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Akiva Weinberger
Akiva Weinberger
00:12
Ah… no pensé en español
Akiva Weinberger
Akiva Weinberger
00:48
Are all simply connected open subsets of the plane homeomorphic
I guess yes by Riemann mapping
Ted Shifrin
Ted Shifrin
Well, Riemann mapping omits the whole plane :)
Akiva Weinberger
Akiva Weinberger
I can also say connected open subsets of the plane whose complement has no bounded connected components, right?
(are all homeomorphic)
Riemann mapping feels like a really big hammer for this
Fails in 3D, though, so we need to use 2D essentially somewhere
25
Q: Riemann mapping theorem for homeomorphisms

JaikrishnanHow do you prove to any two simply-connected domains in the plane are homeomorphic without using the Riemann mapping theorem? An elementary proof would be nice.

gt.geometric-topology cv.complex-variables alternative-proof
Found something
Akiva Weinberger
Akiva Weinberger
01:34
Symmetric linear transformations on Hilbert spaces are necessarily continuous, right?
Something something uniform boundedness
Ted Shifrin
Ted Shifrin
I don’t buy that.
Oh, maybe I do. Over my pay grade. Where’s leslie?
leslie townes
leslie townes
if they're everywhere defined, sure. but sometimes the use of the term 'symmetric' is already a hint that maybe the domain is not all of the hilbert space.
Ajay
Ajay
Would the physics tag be applied to this question?
0
Q: What are the speeds of a bus and bike if they are heading in opposite directions? Unsure on how to go about solving for speed here

KaleCities A and B are 70 miles apart. A biker leaves City A at the same time that a bus leaves City B. They travel toward each other and meet 84 minutes after their departure at a point between A and B. The bus arrives at City A, stays there for 20 minutes, and then heads back to City B. The bus...

linear-algebra
Linear algebra should also be removed right?
Coz I get these kind of questions on my physics papers and math papers in school.
Ted Shifrin
Ted Shifrin
Yes, the mis-tagging is exponentially increasing. Complex-geometry is the worst offender, but — gee — linear equations means linear algebra.
Calling that physics does physics a disservice. Just precalculus-algebra.
Ajay
Ajay
Seems about right, thanks.
leslie townes
leslie townes
01:43
we saw a dead opossum on our walk. cause of death could not be determined. it prompted the question, "why is the opossum sleeping like that."
Ted Shifrin
Ted Shifrin
@leslietownes Interesting. Why?
Cuz he’s realllllly tired.
leslie townes
leslie townes
i dunno, some writers only use 'symmetric' for densely defined but not everywhere defined operators. it might be a physics thing where those are the only cases of interest.
Ted Shifrin
Ted Shifrin
But why does symmetry have anything to do with it? There are plenty of ddo in (geometric) analysis.
leslie townes
leslie townes
what is "it"?
Ted Shifrin
Ted Shifrin
Your jumping to the ddo conclusion.
leslie townes
leslie townes
01:50
what? i only know what i've seen. some people use 'symmetric' to mean 'symmetric and not self adjoint' the way some people use 'rectangle' to mean rectangle that isn't a square. a lot of the people who do that work with densely defined operators that are not everywhere defined.
i don't know what geometric analysts do. this might be a math physics thing.
Ted Shifrin
Ted Shifrin
Oh, I’ve never seen that.
Ajay
Ajay
Speaking of science, I have my science exam at 12.
Ted Shifrin
Ted Shifrin
I see the point. I assumed symmetric to be synonymous with self-adjoint, but maybe the point is symmetric ≠ hermitian.
Nuances …
@Ajay good luck!
dc3rd
dc3rd
Was just rewatching one of your lectures Ted and you said something which was probably not meant to be in depth but I took it there. It was about optimization (max/min problems). You said for the constraint expression that we should re-write it to get rid of one of the variables so we can try and make a compact set.
So my mind started firing.... is this why we place such importance on manifolds?
as in if we can write an implicit expression in terms of a grpah we can then optimize whatever functions are of interest over said set?
as in we can "make" the graph be a compact set and then if we had some functions we wanted to do some work over that set we know we can optimize them on it.
leslie townes
leslie townes
02:09
ted: see my now 10-year-old comment on math.stackexchange.com/questions/38387/…
for an amusing anecdote in this vein.
nate's answer gives the usual formulations. but why answer with helpful definitions when you can paste an anecdote.
per nate, some use 'hermitian' to imply bounded, which i am not sure that i have seen that often. but it might be in reed and simon.
copper.hat
copper.hat
02:23
off looking for where Xander issued the language admonition...
ah crapola, it seems to have been removed :-(
dc3rd
dc3rd
@copper.hat Let's just create a new one: HFYR**!!?**$$$
copper.hat
copper.hat
:-)
did you get the Lebesgue number lemming sorted out?
dc3rd
dc3rd
@copper.hat I did......gonna be a joy when I have to reprove it when I do Munkres topology later in the spring....
:(
copper.hat
copper.hat
I found Munkres hard going, not because it was difficult but because I did not get the motivations behind stuff.
dc3rd
dc3rd
I'm using the donut --> coffee mug as motivation and hoping somewhere down the road I'll be struck by the lightning bolt as to why it pertains to Stats...
I mean I could see it in the sense of Measures and things of that nature why it would be important
but now I'm treading in unchartered territory and thinking things align when they don't. Even though I have googled "topology and stats" and its derivatives to see what is out there
geocalc33
geocalc33
02:50
does anyone care about functions of a random vector?
or just functions of a vector
Calvin Khor
Calvin Khor
03:02
@geocalc33 of course, why do you ask?
@dc3rd a few years ago i was told that the "next hot thing in data analysis" was using topology to do dimension reduction or smth
geocalc33
geocalc33
@CalvinKhor I have a definition that I think is not quite correct but very close
dc3rd
dc3rd
@CalvinKhor Kill off all degrees of freedom.
Calvin Khor
Calvin Khor
@geocalc33 definition of?
geocalc33
geocalc33
@CalvinKhor I have a purely statistical definition of a curve as a function of a random vector, and I want to try to either understand that definition or turn it into a regular vector valued function definition
I attempted to translate the statistical definition into the non-random vector type definition
Calvin Khor
Calvin Khor
im too scared to ask what a statistical definition is :)
geocalc33
geocalc33
03:12
Well here's what I have so far: Let $L^1_+$ the set of all $1$-dimensional nonnegative vectors $X$. Let $C^{\infty}$ be the class of all smooth functions from $\Bbb R_+$ to $[0,1].$ Let $f \in C^{\infty}$ s.t. $f:\Bbb R_+ \to [0,1].$ Define the $\text{PRO}$ curve as $\text{PRO}(X)=\bigg\{\bigg(\int f(x)~dx, \int xf(x)~dx:f \in C^{\infty}\bigg)\bigg\}$
Calvin Khor
Calvin Khor
that's....not what I would call a curve, but ok
geocalc33
geocalc33
maybe I messed up trying to convert the stat definition
Let $L^2_+$ the set of all $2$-dimensional nonnegative random vectors $X = (X_1, X_2)^⊤$ with finite and positive marginal expectations, and let $Ψ^{(2)}$ the class of all measurable functions from $\Bbb R^2_+$ to $[0, 1].$ Then $L(X)$ of the random vector $X$ with joint CDF $F$ is:

$$L(X)=\bigg\{\bigg(\int \psi(x)dF(x), \int \frac{x_1\psi(x)}{E(X_1)}dF(x),\int \frac{x_2\psi(x)}{E(X_2)}dF(x):\psi \in Ψ^{(2)}\bigg)\bigg\}. $$
I really have no idea how this forms a manifold, a convex surface, as a subset of the unit 3-cube, where the surface includes $(0,0,0)$ and $(1,1,1).$
Ted Shifrin
Ted Shifrin
@dc3rd You’ve definitely garbled this.
@leslietownes it gave me a headache, but very interesting post.
geocalc33
geocalc33
03:29
I'm going to email the author
dc3rd
dc3rd
@TedShifrin THen let's not make hay about it anymore
Ted Shifrin
Ted Shifrin
Just dried out straw?
dc3rd
dc3rd
Just made caccio e pepe actually....nothing dry here
geocalc33
geocalc33
03:56
say $f_n(x)=x^n$ is a particular subclass of all smooth functions. and say you have an operation $x^n \times x^k =x^{n+k}.$ If you have a linear functional $x^n \mapsto n$ for $n \in \Bbb R_{\ge 0}$ can you associate the subclass $f_n$ to the monoid $(\Bbb R \cap (0,1),\times)$ due to the fact that the linear functional sends each curve to an element in the monoid?
Ted Shifrin
Ted Shifrin
That ain’t no linear functional.
geocalc33
geocalc33
I guess I could do $\int_0^1 x^n ~dx \mapsto \frac{1}{n+1}$
geocalc33
geocalc33
04:29
*$f_n(x) \mapsto \int_0^1 x^n ~dx$
 
1 hour later…
Calvin Khor
Calvin Khor
05:40
@geocalc33 $\psi=0$ gives (0,0,0) and $\psi=1$ gives (1,1,1). Since $x_idF/E(X_i)$ is a probability, its a subset of the unit cube. Convexity looks like it follows trivially from the definition. $L(X)$ is also clearly closed, and (mumble something about boundary of closed convex set can't be too ugly) so its a (topological?) manifold
@geocalc33 can't understand anything here
Ajay
Ajay
06:22
Exam is over now.
Paper was very easy compared to past papers.
Prithu biswas leftmse
Prithu biswas leftmse
06:49
@TedShifrin Yes you are right! In the later edition spivak does have a proof. I will take a look at it. Thanks Ted =)
 
2 hours later…
porridgemathematics
porridgemathematics
08:38
hi, does anyone knowhow to proceed here? Im trying to prove that for a compact complex manifold $X$, and for a holomorphic line bundle $E$ over $X$, $H(X,E)$ the space of holomorphic sections of $E$ is finite dimensional using montels theorem, so I put some hermitian metric on $E$ and defined the $L^2$ inner product as usual, now I want to show the space of holomorphic sections is closed
So I reached a point where I can show that if $s_k \in H(X,E)$ converge in $L^2$ to $s$, then there is some open set $U$ over which $E$ is holomorphically trivial, and $X \setminus E$ is a null set, and $s$ is equal almost everywhere to a holomorphic section on $U$
I can make this $U$ an open neighbourhood of any point of $x$.. so I reach a weird conclusion where $s$ is kind of holomorphic, but I dont know how to go from here to $s$ is equal to a holomorphic section almost everywhere on $X$
porridgemathematics
porridgemathematics
08:50
maybe this works, to show my holomorphic section extends to all of $X$, I just repeat the procedure for some point $x$ not in $U$, then taking smaller and smaller balls centered at $x$, over which $E$ is holomorphically trivial, these balls have nonempty intersection with $U$ since $U$ is of full measure, and I have that my old section is equal to this new one over the intersections of these balls decreasing to $x$ with $U$, which should mean I can extend it to $x$ with no compatibility issues
since $x$ was arbitrary, its really the restriction of some global holomorphic section to $U$
Mad Spaces
Mad Spaces
09:08
quick question. if $ h $ sesquilinear $ V $ complex vector space $ \varphi \circ \varphi* $ hermetian obviously why is it then that $ \varphi \circ \varphi* $ also positive definit ? say that $h_{\varphi \circ \varphi*}= h((\varphi \circ \varphi*)(v),v) \geq 0 \forall v$\0
porridgemathematics
porridgemathematics
what is $\phi$ and $\phi^{\ast}$?
Mad Spaces
Mad Spaces
endmorphisms. and the adjunct
porridgemathematics
porridgemathematics
if $\phi$ was just the zero map, wouldn't we have $h((\phi \circ \phi^{\ast})(v),v)) = 0$ for all $v$?
Mad Spaces
Mad Spaces
Yes
porridgemathematics
porridgemathematics
oh okay, so just semi-positive definite I guess is enough
Mad Spaces
Mad Spaces
09:20
and why is it that? the sesquilinear form $h$ has no "to my knowledge" restriction to be positive.,here is h_phi circ Phi * also hermetian since the function is hermetian
porridgemathematics
porridgemathematics
i thought positive definite usually refers to a bilinear form satisfying $\langle v,v \rangle > 0$ for nonzero $v$
Mad Spaces
Mad Spaces
the hermetic sesquilinear form here is bilinear. (but anti linear in second component )
porridgemathematics
porridgemathematics
right, if I understand your question correctly, you want to show that $(v,w) \rightarrow h((\phi \circ \phi^{\ast})(v),w)$ is a positive definite hermitian sesquilinear form?
Mad Spaces
Mad Spaces
no it has to be the same vector, but other than that yes
Hermetian is obvious and easy but i am not getting the positive part
porridgemathematics
porridgemathematics
eh, actually im confused, to define $\phi^{\ast}(v)$ i'm assuming you're identifying $v$ with something in $V^{\ast}$, how are you doing that?
Mad Spaces
Mad Spaces
09:26
No we are not talking about the dual space.
porridgemathematics
porridgemathematics
isn't $\phi^{\ast}$ defined on $V^{\ast}$?
Mad Spaces
Mad Spaces
$\varphi^* $ is such that $ h(\varphi(v),w) = h(v,\varphi^*(w)$ )
porridgemathematics
porridgemathematics
09:38
if $h$ is just any sesquilinear form, and you define $h = 0$, by your above definition, can't $\phi^{\ast}(w)$ be anything in $V$ at all?
how can you show that $\phi^{\ast}(w)$ is well defined by that condition without further assumptions on $h$
Mad Spaces
Mad Spaces
this is the definition of an adjunct function to varphi
porridgemathematics
porridgemathematics
okay, so $\phi^{\ast}$ is just any adjunct, it doesn't need to be well-defined?
Mad Spaces
Mad Spaces
there is a proof to why it is one only one with this attribute.
but this is not my problem lol
porridgemathematics
porridgemathematics
well that proof should assume $h$ is not just identically zero, then
Mad Spaces
Mad Spaces
Sure assume that
porridgemathematics
porridgemathematics
09:43
if all $h$ needs to be is sesquilinear, then $\phi^{\ast}$ does not make sense.
and even then ,im not sure if $\phi^{\ast}$ makes sense if you assume $h$ is sesquilienar nonzero
im guessing you need $h$ to at least be non-degenerate
Mad Spaces
Mad Spaces
Nvm..
Emil Haugen
Emil Haugen
09:59
Anyone here an expert on Laguerre-functions? I'm wondering if it's possible to find a self-adjoint (e.g. Sturm Liouville) operator such that the eigenfunctions are the generalized Laguerre functions (https://encyclopediaofmath.org/wiki/Laguerre_functions)?

I did find this result: https://arxiv.org/pdf/hep-th/0109028.pdf (see equation 16). But here they use Laguerre functions that are a basis for the weighted space L^2(0,\infty, dx/x) whereas I would like a similar result in the unwieghted space L^2(0,\infty).
 
1 hour later…
PM 2Ring
PM 2Ring
11:03
There's a recent question on Physics.SE asking about "physics theorems". physics.stackexchange.com/q/707527/123208 There's a good answer there that explains that theorems are actually mathematics, but there are some weird opinions floating around on that page. IMHO. ;)
Adam Warlock
Adam Warlock
11:33
How to calculate the ratio of maximum and minimum value of sum of mod of 4 numbers given sum of their squares is 1.
Any suggestions or links?
Mats Granvik
Mats Granvik
12:21
$$1-x^{\log (2)}+x^{\log (3)}-x^{\log (4)}+x^{\log (5)}-x^{\log (6)}+x^{\log (7)}-x^{\log (8)}+x^{\log (9)}-x^{\log (10)}+\text{...}(-1)^{n+1}x^{\log (n)}=0$$ @AkivaWeinberger
(*Mathematica start*)
x = N[Exp[-ZetaZero[1]/10], 100]
Sum[(-1)^k*x^(Log[k]*10), {k, 1, Infinity}]
(*end*)
49
A: Unexpected examples of natural logarithm

Akiva WeinbergerYour first point can be generalized. Write $[a_1,a_2,a_3,\dots]$ for $\sum a_n/n$. You wrote:$$[\overline{1,-1}]=\ln2.$$(The bar means repeat.) Then we also have:\begin{align}[\overline{1,1,-2}]&=\ln3,\\ [\overline{1,1,1,-3}]&=\ln4,\end{align}and in general:$$[\overline{\underbrace{1,1,\dots,1}_{...

Replace 10 with 100,1000,10000,... and so on, in Exp[-ZetaZero[1]/10] and (Log[k]*10)
10,100,1000,... and so on must be greater than or equal to Im[ZetaZero[j]]/Pi
j=1,2,3,4,5,...
oeis.org/draft/A191904
Lucas Henrique
Lucas Henrique
12:38
i don't want to be rude but you just threw a lot of information with no further information on what you want to communicate...
Mats Granvik
Mats Granvik
12:48
@LucasHenrique I was thinking something along the lines $x^{\log(2)}=x^{1-\frac12+\frac13-\frac14+\cdots}=x^{1} \cdot x^{-\frac12} \cdot x^{\frac13} \cdot x^{-\frac14} \cdots$ and how it would possibly relate to the product recurrence in OEIS table A191904.
Wojciech Kulma
Wojciech Kulma
13:36
Hi folks! Anyone acquainted with Galton-Watson processes here?
Calvin Khor
Calvin Khor
@WojciechKulma many years ago.....
Semiclassical
Semiclassical
@PM2Ring when it comes to settled areas of physics (eg mechanics or thermal physics) then I agree
I’m less sure when it comes to QFT level stuff
(Though I guess from the math POV it would be more “physics conjectures” rather than physics theorems)
Calvin Khor
Calvin Khor
Can I roll a ball at prescribed velocity to 'prove' the fundamental theorem of calculus
with displacement = integral of velocity
Wojciech Kulma
Wojciech Kulma
@CalvinKhor I'm sruggling with an apparently simple excericise. Given is a crtictical GW process $Z_n$ with probability generating function $G(z) = p_0 + p_2z^2 + p_4z^4. The process is critical so E(Z) = 1. For the family trees t1 and t2 as on the picture following holds: $5P(T=t1) = 4P(T = t2 > 0)$

https://imgur.com/a/URmVgZc
now I got to obtain the equalities for p coefficients, $p_0$, $p_2$ and $p_4$

I obtained that $EZ = 1 = G'(1) = 2z p_2 + 4zz^3p4$
and tbh I have no idea where how to proceed :<
PM 2Ring
PM 2Ring
14:07
@Semiclassical My take is that physics has theories, which can (potentially) be falsified. Mathematics has theorems, which can be proven from axioms. The phrase "physics theorem" IMHO only makes sense if it's shorthand for "mathematical theorem used in physics", eg Noether's theorem. The topic of that question, which asks if you can prove mathematical things Pythagoras' theorem using physics seems misguided to me.
OTOH, there has certainly been a lot of fruitful interplay between mathematics & physics over the centuries, and many of the greatest physicists have also been great mathematicians, eg Archimedes, Newton, Euler, Gauss.
And although I'm not entirely happy with what Mozibur's saying, I have to concede that my attitude is relatively modern, and the ancient Greeks (probably) didn't make a strong distinction between physics & maths, or between geometry & geography.
@CalvinKhor I don't think so. You can certainly illustrate mathematics with physical demonstrations & observations, though.
oneofvalts
oneofvalts
14:58
@Semiclassical hey semi!
Calvin Khor
Calvin Khor
15:49
@WojciechKulma sorry, absolutely nothing is coming back to me
Wojciech Kulma
Wojciech Kulma
16:07
@CalvinKhor no probs, thanks for looking into it :)
robjohn
robjohn
@PM2Ring or between astronomy and astrology.
Wolgwang
Wolgwang
Why am I getting $\lim_{x\to0} \dfrac{\tan(x)-\sin(x)}{x^3}=-\dfrac16$ by using expansion?
soupless
soupless
Just some random question. What kind of statistics do you need to analyze an election result to check for anomalies?
Wojciech Kulma
Wojciech Kulma
@soupless I'd start with a T-test
Wolgwang
Wolgwang
@Wolgwang $$\lim_{x\to0} \dfrac{(x-\frac{x^3}{3}+\frac2{15}x^5\cdots )-(x+\frac1{3!}x^3\cdots)}{x^3}$$
Ted Shifrin
Ted Shifrin
16:23
@Wolgwang Why not?
robjohn
robjohn
@Wolgwang series for $\sin(x)$ needs some work
Same for $\tan(x)$
tan does not oscillate, but sine does. Also, check your arithmetic.
PM 2Ring
PM 2Ring
@robjohn Good point. ;)
robjohn
robjohn
@Wolgwang using that limit, I get $-\frac12$, but the actual value should be $\frac12$.
PM 2Ring
PM 2Ring
Speaking of sine, a question from 2014 about calculating arcsin got bumped a day or so ago, so I decided to write an answer. I got a nice solution (adequate for 5 sig figs) using a pair of cubic minimax polys. math.stackexchange.com/a/4447225/207316
robjohn
robjohn
@PM2Ring I'll take a look. I came up with a few rational approximations that were nice. I'll try to find them.
Oh, I believe mine were for sine.
PM 2Ring
PM 2Ring
16:34
@robjohn We had a conversation about sine approximations a few months ago. I think you may have also mentioned arcsin. I've done lots of stuff with arctan (mostly in connection to Machin-like formulae for pi), but I haven't played much with arcsin.
Sine is relatively benign. That vertical in arcsin as you approach arcsin(1) is a killer for poly approximations.
Ted Shifrin
Ted Shifrin
@robjohn I've never observed that before. Do you have an a priori argument that all the coefficients in the $\tan$ series are positive?
PM 2Ring
PM 2Ring
All the Machin-like identities you can eat: machination.eclipse.co.uk COMPUTING PI: LISTS OF MACHIN-TYPE (INVERSE COTANGENT) IDENTITIES FOR PI/4
There's another old-school website for you, @leslie.
Wolgwang
Wolgwang
16:54
@TedShifrin Because while simply solving by trigonometric formulae, I am getting 1/2. :(
@robjohn Oops! I missed a - before x^3 term.
robjohn
robjohn
@TedShifrin It seems pretty simple to see that all the coefficients of $P_n$ are non-negative:
$$
\frac{\mathrm{d}^n}{\mathrm{d}x^n}\tan(x)=P_n(\tan(x))\\
\frac{\mathrm{d}^{n+1}}{\mathrm{d}x^{n+1}}\tan(x)=P_n'(\tan(x))(1+\tan^2(x))\\
P_{n+1}(x)=P_n'(x)\left(1+x^2\right)\\
P_0(x)=x
$$
Since $P_n(0)\ge0$, all the derivatives of $\tan(x)$ at $x=0$ are non-negative
I was actually only talking about in the approximation when I made my comment, but that was a challenge I couldn't refuse.
Ted Shifrin
Ted Shifrin
I'm a bit confused by this. $P_n$ isn't actually the $n$th degree T.P.; you're thinking of the $n$th derivative of $\tan$ as a function of $\tan$, which is unusual.
robjohn
robjohn
@TedShifrin The derivative of $\tan(x)$ is $\sec^2(x)$ which is $1+\tan^2(x)$
Ted Shifrin
Ted Shifrin
Yes, that I am aware of.
So you're observing that, inductively, all the derivatives are in fact functions of $\tan$.
I can guarantee I wouldn't have thought of this.
And I've never wondered (before your comment) if all the Taylor coefficients were positive. Mathematica suggested it (up to order 25).
But, just as a word to explain why I was a bit confused: Your first sentence used $P_n$ for the Taylor polynomial itself of $\tan$.
robjohn
robjohn
17:09
@TedShifrin Yes, but that guarantees that the constant coefficient of $P_n$ is non-negative.
Ted Shifrin
Ted Shifrin
Well, to me $P_n$ is the $n$th degree T.P. Your notation is unusual to me.
robjohn
robjohn
Sorry, to confuse. I intended it to be defined in the first equation. I meant the first statement to be an observation verified after the induction.
I modified the second statement to hopefully be clearer, but maybe not.
It was a spur of the moment proof, so I apologize for the roughness.
HERO
HERO
if
Ted Shifrin
Ted Shifrin
Anyhow, I get it now. Very clever argument. +1 to @robjohn Thanks.
HERO
HERO
if two variables x and y are independent then should you be able to parametrize them at all?
robjohn
robjohn
17:19
@HERO what do you mean by parametrize them?
Ted Shifrin
Ted Shifrin
My question precisely.
robjohn
robjohn
write one in terms of the other?
or write both as a function of one variable?
HERO
HERO
express them as functions of another independent variable t
Ted Shifrin
Ted Shifrin
That makes no sense.
What do you mean when you say "two variables are independent"?
If they're each a function of $t$, then implicitly there is a function relating them, so they can't be independent.
robjohn
robjohn
what he said
Ted Shifrin
Ted Shifrin
17:21
But the whole notion of "independent variables" is murky.
robjohn
robjohn
@HERO so, no ;-)
HERO
HERO
user image
i was studying multivar calc where x and y were denoted as independent variables of f(x, y) but now they go on to define the chain rule for situations where x and y are differentiable functions of t... at this point how can i think of them as independent variables though? there functions of t!
do i take it that x and y arent independent now, then? but look at the wording it says "chain rule for functions of two independent variables"
Ted Shifrin
Ted Shifrin
Think about the $xy$-plane with coordinates $(x,y)$. Now you take a curve in the plane, given by $x=x(t)$, $y=y(t)$, and differentiate $f(x(t),y(t))$.
You're restricting the function to a curve, so on that curve $x$ and $y$ are no longer "independent."
HERO
HERO
oh
OH
i think i get it now
thank you mister
Ted Shifrin
Ted Shifrin
Sure thing.
robjohn
robjohn
17:47
@Ted my random downvoter has struck again.
we all have them, so don't get a swelled head ;-)
Semiclassical
Semiclassical
@soupless it depends a lot on the particulars of the election. see for instance youtube.com/watch?v=etx0k1nLn78
statistical tests depend on assumptions, so if the scenario doesn't satisfy the assumptions of a test then that's it's not useful for that purpose
Ted Shifrin
Ted Shifrin
@robjohn Mine went back into hiding after a few days of activity.
Semiclassical
Semiclassical
so here's something silly that's been bothering me, mostly b/c i feel like the way i'm saying it is too imprecise
so i'll do a boring example
Suppose I want all solutions to the problem $-y''(x)=\lambda y(x)$ subject to periodic boundary conditions on $[0,2\pi)$. (By solution I mean both the choice of $\lambda$ and $y(x)$.)
obviously I can write down the solutions as $\lambda = m^2$ where $m$ is an integer, with functions labelled as $y_m (x)=e^{i m x}$. So the usual eigenvalues/eigenfunctions for this problem.
but i could instead use the real basis $\{1,\sin x,\cos x,\sin 2x,\cos 2x\}$ etc. and in that case i can further split up the basis into even solutions and odd solutions
What i want to say is that i could have done this from the start: namely, solve $-y''(x)=\lambda y(x)$ with periodic boundary conditions and require that $y(x)$ is even
(or, equivalently, periodic boundary conditions + $y'(0)=0$)
then do the same for odd solutions $(y(0)=0)$
Is there a better way to say that?
Derivative
Derivative
18:12
0
A: The number of integer points on the curve $(7x-1)^2+(7y-1)^2=n$

sirousComment: We may use following equation as a particular case: $$(a-1)^2+(a+1)^2=b^2+1\space\space\space\space\space\space(1)$$ This equation can have infinitely many solution; if a and b satisfy this equation then considering following identity we can have subsequent solutions: $$(2b+3a-1)^2+(2b+3...

does anyone want to help me understand this?
I don't see how he answered the question
Semiclassical
Semiclassical
@Derivative looking at it, i wouldn't be surprised if all they've shown is how to find some solutions rather than being certain they've found all solutions
Derivative
Derivative
that's what I'm thinking
Semiclassical
Semiclassical
in fairness, doing this for arbitrary $n$ sees pretty tough
Derivative
Derivative
I have one idea that "solves" it but it's rather horrible
I'll post it as an answer sometime
leslie townes
leslie townes
threats will get you nowhere
Koro
Koro
18:27
:)
Koro
Koro
18:38
is there a function whose third derivative exists everywhere but nowhere continuous?
Ted Shifrin
Ted Shifrin
18:52
Is there such for the first derivative?
Koro
Koro
I meant a function $f$ whose third derivative $f^{(3)}$ exists such that $f^{(3)}$ is nowhere continuous.
Ted Shifrin
Ted Shifrin
I understand. My question stands.
cfp
cfp
Hi. I'd love to draw some attention to this question: https://math.stackexchange.com/questions/4447440/uniqueness-of-the-solution-to-a-certain-set-of-integral-equations-from-moment-c
I will post a bounty once it's eligible.
I've been bashing away at this and related questions for weeks and getting nowhere.
Koro
Koro
I can't come up with an example where f is differentiable but f' is nowhere continuous.
Ted Shifrin
Ted Shifrin
In that case, work on this, not the other. I asked you a question a long time ago which is relevant.
Koro
Koro
19:00
If f' is nowhere continuous then f' is not integrable. I think I should think on this also.
Ted Shifrin
Ted Shifrin
If you take the pointwise limit of a sequence of continuous functions, is it continuous somewhere?
Koro
Koro
I think yes.
Ted Shifrin
Ted Shifrin
Do you see why this is relevant?
Koro
Koro
is it because if $g_n(x):=\frac{f(x+1/n)-f(x)}{1/n}$ and f is differentiable then $g_n$'s are continuous and therefore $\lim g_n$ should be continuous somehwhere?
Ted Shifrin
Ted Shifrin
Yup. So this result is what you should be working on!
Koro
Koro
19:12
and noting that $\lim g_n=f'$ (pointwise), we are done.
fixed.
:)
robjohn
robjohn
@Koro Only in the Riemann sense.
Not necessarily so in the Lebesgue sense
Koro
Koro
I was thinking of integrating Conway base 13 function 3 times but then I realized that it is not integrable (in Riemann sense).
robjohn
robjohn
I am not sure, but Darboux's Theorem might be useful.
unfortunately, I have to go run some errands. BBL
Ted Shifrin
Ted Shifrin
19:31
Darboux tells you that there are no jump discontinuities for the derivative, but ….
Akiva Weinberger
Akiva Weinberger
Is there a name for the graph whose vertices are the integers and the edges join adjacent integers
My instinct is to call it a "doubly-infinite path graph"
(to contrast with just the positive integers which would be singly-infinite)
Ted Shifrin
Ted Shifrin
An (infinite) tree with two ends?
Akiva Weinberger
Akiva Weinberger
In any case, suppose a graph can be edge-partitioned into $n$ doubly infinite path graphs. Is that $n$ unique?
(This is a simpler version of my older question)
(I don't know what the simplest counterexample is)
Ted Shifrin
Ted Shifrin
I know nothing.
leslie townes
leslie townes
19:48
Darwho?
Akiva Weinberger
Akiva Weinberger
20:03
The set of all places on earth experiencing sunset and the set experiencing sunrise join up into a circle
leslie townes
leslie townes
round earther has entered the chat
Akiva Weinberger
Akiva Weinberger
20:31
Decided to post my answer
0
A: Can you tell how many open disks are added together?

Akiva WeinbergerHere's a counterexample. This was as simple as I could make it; I don't know if there's a simpler solution. In the first image, you have two open sets (red and black); in the second, you have four (red, black, blue, and green). Their sums are identical, including where the overlaps are. Note, b...

Xander Henderson
Xander Henderson
20:46
Because this has always bothered me, I worked out why the catenary describes a hanging chain this morning. It is silly that I never bothered to do this earlier in my career, but I feel like I am a better person for finally having figured it out now.
Ted Shifrin
Ted Shifrin
Good. I have that as an exercise in the first section of my diff geom text.
Xander Henderson
Xander Henderson
@TedShifrin It is a fun exercise.
And working through it probably doubled what I know about physics (I don't know very much physics at all).
Ted Shifrin
Ted Shifrin
Not much physics here … just a Newton’s law or two.
leslie townes
leslie townes
21:02
also the thing for a chain with equally spaced weights.
it couldn't be too hard if the ancient greeks could figure it out. phbthbht
Xander Henderson
Xander Henderson
@leslietownes Didn't say it was hard, only that I hadn't ever bothered to work through the details before.
@TedShifrin Yeah, that's, like, all the physics I know. :D
(I took a semester of physics in high school in 1997...)
leslie townes
leslie townes
there's a ton of stuff in that area that is really fun to work out. you should do more.
Xander Henderson
Xander Henderson
@leslietownes Probably.
Oh, I also took a quarter of quantum mechanics in grad school. But that was just funky analysis.
leslie townes
leslie townes
another classic is the pendulum. say the chain is length R, mass whatever. how long does it take to go from one side to the other, when pulled away from equilibrium by angle theta and let go. it's an integral, i think, but you can approximate it pretty well.
yeah, QM in grad school doesn't count.
my unbounded operators class was cross-listed as QM, i think, or something equivalent.
people got lazy.
Xander Henderson
Xander Henderson
Heh.
leslie townes
leslie townes
21:05
at some point the college was just like "whatever lol nerds in this class"
@PM2Ring weirdly one of my first interactions with the web, in an academic setting, was pulling machin-like identities off of sci.math and using them to improve a program i wrote for the ti-85 calculator.
robjohn
robjohn
@XanderHenderson High school in 1997... gads I feel old.
Xander Henderson
Xander Henderson
Heh.
leslie townes
leslie townes
robjohn: the new generation is here, and there's nothing you can do about it, old man!
you don't have to like it! slicks back hair plays loud jazz music
robjohn
robjohn
I can deny and hide in 80s movies...
Gwyn
Gwyn
Is this a correct proof of the pigeonhole principle? "Suppose by contradiction that n boxes contain exactly one object. Then n+1, n+2, ..., n+k objects , with k>=1 integer, must fill n+1, n+2, ..., n+k boxes respectively. This contradicts the hypothesis that there are n boxes."
leslie townes
leslie townes
21:15
time to ride my automobile to the malt shop
robjohn
robjohn
@leslietownes make sure the rumble seat is open
leslie townes
leslie townes
and the running board
robjohn
robjohn
of course!
leslie townes
leslie townes
xander and i actually overlapped in high school. i graduated in 97 :D
robjohn
robjohn
BAH!!
leslie townes
leslie townes
21:17
this is why i use up-to-the-minute slang, like "malt shop"
robjohn
robjohn
I graduated in 77
Koro
Koro
@leslietownes I was born one year before that :D
robjohn
robjohn
gives Koro a binkie
leslie townes
leslie townes
koro: you and my daughter (b. 2018) can have some conversations i guess
Xander Henderson
Xander Henderson
@leslietownes I graduated in 1999.
robjohn
robjohn
21:19
My son was born in 92
Xander Henderson
Xander Henderson
@robjohn My sister was born in 96, I think.
Might have been 95?
@robjohn I have siblings born in 83, 86, 89, and 96.
robjohn
robjohn
Big separation there.
Koro
Koro
haha
my sister was born in 99.
robjohn
robjohn
@Koro You looked up "binky"?
leslie townes
leslie townes
koro how did that job stuff ever work out?
Koro
Koro
21:20
@robjohn yes :)
leslie: I have not resigned from the job yet. I have already shifted to the new location. Saturday is also a working day here and Sunday too sometimes but I get paid for this. I'm awaiting result of an exam. If I pass, I'll resign from the job. If not, I think I'll still resign and devote me more time to prepare for that exam. :)
leslie townes
leslie townes
nice :)
particularly, getting paid for saturday and sunday. this is a must. :D
robjohn
robjohn
@Ted: when you get back, I just ran across this question. I wonder if it can be solved the same way I did for $\tan(x)$.
Koro
Koro
there is office in time but there is no out time. I've been leaving office around 10 pm ever since I shifted to this location 😑
$\lim_{a\to \infty} \int_0^1 a^x x^a\, dx$
@CalvinKhor: I tried MVT here as you suggested for the other limit (involving integral) the other day but no luck :(
Ted Shifrin
Ted Shifrin
21:42
@robjohn Very interesting.
@Koro Have you recalled the context in which I posed you that question a while ago (about the continuity of a pointwise limit of continuous functions)?
Xander Henderson
Xander Henderson
@robjohn I have siblings born in 83, 86, 89, and 96.
The baby was an "oopsie".
Ted Shifrin
Ted Shifrin
My one sister was 8 years after me. I think my parents finally decided that they'd risk a second after dealing with the terror that was I.
leslie townes
leslie townes
i think that ted secretly identifies with munchkin.
Ted Shifrin
Ted Shifrin
I think my parents would have thrown me into the SF Bay if I had behaved like that.
leslie townes
leslie townes
envy.
i forgot to tell you the best. yesterday my daughter yelled at my wife upon being picked up from day care. her crime? not having brought a piece of cake with her.
she had this leftover chunk of cake in the fridge we had all forgotten about, until then.
Ted Shifrin
Ted Shifrin
21:51
One more yell and you get no cake!
The guillotine for munchkin.
leslie townes
leslie townes
the worst of it was, it was my job to offer the cake that morning for breakfast, but i forgot to.
so my wife was catching the heat for my omission.
Ted Shifrin
Ted Shifrin
@robjohn If you feel old, what does that make me?
The daycare people were probably appreciative of your having forgotten. Can you imagine munchkin with that much sugar?
leslie townes
leslie townes
i'm not saying ted's old, but some of his books are out of copyright, and not for failure to comply with the formalities of the 1909 act.
Xander Henderson
Xander Henderson
Ha!
Ted Shifrin
Ted Shifrin
97 was about ten years after I got tenure.
leslie townes
leslie townes
21:54
that's why they call it ten-year.
Ted Shifrin
Ted Shifrin
Yup.
Xander Henderson
Xander Henderson
That tracks.
leslie townes
leslie townes
ted: one time in a cafe we let her eat an entire donut and it was like dealing with a person who was high on something illegal.
Ted Shifrin
Ted Shifrin
The train just smashed.
Yes, I think you told us about the doughnut and I made the same comment then.
leslie townes
leslie townes
she kept dancing. we were very close to being kicked out of the cafe. i could feel it.
Ted Shifrin
Ted Shifrin
21:56
Wow. I'm looking at the exercises I wrote in 1980 when I taught complex manifolds/geometry. I sure knew/figured out a lot of stuff back then ... that I no longer can do.
leslie townes
leslie townes
1980 like when i was born 1980?
ahaha, you really are old.
i had an accelerated version of this when i left math. most of what i did in 2005 is incomprehensible to me now. i have vague ideas, that's it.
it was all gone by 2015.
Akiva Weinberger
Akiva Weinberger
I remember my high school graduation like it was five years ago
which is bad, because it actually was four years ago. I just have a bad memory
robjohn
robjohn
@TedShifrin epic?
leslie townes
leslie townes
boo
robjohn: or monic
Ted Shifrin
Ted Shifrin
Maniac?
leslie townes
leslie townes
22:04
as foretold in the prophecies of hall and oates
Akiva Weinberger
Akiva Weinberger
Is elder the superlative of eld
Ted Shifrin
Ted Shifrin
Comparative, you mean?
Akiva Weinberger
Akiva Weinberger
Ah, yes, of course. That would be eldest
which actually is a word
0
Q: Decomposing a graph into line graphs

Akiva WeinbergerBy a line graph, I mean a graph isomorphic to the graph whose vertices are $\Bbb Z$, the integers, and whose edges connect adjacent integers. (I suppose a more "proper name" would be a doubly-infinite path graph? As opposed to a singly-infinite path graph, which resembles a ray. If there's anothe...

graph-theory
So far the only values of $n_G$ I know of are $\{1+k\}$, $\{2+k,4+k\}$ for $k\ge0$

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