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D.C. the III
D.C. the III
00:09
what Ted said...so not far off...........
Yea I don't know why I wouldn't conceive garlic bagels when I do see onino and don't get onion ones for the reason you just mentioned
When I try a garlic bagel (which I might do tomorrow) I'll see where to place it in rankings, but going on its makeup Akiva's order may be the right order
Akiva Weinberger
Akiva Weinberger
Did you know they make bagels by boiling the dough
D.C. the III
D.C. the III
I realize why I haven't seen a gralic bagel, it's not really a thing in toronto....all the places I go to which are deemed the better places don't offer them. Looks like that may have to wait until I go south of the border again.....
Ted Shifrin
Ted Shifrin
@AkivaWeinberger Yes, and then baking.
leslie townes
leslie townes
i don't know where to get a decent bagel within a short drive of here. although the mediocre place within walking distance of here does sell garlic bagels.
Akiva Weinberger
Akiva Weinberger
Just tried something called "Malta India" that I saw at a shop, out of curiosity
"Non-alcoholic malt beverage"
Initial reaction: ew
@leslietownes If you get the opportunity (and time) to boil and then bake bagels…
Calvin Khor
Calvin Khor
00:36
@TedShifrin if you recall from before, does this make 'smirnov's theorem' any more familliar:
I appreciate this addendum as I was thinking about the solutions to the Laplacian right after this question. Aside from this my intuition comes from the Smirnov's theorem as I apparently failed to explain at the bottom of my question. It tells you how 1-currents can be represented by an integral over currents on curves. It should be able to make me represent the divergence of a vector field with no periodic integral curves as the difference between the starting point of the solution of the related continuity equation and the ending point, which should be the dirac delta for big enough times. — Lolman 15 hours ago
monoidaltransform
monoidaltransform
00:48
If $f\in C^{\infty}(M)$, where $(M,g)$ is a riemannian manifold, now if $f$ is a smooth function on $M$, we can define $\nabla^kf$, where $\nabla$ is connection. Now, I was wondering why $g(\nabla^kf, \nabla^kf)$ makes sense
$g$ is a function $\mathfrak{X}(M)\times \mathfrak{X}(M)\rightarrow C^{\infty}(M)$
Akiva Weinberger
Akiva Weinberger
Wait, what's $\nabla f$?
I thought connections took two vector fields
Oh wait, I vaguely remember this…
not fully, though. Remind me?
monoidaltransform
monoidaltransform
$\nablaf = df$
which is regarded as a $C^{\infty}(M)$ linear map $\mathfrak{X}(M)\rightarrow C^{\infty}(M)$
in this case
Akiva Weinberger
Akiva Weinberger
Then $\nabla^2 f$?
monoidaltransform
monoidaltransform
So, $df(X)=Xf$
$\nabla^2f: \mathfrak{X}(M)\times \mathfrak{X}(M)\rightarrow C^{\infty}(M)$.
Akiva Weinberger
Akiva Weinberger
I guess in Einstein notation you're gonna put as many $g_{ij}$s or $g^{ij}$s as necessary to make the number of subscripts and superscripts match?
It's been a long time since I learned this, I forgot most of it
Oh what I just said doesn't make sense
monoidaltransform
monoidaltransform
00:59
user image
Akiva Weinberger
Akiva Weinberger
The extra input tells you where to directionally differentiate A, I'm guessing
monoidaltransform
monoidaltransform
yes
so $\nabla^kf$ for $f$ smooth function makes sense
Akiva Weinberger
Akiva Weinberger
Not sure how to get $g$ to take arbitrary $k$-forms as inputs
monoidaltransform
monoidaltransform
it's going to be a $C^{\infty}(M)$ linear map $\bigoplus_{i=1}^k \mathfrak{X}(M)\rightarrow C^{\infty}(M)$
Akiva Weinberger
Akiva Weinberger
(tensors with $k$ inputs)
monoidaltransform
monoidaltransform
01:01
yes, that's what I don't get
, if $g$ is a riemannian metric on $M$ and $f$ is a real valued smooth function on $M$, Then, why does $g(\nabla^kf,\nabla^kf)$ make sense? How does one interpret this expression? Can one view $\nabla^kf$ as a a smooth vector field on $M$?
perhaps the answer in the following explains the situation more math.stackexchange.com/questions/3036321/…
Ted Shifrin
Ted Shifrin
01:36
Not $k$-forms at all. But $g$ induces a metric on tensor bundles.
@monoidaltransform you don’t.
Akiva Weinberger
Akiva Weinberger
02:12
@TedShifrin Oh do those have to be antisymmetric?
Ted Shifrin
Ted Shifrin
02:24
Um, yes. Perhaps bad language (cuz multilinear forms need not be).
D.C. the III
D.C. the III
03:07
been pondering for awhile @TedShifrin about finding equation of plane through point#(1,2,2)$ that cuts off smallest possible volume in first octant....
after some doodling.... I have this nebulous idea that I need to minimize the "slope" of my plane. THis idea comes from me picturing that I have my plane at the point and a bearing is attached to it so I am changing the direction of the plane all over to try and maximze the volume of my object.
Alexandru Ionut
Alexandru Ionut
@TedShifrin I have a little result you might find aesthetic that you told me you would like to see a year or so ago (if I ever get it)
CroCo
CroCo
Hi everyone, how can I compute this derivative where a and b are unit dual quaternion?
user image
K is symmetric positive definite matrix.
porridgemathematics
porridgemathematics
03:23
@monoidaltransform there is a natural extension of $g$ to the tensor bundle of $TM$ , and in the context of sobolev spaces, you can choose nice coordinates, say geodesic coordinates, to see that $|\nabla^{k} u|^p$ in these coordinates is basically a sum of the norms of the partial derivatives of $u$ of order $k$ raised to the power $p$
so working locally you get the usual euclidean sobolev norms
and you can show with some extra work that if you defined sobolev norms locally and patched them together with partitions of unity, they would all be equivalent to this one stated coordinate free using covariant derivatives
Ted Shifrin
Ted Shifrin
@D.C.theIII what are the xyz-intercepts of a plane?
@AlexandruIonut I don’t remember, but sure!
porridgemathematics
porridgemathematics
of course there would be some error terms you would need to make small to do what I am saying is possible
Alexandru Ionut
Alexandru Ionut
@TedShifrin Do you remember when we spoke about clothoids?
D.C. the III
D.C. the III
$(x,0,0), (0,y,0), (0,0,z)$
Ted Shifrin
Ted Shifrin
Vaguely, Alexandru.
porridgemathematics
porridgemathematics
03:27
@monoidaltransform the actual extension to higher form is as you would expect, you first define it on pure tensors by $g(a \otimes b, c \otimes d) = g(a,c) g(b,d)$ and then extend linearly
Ted Shifrin
Ted Shifrin
@D.C.theIII But you need an equation of the plane. Then tell me.
CroCo
CroCo
user image
correction.
any help.
D.C. the III
D.C. the III
So my method for "thinking" out things revovles around me writing out "Key questions", I am literally right now trying to answer the key question of "how canI characterize the plane?"...funny. ..gimme a minute
Alexandru Ionut
Alexandru Ionut
There is a 3d generalisation that was studied by a few folk in the 80s and 90s. Mehlum found expressions for parameter functions using some pretty exotic multivariable hypergeometric series and Ron Resch found a geometric construction for it. It is the spherical clothoid, the spherical curve with geodesic curvature a linear function of arc length (term coined by Ulo Lumiste in unrelated work). Well after 2 years of work, I found MUCH simpler expressions for the parameter functions :)
It's the best result of my life
arxiv.org/pdf/2203.07963.pdf
Ted Shifrin
Ted Shifrin
Congrats! This is stuff I don’t know at all, but I’ll read through it tomorrow.
Alexandru Ionut
Alexandru Ionut
03:32
I appreciate it greatly. I need to fix a sentence or two in my derivation (the way I worded some of the reasoning is wrong but the results all hold)
but you can still appreciate 99% of the paper I think
Ted Shifrin
Ted Shifrin
I know nothing about special functions …
Alexandru Ionut
Alexandru Ionut
I know it is quite niche.
D.C. the III
D.C. the III
let $v_1 = (x,0,0) - (0,y,0) = (x,-y,0)$ and $v_2 = (0, - y, z)$. I now have two direction vectors. Take the cross product, get a normal, call it $A$. Then eqn of plane will be $A \cdot \mathbf{(x - x_0)}$. where $x_0 = (1,2,3)$. Now this would be my constraint.
Ted Shifrin
Ted Shifrin
You’re destined to fail using xyz as two different things.
D.C. the III
D.C. the III
did I inadvertently use xyz as a set of points and as legths?
Ted Shifrin
Ted Shifrin
03:47
As intercepts and as coordinates of a generic point.
D.C. the III
D.C. the III
hmm.
Ted Shifrin
Ted Shifrin
It’s ok to use xyz for intercepts, but then you need XYZ for coordinates.
You want to maximize what?
D.C. the III
D.C. the III
so using a,b,c as the intercept pieces....I see the reasoning, but right now what is making me feel weird is that wouldn't this mean I'm fixing some arbitrary values for my vectors. In essence already giving the elevation of my plane?
I want to maximize volume of the octant
Ted Shifrin
Ted Shifrin
So that’s some constant times xyz.
D.C. the III
D.C. the III
Ok yeah, that makes sense thinking about it for a minute. Going to eat then write out my solution
copper.hat
copper.hat
04:13
those actions do not commute
at least when hungry
one potato two potato
one potato two potato
05:02
How many cups of coffee do you drink a day?
copper.hat
copper.hat
05:31
typically zero
D.C. the III
D.C. the III
05:42
only merlots or sauvignons for you
copper.hat
copper.hat
just had some NZ sauvignon blanc before hitting the sack
:-)
buenos noches...
 
1 hour later…
porridgemathematics
porridgemathematics
06:54
never dink and derive!
drink*
porridgemathematics
porridgemathematics
07:26
im wondering if the nowhere differentiable continuous (real valued) functions on an open interval of $\mathbb{R}$ form a co-meager subset of the continuous functions as they do in the compact interval case, lets say we use the topology of uniform convergence on compact subsets, anyone know?
maybe the answer affirmative or not could be an easy corollary to the compact interval case? My concern is that the proof im familiar with in the latter case involves exploting uniform continuity
user 1591719
user 1591719
07:51
@robjohn Hi. How is it going? Are you around?
I was thinking to ask you a question, but maybe I think more over it before, as regards the existence of a class of integrals in literature with a certain property (they certainly exist, but not sure if the bear a name).
Wave
Wave
08:38
Hello. If $X$ is a continuous real valued random variable and we denote $f_X$ it's density function. Then I know by definition that $E(X)=\int_\Omega X(\omega) dP(\omega)$. But is it true that from this definition it does NOT follows immediately that $E(X)=\int_\Bbb{R} x f_X(x)dx$ where $dx$ denotes the lebesgue masure?
I think there is a lot going on there, which can't be easily proven but maybe it's trivial and I don't see it. So what I know is that if $Y:\Omega\rightarrow M$ for some set $M$ and $g:M\rightarrow \Bbb{R}$ is a measurable bounded function then $E(g(Y))=\int_M g(y) f_X(x)dx$…
Am I correct or is it really an obvious fact and I'm to stupid to see it?
robjohn
robjohn
08:53
@user1591719 once in a while
porridgemathematics
porridgemathematics
@Wave in general, one has that for $g : X \rightarrow Y$ a measurable function, and $f : Y \rightarrow \mathbb{R}$ measurable, $\int_{Y} f d(g_{\ast}(\mu)) = \int_{X} f \circ g d\mu$ where $\mu$ is a measure on $X$, with equality in the sense that both are well-defined simultaneously. So in general for your case, $\int_{\mathbb{R}} x dX_{\ast}(\mathbb{P})(x) = \mathbb{E}[X]$ whenever the expectation exists
$X_{\ast}(\mathbb{P})$ is a regular Borel measure on $\mathbb{R}$, induced by the cdf of $X$, lets call that $F_{X}$, then in general the induced measure is a Stieltjes measure, and need not be absolutely continuous with respect to the Lebesgue measure
the induced measure will be absolutely continuous with respect to the lebesgue measure precisely when it is equal to $f_{X} dx = F' dx$
Wave
Wave
@porridgemathematics Ah but the fact that for $g : X \rightarrow Y$ a measurable function, and $f : Y \rightarrow \mathbb{R}$ measurable, $\int_{Y} f d(g_{\ast}(\mu)) = \int_{X} f \circ g d\mu$ where $\mu$ is a measure on $X$ is a fact from measure theory? Because I have never seen this.
porridgemathematics
porridgemathematics
indeed, just a general fact from measure theory
you can prove it by a standard procedure, start with indicator functions, use standard limit theorems
Wave
Wave
Ah okey and using this we are done.
porridgemathematics
porridgemathematics
not quite, what you want to get your result is a little stronger than $X$ just being continuous, i.e. your CDF just being continuous, you actually need your CDF to be absolutely continuous to get that it is the integral of its derivative
if you know additionally that $X$ is compactly supported , continuous AND that $F$ (its CDF) has derivative defined everywhere besides a countable set, THEN you can get exactly what you want
but this is not a trivial fact
Wave
Wave
09:11
okey I see thank you I think this goes a bit too far for this course but thanks
porridgemathematics
porridgemathematics
for instance, take the cantor function, this is the CDF of some continuous random variable
but the density is $0$
so integrating the density does not give you a probability measure
@Wave en.wikipedia.org/wiki/Cantor_distribution , see here, the expectation is $\frac{1}{2}$, but the density is $0$, so we do not obtain the expectation by integrating against the density
Wave
Wave
09:31
@porridgemathematics sorry but like a stupid question. If I have $X:(\Omega,F,\Bbb{P})\rightarrow (\Bbb{R},B(R))$ and then take $id=f:\Bbb{R}\rightarrow R$. Then id is clearly mesurable with respect to the borel measure. And in addition $f(X)$ is integrable. So I would say I don't need explicitly boundedness of $f$ but then applying the statement from above i have $E(f(X))=\int_\Bbb{R} x P_X(dx)$ where $P_X(dx)=g_Xdx$ if $X$ has a density. WOuldn't this work?
porridgemathematics
porridgemathematics
im assuming you mean that $X$ having a density in this case just means $\mathbb{P}_{\ast}(X) = f_{X} dx$ where $dx$ is the lebesgue measure, in which case yes, $\mathbb{E}[g(X)] = \int_{\mathbb{R}} g(x) f_{X}(x) dx$ holds in the sense that one side of the equality is well-defined precisely when the other is
so if $X$ has a finite expectation, or is a positive random variable so $\mathbb{E}[X] \in [0,+\infty]$ still makes sense, your formula (i.e. with my $g$ equal to the identity) will hold
you do not need boundedness of $f$. But boundedness of $f$ will guarantee that $\mathbb{E}[f(X)]$ is finite, so that this is a special case in which the formula I wrote is valid
Wave
Wave
Yes I mean this. But then the claim also follows from my statement here:if $Y:\Omega\rightarrow M$ for some set $M$ and $g:M\rightarrow \Bbb{R}$ is a measurable bounded function then $E(g(Y))=\int_M g(y) f_X(x)dx$ if $f_X$ is the density of $X. But without the boundedness. But then both sides are equal in sense of [0,\infty]$
porridgemathematics
porridgemathematics
yes, you can omit the boundedness, with equality being understood in the sense that either side is well defined in $[-\infty,+\infty]$ whenever the other side is, and in this case both sides are equal
Wave
Wave
ah okey now it makes sense thanks a lot!
porridgemathematics
porridgemathematics
this just follows from the general measure theory fact stated before
np
 
1 hour later…
user 1591719
user 1591719
10:59
@robjohn Okay
AncientSwordRage
AncientSwordRage
What's the best way to get more eyes on this: math.stackexchange.com/q/1576115/20792 ?
The Bounty is expiring soon and I didn't want it to go to waste
Akiva Weinberger
Akiva Weinberger
11:26
Neat puzzle
You and your friend separately shuffle two decks of 52 cards, then you go through them together, with both of you revealing successive cards in the shuffled order.

On average, how many times would you expect both of you to reveal exactly the same card?
Another way to phrase the question: what is the expected number of fixed points in a random permutation of 52 cards?
Mad Spaces
Mad Spaces
Good day.
If i have two curves, begining in one point and ending at the same point, both of these paths are in a connected space. How do you construct explicity a homotopy between them, given no explicit information about the exact path of the curves?
It is however, to my eyes, very evident, that a homotopy exists. regardless of how the paths look like. Is there some theorem, maybe the axiom of choice?
monoidaltransform
monoidaltransform
@MadSpaces You require convexity
Mad Spaces
Mad Spaces
Does connectedness not suffice?
Example, if we have $ \alpha, \beta : [0,1] \rightarrow \mathbb{D}, \alpha (0) = \beta (0 ) = a,\alpha (1) = \beta (1 ) = b , $ where as $ D $ is a connected , none empty, open set.
porridgemathematics
porridgemathematics
@MadSpaces no because you can draw two paths between the same paths in an annulus with no homotopy between them that stays in the annulus, simply because there are nontrivial loops in an annulus
*points
being able to do that for all paths in some connected space between any two points is the same thing as saying your space is simply connected
Akiva Weinberger
Akiva Weinberger
11:42
The thing to look up is the phrase "simply connected"
Tori and annuli (uh, toruses and annuluses) are connected but not simply connected
@MadSpaces
Mad Spaces
Mad Spaces
So if you have simple connectdeness, you will have a homotopy
Or if you have convex (does star shaped sets here also work?) you also have a homotopy
However if you just have connectdeness, you can not ensure a homotopy
Did i unerstand right?
monoidaltransform
monoidaltransform
yes, in simply connected spaces, all loops are path homotopic
convex subsets of $\mathbb{R}^n$ are simply connected
Mad Spaces
Mad Spaces
Alright thank you.
Calvin Khor
Calvin Khor
11:58
@AncientSwordRage i think the comment probably gives the answer up to change in notation (a power series is obtained in the PDF for the arclength.) You might be able to say, install the free wolfram engine and get it to spit out some human-readable steps...
Koro
Koro
12:18
$\mathbb R^2$ is a vector space of field $\mathbb R$.
$\mathbb C^2$ is also a vector space over $\mathbb R$. Right?
Mad Spaces
Mad Spaces
Well sure why not.
C^2 is also a vector space over C.
Koro
Koro
Or over $\mathbb C$.
$\mathbb C^2$ over $\mathbb C$ has dimension $2$.
I ask because when one sees $\mathbb C^2$, what is immediately understood (to discuss vector spaces)?
is it understood to be a vector space over R or over C?
Akiva Weinberger
Akiva Weinberger
One has strictly more structure than the other, right? So probably the one with more structure
Depends on the context
Like, as a vector space over $\Bbb C$, you can multiply by $i$, and as a vector space over $\Bbb R$, you can't
Mad Spaces
Mad Spaces
You are merly taking a scalar and multiplying the elements with it. If you take scalars from C then you are already including the ones of R
If you restrict yourself on R, you are restricted to changing lenghts of the vectors, if you expand to C, your multiplication will also result into change of angle
Koro
Koro
$\mathbb C^2$ over $\mathbb R$ is of dimension 4.
$(1,0),(0,1), (i,0),(0,i)$ is a basis of $C^2$ over R.
@MadSpaces yes. When I see $\mathbb C^2$, I get confused thinking -should it be considered over R or over C.
I think that $F^n$ is understood to be $F^n$ over F unless stated otherwise.
Mad Spaces
Mad Spaces
12:29
My experiance so far approves your last statement.
porridgemathematics
porridgemathematics
if the only concern is dimensions, you could always clarify by saying $V$ has $K$-dimenison $n$
in the absence of being more specific it should usually be clear from context
Mad Spaces
Mad Spaces
7
Q: Are complex numbers two dimensional or one dimensional?

Random ProgrammerComplex numbers are represented as: z = x + yi This gives the impression that complex numbers are a real component plus an imaginary component. However, when doing math with complex numbers, they are represented as 2-D vectors like in this picture: Complex Vector Are complex numbers one o...

linear-algebra complex-numbers
I think this might be helpful
Vrouvrou
Vrouvrou
12:42
Hello
Akiva Weinberger
Akiva Weinberger
@Koro In Galois theory, we often talk about field extensions (when one field is contained in another)
If we have three fields $F\subset K\subset L$, then $L$ is a vector space in three different ways: it's a 1-dimensional vector space over $L$, as well as a vector space over $K$ and a vector space over $F$
Vrouvrou
Vrouvrou
If i have that $ u"(x) \leq v"(x)$ a.e $x\in (a,b)$ how we can say that $u'-v'$ is monoton nonIncreasing on (a,b)
?
Akiva Weinberger
Akiva Weinberger
The notation is $[L:K]$ for the dimension of $L$ over $K$ (called the degree of the field extension), etc
One of the basic theorems is that $[L:F]=[L:K][K:F]$
@Vrouvrou Well $(u'-v')'\le0$
though, are $u''$ and $v''$ defined everywhere? That is, are $u'$ and $v'$ differentiable?
Otherwise I could imagine introducing a large discontinuity into $u'-v'$ messing with monotonicity if I only care about things almost everywhere
Vrouvrou
Vrouvrou
v" is continuous but u" is continuons a.e
Continuity almost every where is sufficient ?
Koro
Koro
@AkivaWeinberger yes, I agree.
Akiva Weinberger
Akiva Weinberger
12:56
Are they defined everywhere though? @Vrouvrou
@Koro One consequence of $[L:F]=[L:K][K:F]$ is that $\sqrt[3]2$ cannot be written in terms of nested square roots
I can guarantee that, eg, $\sqrt{3+\frac1{\sqrt{2+\sqrt5}}}\cdot\sqrt{\sqrt5}$ doesn't equal the cube root of 2, without bothering to check
The reason is, I can define a large field $L$ containing that number, which has degree a power of $2$ over $\Bbb Q$ (by repeatedly doing that lemma)
and the field $\Bbb Q(\sqrt[3]2)$ (the smallest field containing $\Bbb Q$ and $\sqrt[3]2$) has degree $3$ over $\Bbb Q$
and by that lemma, if $L$ contains $\Bbb Q(\sqrt[3]2)$, then $3$ is a factor of $2^n$
(because $2^n=[L:\Bbb Q]=[L:\Bbb Q(\sqrt[3]2)][\Bbb Q(\sqrt[3]2):\Bbb Q]=[L:\Bbb Q(\sqrt[3]2)]\cdot 3$)
The final gap is, why does $L$ have to have degree a power of $2$? Because if $K$ is some field and $K(\sqrt a)$ is the smallest field containing $K$ and $\sqrt a$ (where $a\in K$), it can be shown that $[K(\sqrt a):K]$ equals either $1$ (if $\sqrt a\in K$) or $2$
so we repeatedly do that for each new square root until we have a monster field containing everything we need
Vrouvrou
Vrouvrou
@AkivaWeinberger i don't understand
Akiva Weinberger
Akiva Weinberger
Like do we know that $u'$ is differentiable everywhere or can it have a discontinuity
because you only wrote that $u''\le v''$ "a.e"
Vrouvrou
Vrouvrou
What i have is u"=f(t,u(t),u'(t)), where f is Carathéodory
u' is continuous yes
u is in C^1
Akiva Weinberger
Akiva Weinberger
I'm also worried about maybe $u'$ being the Cantor staircase or some other weirdness
The Cantor staircase has derivative 0 a.e. but is nonconstant
Vrouvrou
Vrouvrou
13:12
There is no this term in the paper
Calvin Khor
Calvin Khor
@AkivaWeinberger and is continuous
Akiva Weinberger
Akiva Weinberger
In any case, except for possible weird analysis counterexample weirdnesses like that, $u'-v'$ weakly decreasing follows from $(u'-v')'\le0$
Calvin Khor
Calvin Khor
it seems that the solution is supposed to be absolutely continuous
Akiva Weinberger
Akiva Weinberger
(weakly decreasing = nonincreasing but can be constant)
Calvin Khor
Calvin Khor
glancing at the caratheodory existence thm
Vrouvrou
Vrouvrou
13:15
How I can sent you the paper ?
Calvin Khor
Calvin Khor
give the name or url?
if it is semi-new it might have an arxiv link
Akiva Weinberger
Akiva Weinberger
@Vrouvrou You should know that I plan to learn French next semester
I have not yet started
but that is my plan
Koro
Koro
Tres bien @Akiva.
Calvin Khor
Calvin Khor
@Koro et tu?
Koro
Koro
😁
Vrouvrou
Vrouvrou
13:17
The paper is in English
Akiva Weinberger
Akiva Weinberger
(They say French people never appreciate when foreigners try to learn their language…)
Koro
Koro
I find chinese language very difficult.
Calvin Khor
Calvin Khor
@Koro 中文真的非常难，没有办法的👐
Akiva Weinberger
Akiva Weinberger
That only works if we have your computer
Koro
Koro
@Vrouvrou: that's your local storage link.
Akiva Weinberger
Akiva Weinberger
13:20
What's the title of the paper?
Koro
Koro
@CalvinKhor Oh wow! you know Chinese language :)
Calvin Khor
Calvin Khor
you cannot upload pdfs on this chat
Akiva Weinberger
Akiva Weinberger
(Or maybe you can make a Dropbox link?)
Koro
Koro
I think the first one is 'hito' ?
hito =man
Vrouvrou
Vrouvrou
aimsciences.org/article/doi/10.3934/proc.2011.2011.209
Akiva Weinberger
Akiva Weinberger
13:21
That's Japanese, and 人
Calvin Khor
Calvin Khor
中文=chinese真的=really 非常=super 难 = hard，没有办法的=nothing you can do about it
2
Koro
Koro
@AkivaWeinberger oh I remember now.
:D
Calvin Khor
Calvin Khor
ooh open access
@AkivaWeinberger do you know japanese?
Vrouvrou
Vrouvrou
user image
porridgemathematics
porridgemathematics
isnt it because you have absolute continuity? Its like asking if $f,g$ are absolutely continuous and $f' \leq g'$ a.e., then is $f-g$ monotone nonincreasing on some interval, the answer is yes because $(f-g)(x) = (f-g)(x_0) + \int_{x_0}^x (f-g)' $, so $(f-g)(x) - (f-g)(x_0) \leq 0$ for $x \geq x_0$
since everything in that paper seems to be a solution to a ODE in the sense of caratheodory, it seems by definition these solutions are absolutely continuous
basically so that they can be recovered by picard-lindelof in the first place
Akiva Weinberger
Akiva Weinberger
13:38
@CalvinKhor 少しだけ
porridgemathematics
porridgemathematics
yeah @Vrouvrou it says at the start that $\alpha \in W^{4,1}(I)$ and explains what that means, $\alpha'''$ is absolutely continuous and its fourth derivative is in $L^1(I)$
Vrouvrou
Vrouvrou
where is used the absolutly continuity ?
Akiva Weinberger
Akiva Weinberger
Does $f(x)=f(x_0)+\int_{x_0}^xdf$ only hold if $f$ is absolutely continuous?
I mean, I suppose probably, since the Cantor staircase is a counterexample that is not absolutely continuous
porridgemathematics
porridgemathematics
yes , its an if and only if, with $f'$ being in $L^1$ on that compact interval
(assuming this is being formulated on a compact interval)
$f$ is AC on $[a,b]$ iff there is some $g \in L^1[a,b]$ s.t. $f(x) = f(a) + \int_{a}^x g(t) d\lambda(t)$ where $\lambda$ is the lebesgue measure
(equivalently, same thing with $g$ replaced by $f'$)
yeah, the cantor staircase is an example of something that is of bounded variation on a compact interval, so it defines a regular borel measure (in this case finite) on that interval, but this borel measure is mutually singular to $\lambda$
mutually singular is basically the opposite of absolutely continuous
(the measure the staircase induces weights the cantor set with mass $1$)
and is a probability measure
Akiva Weinberger
Akiva Weinberger
Absolute continuity means that it maps small sets to small sets, yeah?
Vrouvrou
Vrouvrou
13:49
please is continuous a.e implies absolutly continuous ?
Calvin Khor
Calvin Khor
@AkivaWeinberger the only word i can read is 少 which is also chinese :) (meaning few or little)
Akiva Weinberger
Akiva Weinberger
@CalvinKhor "Just a little" :)
Calvin Khor
Calvin Khor
@Vrouvrou no, in fact continuous everywhere does not imply AC (see devil's staircase above)
Akiva Weinberger
Akiva Weinberger
Cantor function
In mathematics, the Cantor function is an example of a function that is continuous, but not absolutely continuous. It is a notorious counterexample in analysis, because it challenges naive intuitions about continuity, derivative, and measure. Though it is continuous everywhere and has zero derivative almost everywhere, its value still goes from 0 to 1 as its argument reaches from 0 to 1. Thus, in one sense the function seems very much like a constant one which cannot grow, and in another, it does indeed monotonically grow, by construction. It is also referred to as the Cantor ternary function...
^@Vrouvrou Please see this carefully
porridgemathematics
porridgemathematics
@AkivaWeinberger yeah the usual definition
Akiva Weinberger
Akiva Weinberger
13:53
@CalvinKhor 少 is a Chinese character (in Japanese called a "kanji", written 漢字), but しだけ are all kana (symbols representing sounds - syllables - but not meanings, originally derived from cursive versions of certain kanji used for their sound rather than their meaning)
少し = sukoshi (pronounced more like skoshi or even skosh) meaning "a little", and だけ = dake meaning "just"
Vrouvrou
Vrouvrou
i don't understand how he deduces
that $u'''-\alpha'''$ is nonincreasing
porridgemathematics
porridgemathematics
on compact intervals measure theoretic notion of absolute continuity and the version for functions coincide, on open intervals you can have a regular signed measure which is absolutely continuous w.r.t $\lambda$, but the corresponding function defining it will only be guaranteed to be absolutely continuous on compact sub-intervals
also the function defining it wont be of bounded variation, but it will be of bounded positive or negative variation
Vrouvrou
Vrouvrou
if i suppose that $\alpha $ is $C^3$
and $u^{(4)}$ is continuous a.e
such that $u^{(4)}(x)\leq \alpha^{(3)}(x)$
can i deduce that $u'''(x)-\alpha'''(x)$ is nonincreasing
porridgemathematics
porridgemathematics
I think both $u^{'''}$ and $\alpha^{'''}$ are absolutely continuous
Vrouvrou
Vrouvrou
absolutely continuous is important ?
porridgemathematics
porridgemathematics
14:02
it seems to me like it could be, but i am pretty unfamiliar with this stuff
apparently solutions to a caratheodory type ode are absolutely continuous
so if $u^{'''}$ is one , then it should be.
because you only have fourth derivatives defined a.e, and you want to recover information about third derivatives
you want to use FTC somehow to do this
FTC for lebesgue integrals is equivalent to absolute continuity
Vrouvrou
Vrouvrou
what is ftc ?
porridgemathematics
porridgemathematics
as has been pointed out, if you only know $f'$ is continuous, you cant recover information about $f$ using the ftc
fundamental theorem of calculus
here $f' \in L^1$ of course, so its implicit that this is a.e. defined
because you can have $f $ non-constant, and $f' \in L^1$ equal to $0$
you can even have $f$ strictly increasing, and $f' = 0$ a.e., and in $L^1$
but if you know $f$ is absolutely continuous, these pathologies go away
typo above also, i meant if you only know $f$ is continuous with derivative existing a.e. and in $L^1$
that isnt enough to recover it from its derivative
(also the pathologies I mention are all continuous, i.e. they serve to show why you need something more than just continuity)
it may be worth mentioning though, that if you have $f$ is continuous, $f' \in L^1$ exists everywhere besides an at most countable number of points, then you get $f$ is absolutely continuous (you can recover it from its derivative)
so the pathologies I mention have $f'$ undefined at an uncountable number of points necessarily (but still measure zero of course)
Vrouvrou
Vrouvrou
so there is an error in the paper ?
porridgemathematics
porridgemathematics
i wouldn't assume so, i really just skimmed bits and pieces, i dont understand that paper
i think they assume the reader is familiar with caratheodory solutions to things
robjohn
robjohn
14:41
@porridgemathematics If $f’$ is continuous and $0$ a.e., then $f’=0$ everywhere. Give an example of two functions so that $f’=g’$ is continuous and $f(0)=g(0)$, but $f\ne g$.
porridgemathematics
porridgemathematics
@robjohn I mentioned what you are referring to was a typo
35 mins ago, by porridgemathematics
typo above also, i meant if you only know $f$ is continuous with derivative existing a.e. and in $L^1$
was supposed to be if you only know $f$ is continuous blah blah blah, not $f'$
of course a continuous function equal to some other continuous function a.e. is that other function everywhere
if you know $f'$ is continuous (and integrable), you can use the usual version of the FTC, no lebesgue theory needed
uh, all you need to use the usual FTC is knowing that $f'$ exists everywhere and is riemann integrable, so a.e. continuous and exists everywhere on a compact interval is the same thing
uh, and bounded (sorry)
Koro
Koro
15:17
0
Q: If $ST$ is a nilpotent operator then so is $TS$.

KoroSuppose that $T, S \in L(V)$ where $V$ is a finite dimensional complex vector space. I am trying to prove that if $ST$ is nilpotent, then $TS$ is nilpotent. First I consider the case $ST=0$. From here, I want to show that $TS=0$. Suppose on the contrary that it is not so. There exists a non zero...

linear-algebra eigenvalues-eigenvectors
porridgemathematics
porridgemathematics
that comment answers the question, doesn't it?
he is saying $(TS)^k = TS(TS)...(TS) = T [ST(ST)...(ST)]S = T [(ST)^{k-1}] S$
Koro
Koro
15:33
yes, I understood. Thanks a lot.
user image
porridgemathematics
porridgemathematics
you could just use the definition of nilpotent in terms of $A^k = 0$ for some positive integer $k$
not sure why you need to write null a bunch of times
Koro
Koro
I used the theorem that null T^ dim V = V if T is nilpotent.
porridgemathematics
porridgemathematics
sure, but you dont even need that
Koro
Koro
Another theorem that I used was: If null T^k = null T^{k+1} then null T^k = null T^{k+i}
for any $i\in N$.
porridgemathematics
porridgemathematics
you just need to be able to zero the matrix with some power, and writing $(TS)^k$ in terms of $(ST)^{k-1}$ allows you to see this immediately
i think that is slightly overcomplicating the question
Koro
Koro
15:37
I got your point. :)
If $(ST)^k=0$ for any k>0, then $(TS)^{k+1}=T (ST)^k S=0$.
porridgemathematics
porridgemathematics
yeah
Koro
Koro
Another comment also answers my question.
Characteristic polynomials $ch_{TS}(x)=ch_{ST}(x)$
robjohn
robjohn
@porridgemathematics sorry, missed that.
Koro
Koro
If ST is nilpotent, then $ch_{ST}(x)=x^n$ where n= dim V.
porridgemathematics
porridgemathematics
@robjohn no worries, admittedly it was buried since I caught it so late
Koro
Koro
15:42
So $ch_{TS}=x^n$
Now, by Cayley Hamilton's theorem: $(TS)^n=0$
😁
But this is complicated as I know that characteristic polynomial of AB and BA are same if either of A or B is invertible.
porridgemathematics
porridgemathematics
it holds in general over any field actually
(whether they are invertible or not)
robjohn
robjohn
15:58
@Koro For example
They don't even need to be square
if you don't mind a few factors of $\lambda$
AncientSwordRage
AncientSwordRage
@CalvinKhor 😵
Those are words
porridgemathematics
porridgemathematics
another way could be $Q(x,X,Y) = \det(xI - XY)$ satisfies $Q(x,X,Y)\det(X) = \det(xI - XY)\det(X) = \det(xIX - XYX) = \det(XxI - XYX) = \det(X) \det(xI - YX) = \det(X) Q(x,Y,X)$, where $x \in R$ , $X,Y$ are in $R^{n^2}$ where $R$ is some integral domain, so that you can cancel the $\det(X)$'s from both sides
user193319
user193319
16:16
If two bijections $f$ and $g$ of some set have disjoint support, does this imply they commute?
Vrouvrou
Vrouvrou
@robjohn have you an idea on how they work on the paper i work on ?
porridgemathematics
porridgemathematics
for non square, you could also adapt the proof of this,if $X$ is $m \times n$, then $xI_{m \times m} X = x^{m-n} X xI_{n \times n} $ (for $m \geq n$) so up to a $x^{m-n}$ factor they would still be the same
robjohn
robjohn
@Vrouvrou I have no idea what you are asking.
@user193319 how can bijections of a given set have disjoint support? can you give an example of that you mean?
Vrouvrou
Vrouvrou
16:33
$u^{(4)}$ is continuous almos everywhere and $\alpha^{(4)}$ is AC , how from $u^{(4)}>\alpha^{(4)}$ we can deduce that $u^{(3)}>\alpha^{(3)}$
porridgemathematics
porridgemathematics
oh wait, actually I think $\det(X)$ should be cancellable regardless of whether $R$ is a domain, its never a zero divisor in $R[x,X,Y]$, so this fact should hold over any commutative ring
Wolgwang
Wolgwang
> $x=t^5-5t^3-20t+7,y=4t^3-3t^2-18t+3$ where $|t|<2$. Find max and min. value of $y=f(x)$
$$\frac{dy}{dx}=\frac{6(t+1)(2t-3)}{5(t^2+1)(t+2)(t-2)}$$


\begin{array}{|c|c|}
\hline -& +&-\\
\hline (-2,-1)&(-1,\frac32)&(\frac32,2)\\
\hline \end{array}

$x=-1$ is point of minima and $x=\dfrac32$ is point of maxima
What's wrong?
user193319
user193319
@robjohn Don't the cycles $(1,2)$ and $(3,4)$ have disjoint support?
I think I was able to prove it, actually.
porridgemathematics
porridgemathematics
@Vrouvrou $u^{4}$ is also in $L^1(I)$, right?
Vrouvrou
Vrouvrou
yes
porridgemathematics
porridgemathematics
16:45
then $u^3$ is AC
oh, you also need that $u^4$ is bounded, sorry
do you have that?
Vrouvrou
Vrouvrou
u^4=f(t,u,u',u'',u''') which is Carathéodory and bounded by a function L^1
porridgemathematics
porridgemathematics
okay, but then $u^3$ is automatically AC
$y' = f(t,y)$ in the caratheodory sense means $y' = f(t,y)$ holds a.e. and $y$ is AC.
that is what it means
also note that your $u^4$ is not even defined everywhere
Vrouvrou
Vrouvrou
and why with AC it works ?
porridgemathematics
porridgemathematics
i wrote it down before
if you have $u^4 - \alpha^4 > 0$
and $u^3 - \alpha^3 $ is AC
robjohn
robjohn
@user193319 If you are talking about permutations and you consider $x$ to be out of the support of $f$ if $f(x)=x$, then yes.
But none of that was made clear
porridgemathematics
porridgemathematics
16:52
you are asking to show that if $f$ is AC, and $f' > 0$, that $f$ is increasing
Vrouvrou
Vrouvrou
yes
porridgemathematics
porridgemathematics
you do it by using the fact that when $f$ is AC, $f(x) = \int_{x_0}^x f'(t) d \lambda(t) + f(x_0)$
robjohn
robjohn
can you apply the mean value theorem?
porridgemathematics
porridgemathematics
so $f(x) - f(x_0) = \int_{x_0}^{x} f'(t) d \lambda(t) > 0$ since $f' > 0$ a.e.
so to get what you want, that $u^3 - \alpha^3 > 0$ on $[t^{\ast},t)$, you need some $t^{\ast}$ such that $u^3(t^{\ast}) - \alpha^3(t^{\ast}) = 0$ (which I assume you have)
then you just apply monotonicity
(or s.t. that difference is positive already somewhere)
user193319
user193319
A bijection of a set is a permutation
Vrouvrou
Vrouvrou
17:24
is there a theorem which says that if $u^{(4)}$ is caratheodory bounded by a function L^1 then $u^{(3)}$ is AC ?
copper.hat
copper.hat
a bijection is when you get your second dose
Ted Shifrin
Ted Shifrin
17:46
I guess you need a trifectajection.
Alexandru Ionut
Alexandru Ionut
Hi everyone!
Ted Shifrin
Ted Shifrin
18:47
@leslie This might be down your alley, despite the heavy quantum mechanical bent.
leslie townes
leslie townes
huh. position operator. i guess multiplication by coordinates, on L^2(R^3) or something?
the answer in that case is yes, although i don't know how to translate that into a language that would make sense to the OP.
i used to be able to fake their language but those days are long gone. i realized the other day that i've been a law student then attorney longer than i was a grad student then prof.
should i write "if it's multiple choice bubble 'yes'"?
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