00:00 - 17:0018:00 - 00:00

12:00 AM
did his parent write some treatise or something? i mean, the notion has been around forever. under mild hypotheses the family of solution operators to a differential equation given conditions at time __ is a semigroup. but there are algebraic semigroups that maybe didn't get attention until later
my mathematical great^something grandfather einar hille coauthored a pretty good book on semigroups of operators in the 1950s

let me see

hille's book takes a lot of the mystery out of why e^(d/dx) is what it is

he's responsible for "Lipscomb’s Space" in dimension theory
I'm trying to figure out what that is..

looks ugly
at least we now know who to blame

apparently the methods he constructed helped him solve a decades old problem in dimension theory
I might read his book "Incipit"

12:12 AM
Is munchkin on summer break, or are they stuck with her in summer too?

12:26 AM
oh, they're stuck with her in summer. they have a week off in august.
her teacher got covid and is out. there is a sub. another student got covid and is out. it's only a matter of time.
doing the covid test this morning was a pain in the neck.

I wonder when they go back to masks.

her teachers mask but the students are not so good about it. including my daughter. half the time when i pick her up, her mask is off.

Ah.

12:55 AM
Hi, why does Spivak state that “A straight line is the shortest distance between two points” is “hopelessy muddled terminology”? What about this is muddled or unclear?

mm, in some contexts, 'distance between two points' already has a meaning in context. where there aren't 'shortest' distances and longer distances.
that particular wording also equates a geometrical object (a straight line) with a number (a distance), which is undesirable.
and in some abstract settings a line does not provide the shortest distance between two points. so the context that makes it a true statement is absent (but, often, reasonably assumed)

But Spivak focuses only the occasional R^1, and mostly R^2 and rarely on R^3

'hopelessly muddled' might be an overstatement.
it's just not all that you'd need to know to make sense of it.

And I think he makes statements that often only apply to those, but he points this one out in particular?
It just seemed odd to me, as if there's something more to it
And less of "well, the generalization of this isn't always true"
When in the context of the material in the book, it should be true?

i probably don't have spivak so i can't comment in detail. chat regular ted shifrin wrote a lot of the exercises in at least one volume of spivak. maybe he knows more.
even if one is not reaching for a generalization, he could be trying to focus on something that 'feels' obvious but requires proof. if you give a notion of 'distance' that connects to an extremal problem relating lengths of curves connecting two points.

1:05 AM
No I think I made him too frustrated, because I was making some silly mistakes yesterday in a post I made about something I was confused on

ted's just like that. :)

So I don't wanna exacerbate the situation :)

i'm trying to think about how you'd prove it in R^n, given the integral formula for the length of the curve. it's pretty clear from the integral if you choose your points to be on the same coordinate axis, but the existence of rigid motions that can place any two points in such a position maybe isn't obvious.

1:18 AM
in my high school calculus class we had a project that involved writing up a solution to a minimization problem of the type where a person can travel one speed on land and another speed on (non-flowing) water, and how do you choose a path to get somewhere the fastest. i spent so much of my write-up explaining why the path had to be composed of straight lines that i did not finish various subparts of the problem which presumed this. my calc teacher gave me full credit.
great teacher.
he also lent me all of his college and grad school math books. he was the first math major i'd ever met. got an MA in numerical analysis somewhere.

Grad school math books in high school?

yeah!
i just googled him. it looks like he retired from my high school but still teaches community college. there's a ratemyprofessors page which is just page after page of people saying they love him. great guy
made up for a lot of math teachers at my high school who were volleyball coaches first and math teachers second.
my high school was a really weird mix of the completely incompetent and the overqualified true believers.

1:36 AM
The correct statement is this: The line segment joining two points is the shortest path joining them. A line isn’t a segment, and neither is a distance.

ooh, i came close to that up above.

It’s immediate in $\Bbb R^n$ with some vector tools (and the formula for length of a $C^1$ Curve.
Cauchy-Schwarz does it.

i came somewhat close to that, too.
my brain still kind of works, despite its immersion in the law

But it’s also immediate in Spivak’s context for plane curves just with his definition of the length as a sup of lengths of polygonal approximations.
The line segment is the trivial polygonal approx.

i still giggle about spivak citing himself in cyrillic with the 2d version of inscribed polygonal approximations for surface area. that was a good joke.

1:41 AM
Very clever. I ripped off his idea and repeated his theft in my multivariable book.

oh lord. my daughter was apparently 'screaming and hitting' in day care today.
:(

I finally did my proof. Ted Shifrin would be proud. If only I had done that yesterday

LOL, good, polite. I just get tired of repeating the same thing after three times :)
They should kick her out, leslie.

Could you please have a glance when you're not busy?

After dinner.

1:50 AM
Thank you
See the problem yesterday was I was confused about not being sure what I was confused about. When you're not confused, and the proof works, you're sure of yourself that it should work unless there's a subtle error somewhere.

The proof didn’t work.

I meant yesterday I was still confused
So I wasn't sure of what I was doing at all

My point is that your error was very much not subtle! It’s the whole point of the definition of the integral with all those infs and sups.

Oh I know, I'm saying now, that when I'm more confident in the material.
Before, I wasn't sure why this wouldn't work, so I was really confused.

Working through simple examples is good to do in learning every part of math.

2:19 AM
@MadSpaces Here's how I think about tensor products. Let $f$ be a bilinear map on $V$ and $W$ to some other vector space $X$. What things must be in this codomain vector space $X$?
Well, for any $v\in V,w\in W$, $X$ must contain $f(v,w)$, more-or-less by definition.
It must also contain $f(v,w)+f(v',w')$ (for any $v,v'\in V,w,w'\in W$) because it's a vector space.
Now, here's a question. Say I pick one element $v\in V$ and two elements $w,w'\in W$. Must $f(v,w)+f(v,w')$ be in the image of $f$?
Yes - because $f$ is bilinear, $f(v+w)+f(v,w')$ equals $f(v,w+w')$.
(On the other hand, there's no guarantee $f(v,w)+f(v',w')$, with two vectors chosen from each of $V$ and $W$, is in the image of $f$.)
These expressions are all things we can write down that live in $X$, without having actually ever seen an element of $X$ directly.
Let's thing about these expressions themselves, rather than the elements of $X$ they represent. Say two expressions are the same if we can prove using the bilinearity laws that they must refer to the same element of $X$.
(So I'm considering $f(v+w)+f(v,w')$ and $f(v,w+w')$ to be the same expression. I'm not considering $f(v,w)$ and $f(v',w)$ to be the same expression even if they coincidentally refer to the same element of $X$, because I can't prove it using the bilinearity laws.)
Then we can say the set of expressions is its own vector space.
This set of expressions is the vector space $V\otimes W$, just in different notation.
Instead of talking about $v\otimes w$, I was talking about the expression $f(v,w)$.
But they behave exactly the same. $v\otimes w+v\otimes w'=v\otimes(w+w')$, exactly like how $f(v,w)+f(v,w')=f(v,w+w')$ for a mystery function $f$.
When is $v\otimes w+v'\otimes w'$ not a pure tensor? Exactly when you can't prove $f(v,w)+f(v',w')$ is in the image of $f$ for a mystery function $f$.
I probably should have led with that phrasing. Tensor products describe the behavior of a mystery bilinear function.

2:44 AM
ted: apparently there was another girl who hit her first. so they were reluctant to punish too much.

Girls are vicious. Wait for teenage years!

@MadSpaces I'm sorry that was so long. It would have been shorter if I had thought it fully through at the start.
(But then I might not have written it.)

you did not have the time to make it shorter.
some guy said that once.

Pascal et al

3:17 AM
The parenthetical condition supports a lot of concise writing.

george bernard shaw

> Je n’ai fait celle-ci plus longue que parce que je n’ai pas eu le loisir de la faire plus courte.
Mark Twain, Voltaire, Johann Wolfgang von Goethe, Winston Churchill, Pliny the Younger, Cato, Cicero, Bill Clinton, and Benjamin Franklin.

3:43 AM
like i said, some guy
some guys, i guess

I almost forgot math was a thing. I'm gonna be so rusty when I start uni in October

@copper.hat GHOTI

the key is to remember that math is a thing
speaking of fish, we had sushi again tonight.

Or two … just say “sup?”
Place that good?

not the restaurant this time. whole foods grocery store sushi.

3:52 AM
Ah

we were originally gonna cook a lentil soup but both of us had unexpected work in the mid afternoon.

Don’t forget leeks agridolce!

it's on the list.

Yeah, sure.

most of the folks who make grocery store sushi in long beach are immigrants from burma. it's this whole thing they've got going.

3:56 AM
You have proof?

n is fixed and $\sup\{\sum_{k=1}^n\frac {a_k}{a_k+a_{k+1}+a_{k+2}}: a_1,a_2,...,a_n\in \mathbb R^+\}$ is to be found.

well i haven't done a formal survey. this is anecdotal based on the ones i talk to.

here $a_{n+1}:=a_1, a_{n+2}:=a_2$

oh haha i just googled it. apparently there was a burmese guy who got in very early on supermarket sushi. nytimes.com/2017/10/29/us/burmese-refugees-sushi.html
from 5 years ago. it's not a new thing

Your indices aren’t quite right, but no clue.

3:58 AM
Book solution first establishes that n-2 is an upper bound.

Ob, never mind.

what's wrong with indices? :(

Then, the solution takes $a_k= t^k$ and considers $t\to \infty$.
and the $\lim_{t\to \infty} \sum_{k=1}^n \frac {a_k}{a_k+a_{k+1}+a_{k+2}}=n-2$
so supremum is n-2.
I'm having difficulty understanding this last step.

In problem 13-25 (g) of Spivak, we say $f'$ is continuous at $a$ and want to show that $\lim_{x\to a^+} \frac{\int_a^x \sqrt{1 + (f')^2}}{\sqrt{(x-a)^2 + (f(x) - f(a))^2}} = 1$. I see now way around using a version of the MVT for integrals, which states that if $f$ is continuous on $[a,b]$, then $\int_a^b f = (b-a)f(z)$ for some $z \in [a,b]$.
However, in the problem, we don't necessarily have that $f'$ is continuous on $[a,x]$. Obviously, as $x$ approaches $a$, it will be continuous on $[a,a]$, but why don't we have to worry about this issue while applying the theorem?

4:03 AM
sorry, the limit is to be taken as $t\to 0+$ and not as $t\to \infty$.

If $A = \det\left(\left[C_{1}\middle|\frac{I}{0\dots0}\right]\right), B= \det\left(\left[C_{2}\middle|\frac{I}{0\dots0}\right]\right)$ , where $A$ and $B$ have different first columns..the entries in matrix $A$ and $B$ are real. Suppose the eigenvalues of $AB^2$ and eigenvalues of $A^2B$ are less than $1$. Then I think $\det(AB+A+I) < 0$ and $\det(BA+B+I)<0$.
@robjohn I thought of adding impurities to $\det(A^2 +A + I)$, we had a chat last time

@politeproofs Try using the proof of the FTC …

@TedShifrin This problem precedes the chapter which introduces FTC
I try not to use "future" knowledge, since that's not how Spivak intended it(?)

Well, just obvious bounds on the integral .
Max and min of function times $x-a$.
And MVT for denominator.

Ah, I see, so then the bounds will go to $0 \le \mu \le 0$ as $x \to a$
Wait

4:14 AM
Not so fast.

I think I understand now why $a_k=t^k$ works.

Oh wait
We need $m = \text{min}(\sqrt{1 + (f')^2}, M = \text{max}(\sqrt{1 + (f')^2})$, and we're told that $\sqrt{1 + (f')^2}$ is integrable on $[a,x]$, then $\int_a^x \sqrt{1 + (f')^2} ~ dx = (x - a)\mu$, for some $\mu \in [m,M]$

Better.

Professor Ted: I saw your name in an another book.

koro: 'serial killers who were never caught' ?

4:20 AM
So I guess $m(x-a) \le \int_a^x (...) \le M(x-a)$

Yes. Now do the denominator

@leslietownes haha no.

Yeah, denominator is $(x-a)\sqrt{(f'(z)^2+1}$ for some $z \in (a,b)$

That’s not a book :)
Now divide and do the limit.

the shadow of athens, they called him
harvesting souls for geometry

4:22 AM
@Koro There are other actual books.
Well, one, at least.

i'm thanked in the introduction to a book. i showed my wife one time when we were in a bookstore. she was like "ok? ...?"

So we have that $\frac{m}{\sqrt{1 + (f'(z))^2}} \le \frac{\int (..)}{(x-a)\sqrt{1 + (f'(z))^2}} \le \frac{M}{\sqrt{1 + (f'(z))^2}}$.

And so?

Is there any particular way of writing as $x \to a$, $(a,x) \to (a,a]$?

Don’t do that.

4:26 AM
^

Anyway, the value of $m$ on the interval of $\sqrt{1 + (f'(x))^2}$ on the interval $[a,x]$ as $x \to a$ becomes the same as the minimum value of $\sqrt{1 + (f'(x))^2}$ on $[a,a]$. Same with $M$, so this value will go to $\sqrt{1 + (f'(a))^2}$. Then also, of course, $\sqrt{1 + (f'(z))^2}$ for $z \in (a,x)$ will approach $(a,a)$ so it'll approach $\sqrt{1 + (f'(a))^2}$, which implies that we have $1 \le \int/\dots \le 1$
Don't do what?
I'm quite uncomfy about (a,a), by the way.

doctor, it hurts when i do this

Haha

You should be explicit about the use of the continuity hypothesis.

You're right, those values are achieved because $f'$ is continuous at $a$, and therefore the limit is equal to just plugging in the values
Why can't I write it less... like that though?
Surely there has to be a better way to write about the intervals approaching a single point as one of the endpoints approaches the other one in an interval

4:39 AM
@TedShifrin I know :)
@politeproofs you can use limit definition to prove the result you're trying to prove.
Try writing the denominator in integral form.

The denominator has an integral form?

$\sqrt{(x-a)^2+(f(x)-f(a))^2}=\int_a^x \sqrt{1+\left(\frac{f(x)-f(a)}{x-a}\right)^2}\, dt$

I would rather just write the proof in a better way than solve it another way?

Because you said you didn't want to use FTC, the way above uses only 'continuity at a' and limit definition. :)

4:49 AM
Well, Dr. Shifrin's suggested proof (the one I wrote above) does not use FTC?

@politeproofs Did you complete this solution?

@politeproofs it's not clear why " $m$ on the interval of $\sqrt{1 + (f'(x))^2}$ on the interval $[a,x]$ as $x \to a$ becomes the same as the minimum value of $\sqrt{1 + (f'(x))^2}$ on $[a,a]$".
Do you want to say that $m$ is attained by $\sqrt (.)$? If yes, then why?

koro, how is your new life as a mathematician? when does the course start?

leslie, the classes start next month. :)
And thanks a lot for asking : )
I'm excited to go to college again :-)

4:57 AM
@Koro What do you mean $m$ is attained by $\sqrt{\dots}$? $f'$ is bounded, so it has to be attained on a closed interval

f' is bounded but given to be continuous only at a.

That's true. But we only care about the fact that it is bounded for such an $m$ to exist

it's great that you were able to get back into math after some time off. it must have been a long journey. one of my advisors from undergrad said 'when people go away even for a year, they never come back.' i proved him wrong. :) you did too.

@leslietownes The recent Field's Medal winner June Huh was a HS dropout

really? that's funny. one of my colleagues at work is an hs dropout.

5:06 AM
@leslietownes :-)
My UG is in civil engineering.

haha, so you did not get 'back into math.' you broke into math.

haha :D

when i was an undergraduate, the civil engineering department was the best department on campus. they threw wild parties on fridays on the roof of their building. usually with multiple DJs. i never figured out where the tradition started but most of them were from india.
it wasn't all bridges and concrete with those people. they could throw a party.
and probably design some bridges too.

@leslietownes nice :). It wasn't the case at my college though.
Many of the roofs were made conical at the college as there is lot of rain in the region.

well, that's no fun. where is the DJ supposed to set up?

5:13 AM

the math department had lame parties. except one of my friends, who would invite us over on thanksgiving and we would leave at 3am.
my wife would be like "how are they still partying, it's 3 am." older people don't need as much sleep. that's the only explanation.

haha

I meant to say that $\det(AB+A+I) < 0$ and $\det(AB+B+I) < 0$ are not possible

@TedShifrin Interesting individual.

5:20 AM
there don't seem to be a lot of patents relating to civil engineering. that's where you guys miss out. gotta get patents to make money.

Is there any way to write what I want to write more rigorously?

I'm so out of civil engineering now. How I got into civil engineering in the first place is a very long story.

was it just family expectations? a lot of the people i know, it was just very important to the family that they be an engineer.

it was lack of knowledge of options available at the time.

oh, that's interesting.
my parents didn't have a lot of ideas about college. they were just happy that i went.

5:31 AM
@leslietownes same here :)

i wonder what my daughter will do.

5:43 AM
How does this 'silly example' on a non-separable hilbert space $\ell^2([0,1])$, $(Au)_x = u_x$ have every $x\in[0,1]$ as an eigenvalue? Maybe it should be $(Au)_x = x u_x$?

If $A,B \in M_{n}(\Bbb{R})$ such that $A = \det\left(\left[C_{1}\middle|\frac{I}{0\dots0}\right]\right), B= \det\left(\left[C_{2}\middle|\frac{I}{0\dots0}\right]\right)$ , where $A$ and $B$ have different first columns (represented as $C_{1}, C_{2}$).

The entries in matrix $A$ and $B$ are real.

Suppose the eigenvalues of $AB^2$ and eigenvalues of $A^2B$ are less than $1$.

Then I have an intuition that $\det(AB+A+I) < 0$ and $\det(BA+B+I)<0$ is not possible.

The approach I am thinking is if $\det(AB + A + I) < 0$, then the matrix $AB+A+I$ should have odd number of real negative eigenval

calvin: yeah, that's garbage. very bad notation. i'm not sure if their 'A' is not the identity operator.
using $u_y$ both for a sequence and a value of a sequence. that's a notational car crash.

Sorry I messed up the last paragraph in latex
It should be:
The approach I am thinking is if $\det(AB + A + I) < 0$, then the matrix $AB+A+I$ should have odd number of real negative eigenvalues (say $n_{A}$). Similarly, the matrix $\det(BA + B + I) < 0$ should have odd-number of negative eigenvalues (say $n_{B}$). That means we will always have $n_{A}n_{B}$ as an odd number and if the intuition is true then $n_{A}n_{B}$ cannot be an odd number, it will always be even.

@leslietownes shh don't poo poo mathoverflow

math overflow doesn't know the difference between operator theory and a hole in the ground.

5:50 AM
k tyvm!
jumping into the middle of functional analysis is not pleasant. so much i absolutely do not know

waaay back when i answered a post on math overflow with a number of references. the question made a lot of sense, but was just very ignorant, so the references were all to books with titles like 'a first course in operator theory.' the OP, who i think was one of the founders of math overflow, deleted the question.
i ceased participating after that point.

loool

the main reason for considering non separable hilbert spaces is that they allow you to construct representations very easily/cheaply. which is important if you are not already given an algebra of operators as acting on something. you can almost always represent it on a non separable space.

sounds reasonable

some of the people who practice in this area are very bad about details. there are some wrong results out there. i won't name names. people who work in the area 'just know' and you don't cite them.

5:55 AM
i came to the example from a very basic question...i wanted to know if an isolated point in the spectrum must be an eigenvalue. the wikipedia definition of the discrete spectrum makes me think its not enough

that's a good question. i don't have an example offhand but it would surprise me if the answer were positive.

@leslietownes functional analysis version of that italian school of algebraic geometry story? that one has names out though

oh there are some names in functional analysis too. there's a whole chapter of naimark's "normed rings" that vanished.

yikes

if you assume the operator to be self adjoint you might be able to get a proof from the functional calculus.
i just doubt that it's true for a general operator

6:00 AM
In "Duistermaat, Kolk, Multidimensional Real Analysis 1. Differentiation", at some point the authors say that for $(x_1,x_2) \in \mathbb{R}^n \times \mathbb{R^n}$ it is $\| (x_1,x_2) \|^2=\|x_1 \|^2+\|x_2 \|^2$. I don't understand if this is a definition (since they never define norm on cartesian product) or it can be deduced from the definition they gave before, that is $\|x \|^2= \langle x,x\rangle$.

if they define a norm on R^n then that's what you'd get if you looked at the norm on R^2n as a cartesian product.

@leslietownes right yeah. I find myself with some bounded perturbation of a self-adjoint, unfortunately

an arbitrary bounded perturbation? do you have kato's book?
is the underlying 'self adjoint' operator unbounded?

well its from a pde so not too ugly. yes, though i'm rather afraid of kato now
yes, unbounded. but on nice L2 sobolev spaces so not CrAzY

i overlapped briefly with kato at berkeley. i saw him in the elevator
if i had a time machine, now seems like a good time to use it

6:05 AM
next thing i know ur gonna say ur a coauthor of simon

no. my former officemate might be, but i'm not

@Gwyn it is both true. You can define the norm on the cartesian product of hilbert spaces as that sum, and also it follows from the definition of a norm on $\Bbb R^N$ with $N=2n$
ooh. on the wikipedia page for discrete spectrum which i just finished reading, they give an example of an operator with no eigenvalues, and the spectrum is {0}
So, no, its not enough. Further the operator is bounded on separable hilbert
for reference its right shift with decay: on $L^2(\Bbb N)$ send $(a_1, a_2, ... )$ to $(0, a_1, a_2/2^2, a_3/2^3, ... , a_{n-1} / 2^{n-1}, ... )$

6:39 AM
@Calvin Thor Thank you!!

6:59 AM
@Gwyn, you're welcome! I should make a minor correction, it is the definition of the euclidean norm on R^N that you need to use. You can't just use any norm to make it match

2 hours later…
8:34 AM
hi, this is more of a sanity check than a question, usually when I come across greens theorem in the plane being stated, its at most for multiply connected domains whose boundaries consist of finitely many disjoint recitifable jordan curves , im trying to digest this version:
as I understand it, is this basically just a routine application of greens theorem viewed as as special case of stokes theorem for manifolds with boundary? Where the manifold in boundary in consideration here is a $C^1$ manifold with boundary that happens to be the closure of an open connected subset of the plane?
is what I just said essentially how one can prove the above?
here im being a little loose with what I mean by 'greens theorem' exactly, assuming $R$ is some domain with boundary that is a disjoint union of say piecewise $C^1$ jordan curves, I know I can use the usual greens theorem for such domains to derive the result in the image
so im referring to the thing in the image as 'greens theorem' as well (although its really a consequence)
hopefully my question makes some sense..

@porridgemathematics yes, its 'just integration by parts'

okay great
so nothing too sophisticated is being said in that image right?
what got me confused is the term $C^1$ boundary, ive not come across it before
and they dont define it before in the book, i just assumed it means the thing is a $C^1$ manifold with boundary
(embedded in $\mathbb{R}^2$)

locally a C^1 graph

right, I think thats equivalent to it being an embedded $C^1$ MFB right?
you can use the charts at the boundary to get the locally $C^1$ graph condition afaik

yes, sounds right. but know that i am blagging and don't know how to actually prove that

8:42 AM
yeah okay, just wanted to make sure im not talking crazy
havent written it down proper either

thanks for the sanity verification :)

np m8

Hi @CalvinKhor,

@opio ello

8:47 AM
Any ways to naturally generalize this $\hat X=\langle x\log x,-y\log y\rangle$ to 3-space?
that's the form of entropy in the components

$(x,y)\mapsto (\hat X;0)$

can I ask why for functional derivative (of $F$) at say $\rho$, people write $\frac{\delta F}{\delta \rho}(\rho)$
have you seen this notation?

just to differentiate it from the other million derivatives
yeah, its basically a directional derivative but you replace vectors with functions which after all, belong to vector spaces

how do you express the notion of flow lines converging to a singular point?
is that the correct verbiage I mean: "flow lines converging"
I'm thinking "sink" possibly
yes it's "source" or "sink"
im not sure if "source" or "sink" imply the vanishing of the vector field though

but look at this definition i.stack.imgur.com/zpUDX.png
they take differential at $\rho$
but why does $\rho$ appear twice in the notation $\frac{\delta F}{\delta \rho}(\rho)$
seems unnecessary
unless im missing something

9:11 AM
@opio its the same as writing $df/dx$ for $f'$, then evaluating at the point $x$. You end up with $\frac{df}{dx}(x)$
e.g. if $x^2$ is the function $f(x)=x^2$, then $\frac{d(x^2)}{dx}(x) = 2x$
and each of the xs on the left hand side is used in a different way

9:24 AM
yes thanks :) @CalvinKhor
I just thought of it in that way, cheers

np :)

0

Suppose $A,B \in M_{n}(\Bbb{R})$ such that $A = \left[C_{1}\middle|\frac{I}{0\dots0}\right], B= \left[C_{2}\middle|\frac{I}{0\dots0}\right]$ , where $A$ and $B$ have different first columns (represented as $C_{1}, C_{2}$). Let $\lambda_{i}, i=1, \ldots, n$ denote the eigenvalues of $AB^2$. Then ...

9:41 AM
Is there atleast a formal way to see why for a curve $\rho(t)$, $\frac{d}{dt}\Big|_{t=0}F(\rho(t))=\int \frac{\delta F}{\delta \rho}(\rho(0))\partial_t|_{t=0}\rho(t)$

what's F?

10:04 AM
name for a mapping that maps flow lines of a vector field X to flow lines of X?
im guessing that if the local flow of $X$ patches toghether to form a foliation of some manifold, then you could view such a mapping as a diffeomorphism

10:45 AM
Hello there, just want to share a video on proving an inequality:

$$\sqrt{x^{3}+y^{3}+1} + \sqrt{y^{3}+z^{3}+1} + \sqrt{z^{3}+x^{3}+1} \ge 2 + \sqrt{2(x^{3} + y^{3} + z^{3}) + 1}$$

1 hour later…
11:47 AM
Since $\sqrt{x}$ is concave, \begin{align}\sqrt{1+a}&\ge\frac{b+c}{a+b+c}\sqrt1+\frac{a}{a+b+c}\sqrt{1+a+b+c}\\\sqrt{1+b}&\ge\frac{a+c}{a+b+c}\sqrt1+\frac{b}{a+b+c}\sqrt{1+a+b+c}\\\sqrt{1+c}&\ge\frac{a+b}{a+b+c}\sqrt1+\frac{c}{a+b+c}\sqrt{1+a+b+c}\end{align} Add.

sqrt is concave! (which is what you use)

fixed
I hate having to wait for video proofs.
does the video show it this way?

12:10 PM
The video shows 2 ways, one uses algebraic manipulation, the other uses Karamata's.
Would you watch and subscribe to help my friend?
@robjohn what do you mean? there is it the video.

1 hour later…
1:41 PM
@robjohn this is so fantastic :-)
Using Jensen’s there. : )

2:00 PM
Hey, I think I've asked this before, but I either forgot or never got an answer: how do you find, or is there a general method to find, $x(t), y(t)$ such that $f(x) = (y\circ x^{-1})(t)$ and the distance along the curve of $f$ increases linearly with $t$?
Ah, I just remembered it was about solving for one of the bounds of integration in an arc length integral.
How do you solve such problems? e.g. solve $s = \int_0^{x(s)} \sqrt{1+(f'(x))^2}dx$ for $x(s)$.

2:58 PM
@Koro actually, just the definition of concavity.
which is Jensen for this situation

jensen's with a sum of diracs, if one must :)

3:15 PM
@CalvinKhor Yes, Jensen on a space of two point masses (that sum to $1$) reverts to the definition of convexity.

is it Jensen like Jacks or Jensen like yaks?

I've always heard it as in Jeep, Jacks, Jungle, Giraffe...

what kind of isomorphism is defined on :
$< v_1 \otimes v_2 +v_2 \otimes v_1 > \to V \otimes V / <v_1 \otimes v_2 -v_2 \otimes v_1 >$
A similiar question, what kind is defined on:
$<v_1 \otimes v_2 - v_2 \otimes v_1 > \to V\otimes V / <v\otimes v>$
Where it is said, that the left side sets split the tensorproduct space into alternating and symmetrical parts in a direct sum, and these are isomorphic to $Sym^2(V), Alt^2(V)$
i was thinking of something like a projection because a projection will always be atleast surjactive on $V \otimes V/ anything$
So for it to be isomorphism we need that the only zero in the kern, so basically, no elements of the set are in the form of the quotient. (which is true to these sets)
I guess i am generally asking how does the isomorphism look like, just like send the element into its aquivalence class and tats it?

3:41 PM
How to show that $\sum_{k=1}^{2n+1}\frac 1{3n+k}< 2/3$?
oh wait, I'll try with integrals first.

@Koro For $n=0$, it is $1$, but I would show it is decreasing and evaluate at $n=1$

if $V$ is a vector space and $U,W$ are subspaces, you can always consider the composition $U\rightarrow V\rightarrow V/W$ of the inclusion followed by the quotient map
this composite is injective iff $U\cap W=0$ and surjective iff $U+W=V$, so an isomorphism iff $V=U\oplus W$
i dont think what you've written is entirely correct; the relevant direct sum decomposition of $V\otimes V$ you wanna look at is the eigenspace decomposition of the involution $V\otimes V\rightarrow V\otimes V$ given by $v\otimes w\mapsto w\otimes v$
you probably wanna work in characteristic $\neq2$ for this

@robjohn $n\in \mathbb N:=\{1,2,3,...\}$
The book has given a very strange solution to this.
$\sum_{k=1}^{2n+1}\frac 1{3n+k}\le\frac 1{3n+1}+\frac{2n}{3n+2}<2/3$
I understand this solution but it seems unnatural.
is there any systematic way to arrive at this?
I got it now. $\sum_{k=1}^{2n+1}\frac 1{3n+k}< \int_0^2 \frac 1{3+x}\,dx\le \int_0^2 \frac 13\,dx=2/3$

3:58 PM
are you sure about the first inequality?

I think yes. I concluded that by plotting graph of $x\mapsto \frac 1{3+x}$
n is fixed.

@Koro It is false for $n=1$
and for $n=2$
The integral is $\log(5/3)$
It is false all the way to $n=100$ and maybe for all $n$

@robjohn no?

Evaluate the sum and the integral.

For $n=1$, we have LHS=$1/4+1/5+1/6$
and this is < 2/3

4:05 PM
the first inequality

Oh

You can't build a valid proof of a true fact on false premises

yes, the first inequality is wrong. :(
$\sum_{k=1}^{2n+1}\frac 1{3n+k}<\int_0^2 \frac 1{3+x} \, dx+ \frac 1{5n+1}$
i.e $F(\cdot,\cdot)$ is the functional, a
how do I write the variational derivative of $F$ at some $\mu$?
$\frac{\delta F}{\delta \mu}(\mu,\nu)$