@leslietownes While in graduate school in the 70s, my father was TAing a class, and made reference to the Beatles. The class had no idea who the Beatles were, until one of the students said "Oh, yeah! That was, like, Paul McCartney's band before Wings!"

"wings? they're only the band the beatles could have been" - alan partridge

Ugh... I don't want to start teaching again next week.

Last semester was brutal, and I still don't feel the least bit rested or ready. :(

OK, we'll make an exception for you. You may start this week.

have you considered starting, but just doing a really bad job?

As opposed to the usual?

they can make you drive to the room and say stuff or whatever, but they can't make you 'start teaching.'

@XanderHenderson what have they assigned you to this semester?

@Semiclassical Two sections of Calc II, and three sections of Precalc II (mostly trig).

godspeed

All stuff I have taught before, nothing revolutionary.

But... ugh.

i mean, new stuff at least is novel

@TedShifrin inorite?

start teaching revolutionary. your first lecture should be on the illusion of capital.

Das Kapital, you mean?

@leslietownes Followed by a lecture on lesliecoin?

yes. i can get a whole curriculum together.

My lesliecoin account is still blocked.

ted, you scare the markets every time you say that, and you know it isn't true.

I do? Why am I stymied every time?

Man, last semester, I had a student who was angry that I taught things which are not in the book. We spent two days on complex numbers so that I could show them some Newton fractals (and I didn't even test them on that shit!). We did a little bit of analytic geometry so that we could look at functions from a different perspective. And, god forbid, we spend two days sequential convergence because epsilon-delta is (a) rather technical and (b) not super intuitive at first glance.

But that's not in the book!

i won't pry into your private life. all lesliecoin servers are fully operational.

:(

@PenAndPaperMathematics have you seen this paper before? arxiv.org/abs/0903.0340

The response from one very angry student in evaluations was "He didn't teach to the national standard."

i taught out of ross once. he was all about sequential convergence. i grew to like it.

i've glanced at it in the past but sorta forgot it till now

Are any of you familiar with a national standard for college calculus?

i like the appeal to a nonexistent authority. make education great again.

Where can I find this document?

@robjohn You'll be amused to know that after your comment, DrPotato admitted I'd been correct all along (re integration by parts with a non-well-defined function).

@Xander I think lower-level students (no offense intended) tend to be more wedded to "following the book" than more advanced students. I admit that in precalc and calc I tried to conform to the book's conventions, etc., to minimize Angst.

Yes, I think sequences are pedagogically quite good, but since we don't usually teach them until the end of Calc II, there is a logistical issue there.

when i taught out of a book i absolutely hated, i made sure to have all homework and exam problems clearly linked to assigned reading in the book. and sometimes even exercises in the book.

@TedShifrin Yeah, that is very much true. Which is why I need to start teaching out of my own book. :)

"to the national standard"? Sounds 1000% like an education major.

to avoid precisely that criticism.

@Xander As soon as you write that book, you'll find yourself still complaining about the stupid author who did such-and-such ... :D

@TedShifrin No doubt.

I found that quite a few times with my own books. The students found it amusing, though.

Given a linear ODE with constant coefficients of this form: $y^{(n)}+a_1 y^{(n-1)}+...+a_1 y^{(1)}=\sin (ax+b)$, I want to know how to find its particular solution using inverse operator. How can the particular solution can be found using inverse oprator method?

@Xander Students who are struggling probably need more time on the material they need to master.

Explain the inverse operator method, @Koro.

something something $y''+y=(D^2+1)y=f$ shenanigans?

You haven't gotten to inverse with your mumbles.

It develops like this: If we have $dy/dx=f(x)$, say, then we re-write it as $Dy=f(x)$. Clearly, $y=\int f(x)\, dx$. Based on this we can define: $\frac 1D. f(x):=\int f(x)\, dx$ so that $y=\frac 1D. f(x)$.

OK, where $1/D$ formally means $D^{-1} = \int\,dx$.

yes.

So what are you going to do in general?

Say with Semiclassic's example?

@TedShifrin i didn't want to give away the game entirely :P

Now the earlier equation can be written as: $p(D)y=\sin (ax+b)$. I want to say that $y=\frac 1{p(D)}\sin (ax+b)$. But not sure how to evaluate RHS.

What can you do with $1/p(D)$?

Suppose you had $1/p(x)$. What would you naturally do to work with that?

the usual way i handle this is to do Fourier transform but that might be outside of the inverse method

Not germane (or german) here, Semiclassic.

I can factorise it into linear factors and then use commutativity of the inverse operators to evaluate.

But, but

How do you factor $1/p(x)$ into linear factors?

You mean ...

there is only one concern: I am not sure what happens if p(x) has complex roots.

Use them.

@TedShifrin eh, kinda is. it lets us write $\tilde{y}=p(-ik)^{-1}\tilde f(k)$

and then invert that to get an integral representation for $p^{-1}(D)$

We'll end up at the same place, I think, Semiclassic.

Ugh... I've been on hold with Delta for more than an hour. SO FRUSTRATING.

Never gonna fly Delta ever again.

i agree, i just prefer going via FT

Oh, I've heard stories of over a dozen hours ....

All the airlines are suffering.

airlines live and die by the margins

Nationalize air travel.

And fund the trains.

the only choice they've given themselves is to screw their customers

or that, yes

but that'd go against the word of our lawd and savior Ronald Reagan

Yeah, it's, like, socialism.

I've been reading endless stories about all the disasters with Amtrak, especially with the recent snow. Definitely the case that the generic customer is very dissatisfied.

I'll explain with an example. Suppose we have $y''-3y'+2y=xe^x$ so we have $(D-2)(D-1)y=xe^x$. From here, $y=\frac 1{(D-2)}\frac 1{(D-1)}(xe^x)=\frac 1{D-2}. e^x\int xe^xe^{-x}\,dx=-1/2(1+x)^2e^x$

Which makes me sad. I love the train lines in the west. I just wish they ran more often.

Similarly, what should I do if I have sin (ax+b) on RHS?

(in RHS or on RHS?)

The trip from the LA area to northern Arizona is beautiful. I love that ride. And it only takes an hour longer than driving.

@Koro you seem to have something specific in mind for $1/(D-a)$

@Koro "on the RHS"

what rule are you using?

Yes, you are using shortcuts/memorized facts.

I would have used partial fraction decomposition. That's what I was hunting for earlier.

But you still need to know how to do $1/(D-a)$.

yes, semi. $\frac 1{D-r} e^{ax}=e^{rx}\int e^{at}e^{-rt}\,dt$

So what's your shortcut rule for $1/(D-a) \sin bx$?

that's the consequence of integrating factor in order one linear ODE.

@Koro that looks suspicious to me: x as both a free variable and integration variable

okay, that looks more plausible

earlier version was also correct though.

$\frac 1{D-a}\sin bx=e^{ax}\int (\sin bx) e^{-ax}\,dx$

i'll yield the point

OK, and that's a well-known integral.

while RHS can be simplified. I have one problem: We may not that we'll have (n-1) more such factors. So isn't that going to complicate the calculations?

it'll be tedious

I return to my partial fraction suggestion?

side note: if we were given the initial conditions (say, at $x=0$) i'd probably insist on something like $(D-a)^{-1} f(x)=\int_0^x f(t)e^{a(x-t)}\,dt$

I concur that this will lead to clearer answers.

it has the nice bonus of being equivalent to convolution

though i guess the other version was too?

partial fraction suggestion seems nice. However, that will also give complicated final solution, I think. I mean -using partial fractions-are we able to reach this type of nice closed form: if p(D)y=sin (ax+b) then $y=\frac 1{p(D=\text{something})} \sin (ax+b)$?

Yeah, but now that you mention it, why is it obvious that convolution should show up here?

Huh? @Koro

hmm

You mean a linear combination of such things, @Koro?

maybe if one views this through the Laplace transform?

yes, combination of the partial fractions.

Let me write the equations to record what's going on.

Oh, I guess we just need to think about the fundamental solution, @Semiclassic.

sanity check: $(D-a)\int_0^x f(t)e^{a(x-t)}\,dt = f(x)+\int_0^x f(t)(a-a)e^{a(x-t)}\,dt=f(x)$

$p(D)y=\sin (ax+b)$. Suppose that $p(D)=(D-r_1)(D-r_2)...(D-r_n)$, where $r_i$'s may be complex or real. $y=\sum_{i=1}^n \frac {a_i}{D-r_i} (\sin (ax+b))$, where $a_i$'s are constant obtained by converting $1/(D-r_1)(D-r_2)...(D-r_n)$ into partial fractions.

It follows that $y=\sum a_i e^{r_ix}\int e^{-r_ix}\sin(ax+b)\,dx$

And, in complete generality, that's all you can say.

gonna follow my nose with Laplace transform. If $y'(x)-ay(x)=f(x)$, then $(s-a)Y(s) = F(s) + y(0^-)$. for a particular solution, we can choose $y(0^-)=0$. Then $Y(s)=(s-a)^{-1}F(s)$

Semiclassic: So if $\phi$ is the fundamental solution, i.e., solves $\phi'-a\phi = \delta_0$, then the solution is given by convolution with $\phi$. I think this gets us to the same place. So what is $e^{ax}\int e^{-ax}\delta_0(x)\,dx$?

well, the one thing i'm noticing that seems problematic here

one reason to expect convolution to get involved is that the solution operator commutes with translations

But the book did something along these lines: note that $D^2\sin (ax+b)=-a^2\sin(ax+b), D^4\sin (ax+b)=a^4\sin(ax+b),..., (D^2)^r\sin(ax+b)=(-a^2)^r$ so $p(D^2)\sin(ax+b)=p(-a^2)\sin (ax+b)$ so $\frac 1{p(D^2)}\sin (ax+b)=\frac 1{p(-a^2)} \sin (ax+b)$.

That's an interesting remark.

if $p(-a^2)\ne 0$.

But I don't know how that gives me the desired solution.

They're separating out even and odd powers of $D$?

I want $\frac 1{p(D)}\sin (ax+b)$ and not $\frac 1{p(D^2)}\sin (ax+b)$

is that the usual rule one has is $f(t)=e^{-\alpha t}u(t)\implies F(s)=1/(s+\alpha)$ valid for Re(s) > $-\alpha$

Write $p(x)=q(x^2)+r(x)$, where $r$ is an odd polynomial.

yes, then what do I do to r(x)?

so here $f(t)=e^{at}u(t)$ which doesn't have a well-defined Laplace transform unless Re(s)>a

I used to know Laplace Transform. @semi.

So Laplace is not wonderful here, @Semiclassic.

I would have to think from scratch about that.

yeah, it seems too limited

i mean, formally it's fine

(Laplace transformation needs revision at my end.)

You do $r(x)$ as we were discussing in the first place. Or you replace $\sin$ with $\cos$. Aha, so write $r(x)=xQ(x^2)$.

but going to Laplace transform was supposed to make this rigorous, not introduce more appeals to formality

I was happy with fundamental solutions, but I'm rusty.

I last taught this stuff in 1987.

lol what even is 1987?

lemme try Laplace transform method from scratch

the year i was born, lol

I just need to remember what $\int e^{-ax}\delta$ means.

Wow, I was almost tenured in 1987 ... close.

if it's really $\int^x e^{-at}\delta(x-t)\,dt$, then that's just $e^{-ax}$

r(x) as x times even polynomial sounds nice.

@Semiclassical I know that !! Diract delta!!

though i guess that does assume $x>t$

No, I mean $\int_{-1}^x e^{-at}\delta_0(t)\,dt$.

hmm

Looks like $1$ to me, independent of $a$. But ...

@Semiclassical infact $\int_{-\infty}^\infty f(t) \delta (t-2021)dt=f(2021)$

I think we need to interpret everything distributionally to make sense of it, so I have to integrate by parts.

right, semi?

that's the usual meaning, yes

I took a course on probability. I met it there along with spectral density, autocorrelation functions etc.

$\int_{-1}^x e^{-at}\delta(t)\,dt = e^{-ax}H(x) - a \int_{-1}^x e^{-at}H(t)\,dx$.

@TedShifrin Thanks a lot. I think this should work :-)

Yeah, seems the logical thing, @Koro.

The last $dx$ should be $dt$, and of course the first term should be evaluated also at $-1$, but is $0$ there.

:)

in physics you usually run into Dirac delta in junior-level E&M

b/c there you start from Maxwell's equations, and somehow need to recover the notion of a point particle

whereas in intro courses you take point particles as 'obvious' and derive stuff via integrating Coulomb's law

Maybe FT is the easiest way to do the fundamental solution thing I was looking for. Agh, I've forgotten so much.

yeah, let's give that a shot

So we want $(D-a)\phi = \delta$.

FT the whole thing.

$(k-a)\hat\phi(k) = 1$.

shouldn't it be $\delta(x-x')$ or something

Um, my memory is that for fundamental solution we do $\delta_0$.

Think about Gauss's law, etc.

hmm. i mean, translation symmetry should mean it doesn't matter

So, $\hat\phi(k) = \dfrac1{k-a}$. Oops. I missed an $i$ when I differentiated.

This should be $(ik-a)\hat\phi(k)=1$, right?

i want to say it's $-ik$

but that may depend on the convention

I think I'm right.

$\hat{f'}(k)=ik\hat f(k)$.

yeah, i think you're right

Hmm, $\widehat{f'}(k)$ ...

Better.

I'll proceed a little differently but with the same goal. $$(D-a)\phi(x) = f(x)\implies (i k -a)\tilde{y}(k)=\tilde{f}(k)\implies \tilde{y}(k)=(ik-a)^{-1}\tilde{f}(k)$$

So $\hat\phi(k) = \dfrac1{ik-a}$. So $\phi(x) = \begin{cases} 0, & x< 0 \\ e^{ax}, & x\ge 0\end{cases}$ or something.

That seems right.

So now we convolve with the general $f$.

hmm. Wikipedia lists the Fourier transform of $e^{-ax}u(x)$ as $1/(ik+a)$, but only if $a>0$.

so again running into weirdness at $x=\infty$?

Oh right, $a>0$.

We need functions decaying at infinity.

maybe the point is that the inverse doesn't work if $f(t)$ diverges too fast?

Yeah, I have notes in my lecture notes from 1987 on issues like this. Solving $u'' = a^2 u$ doesn't work with FT because we need boundary conditions at $\pm\infty$.

yeah, whereas $u''=-a^2 u$ has no such issues

Right, although we're still not in $L^2$.

yeah

lol lecture notes from 1987. week 1, the geocentric model. week 2, how to identify a witch. week 3, linear differential equations [guest lecturer: leonhard euler]

reminds me, i still want to get a better understanding of how tf people did astronomy back then

leslie is so proud of being only half-antique.

how do you come up with an ellipse model if all you can see is where the planet is in the sky and not how far away it is

heck, how do you come up with -any- model

ted: if i do enough of this, i get younger

Oh, I once knew about this, @Semiclassic. They did take measurements :)

lol

i know it's something something Keplerian elements

but i do have a lot more regard for geocentrism than flat earth

esp. since people have a tendency to think stuff like "oh, earth is closer to the sun during the summer and that's why it's hot"

some species of bird disappear in the winter because they tunnel underground and hide until springtime

which connects two valid facts---"earth's distance to sun changes over the year", and "North America is hot during June / July August"---via invalid reasoning

the fact that aphelion, farthest distance, occurs in July is testament to that :P

the distance is less relevant than the angle

exactly

seasons are about the tilt of the earth

copper: sigh that's not what she said

@Thor Re your intersection theory question. I don't normally think about open embeddings, but what if we embed a torus into an infinite-genus torus and push the intersection off to infinity?

the torus doesn't embed into the infinite-genus surface. but I don't think I understand the idea either. we're looking at intersections coming from the space we're embedding, so how do you mean "pushing off to infinity"?

Well, I need a twice-punctured torus, I guess, but that's still enough to have intersections. I was going to push the torus off to infinity in the infinite one. I dunno. I don't think about these things.

We just passed through perihelion. Here's a plot of the Sun-Earth distance, using my program here: astronomy.stackexchange.com/a/47345/16685 Data points are for 0:00 UTC

i knew the astro talk would bring evening turing out of the woodwork

That "mean" line is only the mean distance over that time period, computed by integrating the Bézier curves passing through the data points.

nice

i mean, the percent variation is tiny

Yeah. The eccentricity is 0.0167.

I've got an embarrassing math's q: stackoverflow.com/questions/70598736/…

The maths bit is at the bottom

so perhaps ironically, i feel like it's easier to get the origin of the seasons correct if you think of earth's orbit as circular

since it forces your attention away from a tempting alternative explanation

But it still does have a measurable effect on the climate. So southern hemisphere summers are a little hotter, and shorter. But other factors are more important. We have less land mass down here, and a big circumpolar current. So it gets pretty cold below 40°S, and not many people live further south than that. Whereas 40°N is still fairly temperate.

right

@Semiclassical Oh, sure. In terms of the seasons, that 23° angle between the ecliptic & equatorial planes is way more important than the eccentricity. But they're both important in determining the difference between apparent solar time & mean solar time, aka the Equation of Time. physics.stackexchange.com/a/469884/123208

nice

actually, I think the answer to my question is positive, I just had an insight

Interesting.

@Semiclassical Kepler was a pretty amazing dude. And a great writer, although he does tend to ramble all over the place. And get a bit mystical. :) He took a couple of years off astronomy & geometry to defend his mother against charges of witchcraft. And they won the case, which was quite unusual.

Kepler's mathematics is even more impressive when you remembervhe did that stuff before calculus was invented, and analytical geometry was still in its infancy. Of course, there were techniques for finding areas under curves, and rates of change, but they weren't codified or well-organised.

yeah

Yup.

enjoysmath.blogspot.com/2022/01/definition-of-monomorphism.html

The notation for interpreting the diagram in terms of $\forall, \exists$ will go into a post about notation

enjoysmath.blogspot.com/2022/01/…

@leslietownes What do you mean?

@AkivaWeinberger Sure. So if you run the Euclidean algorithm on a pair of adjacent Fibonacci numbers, you descend through the Fibonacci sequence. Each quotient is 1, so the algorithm descends at its slowest possible speed. This is reflected in the continued fraction for the golden ratio. It's sometimes said that it's the most irrational number because its continued fraction has the slowest possible convergence.

@PM2Ring i get why this is said, but i do find it a bit strange to elevate the continued-fraction representation in this way

if you use Newton's method, you get the usual rapid convergence via rationals

(and the way is sorta cute: Newton's method ends up iterating $F_{n}/F_{n-1}\mapsto F_{2n}/F_{2n-1}$)

@Semiclassical Right. It's connected to Cassini's fast algorithm for Fibonacci numbers, which can be implemented in matrix form by exponentiation.

yeah

basically just exploiting $A^{2n}=A^n A^n$

@Semiclassical Well, quadratic irrationals are nice because their continued fractions are periodic. Continued fractions have a long history, and a lot of the "big names" in early modern number theory put a lot of time into studying them. At that time, they were one of our few tools for tackling irrationals. The continued fraction is a compact way of representing the recursive solutions to Pell's equation.

tru

Continued fraction convergents give you the best rational approximations of a number, which gives them an advantage over representations using some arbitrary base like 10, 2 or 16. Sadly, they're a bit painful when you need to do arithmetic on the number. ;)

There are families of cute continued fractions for powers of e. Last week, I learned there are nice continued fractions for the incomplete gamma functions. math.stackexchange.com/a/4339669/207316

But getting back to orbit stuff... Yesterday, I was playing with the vis-viva equation, which gives the speed of a body in an elliptical orbit. It's usually written in terms of $r$, but it's easy enough to write it in terms of $\theta$, as I mention here: physics.stackexchange.com/a/675868/123208

Hi, I am trying to prove that the operator $X\mapsto \mathcal{L}_{X}g$ is a closed operator when working on a compact manifold. I know I need to limit my domain so that I will have a Norm space into a norm space, since the space of smooth sections over a manifold is not a norm space. Is there a way to do so?

We get $v^2 = GM\left(\frac{1+2e\cos\theta+e^2}{a(1-e^2)}\right)$, where $e$ is the eccentricity. That's a family of limaçons, with a circle for $e=0$, and approaching a cardioid as $e$ approaches 1 (where the orbit becomes unbound).

The following script plots the curves (in green), along with circles showing the minimum & maximum (in red & blue).

@Myprofileissaved What is $g$?

sagecell.sagemath.org/…

&g& is the metric of the Riemannian manifold

Ah, something you forgot to mention. So the Riemannian metric gives you all sorts of norms.

Does it makes sense to take $Sup_p \langle X_p , X_p \rangle_{g_p}$ as a norm over vector fields?

Sup is max, but that is not quite a norm.

This script makes an interactive 3D plot of the limaçon family in cylindrical coords, with the eccentricity on the Z axis.

$max_p |\langle X_p , X_p\rangle_{g_p}|^{1/2}$

sagecell.sagemath.org/…

Aw come on.

Better!

No abs value needed, of course.

Yes, ofcourse.

Ok. So for $T\in \Gamma{T^2 T^\ast M}$ can I take $Sup|T(X_1,X_2)|$ going over all $|X_i|=1$ $i=1,2$ as a norm? does that makes sense?

with $\Gamma(T^2 T^\ast M)$ being the space of covariant 2-tensor fields?

* sup going also over all $p\in M$

more generally, if $M$ is compact and $E\rightarrow M$ is a vector bundle with a bundle metric, then this constructions defines a norm on the space $\Gamma(E)$ of global sections

the Riemannian metric induces a bundle metric on $T^2T^{\ast}M$, so you also get a norm on $\Gamma(T^2T^{\ast}M)$ that way

Is it possible to show that with such norms $\mathcal{L}: \mathfrak(M) \rightarrow \Gamma(T^2 T^\ast M)$ taking $X\mapsto \mathcal{L}_{X}g$ is closed?

$\mathfrak{X} (M)$ - domain of $\mathcal{L}$

smooth vector fields

@TedShifrin that’s one way to look at Euler Maclaurin: $\frac{D}{1-e^{-D}}D^{-1}$

@TedShifrin I’m glad I was able to evince some correctness there.

@robjohn I was asking for Koro to verbalize. I know the basics. Yes, thanks again.

Sorry. I just hopped in on mobile and it’s hard to get a whole picture.

Here's a really clean proof in Topos Theory: enjoysmath.blogspot.com/2022/01/…

I'm blogging to create an online reference with nice proofs / short & sweet page content

That proof is missing some links to its definitions and properties used, but I have create those first :)

Any idea how to compute $\displaystyle \int_0^1 \frac{x^3-1}{\log(x)}dx$

They’ve messed up more on the mobile activity page: now the navigation pop-up menu takes up a whole row and I can hardly see any of the content past the headers. Ugh!

Web dev is hard man

@Odestheory12 that will require exponential integrals (non-elementary functions).

I've considered $I(t) =\displaystyle \int_0^1 \frac{x^t-1}{\log(x)}dx$ for $t > 1$ and $\dfrac{\partial}{\partial t} I(t)$. It works :D

@Odestheory12 does that give $\log(4)$? I’m on my phone so I’m trying this in my head.

Yes

$\log(1+t)$

If $I(t) =\displaystyle \int_0^1 \frac{x^t-1}{\log(x)}dx$ then $\dfrac{d}{d t} I(t) =\displaystyle \int_0^1 \dfrac{\partial}{\partial t} \frac{x^t-1}{\log(x)}dx = \int_{0}^{1} x^{t} dx=\frac{1}{t+1} $

Often, definite integrals are definitely easier.

Since $I(0) = 0$ then $I(t) = \log(t+1)$ as you already noted :P

I really hate $a^x$ type functions

They make everything too complicated

Depends... sometimes not.

The other day i was solving this: Determine an odd function $f:\Bbb R \to \Bbb R$ that holds $f(1) = 1$ and the relation $\displaystyle \int_{-a}^{a} (a^x - x^2)f'''(x) dx = a^3$

They’re just exponentials really

That $a^x$ ruined essentially every try I did

(Then I got told there was a typo and was meant to be $a^2$ instead of $a^x$)

Ouch

It looks as if integration by parts would be the way to go, especially with a constant $a^2$

Yeah I guess it would work that way.

I just differentiated w.r.t $a$ and solved the ODE I got

Doesn’t that still leave a $\int_{-a}^a2af’’’(x)\,\mathrm{d}x$ in there?

Yeah but the key is to note $f'''(x)$ is even, and hence it cancelates with the other integral

And you are left with $4a f''(a) = a^3$ iirc.

Can you get convergence for some values on a natural boundary?

@geocalc33 this Q&A may be relevant: math.stackexchange.com/questions/1320321/…

Hey, how do the rules of modulus fair for inequalities?

I found an identity for expressions of the form $\lfloor\frac{ab}{c}\rfloor$ that I might be able to use because the conditions ever so conveniently work out for doubling the precision of a given quotient approximation.

Second to last here: wolframalpha.com/input/?i=identities+floor%28ab%2Fc%29