« first day (2560 days earlier)      last day (2757 days later) » 

00:00
We can also check that each set in the trivial topology is both open and closed hence clopen by taking their complements and noting they are elements in the topology
The boundary of some set $B$ is defined to be points $x$ such that for any open set $U$ in a topology $\tau$, it contains at least one point in $B$ and one not in $B$. i.e. $\partial B = \{\forall U \in \tau, \exists x,y \in U \subset B | x \in B \wedge y \in B^c\}$
Correction: $\partial B = \{z \in U|\forall U \subseteq B \in \tau, \exists x,y \in U, x \in B \wedge y \in B^c\}$
The interior of $B$ are the union of all open sets $U$ in $B$ i.e. $B^{\circ} = \{z|\forall U, z \in U \subseteq B\}$
00:31
The limit point of $B$ is some point $x$ such that any open set containing $x$ contains a point in $B$ that is not $x$. i.e. (Not going to write the set builder notation as there are no well accepted notation for a limit point, though $\lim (B)$ is tempting...)
@Secret: I often used to write $A'$ for the set of limit points of $A$.
Ah I see
Of course, people who use $A'$ to mean the complement of $A$ cannot do that.
I tend to use ${stuff}^c$ for complement
I usually wrote $X-A$. :)
00:33
$X - A$ is in some sense clearer as you can convey the information of a relative complement of some set $X$ easily if needed
I prefer $X\setminus A$ for that
using $-$ for set difference seems weird to me. I don't know why.
@Daminark barbarian.
Since I'd want $X - A = \{x-a: x\in X \land a\in A\}$
00:34
I always have ... I'd rather save $\setminus$ for group quotients.
isn't group quotient / ?
Demonark: That only makes sense in a vector space.
Depends which sided cosets you use, Semiclassic. Often you have to do both at once.
\ is too close to /
What is the problem with $-$?
00:35
Well, I've often found myself in a vector space, and usually prefer / for quotients (I've often stuck to one side)
Other than Demonark's objection, and that almost never occurs?
Steinhaus tho
Demonark: If your group $G$ acts on $X$ on the left and not on the right, you need to write $G\setminus X$.
I mean, I get what you mean. But having $A\fslash B$ as a quotient
When you have something with both \ and / it looks terrible.
00:36
But the spacing is wrong with that LaTeX.
(X\A) /Y
I'm little surprised that \fslash and \bslash aren't a thing
That happens a lot with homogeneous spaces and Lie group theory.
I reserve / for quotients and left division in quasigroups, and \ for right division in quasigroups
No, you write $H\!\setminus \!G/ K$. Bah, I forget how to balance these.
00:38
$H / G \backslash K$
so when I denote relative complements, I use setminus
ugh, that looks terrible.
No, no, that's the wrong way, Semiclassic.
Oh lord this left and right stuff is messing with me
00:38
Go back to diff geo exercises, Demonark.
$H\backslash G / K$
Or just be apolitical. Oh, you already are.
Now balance the two slashes, Semiclassic.
ugh, yeah. that's the problem
$H\!\backslash\!G\big/K$.
Ugh, no good either.
ironically, this is easier without using tex
00:39
ROFL, yup.
That's cheating, boy.
even better if you do it in fixed-character font, e.g. H\G/K
...I say that, but I like the previous one better
I punt.
works for me
00:41
asdf
er
omgwtfbbq
I know I figured out how to do it, because I used it in diff geo problem sets. Maybe I'll go back and look ...
hey ted
Hi @Faust :)
00:42
(cont.)
The closure of $B$ is $B$ with its limit points,i.e. $\text{cl}(B)=\bar{B} = B \cup B'$. It is the smallest closed set that contains $B$
See, Secret, you like that ' notation!
tried to find that integral geometry thing its really intresting!
When you learn more, I'll send you one of my papers to read, @Faust. All sorts of cool stuff with projections and slices.
If you want to see a classic exposition of that stuff, it's a book by Santaló, called something like Integral Geometry and Geometric Probability.
My math club wanted to go over it so im trying to find some information or a book on it atm =)
I'm still not exactly sure what you were talking about with Möbius strips, but, nevertheless ...
00:44
yeah, especially when you realised $\lim (stuff)$ can appear in the context of nets (which is going to be daily given how I love highly pathological topologies that are not metric spaces)
In mathematics, more specifically in general topology and related branches, a net or Moore–Smith sequence is a generalization of the notion of a sequence. In essence, a sequence is a function with domain the natural numbers, and in the context of topology, the codomain of this function is usually any topological space. However, in the context of topology, sequences do not fully encode all information about a function between topological spaces. In particular, the following two conditions are not equivalent in general for a map f between topological spaces X and Y: The map f is continuous (in the...
when is it normally taught non of us had heard of it before but we are all undergrads
I've given numerous talks on integral geometry to undergrads. The first part of Santaló's book you can read with just multivariable calculus background.
yeah im not worried about i cant find the notes i had taken on it and honestly i didnt really understandit anyway i just want to learn more about it
Fenchel's Theorem and Fary-Milnor (theorems about curves in $\Bbb R^3$ relating curvature to topology) can be done with Crofton. See my notes :)
You guys can fly me up and I'll give you a lecture :P
what notes? :p
where do you live?
00:46
the diff geo notes ... section 3 of chapter 1. I thought you'd looked at those before.
Anyway, what caused me to suddenly ramble about topology today is because last night dream I saw members from $\{0,1\}^{\Bbb{R}}$ flashing before me in 0.1 seconds each
you have given a few sets of notes
i wanted to go over mulivarable calc before i took anthor round at that book
I'm not close ... San Diego, CA. If I didn't have to teach every Sunday morning, I'd love to spend a while visiting Canada.
Dif geo is the most intresting subject i have encountered as an undergrad
Some people here will accuse you of kissin' up :P
00:48
though i feel most of the class was a bit above my skillset
It shouldn't be, but doCarmo's book is a bit too sophisticated.
i mean given enough time, i could eventually figure out an example
I like the boat on the front.
That's the one with a boat right?
@Astyx: Apparently the correct French title is Courage Mon Amour. It's very much in the weird style of Jacques Tati, but I loved it.
anyway im going to go and try and finish all those multivarable lectures
00:49
Boat? Whatcha talkin' about?
That's a lot, @Faust. It'll take time.
The flashes is due to the actual object in question is a probability distribution $f_{\Lambda}(x)$ which assigns a probability to each member in $\{0,1\}^{\Bbb{R}}$, thus as $f_{\Lambda}(x)$ is written into the white void, it literally morphs very quickly and randomly across all possible indicator functions in the form of black and white specks on an interval
thanks for the info though =) and if you dont mind crashing in a spare room your welcome to come stay with me in Canada would love to pick your brain :p
Wasn't there an edition of DoCarmo DG with like a weird painting of a boat as the cover?
Oh, @PVAL, the linear algebra book has a building design that looks sort of like the mast of a boat.
Nevermind that's spivak
00:51
I've never seen that on a doCarmo.
Right, Spivak has a biblical cast on his.
My linear algebra book has something that's sorta rigging-like.
What's with all this nautical geometric stuff?
So, as I woke up back in reality, I become curious on the subspace topologies of each indicator function (because topology is such a nice tool in helping me to map pathological things) and thus started to ramble about topology
@Secret: Thou dothst ramble.
oh, apropos of that
00:53
People have weird ideas about what topology is.
on one of the youtube videos I was watching, I kept seeing ads for a certain university program
for one it was a nursing program, so I dunno what in my internet habits is suggesting that
you're looking for any job? :P
lol
well, the funnier bit
Nursing is a really good job.
Won't be when only 10% of the population can afford healthcare though.
there were three bullet points on the ad. two of them were fine; I forget the first, but the other was 'M.D.N.S studies' which is fine
but the last one: "Biblical integration"
00:55
Biblical integration?
No joke, @PVAL.
yeeeah.
That's literally what the third bullet point was.
Is that more like Lebesgue or Riemann integration?
And if I google the program now, their website has the first sentence on their School of Nursing page: "The goal of the Cedarville University School of Nursing is to prepare excellent professionals to use nursing as a ministry for Jesus Christ."
It's just totally antithetical.
Well, that's what Pence will mandate for everything.
It's in Ohio, which is a little ironic given Kasich
...but then that's right next to Indiana, sooo
00:58
I'm sure there are plenty of things Pence won't do that for.
He won't do that for public education for instance.
He'll just get rid of that entirely.
No, he'll just mandate vouchers.
or that.
Or anything GLBT- or woman-sympathetic.
But we digress.
For reference, the page I'm referencing is this one: cedarville.edu/Academic-Schools-and-Departments/Nursing.aspx
Hi yall
Anyone know about generating functions and is keen to help me out ?
And while there's not a lot more on that page, there's enough there to make my head hurt.
01:00
At least they don't pretend to have diversity.
@Jacksoja As in, combinatorics?
@Semiclassical Yes, there are lots of weird questions solved by that method
Yeah, GFs are fun.
3
What the idea behind it tho
(There's also GFs in classical mechanics, which are completely different and more annoying)
01:02
I mean take a question like : a+b+c+d = 20 where A is even , B is odd , C is less than 5 and D is at most 11
Should I just make the generating function of each and multiply ?
@semi I don't think any of the certificates they offer are legit.
I see it working but I don't get why is that normal ?
You should understand why it works.
01:04
"The faculty of the School of Nursing supports the mission of Cedarville University and the profession of nursing by offering a biblically based baccalaureate nursing program that emphasizes godly living as a foundation for nursing practice grounded in biblical truth."
How does transposing that problem into infinite series and multiply them have the solution , its like black magic
No, think about exponent rules.
If the foundation for your nursing practice does not include scientific practice, I don't want you as my nurse.
(i'm exaggerating a bit, but only a bit)
Then we have all the doctors in congress who don't believe in science or evolution.
@TedShifrin when we multiply polys we add the powers
01:06
@ted an old reference but a good one: xkcd.com/154
Maybe start with a simple example, @Jacksoja. Suppose I want to make $1.00 out of nickels, dimes, and quarters. How many ways can I do it?
Actually I probably lied.
Ugh, @Semiclassic.
Those are probably all good certifications to have.
There's just very few of them.
the example I like as far as GFs go is linear recurrence relations. but that doesn't help a lot here.
01:07
I'm trying to recall conversations with family members about these things.
You have family issues, @PVAL?
I have family members who have worked in nursing.
and issues as well
oh oh ... I thought you meant politics and anti-science
well I've got those too, but they aren't related enough to me for me to worry about it.
My close family is really super medical though.
not the Paul Broun sort of medical, I hope
01:10
@TedShifrin In that example I look for the power on x^100 ?
coefficient ofc
ah neet
Thanks @TedShifrin @Semiclassical
YOu get it now, @Jacksoja?
I get why it works but it's neat trick , I need to do some more examples
That's the best way for it to make sense, @Jacksoja.
01:12
(x^0+x^3)(x^1+x^5)=x^(0+1)+x^(3+1)+x^(0+5)+x^(3+5)
as an example of how things work.
@Semiclassical that was meant for me ?
Right I see the structure now :)
@Semiclassical Well, we now have the US president that fit this xkcd
So what does that mean, it means: $$\lim_{t\to \infty}person = Endoftheworld$$
...
Did the chat just died suddenly......?
It died a while ago.
01:27
I see
@Ted I might have to hold off on the car problem since it's not responding too well to me and I need to push on, but just wondering, do you have tips for visualizing here?
It's more concrete and computational than you're used to, Demonark, but I think it's cool. You end up solving a differential equation to find the curve, of course.
The next one in section 1 I marked is more of an analysis question, so you might find it more to your taste.
Oh I mean I guess by "here" I mean "in life"
But there's a lot to learn in section 2.
LOL
My visualization skills as of yet are near non-existent
01:29
oh, "in life"?
This doesn't strike me as a visualization question. It's a compute and think to be tricky question.
As I told me diff geo students repeatedly, as Chern used to say, "When in doubt, differentiate." If you have an arclength parametrization, the speed = 1 equation means you can differentiate and get ... = 0. That will come up hundreds of times.
A proper multivariable calculus/analysis course would have had lots of pictures, Demonark.
Well, what I had reached on that line was $\langle PC,QC\rangle = 1$
That doesn't sound right.
Really? How'd you arrive at that?
I expected you to write out the equation $\overrightarrow{OC} = \overrightarrow{OP} \dots$ in terms of the parametrization $(x(s),y(s))$.
Well, we know that $\langle C-Q,C-Q\rangle = 1$, and that $\langle P-Q,C-Q\rangle = 0$, so that their difference is $\langle C-P,C-Q\rangle = 1$
That unless I had a brain fart, which isn't at all unlikely
Interesting. I didn't think about that.
Does that get us anywhere?
What I was hoping when I asked you a while bak whether we could assume $C$ was immediately above $P$ always was that this would imply that $C-P$ is orthogonal to $C$ since $C$ isn't moving vertically
01:36
But what do you mean when you say $C$ there? Oh, I see, the vector from the origin to $C$, which is horizontal. So, yes, that's right.
Because that would just leave us at $\langle PC,Q\rangle = 1$. Then... I'm still not sure, really
With the equation I gave, that's precisely what I had in mind. Use the fact that the vertical component is constant.
You gotta differentiate and do something with the parametrization.
Oh yeah maybe $OC$ would be better to say there
Okay at this point my guess is that $Q$ should be the point that somehow measures vertical change in $C$
I guess you have to compute more and try to second-guess less.
True
01:49
Wait what's with "black holes," aren't most holes black
Usually dark in there
hi, DogAteMy
when are you done in Argentina and returning to the grind?
Hi
Saturday night
ah, one more week of crazy liberty :P
Have you learned a lot and had a good time?
01:50
Yes and yes
Great :)
Okay so it'd be nifty if $QC = (stuff, -x(s)-y(s))$
If $f$ is even and strictly concave, then the minimum of $f(x-a)+f(x-b)$ is at $\frac{a+b}2$
Like, the minimum of $(x-a)^2+(x-b)^2$ is at the average of $a$ and $b$
Wait merp nevermind me
Are you doing the square-wheel-weird-road thing's-above-the-other-thing thing?
01:53
$\overrightarrow{QC} = (-y'(s),x'(s))$.
Yes, DogAteMY :D
It's a unit vector, Demonark.
If I were to BS my way into this, I'd start by looking at an infinitesimal time increment or something
comparing times $t$ and $t+dt$
Oh... that comes as a bit of surprise to me
No idea if that's useful
Well, try it, DogAteMy.
What's a surprise, Demonark?
It's a unit vector orthogonal to the tangent vector to the curve. Ergo.
OH QP IS TANGENT TO THE CURVE
01:56
duh
What's Q and P
In mathematics, a concave function is the negative of a convex function. A concave function is also synonymously called concave downwards, concave down, convex upwards, convex cap or upper convex. == Definition == A real-valued function f {\displaystyle f} on an interval (or, more generally, a convex set in vector space) is said to be concave if, for any x {\displaystyle x} and y {\displaystyle y} in the interval and for any α ∈ ...
Download the damn text, DogAteMy. :) Problem 10 in the first section.
Is that slanted parabolic curve strictly concave?
Parabolas are always strictly one or the other.
01:57
cause its max does not seemed to lie right in the middle?
Oh wait yeah I knew that yeah
Who needs a middle, Secret?
Wait I might have meant convex @Secret
Whichever one $x^2$ is
Oh I guess we're in the plane
That's convex. $-x^2$ is concave.
glowers at Demonark
01:58
Right yeah the line orthogonal to something has a unique generator
Oh. So I meant convex
Okay I'm dumb
Sorry
Either that or swap out max for min
I'm not calling names (yet), Demonark.
I'm glad you're working at it.
BTW, my comments about differentiating things was a really important comment. Even for this problem (in a bit).
But I'm going to cook/eat dinner now. Bye.
something like this does not have min right in the middle
02:00
If you're referencing the thing I wrote earlier, I did say "even"
O nvm then
Also, it's kinda weird how, even if you add $x$ to a parabola (equivalent to shearing the graph), it still ends up symmetrical
Just shifted
In fact, should be true for any shear applied to the graph, not just vertical ones. You just end up with a rotated (but still symmetric) parabola. Right?
Same for ellipses
and I'm guessing hyperbolas because that would make sense
Well for ellipses and hyperbolas it's weird because not all ellipses are similar to each other, and same for hyperbolas
but all parabolas are similar to each other
^That's also a kinda cool fact
$x+x^2 = (x+\frac{1}{4})^2-\frac{1}{2}$
complete the square saves the day
02:04
Also makes it make sense why it wouldn't work for things like $x^4$
$x^4+x$ is all lopsided and not symmetrical at all
attempt to complete the tesseract and you end up with too many cross terms, they skewed the graph nonlinearly
I suspect similar shifting effect happens for graphs of the form $x^n+bx^{n-1}+\cdots + nx$
for suitable $(b,c,...,n)$
@Secret I threw together a thing desmos.com/calculator/7czqtzmirt
Mess with the slider to shear them
I also put equations serving as grid lines on there but turned them off
so you can turn those on to see the shearing better I guess
Parabola retains its parabolalality, quartic curve does not
The trajectory/locus of the shifting also seemed to trace out a parabola
Oh, you mean the locus of lowest points of the parabolas? @Secret
Yeah it does
We want to look at the lowest points of $x^2+ax$
which equals $(x^2+\frac a2)^2-\frac{a^2}4$
The lowest point is thus at $(-\frac a2,-\frac{a^2}4)$
which lies on the parabola $y=-x^2$
Yeah?
yup
02:16
@Secret Also the lowest point of $y=x^2+\frac ax$ traces out $y=3x^2$
Reminds me of the following observation
NB: Shifting can be generalised to $x^3+bx^2+cx+1$ for some tuple $(b,c)$. Pattern to be found
Gonna predict Semi's about to talk about the extrema of $\sin x/x$ thing
Nah.
Consider the second-order Taylor series of $e^x$ about the point $x=a$: $T(x;a)=e^a+e^a(x-a)+\frac12 e^a (x-a)^2$.
That'll have a unique minimum at $x=a$, obviously, corresponding to the point $(x,y)=(a,e^a)$.
Which means that the vertices of that family of quadratics traces out an exponential curve.
And they're all the quadratics that fit the closest to the exponential at that point.
Cool.
02:26
I wonder if there's a more general result, such as where the criteria for single minima functions $f$ with parameter $a$ such that the minima is $(a,f(a))$, I guess $f$ may be the second order taylor series of any analytic function $f$
What's more, that property is actually valid if we consider the $2n$-th Taylor polynomials instead of the quadratic specifically
Wait
They don't have a minimum there
Not at $a$, no.
They have a minimum at $a-1$
The quadratics
...huh.
you're right. not sure why I thought otherwise.
02:29
Which, coincidentally, is the x-intercept of the closest line to $e^x$ at the point $a$, IIRC
the claim still holds, though. at $x=a-1$, we have $T(x;a)=e^{a}-e^{a}(1)+\frac{1}{2}e^{a}(1)^2=\frac{1}{2}e^a$
so you've got the set of points $(a,\frac12 e^a)$.
$(a-1,\frac12e^a)$, no?
hng. yes.
Yeah, so it traces out $\frac e2e^x$ or $\frac12e^{x+1}$
Which is still exponential in $a-1$ since $\frac{e}{2}e^{a-1}$
Right.
The claim goes like this. Let $x_a$ be the location of the minimum of $T(x;a)$, and $y_a$ its value there. Then $(x_a,y_a)$ traces an exponential as $a$ is varied.
02:33
Cool
And, moreover, that remains true if we do quartics instead
and so forth for all even powers.
Easiest way to see that in practice is to do a log-scale plot of those polynomials:
$x^3 + bx^2 + cx + 1=x^3 + \frac{3b}{3}x^2 + \frac{3c}{3}x+1 = (x^3 + \frac{3b}{3}x^2 + \frac{3c}{3}x + k^3) +1 - k^3$

Shifting holds if $$\frac{b}{3} = k^3, \frac{c}{3} = k^2$$ i.e. $3b^2=c^3$
@Secret Have I ever told you that the extrema of $\dfrac{\sin x}x$ lie on the curve $y=\cos x$?
If not, graph it, it's cool
nope
and it's actually a special case of the fact that, as $a$ varies, the extrema of $\dfrac{\sin x+a}x$ trace out the curve $y=\cos x$
It looks cool with how a local maximum and local minimum pair can cancel each other out as $a$ varies, and how that corresponds to the curve no longer intersecting the cosine curve there
Hm, experiments suggest that as $a$ varies, the extrema of $\sin x+a\cos x$ trace out the curve $\csc x$
And calculus confirms it
02:41
Here's a plot on log-scale of the quartic case.
Since it's plotted on log scale, the minima line up in a straight line.
(It's always of the form $(x_a,y_a)=(a-t_0,be^a)$)
Cool
So it's a consequence of the fact that, on the log plot, those quartics are all congruent to each other
They're all translations of each other
Which is true, right?
I'm not sure that's right.
But I should admit I no longer remember the solution off the top of my head...
but I do know where to find my solution :)
$(\frac{\sin x}{x})' = \frac{\cos x}{x} - \frac{\sin x}{x^2}$

Extremas where $(\frac{\sin x}{x})' = 0$, thus $x \cos x - \sin x = 0, x =\tan x$

and tan x intersect cos x at $\sin ^ x + \sin x -1 =0$
@Secret If we flip our earlier thing, we find that the extrema of $ax^2+x$ trace out a line, $x/2$
@Secret $x=\tan x$ implies $x=\frac{\sin x}{\cos x}$
implies $x\cos x=\sin x$
implies $\cos x=\frac{\sin x}x$
QED
"Note that $T(x; a) = e^aT(x - a; 0)$ achieves its minimum where $T(x - a; 0)$
does, namely at the unique $x$ value with $x - a = t_0$. Thus $t_a = a + t_0$ and so $$ T(t_a; a) = e^aT(t_a -a;0) = e^{t_a-t_0} T(t_0; 0) = [e^{-t_0}T(t_0; 0)]e^{t_a}$$ which is exponential in $t_a$, as desired."
02:48
I wonder for odd power taylor series, the point of inflection also follow the same trend?
Good question. I don't know.
If you're wondering how I could dredge that up so fast...well, this problem has a certain significance to me.
The extrema of $x^3+ax$ follow the graph $-2x^3$
and if you graph that, it looks weird
because it only has extrema for $a$ negative
Otherwise it's increasing
Namely, I ran into this problem in Mathematics Magazine during my undergrad, and (with assistance of one of my math profs) submitted a solution to it.
And it was ours which got featured :)
Some Correction:
The extrema of $x^3+ax^2$, on the other hand, follow the graph $-\frac12x^3$
02:52
So all I had to do was look up our old published solution :P
which also looks weird, as $0$ is always an extremum of that, so the other extremum "passes through" it
Ah, that might explain how the cancellation occurs that result in just shifting for suitable coefficients of $x^2$ and $x$
and it kinda looks like the local maximum and minimum are trading places
$x^3 + bx^2 + cx + 1=x^3 + \frac{3b}{3}x^2 + \frac{3c}{3}x+1 = (x^3 + \frac{3b}{3}x^2 + \frac{3c}{3}x + k^3) +1 - k^3 = (x+k)^3 +(1-k^3)$

Therefore shifting condition is:

$$\frac{b}{3}=k,\frac{c}{3}=k^2 \implies 3c=b^2$$
@Secret Pretty sure I agree with your comment re: the inflection points now
02:57
Because I can: the extrema of $x^3+ax^2+x$ follow the curve $-\frac12(x-1)(x)(x+1)=-\frac12x^3+\frac12x$
That doesn't even have extrema for $a$ between $-\sqrt3$ and $\sqrt3$

« first day (2560 days earlier)      last day (2757 days later) »