\in M_{n\times n}(\mathbb{R})$ be a matrix. Is it true that $AA^{T}$ is positive-definite? Clearly $AA^{T}$ is symmetric. I have shown that a symmetric matrix $S\in M_{n\times n}(\mathbb{R})$ is positive-definite if and only if $S$ has only positive eigenvalues. Can this be helpful?

@Ramanujan Can you prove that the derivative of x^(1/3) is (1/3)x^(-2/3). This should be slightly harder.
 
@DHMO first iam trying now :P
 
@Aresloom in what context, matrices?
 
So A^TA is semipositive definit if A hasn't full rank and is positive-definit with full rank, is it true that A^TA or A*A^>T has the same rank as A?
@s.harp yes
 
@DHMO idk how to do it :( now reveal some hint
 
2:17 PM
@Ramanujan You use the same trick.
Hint: a^3 - b^3 = (a-b)(a^2+ab+b^2)
 
The formatting on your post is messed up so I can'T tell what you are asking, but the ranks of $AA^T$, $A^TA$ are the same as the rank of $A$
 
ok my second statement was true: http://math.stackexchange.com/questions/349738/prove-rank-ata-rank-a-for-any-a-m-times-n

I' would appreciate if one could confirm my first one about psemi-/positive definit
 
hmm, it seems $\mathbb{RP}^2$ is basically a saddle folded up onto itself after a twist
Briefly, referring to your first diagram: The circle $AB$ looks just like the circle $AC$, and the center pinch point $A$ is a saddle for "height" measured along the axis $CB$, but $A$ is a smooth point since the projective plane is a manifold. — Andrew D. Hwang 1 hour ago
 
Hmm @DHMO now I got
Shall I give you one task?
 
Sure
 
2:23 PM
@s.harp ok, thanks for that, my last question is: Is it true, that $AA^T$,$A^TA$ is semi-positive definit if A has not full rank otherwise if A has full rank it is positive definit?
 
yes
 
@s.harp perfect! Thank you
 
You know how projection (onto the $x$-coordinate) takes a subset of $\Bbb R^2$ and returns a subset of $\Bbb R$?
Is there a name for a sort of "weighted projection", where
 
Show that lengthy of the tangent for a curve (represented by y=F(x) ) is = {y√(1+(y')^2)}/{y'} @DHMO
 
given a subset of $\Bbb R^2$, it returns a function from $\Bbb R\to\Bbb R$ that maps $x$ to the measure of the part of the set that has it as the first coordinate
 
2:27 PM
So you're measuring the slices?
 
It's also known as zero
 
Not the measure in $\Bbb R^2$
The measure in $\{x\}\times\Bbb R$
 
okay okay but I really wanted to say that
 
I'm curious what we can figure out about a set $X\subset\Bbb R^2$ given the "weighted projections" of the rotations of $X$, $R_\theta(X)$
 
2:29 PM
@Ramanujan ?
 
@MartianCactus you got that ratio now?
 
@Ramanujan tangent is a line
how can it have a length?
 
i didnt calculate those
 
Lenght from line touching points (viz point on x axis and point on curve)
 
@AkivaWeinberger There is no closed subset of $\Bbb R^2$ in the kernel of that map, I believe.
 
2:33 PM
I don't know if there is a specific name for your function but slices of measurable set arise often in measure theory @Akiva
 
That is to say every closed subset of $\Bbb R^2$ which has measure zero slices is measure zero itself
It's a special case of Fubini's theorem.
 
@Ramanujan what is that book?
 
@Akiva do you know what $\mu_{proj}(R_\theta(X))$ is for all $\theta$ or do you know what $\mu_{proj}(\bigcup_{\theta} R_\theta(X))$ is?
 
@BalarkaSen Wouldn't $S^1\subseteq\Bbb R^2$ work?
 
@euclid my high school textbook
 
2:35 PM
That's measure zero
 
@s.harp The first one
 
@BalarkaSen Yeah. Not all measure-zero sets are in the kernel (take a vertical line segment).
 
@Ramanujan is there a file for it?
 
@Ramanujan ^
 
2:36 PM
@MartianCactus
 
@Akiva so clearly you should be working with getting maps Möbius band = affine geassmannian -> R
 
@BalarkaSen I didn't see this comment, sorry
 
@AkivaWeinberger I am not sure what's the point you're trying to make. A vertical segment hits it at two/one points, which is measure zero in R
 
If $E$ is measurable in $\Bbb R^2$ almost all of it's slices are measurable in $\Bbb R$ if I remember correctly right?
 
@Ramanujan yeah :)
 
2:38 PM
@BalarkaSen I thought you were claiming that it's in the kernel iff it's measure zero, sorry
 
@Balarka He's saying that if you apply your operation to a vertical line segment you get a function which is nonzero st a poin.
 
@MikeMiller …What?
 
Oh.
I see. Yeah, that's not what I meant.
@AkivaWeinberger Space of lines passing through a single point is the moebius strip.
 
@euclid unfortunately its PDF is not available , in which class are you?
 
No, space of all lines
 
2:39 PM
@DHMO there is some good proof in my book
 
$\mu_{proj}(R_\theta(X))=0$ for any $\theta$ implies $\mu(X)=0$, on the other hand $\mu(X)=0$ does not imply that ther exists a $\theta$ so that $\mu_{proj}(R_\theta(X))=0$ as the circle shows
 
Ah, I see. Open Möbius strip?
 
@Ramanujan ok
 
Ah, ok, yeah.
 
That is, projective plane minus a point. I see.
 
2:41 PM
@Ramanujan i am graduate, but it seems it is a good book.
 
If the subset is measurable is the induced function measurable? I guess so.
This is a really nice operator.
 
Exercise: prove that {{a},{a,b}} = {{c},{c,d}} iff a=c and b=d.
 
Yeah of course it's measurable.
 
A person is standing at the junction ( crossing) of two straight paths represented by equation 2x-3y+4=0 and 3x+2y-5=0 wants to reach the path whose equation is 6x-7y+8=0 in the least time. How to find the equation of the path that he should follow?
 
$\exists \theta, \mu_{proj}(R_\theta(X))\neq 0$ are we projecting a circle to a point for this particular case?
 
2:44 PM
@Ramanujan use vectors lol
i'm just kidding
find the intersection first
 
Actually isn't that the function sending the a point in $\int_{\text{slice}}1dL^1$ so it's a special case of Fubini's theorem?
 
@DHMO eh, {{a},{a,b}}={{c},{c,b}}={{c},{c,d}}?
 
@Null prove the first and the second equality
maybe you misread "iff" as "if"
 
@DHMO yeah
 
@Secret the way I have understood it $\mu_{proj}(X) = \mu_{\mathbb R}(\{ x\mid (x,y)\in X\}$
 
2:45 PM
@Akiva The only thing that makes me sad is it doesn't record when the fiber is nonempty but measure zero.
 
@AlessandroCodenotti Can't you take a nonmeasurable subset of $\Bbb R$ and cross it with $\Bbb R - \Bbb Q$ to get a subset of $\Bbb R^2$ only countably many slices of which are measurable?
 
@DHMO (x,y)=(12/17 , -16/17) ?
 
@DHMO but can we agree that if a=c... is not the hard part?
 
@Ramanujan then use the distance formula
@Null yes we can
 
nevermind.
 
2:47 PM
@Balarka in that case you have a non-measruable subset of $\mathbb R^2$ though.
 
there's no reason that subset of R^2 is measurable
 
@Ramanujan I mean, distance formula from point to line
 
Yeah, I don't have my notes here now but I think that stuff about slices should be true
 
hi chat
 
the distance from the line Ax+By+C=0 to the point (X,Y) is given by |AX+BY+C|/sqrt(A^2+B^2) @Ramanujan
 
2:48 PM
The fact that the induced function is measurable is a definition push
 
You can get some "bad" examples with stuff like a Vitali set times a singleton but they have a small set of bad slices
 
We should add "no random starring" to the chatiquette.
12
 
@DHMO yes we should
 
@DHMO I got distance = 44/17√85
 
Really...? @starrers
 
2:51 PM
@Ramanujan does that match with the answer?
 
@DHMO :/
 
@MikeMiller hi. i have a question
 
9 mins ago, by Ramanujan
A person is standing at the junction ( crossing) of two straight paths represented by equation 2x-3y+4=0 and 3x+2y-5=0 wants to reach the path whose equation is 6x-7y+8=0 in the least time. How to find the equation of the path that he should follow?
equation
 
oh
in that case
 
2:53 PM
@Ramanujan you find the slope and a point of the required line
 
I don't think proving from a=c and b=d back to the sets is that simple, but I am not sure what I miss
 
@DHMO and what about leat time ?
 
@Secret how does the first step follow from the definition of set equality?
@Ramanujan you just found it
 
well, {a}, {a,b} are elements of the big set given
even though they are sets
 
@Ramanujan you know, speed x time = distance
 
2:55 PM
This is valid considering that $\{\emptyset\}$ is also an element of some set
 
@Secret what is the definition of set equality that we're using?
 
I am being stupid today.
 
Sets are equal iff all elements it contains are
 
@BalarkaSen I only know that I know nothing.
@Secret why cannot it be the case that {a} = {c,d} and {a,b} = {c}?
 
@Ramanujan Since they don't tell you anything about the speed, I presume you can assume that you're walking at a constant speed.
 
2:57 PM
@DHMO But do you know that you know that you know nothing?
 
@BalarkaSen No
 
because {a} has 1 element and {c,d} has 2 elements, thus the 2nd iteration of using the set equal definition will fail...??
 
In which case the path of least time is the same as the path of least distance.
 
@BalarkaSen my set of knowledge is only that I know nothing
 
Hmm, now I shall prepare for tomorrow chemistry exam, I will discuss later :)
 
2:58 PM
@BalarkaSen that I know that I know nothing is not included in my set
or I should say, class
 
But what about {a} vs {a,a,a,a,....} I have no idea, I must be missing somehting important
 
Too much ignoramus et ignoramibus for my taste
 
@Secret you're jumping too fast
@Secret do you understand that {a} is just a notation
this isn't even part of the ZFC axioms
 
But then how do we define $\{\emptyset \}$?
 
@Secret Define Z such that forall x: x in Z equiv x = emptyset
then Z is {emptyset}
 
3:01 PM
@Secret a set whose cardinality is 0 ?
 
$\forall x:[x \in Z \equiv x = \varnothing]$
@Ramanujan its cardinality is 1, unfortunately
 
Ramanujan: Nothing is something. It is a type of abstract existence
 
In mathematics, and more specifically set theory, the empty set is the unique set having no elements; its size or cardinality (count of elements in a set) is zero. Some axiomatic set theories ensure that the empty set exists by including an axiom of empty set; in other theories, its existence can be deduced. Many possible properties of sets are trivially true for the empty set. Null set was once a common synonym for "empty set", but is now a technical term in measure theory. The empty set may also be called the void set. == Notation == Common notations for the empty set include " ...
 
@Ramanujan {} is not {{}}
 
Just read first two lines
 
3:03 PM
{emptyset} is {{}} not {}
because emptyset is {}
 
the Z that DHMO defined above was not the empty set, but rather a set containing only the empty set
 
@Secret the notation {{}} is a language
(language in the context of mathematics)
 
like a folder containing within it only another folder.
 
@Semiclassical folder analogy is good
in the sense that we can actually create that in our computers
 
Or a box in a box
 
3:04 PM
But in wikipedia article ( in first two lines it says)the empty set is the unique set having no elements; its size or cardinality (count of elements in a set) is zero
 
yeah. any abstract notion of 'container' will do.
 
@Ramanujan the set containing the empty set is not the empty set
a set is still something
@Semiclassical until you create a shortcut in the folder to itself
 
@Ramanujan The empty set is like an empty box. It contains nothing, hence it has no elements.
But I could put that empty box in another box.
 
@Semiclassical and what will be it's cardinality?
 
@MaryStar hi
 
3:06 PM
The empty box has no elements and so has cardinality zero.
 
@Secret anyway, back to our discussion?
 
|{{}}|=1
 
Actually, considering this extra point, what would happen if we do proceed this proof {{a},{a,b}}={{c},{c,b}}={{c},{c,d}} and getting the proposition {a}={c,b} & {a,b}={c}. Will we get into trouble?
 
If I put an empty box in another box, then that bigger box contains 1 element. By contrast, the box it contains is still empty and so still has cardinality zero.
 
So putting 10! Empty boxes in empty box,still cardinality is 0? @Semiclassical
 
3:06 PM
Cardinality of what?
 
@Secret we will not, as long as a=b=c=d
@Ramanujan and this is where the box analogy fails
 
If you mean the individual empty boxes, then yes---none of them contain anything, and so will have cardinality zero.
 
@Semiclassical big empty box in which 10! Empty boxes are put
 
|{{}}|$\not=$|{}|
 
That has cardinality 10!. It's not empty once I put something in it!
 
3:07 PM
because we only count same elements once
@Semiclassical no, still 1
{{},{}} = {{}}
 
:/
Point.
 
@DHMO This is kinda an unexpected pathway the proof of this simple looking proposition can go. Hmm I wonder if other proofs also have such unexpected pathways...?
 
@Semiclassical then why putting 1 empty box counts?
 
@Ramanujan because you don't have other empty boxes in your bigger box
@Secret how do i know
 
Eh, DHMO's got the right of it. The box analogy doesn't work at that point.
 
3:08 PM
@Secret so, how do you define XU{X}?
assuming that X is already defined
 
What is U, some mophism?
 
By contrast, the computer folder analogy suggested earlier works a bit better.
 
Suppose I have an empty folder in a computer. Let me label it as A.
 
@Secret union
 
3:09 PM
This set theory is pretty fancy word and not that i understand it, but it seems a rather simple concept.
 
@euclid Hello!!
 
@LuBu if you think you understand set theory, then you don't understand set theory.
 
I then create an empty folder inside A. I'll call it B.
 
@DHMO Good to know.
So if id try to answer this question XU{X} and U would mean union XU={X,X}
 
@MaryStar i have question that it is about algebra,i think it is simple for you
 
3:11 PM
@LuBu it is very simple,i just have some last year knowledge iam debating with these 0 mathematicians of internet :P
 
In my mind.
 
If both X and {X} are sets, then there is no problem. But if X is an element, then it has to be put into a set in order to be compatible with unions, that is, for X is an element, we need to take the convention that under union, X act like a singleton {X}. But otherwise I have no idea. I think something in the formal language of set teory will help me to fill in the gap
 
Im still a noob though.
 
...actually, no, even this approach doesn't help so much :/
 
@Secret element and set are not mutually exclusive
you can assume that X is a set
 
3:11 PM
The only thing i know about set theory is that i've heard the name.
 
@Secret if you want formal language, study the ZFC axioms
 
Where I was going was that if you try to create two empty folders in the same folder, then the computer will require you to give them different names.
 
@Secret I think defining {X} would be an easier task
 
But I think this analogy only works if you say that the 'label' on each folder is itself an element of said folder :/
 
@euclid Tell me
 
3:13 PM
Yeah {X}={x|x=X}
 
Oh well.
 
@Secret true but not what I expected
although I can go with this
now proceed to defining X U {X}
 
@MaryStar please look at this link,the first answer
 
Anybody got an example of a continuos function $(0,1)\times (0,1)$ so that the integral over the first component always exists but is not continuos? (ie $f$ is continuous but $\int_0^1 dt f(t,x)$ is not continuous)
 
@euclid Which link?
 
3:16 PM
@s.harp i don't think it exists
 
I have a hard time seeing how that could be possible either.
 
$X \cup \{X\}=\{x|x \in X \wedge x \in \{X\}\}$
 
$f$ has to be unbounded in some manner in the first variable
 
So unbounded at zero or at one.
 
@Secret a slight mistake
 
3:17 PM
Yes, I would be surpised if no such thing exists
 
$\frac{x}{t-1}$ should do?
 
continuous on $[0,1]\times (0,1)$ implies continuity of integral sure
 
11
A: Is an integral always continuous?

Pedro TamaroffOne can prove the following THM Let $f:[a,b]\to\Bbb R$ be Riemann integrable over its domain. Define a new function $F:[a,b]\to\Bbb R$ by $$F(x)=\int_a^x f(t)dt$$ Then $F$ is continuous. That is, the map $$f\mapsto \int_a^x f$$ sends $\mathscr R[a,b]$ to $\mathscr C[a,b]$. PROOF Let $c\in[a...

An integral is always continuous.
 
That's continuous as a function of $x$, though.
Not as a function of some other parameter $t$.
 
so we want a function whoose integral between [0,1] discontinous in both x and t?
 
3:19 PM
@Secret the integral of that is infinte
 
I can't see an example either, but I don't think that particular result directly closes the door
 
DHMO: o sorry, should be $\vee$
 
@DHMO the situation here is different, they are showing continuity of $\int_0^x f(t)dt$ for some integrable $f:[0,1]\to\mathbb R$
 
@Secret nice, and $x \in \{X\}$ can be simplified to $x = X$ since $x \in \{X\} \equiv x = X$.
 
What I am intersted in is continuity of $F(x)=\int_0^1 f(t,x)\,dt$ provided $f:(0,1)\times(0,1) \to\mathbb R$ is continuous
 
3:21 PM
Why don't I see all this math showing up as math?
 
What worries me is that, if said definite integral is discontinuous at some $t\in (0,1)$, then that'd seem to mean that the integral can't exist there.
@wyattbergeron1 Use the "Latex in chat" link the room description.
 
@wyattbergeron1 You need to install mathjax to see it formatted.
 
Oh, got iut
Someone wrote out all of MathJax into one function
 
3:24 PM
If an integral is discontinous, then its derivative must be discontinuous at those points, no? (Converse not true, because the differentiable function can have vertical tangents)
 
hey @AntonioVargas
 
@Semiclassical yo
 
@Secret Have you read the ZFC axioms?
 
O, so axiom of infinity is kinda like induction
 
@MaryStar i dont understand first paragraph completely.
 
3:26 PM
@Secret basically it defined $\Bbb N$ in terms of von-Neumann ordinals
 
yup that's the von neumman way of defining numbers (better than the other way imo at least you don't need to track of so many types of nested boxes)
 
what is the other way?
 
@AntonioVargas What math are you up to these days?
 
@Semiclassical took a couple weeks off for christmas, just starting up again today
 
Gotcha.
 
3:29 PM
Continuing on with the big RH project
 
O sorry, mistake, that way I don't like is von neumman (because of nested boxes). The other way is Zermelo's construction
 
@Secret I don't see any difference.
 
@Semiclassical actually atm I'm selecting exercises for the complex analysis class I'll be teaching in the spring
 
You only ever deal with one type of nested boxes $\{\cdots \{x\}\cdots\}$. They are equivalent, so the choice does not matter
 
Oh, nice.
 
3:30 PM
Ok, I think the result is that it is always continuous
 
@Secret oh, I confused the two definitions.
I wrongly said that it is von Neumann
 
@s.harp Is this a 'prove or come up with a counterexample' problem?
 
@Semiclassical its a "students used it in an exercise sheet and I want to tell them they can't" problem
 
@MaryStar are you there?
 
@s.harp I'm imagining as a three-dimensional grid, you know
 
3:38 PM
hmm
 
if the surface is unbroken, I do not see how your integral could be discontinuous
 
If it's discontinuous in $t$, then there's some $t\in(0,1)$ where the discontinuity occurs.
The only way I can see that happening in the present case is if $f$ develops a non-integrable singularity at that specific $t$
 
A discontinous integral for a continous function will be a very bizarre scenario where for some particular values of t, the total volume under the surface becomes undefined as if all the prisms suddenly vanished only for those points
 
actually
$f(x,t)=\frac{1}{x^2+(t-1/2)^2}$.
 
discontinuous at $t=1/2$
 
3:41 PM
Not as a function of $t$ for finite $x$.
 
oh yes
sorry of course
 
I think that works as an example, actually.
 
Do you guys have some sort of LaTeX plugin that works in chat?
 
Not a plugin. See the latex in chat link in the room description.
 
@euclid Ok, I will read it.
 
3:43 PM
thanks
 
Right, thanks.
 
@Semiclassical it integrates to (2 ArcTan[1/(2 Sqrt[x])])/Sqrt[x]
which is continuous
 
Mathematica gives $\frac{1}{1/2-t}\tan^{-1}\left(\frac{1}{1/2-t}\right)$.
Which is certainly not continuous at $t=1/2$.
 
one of our mathematicas is lying!
2
 
you did x + blah, not x^2 + blah
 
3:46 PM
That gives me $2\, arccot(2x)/x$ which is continuous
 
Also, you're integrating in t not in x.
...which you said you were doing above. woops
 
oh, you wanted to integrate in $x$ :)
 
yeah. but it doesn't make a difference: just do f(x,t)=1/(t^2+(x-1/2)^2).
 
yes I know how to rename variables ^^
 
right, right.
 
3:47 PM
it works
 
the key is that the nonintegrable singularity at t=0 only occurs at x=1/2.
once you have that idea in play, it's easy to construct examples.
 
won't you just blow up the integral towards infinity near t=1/2?
 
the integral will, sure.
that's the point.
but the function itself is still continuous on (0,1)x(0,1)
 
no at $t=1/2$ the function is $1/x^2$ which... is not integrable :/
I asked a question at the main btw: math.stackexchange.com/questions/2090411/…
 
well, if the definite integral in $t$ is not going to be continuous in $x$, something weird had better happen at that point in $x$.
This example gives that by having the definite integral diverge at precisely x=1/2.
If you're wanting something like a jump discontinuity, then this isn't an example.
 
3:52 PM
note that the function is still "continuous" as a function to $\mathbb R^+$, as however $t\to1/2$ you have the integral converges to $\infty$, the same as the value at $t=1/2$
 
But I see that this does run afoul of your 'integrable' prescription.
 
RIP hats :(
 
Sorry to hear that
 
See ya next year
 

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