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12:26 AM
another version that avoids the notational issue of 'n' in an interesting way. i don't think it would make sense on an in-class quiz. math.stackexchange.com/questions/2775599/…
one from 2012. math.stackexchange.com/questions/226347/…. i like distinguishing the index in the affine combinations from the dimension of the space. i don't like only implicitly defining 'affine combination' in the problem statement.
what a weirdly rich tapestry of different ways of posing essentially the same problem.
 
Interesting that I missed all the previous ones.
Talk about duplicate posts!
Lee’s proof is identical to mine, of course.
 
1:11 AM
@leslietownes yeah, I got that, Thanks :)
 
koro: can you prove that there are uncountably many non-isomorphic subgroups? :)
some infelicity of language there but i hope the meaning is clear
 
You mean there are subgroups that are not finitely generated?
 
i sure hope so.
 
Darn.
 
1:59 AM
@leslietownes on an entirely different note, i'm pleased that "state" means the same thing (when translated into C* algebras) in physics as math
it's nice to not have to use different words
 
You’re in quite a state, Semiclassic.
 
heh
an excited state, perhaps
it looks like i may be teaching a class of my own next semester on quantum computing
 
Oh, wow. Not just a standard upper division course.
 
which is simultaneously a very exciting opportunity and a waterfall of "oh shit" moments
well, this one would actually be aimed at non-physics people
 
I know nothing thereabout.
 
2:02 AM
college of science and engineering
the main challenge is how tf to pick the math level
 
Ah, interesting. Better pay than lab TA?
 
Grrr
 
it's not something i'm doing for the money
 
Still a challenge and good on your resume.
 
2:04 AM
yup
 
What are the prerequisites?
 
that's the big question right now
it's not a pre-built course
we're having to put it together before spring
 
You get to decide.
 
right
 
At least some discrete math? Hmm @ linear slgebra
 
2:05 AM
yes. it's hard to envision teaching a class like that without "some" linear algebra, if only at the level of "matrices multiply each other and vectors"
one would obviously prefer more than that
 
That’s usually in discrete math.
 
@leslietownes does this require a proof? I thought that since every finitely generated subgroup of Q is cyclic and total subgroups of Q are uncountable so set of non finitely generated subgroups is uncountable - countable which is uncountable.
 
i probably can't assume most people will come in knowing how to diagonalize matrices
 
Nope
 
not a small task
one of my TA assignments this semester is to assist / grade for the physics version of a QC course
so i'm definitely going to be mining that to some extent
but have to go much slower
 
2:07 AM
Unless linear alg is a prerequisite, which means they have to go through several calc. Our quantum computing course, joint math/Cs/phys,did have serious prereq.
 
right. that to me is the big point to hash out
 
@Semiclassical hi
 
have noticed my ping related to a question/
 
saw it, forgot about it
 
2:11 AM
$(x^2.d^2y/dx^2+x.dy/dx-y)=x^2$
$y=\frac{1}{D^2-\frac{1}{x}D+\frac{1}{x^2}1}$
 
Also, showing $Q$ and $Q\times Q$ are not isomorphic turns out to be bit tricky. For showing isomorphism, I look for cardinality (for example, R and Q can't be isomorphic), or that domain group has elements of certain order but the codomain group has no element of that order etc.
 
i'd probably not try to use that method here
 
6 mins ago, by Koro
@leslietownes does this require a proof? I thought that since every finitely generated subgroup of Q is cyclic and total subgroups of Q are uncountable so set of non finitely generated subgroups is uncountable - countable which is uncountable.
 
ok..?
 
to the extent that i try to use formal operator methods, it's on ODEs with constant coefficients
that said, you should be able to bring your ODE into that form
 
2:13 AM
I see variable coefficients and series solution of ode comes to mind.
 
one thing to notice about the left-hand side is what $x^2 d^2/dx^2$ and $x \,d/dx$ do to monomials $x^k$
So, what's $x^2 \dfrac{d^2}{dx^2}x^k$?
 
@copper How is your comment on that affine subspace question relevant?
 
Here's a combinatorics terminology question that's been annoying me
 
@TedShifrin If I understood correctly, you just need $n \ge 2$ (not $3$).
 
I say, suppose on the contrary that $Q\sim Q\times Q$, there exists a bijection $f: Q\to Q\times Q$ so let $f(p/q)=(1,0)$ and $f(r/s)=(0,1)$. It follows that $f(p/q)=f(ps. \frac 1{qs})=ps f(\frac 1{qs})=(1,0)$ and similarly $f(r/s)=rq f(\frac 1{qs})=(0,1)$. These two relations force $f(\frac 1{qs})=(0,0) $ whereby we get $f(p/q)=f(r/s)=(0,0)$ which is a contradiction.
6 mins ago, by Koro
Also, showing $Q$ and $Q\times Q$ are not isomorphic turns out to be bit tricky. For showing isomorphism, I look for cardinality (for example, R and Q can't be isomorphic), or that domain group has elements of certain order but the codomain group has no element of that order etc.
 
2:17 AM
Oh. How do you deduce the actual question with $n=2$?
 
no
$d^2/dx^2$ carries out two derivatives
what's the second derivative of $x^k$?
 
@TedShifrin If you have $\lambda x + (1-\lambda)y \in A$ for any $x,y \in A$ and $\lambda \in \mathbb{R}$ then $A$ is affine. Am I missing something?
 
oh sorry $x^2.(k)(k-1).x^{k-3}$
 
still no.
 
We’re trying to get to the other definition.
 
2:21 AM
how much does differentiation lower the degree of a monomial?
stop
you're doing nonsense.
what's the power rule in calculus?
or, more simply: what's the first derivative of x^k?
 
$\frac{dx^n}{dx}=nx^{n-1}$
$k.x^{k-1}$
 
yes. so if the first derivative of $x^k$ is $k x^{k-1}$, and you differentiate again, what do you get?
 
$k.k-1.x^{k-2}$
 
yes
(i mean, should have parantheses so k(k-1)x^(k-2))
 
I just woke up
 
2:26 AM
me too
 
sorry for real stupidity
 
but i'll set that aside. so if $d^2/dx^2(x^k)=k(k-1)x^{k-2}$, then what's $x^2 d^2/dx^2(x^k)$?
 
$k.(k-1).x^k$
 
right. so (switching to the D notation b/c i'm tired of writing d/dx) we see that $x^2 D^2$ maps monomials back to themselves, up to the factor $k(k-1)$
What about $xD(x^k)$?
 
$(k)x^k$ @Semiclassical
 
2:30 AM
not quite.
you need a coefficient
right
 
@Koro the 'up to isomorphism' question is a little subtler and not just a subgroup count. it is generally possible for large numbers of distinct subgroups of an abelian group to be isomorphic to one another.
 
so if we look at our original ODE, consider what the operator $x^2 D^2+xD-1$ does to $x^k$
 
consider e.g. the group G of all functions [0,1] to R, componentwise addition as the operation, and for x in [0,1] let G_x = {f in G: f(t) = 0 for all t \neq x}. the subgroups {G_x: x in [0,1]} are mutually distinct, there are uncountably many of them, and yet they're all isomorphic to one another.
 
i.e., what's $(x^2 D^2+xD-1)x^k$?
 
($k(k-1)x^{k}+k.x^k-x^k)$
 
2:33 AM
watch out. the first term also has x^k
 
@leslietownes hmm, I probably didn't understand your question clearly. But is my understanding here correct?
29 mins ago, by Koro
@leslietownes does this require a proof? I thought that since every finitely generated subgroup of Q is cyclic and total subgroups of Q are uncountable so set of non finitely generated subgroups is uncountable - countable which is uncountable.
 
yeah. and x^k factors out
try factoring it out and simplifying the result
 
@leslietownes non isomorphic
 
$x^k(k^2-k+k-1)$
 
simplify more :)
 
2:36 AM
$x^k(k^2-1)=x^2$
k=2
 
pretty much, yes.
 
Doesn’t work?
 
$k=\sqrt2$
 
not quite yet, no.
the more general lesson is that $x^2 D^2+xD-1$ sends monomials $x^k$ to themselves, along with a factor $k^2-1$
So if you want to arrive at $x^2$, then using $x^2$ almost works
problem is, you get $3x^2$ instead. that is, $(x^2 D^2+xD-1)x^2=3x^2$ not $x^2$
But what would you do to both sides to make the right-hand side $x^2$?
 
$k$ is an integer?
 
2:41 AM
...
i misunderstood what you were saying
you do need $k=2$, otherwise $x^2D^2+xD-1$ would give you an $x^k$ different than $x^2$
but what would you do to $3x^2$ to get to $x^2$?
 
yes
divide it by 1/3
 
Um, almost
 
oh. yeah, not quite
 
@JackRod multiply by 1/3
 
$\frac{1}{3}(x^2D^2+xD-1)$
 
2:45 AM
sure. and then you might as well move that to the right of the operator
so $(x^2 D^2+xD-1)(x^2/3)=x^2$
Therefore, $x^2 D^2+xD-1$ sends $x^2/3$ to $3x^2$. so that's a particular solution to your ODE
this same idea works whenever your operator only involves terms like $x^n D^n$
 
one more question I have is this one
 
namely, such an operator always sends monomials to monomials. so if you've got something like $(ax^3 D^3+bx^2D^2+cxD+1)y=x^n$, then the answer will still be some multiple of $x^n$.
another cute operator observation one can make here: $x^2D^2+xD=(xD)^2$
 
$\frac {dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz}$
 
method of characteristics?
 
no
 
2:50 AM
then i have no earthly idea what you're doing
 
It is a different question
 
i understand. but the only source i know for equations like that, is method of characteristics / Lagrange-Charpit
 
@Semiclassical
 
so what's the actual question?
 
this the actual question
 
2:51 AM
No, that's an equation.
 
it is equation of sphere
I mean differential equation of it
 
that seems very unlikely
 
Huh?
 
the x^2-y^2-z^2 seems wrong for that
 
earthquake
 
3:00 AM
answer is $x^2+y^2+z^2=x-c$
 
i think you have a typo in your original equations.
hmm
no, scratch that
 
It is correct
@Semiclassical
 
yeah, it does work out in terms of equations
i can't say i know much to say about this, though
the only thing which comes to mind is to make the substitutions $x=r \cos \theta, y=r\sin \theta$
and hope that that simplifies the system
 
If i take pair at time like last two i get
y=cz
 
yeah
but then it's not apparent how one subsitutes that back into the rest
err. i goofed on what i said: i meant $y=r\cos \theta,z=r\sin \theta$
in which case one does have $z/y=\tan\theta$, so $y=cz$ would mean $\theta$ doesn't matter
which fits for a sphere
hmm, maybe what you wrote is enough tho
substituting $y=cz$, we have $$\dfrac{dx}{x^2-y^2-z^2}=\dfrac{dx}{x^2-(1+c^2)z^2}=\dfrac{dz}{2xz}\implies \frac{dx}{dz}=\frac{x^2-(1+c^2)z^2}{2xz}$$
the substitution $x=zu$ here now should do some damage
$$1+\frac{du}{dz} =\frac{z^2u^2-(1+c^2)}{2uz^2}=\frac{u^2-1-c^2}{2u}$$
 
3:15 AM
I did ask robjohn the same question and what he said is this....before he left
 
that doesn't simplify a ton, but it can be solved
 
$\frac{\mathrm{d}x}{x^2-\left(c^2+1\right)y^2}=\frac{\mathrm{d}y}{2xy}$
 
yeah, that's the same as what i'm doing
 
$\frac{2x\,\mathrm{d}x}{x^2-\left(c^2+1\right)y^2}=\frac{\mathrm{d}y}{y}$
$\frac{2x\,\mathrm{d}x}{x^2-\left(c^2+1\right)y^2}=\frac{2y\,\mathrm{d}y}{2y^2}$
dv2v $u=x^2$ $v=y^2$ $\frac{\mathrm{d}u}{u-\left(c^2+1\right)v}=\frac{\mathrm{d}v}{2v}$
now what i feel we can take u/v=k
or u=vk
 
hmm, yeah, that substitution $u=x^2,v=y^2$ seems sensible
 
3:18 AM
@robjohn hi
 
get rid of $u$ with that substitution and you get a seperable ODE
 
@Semiclassical you and robjohn discussed about operator question?
 
not that i remember
but this stuff is familiar from physics
you use $D(xy)=y+x Dy=(1+xD)y$ a ton there
(in a slightly different form)
 
are u agreeing with $\frac{e^{ax}}{F(d)}=\frac{e^{ax}}{F(a)}$
 
what?
 
3:25 AM
yes
 
I feel except me all the guys on physics site use this
 
3:43 AM
Huh?
 
@JackRod that is the right way to go. That leads to a separable equation.
 
@robjohn thanks sir
 
$(ax+by)\,\mathrm{d}x+(cx+fy)\,\mathrm{d}y=0$ is separable using $y=ux$
 
 
1 hour later…
4:55 AM
@TedShifrin I added a long winded comment (in an answer) illustrating why $n=2$ suffices.
 
 
6 hours later…
10:33 AM
hello, is there some resource where one can learn to use generating function to find sum of series.. series like partial series, for example $\sum_{i=1}^{m}\binom{n}{i}i^p$ where n may be greater or lesser than n
 
 
1 hour later…
11:35 AM
hi, could someone help me out with this ? Let $\Omega \subset \mathbb{C}$ be a region( open + connected), and suppose $p dx + q dy$ is a locally exact differential in $\Omega$, suppose $f(z) = \int_{\gamma_{z_0 \rightarrow z}} p dx + q dy$, where $\gamma_{z_0 \rightarrow z}$ is any path from $z_0$ to $z$, and assume $f$ is well-defined in the sense that for any two such paths, the difference of that integral is $2 \pi i k$ for some integer $k$ depending on the paths.
Moreover , suppose there is a harmonic function $u$ in $\Omega$, such that locally, $du + i (pdx+qdy)$ is of the form $g dz$ for some holomorphic $g$, (i.e. $pdx + q dy$ is the conjugate harmonic differential of $u$), then why is $exp(f)$ holomorphic in $\Omega$?
I can see that $exp(f)$ is now unambiguously defined, but I fail to see why it is holomorphic
sorry i mistyped the question
Let $\Omega \subset \mathbb{C}$ be a region( open + connected), and suppose $p dx + q dy$ is a locally exact differential in $\Omega$ ($p,q$ real-valued and continuous, say), suppose $f(z) = \int_{\gamma_{z_0 \rightarrow z}} du + i( p dx + q dy)$, where $\gamma_{z_0 \rightarrow z}$ is any path from $z_0$ to $z$, and assume $f$ is multi-valued in the sense that for any two such paths, the difference of that integral is $2 \pi i k$ for some integer $k$ depending on the paths.
Moreover , suppose $u$ is a harmonic function in $\Omega$, such that locally, $du + i (pdx+qdy)$ is of the form $g dz$
 
 
1 hour later…
1:13 PM
@porridgemathematics is it not the case that $f(z)=g(z)-g(z_0)+2\pi ki$, depending on the path we follow to compute $f$?
So, do we not then have $e^{f(z)}=e^{g(z)-g(z_0)}$, no matter the path taken?
 
1:32 PM
Hello
I am having a doubt
Are Trig functions empirical ?
 
1:46 PM
@Ishwaran What do you mean by "empirical"?
 
@Ishwaran trig functions are based on geometric investigation/experience, but can also be derived totally by theoretical means. So they are and they are not.
it depends on how you approach them, but I'd say they are mostly theoretical with a bit of empirical background to them.
 
 
1 hour later…
3:13 PM
c++ is the best language
why? it gives this output:
JUST BECAUSE I FORGOT THE RETURN STATEMENT
I want to know why I think I should go to stackoverflow chatrooms
 
@XanderHenderson Something proven by observation rather than a idea or logic
@robjohn What is the Theoretical approach to it ?
 
3:56 PM
and
can anyone help with my math problem ?
In the given figure, PS and QT are perpendiculars to QR and PR respectively. Then, prove that triangle RST is similar to Triangle RPQ
 
4:40 PM
@Euler2 at least you didn't forget a semicolon.
 
the great sin
 
@Ishwaran almost all of the formulas for trigonometric formulas are theoretical. That sine is increasing from $0$ to $1$ in the first quadrant is an empirical observation, but to actually derive that $\sin(x)=\sum\limits_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}$ needs more than simple observation.
 
Why do they draw these things with equilateral triangles? That's revolting.
Hi, @robjohn. Hope you and the pup are doing better!
 
@robjohn ^
 
@TedShifrin I am doing a bit better, the pup ate quite a bit yesterday, so we are hopeful. The thing is that she will like something one day and not want to look at it the next.
 
4:56 PM
How very human of her!
 
@TedShifrin Do you mean the diagram ?
 
@Ishwaran Absolutely.
 
You can expect a lot like this in India :P
 
I have a question from India in limits
 
Of course. You should draw a more generic picture ... no false parallel lines, etc.
 
5:06 PM
real quick, if $(A_n)$ is a sequence of events such that $P(A_n) \rightarrow 0$, and $X$ is a random variable, then for $E(X 1_{A_n}) \rightarrow 0$, we need $X < \infty$, don't we?
that condition is missing from the problem statement, and I seem to have a counter example without it
 
Is a random variable allowed to be infinity? Even so, if it's on a set of measure zero, do we care?
 
When we try to find dy/dx then we take delta fx/deltax and make it smaller and smaller. We get better rate. But that rate never stops at a fixed value. So we take limit and and we assign dy/dx a value which it can never have ie the limit of delta fx/deltax.
 
@TedShifrin I think so, why not. An rv is just a measurable function
and for the second thing you said, let $P$ be the uniform probability measure on $[0,1]$, and consider the sequence $A_n = (0, \frac{1}{n})$, with $X(\omega) = \frac{1}{\omega}$
 
@TedShifrin Yes, they can, but they won't, they just want to trick us in all possible ways. I was asked to give 2 possible and unique proof to the solution. And yes, I got two proofs
 
then $E(X 1_{A_n}) = \frac{1}{n} \int_0 ^{\frac{1}{n}} \frac{1}{\omega} d\omega = \infty \not\rightarrow 0$
 
5:12 PM
Oh, you mean you need $E(X)<\infty$.
That makes more sense.
@Ishwaran I haven't seen this before. It's cute. I don't have a proof yet.
 
I think the intention here was that $X < \infty$
but yeah I need some boundedness condition
 
I think the issue is that the integral diverges. The value at $0$ can certainly be finite.
Just saying $X<\infty$ won't fix your counterexample.
 
infinite, yes
sorry I meant $X < c$
 
No, change the value at $0$ and then it's everywhere finite.
That seems too stringent. Just make it $L^1$.
 
oh, shit it is L^1
durrrrrr
 
5:16 PM
smacks joe shmo
 
@TedShifrin I derived myself by proving similarity btw triangle pto and qso where o is the point of intersection of line segments tq and ps, another one is given by Paramanand in Basic Maths room by taking pq as diameter
 
I'm just playing with angles, @Ishwaran. I think that'll get it.
 
@TedShifrin :D
The worst part is I have to take class to my juniors on this problem and explain the 2 proofs I got . The more points they give for me, the more internal marks I get
 
For example, $\angle SPQ = \angle TQP$.
 
How ?
 
5:25 PM
Let me see if I can finish. I just want to prove angle congruence. So call the right bottom angles $\alpha,\beta$, left ones $\gamma,\delta$, and the two angles in our little triangle $x,y$. Relate them all because of right angles and sum of angles, etc.
Hmm, no, I must have a mistake. Let me go back.
Found the mistake.
What is correct is $\angle SQT = \angle TPS$. That should do it, I think.
Hmm ...
 
but... I guess there is a mistake again, I think We must forget the line TS
 
Huh?
 
how do you got angle SQT = TPS ?
 
Writing down a bunch of equations (with Greek letters).
So when you say to forget $TS$, are you saying the problem statement is incorrect?
I suppose we can use some arcane facts about the orthocenter, but I was trying to avoid all that.
 
wait.. I found a mistake in my proof too
This problem is not going to let me sleep.. brrrrr
 
5:39 PM
LOL ... I spent many months like that in graduate school :P
Oh, actually, it's really easy to see those angles are equal. Just look at the two right triangles with vertex at $R$.
 
oh yess
then angle rqt = rps
therefore angle sqt = tps
@TedShifrin If we get these type of sums, who doesnt ? :P
 
I can't do it with all those letters. I have to look at angles in my picture.
 
yes
 
5:54 PM
@TedShifrin Better?
 
I have my own better picture, but I still am not seeing the proof. Close. Sorta amazing.
 
you know you're wrong with that font.
 
So it boils down to seeing that $\angle SPQ = \angle STQ$. Makes it feel circle-ish.
Want inscribed angles at $P$ and $T$ for that. Hmm.
So is this a concyclic quadrilateral?
 
@leslietownes Are you talking about my diagram?
 
He's channeling his troublemaker daughter.
 
6:00 PM
@TedShifrin can I throw a blanket over him when he turns into a cat at 6:45?
 
Please do.
 
@robjohn yes.
 
@leslietownes I had meant to xkcdConvert it, which I now have. Look again.
 
haha. the metamorphosis is complete.
 
Lovely. Let's have mimosas.
 
6:04 PM
xkcd or late-era 'peanuts,' i can't tell.
no problem of false parallel lines in the xkcd version.
 
@leslietownes I never saw a Peanuts strip with a geometric diagram in it
But the font is about the same
 
that's the nice version, although i was also making a somewhat tasteless joke about charles schulz's inability to draw non-wobbly curves in his last years.
peanuts had sort of come and gone by the time i was reading comics as a kid. i knew him as the guy who owned the only ice skating rink i had ever seen in northern california.
 
6:17 PM
not a lot of good matrix questions on math.SE today.
 
@Ishwaran The simplest thing is to note that $\triangle PSR\cong\triangle QTR$
since they are right triangles with an equal acute angle
 
That makes them similar, not congruent.
 
@TedShifrin that gives you the ratios you need
it only asks for similar. The one is obviously smaller
That means that $\frac{RS}{RP}=\frac{RT}{RQ}$ which gives $\frac{RS}{RT}=\frac{RP}{RQ}$
 
I was working with angles only, but I'm stuck. Are you getting it just from lengths?
 
yes
 
6:23 PM
The concyclic quadrilateral thing is appealing.
Hmm, that was sure too easy. I saw those triangles right off the bat, but was addicted to angles only.
Shame on Ted!
 
6:38 PM
Hello all. Officially left academia yesterday.
 
huzzah!
 
Mazltov!
 
Still hasn't hit me yet, but I think it will hit me on Monday.
I'll be taking 2 weeks off between jobs, but will have a short meeting with my new supervisor on Monday
 
6:55 PM
@Clarinetist can you tell us what you are going to be doing, without having to kill us?
 
closes ears
 
Lol. I'm taking on a data scientist role, going the food science route. Won't be revealing the company name, but anyone who knows where I live should be able to guess which one.
 
Is cereal a hint?
 
give my regards to mrs. boyardee.
 
That's not food, leslie.
Speaking of food, have you done the leeks yet, or is munchkin's mishap derailing it permanently?
 
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