Mathematics - 2022-01-05 (page 2 of 3)

AMDG

15:00

How does one vector tell another vector apart? :P

That would make no sense if you instead said two planes (no pun intended).

Semiclassical

we're not talking about vectors, though. we're talking about orientations of objects

for orientations of objects, palm facing up is different than palm facing down

Thorgott

glad to have helped

AMDG

Well all I've described is a system for describing the orientation of a thing in space, but it obviously can't consider the subjective orientations of each individual type of object. Those you have to add data for yourself.

I mean you could always add an imaginary component to describe such a thing.

Semiclassical

i mean, it's not exactly subjective

AMDG

$(\alpha, \beta, i\gamma)$

Semiclassical

15:03

the way one usually does it is one of two ways

AMDG

It is subjective by definition. That orientation derives its meaning (e.g. palm up, palm down) from the object itself.

Semiclassical

i don't see what you mean. how is my palm facing up any less objective than the direction my arm is pointing?

AMDG

My apologies, I misunderstood what exactly you were referring to by orientation in space. Well the problem is solved by just adding another component $\gamma$ (real or not) to describe orientation, but my intent was just a system to describe a vector in $\Bbb{R}^3$.

Semiclassical

sure. if all you're interested in is direction, then unit vectors indeed suffice

AMDG

I mean you could define an axis ad infinitum, recursively, and just keep slapping on components.

Semiclassical

15:08

the reason quaternions are useful is because they address orientation as well

they're not required, but they do serve that purpose

AMDG

I see

Semiclassical

the other way is Euler angles, and they're not so different from what you were saying.

with those you proceed like this

AMDG

Yeah, but I don't like those systems because of the problems I've heard about them, namely gimbal lock and whatever else.

Semiclassical

ya

i won't go into that then

AMDG

Well I wouldn't mind if you did

Semiclassical

15:10

lol, ok

you start out with your object in some definite orientation, e.g., my hand pointing in the (0,0,1) direction with palm facing in the (1,0,0) direction

AMDG

I don't really understand those systems and I'll have to write subroutines that use them anyways

Semiclassical

you first rotate your hand keeping your arm fixed, so a rotation around the z-axis

i should have also specified that your upper arm is initially horizontal (in the (0,-1,0) direction, say) and forearm is pointing straight up along with your hand

AMDG

I should note that I don't do well with analogies. I do better seeing the thing for what it is.

Semiclassical

fair. lemme see if i can find one

313 rotation sequence (Euler Angles)

►

that's one version of Euler angles. there's a few different conventions, which is a headache when learning it

different choices of axes

AMDG

That's the only thing I really find lacking about geometry. It lacks algebra for describing the objective apart from the subjective in an objective manner.

This is in spite of the fact that one can easily define, for instance, n-dimensional planes recursively by creating a new 1D axis perpendicular to the rest, and then procedurally assign unique identifiers for axes relative to the order in which they were created so that there is no ambiguity since any person repeating the process ends up with the same relative structure, thus it is objective by definition.

Semiclassical

15:19

i had to look at some of these conventions again during the semester when trying to remember how to describe gyroscopes

AMDG

Conventions are pain :)

Semiclassical

which are fun to play with, but understanding them is...oof

AMDG

Standards are based.

Semiclassical

gyroscopes do give a good instance of having 3 degrees of freedom for describing rotations, though

spin, precession, and nutation

AMDG

The degree of constancy of a thing with respect to everything it upholds is directly proportional to its degree of perfection.

So conventions are less perfect than standards.

@Semiclassical Seems like this is just a less perfect version of the system I described. For whatever reason, he has defined two of the axes to be parallel. Psi and phi are both rotating in the same direction on different planes. That's inefficient, too, because this system clearly has redundant values outside of mod pi for any of its angles making it waste values.

Semiclassical

15:29

yeah, proper Euler angles always have first rotation same as third

AMDG

._.

cringe

Semiclassical

in aeronautics they apparently use so-called Tait-Bryan angles

AMDG

Ok, hot take: I have now decided I will not be implementing subroutines for euler angles.

Semiclassical

in which case the rotations are all different apparently

c.f. en.wikipedia.org/wiki/Aircraft_principal_axes

that one might be closer to what you had, but tbh trying to think that through is a headache

AMDG

Gotta love the needless complexity. It is things like this which make things like web browsers take up more than 1% CPU usage. Bruh, I have a high end gaming laptop. I kid you not: I watch YT on this thing and I get stutters. Granted, it's older and the components are probably losing stability, but it's inconsistent and tends to be fixed by restarting the app or the computer, so I blame the software.

Semiclassical

15:32

that said, if all you're concerned with is direction, then one of the rotations can be cut out (either first or third, i forget which)

AMDG

Ah, well with my system you never forget :)

Semiclassical

for pure directions in space you really can just do spherical coordinates

AMDG

(Always cut out the last)

Semiclassical

i think it's just a matter of which perspective you take though

AMDG

I suppose a good way to describe this orientation system I've defined is a subjective 3D plane where all the 1D axes are not perpendicular to each other.

Or at least not all of them

Semiclassical

15:35

looking at that video, something does confuse me. hmm

AMDG

So how do you get gimbal lock? By not using Tait-Bryan angles or something? :P

Listen, aight. I prefer to have a domain of mod pi, not mod pi with restrictions, k.

Semiclassical

i doubt it, i think any 3-rotation system always has to fear gimbal lock

AMDG

How does gimbal lock occur? In what context?

That's the thing I still don't yet understand.

Semiclassical

yeah, i'm not great with it either. there are visualizations ofc

AMDG

Clearly gimbal lock, like all other confusing things, were invented by the government.

mohan10216

15:40

What are you guys talking about?

Semiclassical

what's true about gimbals: you can use them to put an object into an arbitrary orientation

the trouble is that, if you get your gimbals in certain configurations, then instead of having three degrees of freedom you end up with just two

AMDG

Also true about gimbals: every part can be moved independently of the other, hence my confusion about gimbal lock.

Semiclassical

which...i say that as though i properly understand it

yeah, the wiki page points out that the wording is misleading

"The word lock is misleading: no gimbal is restrained. All three gimbals can still rotate freely about their respective axes of suspension. Nevertheless, because of the parallel orientation of two of the gimbals' axes there is no gimbal available to accommodate rotation about one axis."

AMDG

From reading the wiki page, it seems to be an issue of the choice of rotation transform.

This would never be a problem if one simply used two vectors on the surface of a sphere. :)

Semiclassical

it's not a lock in the sense of "gimbals aren't moving" but of being locked into certain kinds of rotations

AMDG

15:44

Yes, but my question is how. The mechanism of rotation seems to be the cause.

Semiclassical

well, looking at the gif above: in the final situation, turning the outer gimbal around its axis has the same effect as rotating the inner gimbal around its axis

and so long as the middle gimbal stays parallel with the outer, that's always going to be the case

AMDG

Ohhh, I see now. The issue is the design of the gimbal itself.

Semiclassical

yeah. gimbals are simple but have problems because of it

AMDG

The solution is to make a gimbal whose connection points change which is probably impossible in the usual sense of a gimbal.

Semiclassical

well

AMDG

15:46

Mathematically, that's as easy as swapping two components

The two have become parallel, so...

Semiclassical

"This problem may be overcome by use of a fourth gimbal, actively driven by a motor so as to maintain a large angle between roll and yaw gimbal axes. Another solution is to rotate one or more of the gimbals to an arbitrary position when gimbal lock is detected and thus reset the device."

AMDG

This sounds like the mythical "centripetal force"

Semiclassical

but modern systems tend to avoid gimbals in the first place

AMDG

i.e. gimbal lock doesn't exist; it is purely conceptual.

Semiclassical

i mean, gimbal lock certainly exists if you're trying to fly using a gimbal-based system :P

AMDG

15:48

Well you see it says right there that the other solution is to rotate one or more of the gimbals to an arbitrary position when gimbal lock is detected.

You can just do that at the start of any rotation...

If you can move all the gimbals independently of each other, then there can't be gimbal lock... but this is saying that if you can only move two axes of three at a time, then there is the possibility of gimbal lock.

so... make a gimbal in which all of the parts can be moved independently.

Semiclassical

well, if you can move the orientations of your gimbals, yeah. not sure how that works in practice

AMDG

I mean I can come up with several ideas off the top of my head that would produce such a functionality.

Semiclassical

but for instance gimbal lock was a (minor) problem on apollo 11

b/c they stuck with a three-gimbal system, and relied on the astronauts to manually get away from gimbal lock if the system ended up there by accident

AMDG

If the gimbal is small enough, you can transmit signals wirelessly to small motors which are also sufficient to drive the motors as a form of wireless power.

Semiclassical

ah, this version of Euler angles is more what i'm used to in conventions

ZY'Z'' rotation sequence (Euler Angles)

►

AMDG

15:55

@Semiclassical Yeah, so if I'm understanding correctly, it sounds like the issue is one of the gimbals being held stationary with the other two free. That effectively creates the 2D vector system I initially described. A four-gimbal system would be required to avoid gimbal lock if one is held stationary. Simple and cheap, and I believe you just need to modify the transform matrix a bit.

Which would explain why we use 4D matrices for rotations I suppose.

Semiclassical

@AMDG sounds right. of course on apollo 11 an issue was that introducing more complexity meant introducing more failure points

hence them sticking with a system which, while it did have a failure point, was one which could be manually corrected if necessary. (not that the pilots necessarily agreed!)

but yeah, quaternions are basically just another way of saying 4D matrices

AMDG

Well the issue is solved if one can freely suspend the gimbals.

@Semiclassical And this expense makes no sense to me. 4D matrices is a bunch of FMA. That's slow compared to the cost of computing $e^{i\theta}$ on a GPU these days, especially when one considers you can do these operations with vec4 data types.

Yeah, sure, there's specialized hardware for computing these matrices on vectors; that doesn't change the fact that you can still get something that's faster which doesn't involve these particular matrices per object that you wish to reorient.

Semiclassical

well, when i say quaternions = 4D matrices i mean they're mathematically equivalent. computationally quaternions are a bit nicer b/c you don't end up doing any actual matrix multiplication

AMDG

Mathematically, the gimbal system, if we're applying it to the system I described, is three gimbals existing in the same space and operating independently of each other, something which is physically impossible but not mathematically impossible since these surfaces are infinitely thin.

Semiclassical

writing ij=k can be represented in terms of 4-by-4 real matrices, but those matrices aren't needed if you take ij=k as a definition

AMDG

16:06

Sure, but I'd rather just avoid matrices altogether if possible. Those madds are expensive.

Semiclassical

that's what i mean, though. quaternions don't require matrix multiplication

AMDG

Ok, so how do quaternions work exactly?

Semiclassical

well, there's two questions there. one is how they represent rotations and i'm not so familiar with that---see the wiki article i guess. but that's not needed to say how quaternions work algebraically

you write a quaternion as such: $q=t+x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$

so it's a 4-component object

the rule for adding quaternions is just what you'd expect: $q+q'=(t+t')+(x+x')\mathbf{i}+(y+y')\mathbf{j}+(z+z')\mathbf{k}$

the harder bit is how multiplication works

AMDG

If I remember correctly, each imaginary unit is defining a new axis. If I wanted to intuitively define the notion of a vector describing direction in $R^3$ using imaginary numbers, the point on a sphere would be represented by $x + iy +jz$, but this doesn't define orientation, so we add another axis k to describe an orientation $x + iy + jz +kw$

Semiclassical

$ij=-ji=k, jk=-kj=i, ki=-ik = j$

it's something like that, yeah. there's an Euler formula which makes that precise

oh, and $i^2=j^2=k^2=-1$

AMDG

16:12

Yeah except I don't think I'd do any operations involving i,j, nor k themselves. Instead, I would only specify x, y, z, and w.

Semiclassical

yeah

AMDG

Hm, I should be able to describe this as a multi-valued function of the exponential.

Semiclassical

you have $qq'=(tt'-xx'-yy'-zz')+(yz'-y'z)i+(zx'-xz')j+(xy'-yx')k$ (might be off by a minus sign or so)

AMDG

Ah, yes, the lovely mystery algebra. :P

Semiclassical

so therefore you can think of it as taking 4-vectors $(t,x,y,z), (t',x',y',z')$ to $(tt'-xx'-yy'-zz',yz'-y'z,zx'-xz',xy'-yx')$

so no matrix multiplications, just simple multiplications and additions

AMDG

16:14

Noice. Remind me again why we care about matrix transforms for rotations? LOL

Semiclassical

depends on context, really

AMDG

Expressions of that form are far cheaper to compute compared to a 4x4 matrix multiply.

Imagine all the frames we could compute!

Semiclassical

a lot of the times in physics we do stuff like $e^{i\theta/2 \sigma_x}=\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\sigma_x$

and then just make use of the algebra rules for Pauli matrices, which are pretty much just the quaternions in another form

AMDG

Yes, so I'm wondering what the quaternion version of the exponential is

Semiclassical

same as the physics one, actually

AMDG

16:17

Which is that?

Semiclassical

suppose you want to make a rotation of $\theta$ around the direction $\vec{u}=(u_x,u_y,u_z)$

AMDG

That would be great if I could just use the quaternion exponential to compute orientations. It would be stupidly cheap.

Semiclassical

then the corresponding Euler formula is just $$q=e^{\frac{\theta}{2} (u_x \mathbf{i}+u_y \mathbf{j}+u_z \mathbf{k}) }= \cos\frac{\theta}{2}+(u_x \mathbf{i}+u_y \mathbf{j}+u_z \mathbf{k})\sin\frac{\theta}{2}$$

the way you apply it to a vector $\vec{p}=(p_x,p_y,p_z)$ is to write it as $p=p_x\mathbf{i}+p_y\mathbf{j}+p_x\mathbf{i}$ and write $p'=qpq^{-1}$

i am stealing this blatantly from the wikipedia page, mind

this becomes the same as the physics one if one identifies $\mathbf{i}=i\sigma_x$

AMDG

Yeah, but I'm meaning more in the form of something that combines $e^{i\alpha}$, $e^{j\beta}$, and $e^{k\gamma}$ into a single function such that $f(\alpha, \beta, \gamma) = e^{g(\alpha, \beta, \gamma)} = x + iy + jz + kw$.

(and what the generalized formula is)

Semiclassical

fair

AMDG

16:27

so basically we have $g(\alpha,\beta,\gamma) = \ln(x + iy + jz + kw)$ and I want a formula which describes this for all branches of quaternion logarithm.

Semiclassical

16:37

well, if you have $g(\alpha,\beta,\gamma)=i\alpha+j\beta+k\gamma$ then what i have above will suffice

but if you mean smething like $e^{i\alpha}e^{j\beta}e^{k\gamma}$ then yeah

AMDG

Well when I say quaternion exponential and logarithm, I'm not meaning for quaternion arguments, but for quaternion outputs.

Quaternion arguments is pretty straightforward to compute the exponential of.

Semiclassical

yeah, Euler makes it simple

Alex

Hi, how can I calculate $\displaystyle \int_{0}^{+\infty} \frac{\cos(xt)}{{\rm cosh}^{2}(x)}{\rm d}x$?

Semiclassical

also, if your quaternion happens to be a imaginary unit quaternion (i.e., $x=0$ and $y^2+z^2+w^2=1$) then it's $iy+jz+kw=e^{\pi/2 (iy+jz+kw)}$

AMDG

For describing my system, it's just nested rotation transforms, so the same will be the case here for computing a quaternion representation for orientation.

That's a convenient trick for unit quaternions.

Semiclassical

16:44

or $\ln(iy+jz+kw)=i(y\pi/2)+j(z \pi/2)+k(z\pi/2)$, with the only branches occuring by replacing $\pi/2$ with $2\pi n+\pi/2$

AMDG

I see

Semiclassical

i think that's right anyways

you do see remnants of this in physics, but we've collectively moved away from describing them in terms of quaternions

AMDG

Yeah, it's all differential equations if I remember correctly.

Semiclassical

in special relativity you usually implement the pluses and minuses via the metric tensor

while in QM it's by means of the Pauli matrices

Alex

$\cos(xt)$ I think we can to expand in series but what's about $\cosh^{2}(x)$ ?

robjohn

16:55

@Semiclassical Aren't there $i(tx'+t'x)$ etc terms in there?

or am I misremembering?

if it is like this, then there are just as many multiplications as multiplying a vector by a matrix in $\mathbb{R}^4$

Alex

$\cos(xt)=\frac{e^{ixt}+e^{-ixt}}{2}$ and $\cosh(x)=\cos(ix)$ hence $\cosh^{2}(x)=\cos^{2}(ix)=\left(\frac{e^{i^{2}x}+e^{-i^{2}x}}{2}\right)^{2}$. So I'm looking for $\displaystyle 2\int_{0}^{+\infty} \frac{e^{ixt}+e^{-ixt}}{(e^{-x}+e^{x})^{2}}{\rm d}x$. Is it sense?

robjohn

@Alex You might be able to proceed as in this answer

I think that should work

AMDG

@robjohn He's referring to the product of two 4-vectors. The equivalent in matrices would require that many perhaps as two 4x4 matrices that have been reified from row and column vectors. A number of terms have zero coefficients, so in the verbatim implementation, yes, there are that many multiplies, but in the end, it is fewer multiplies this way using $q\cdot q'$.

robjohn

@AMDG you multiply two quaternions to get the result of multiplying a matrix by a vector, at least as I remember

same number of multiplications, I thought

Semiclassical

@robjohn oh ugh, you're right

user193319

17:07

Problem: Let $f$ be entire such that $\lim_{|z| \to \infty} |f(z)| = \infty$. Show that $f$ is a polynomial.

Semiclassical

yeah, this is the problem with trying to multiply it out in my head

user193319

Here is my attempt. Consider the analytic function $g : \Bbb{C} \setminus \{0\} \to \Bbb{C}$ defined by $g(z) = f(\frac{1}{z})$. Note that $$\lim_{|z| \to 0} |g(z)| = \lim_{|z| \to 0} |f(\frac{1}{z})| = \infty,$$ which tells us that $z=0$ is a pole of $g$, say of order $m$. This means that the Laurent expansion of $g$ at $z=0$ in $\Bbb{C} \setminus \{0\}$ looks something like:

$$g(z) = a_m z^{-m} + a_{m+1} z^{-m+1} + ... a_1 z^{-1} + a_0 + a_{-1} z + a_{-2} z^2 + ...$$
which gives us Laurent series expansion of $f$ after replacing $\frac{1}{z}$ by $z$: $$f(z) = ... + a_{-2}z^{-2} + a_{-1} …

Does that seem right?

Alex

@robjohn Thank you. I'm going to read the solution

robjohn

@user193319 this question might help

AMDG

Well the important thing is that I'm not trying to go for quaternions as presented by Hamilton himself. If anything, it is a sort of inverted quaternion in the sense that an orientation is a single real axis oriented arbitrarily in $\Bbb{R}^3$ with an angle associated with it.

user193319

17:11

@robjohn Which problem did you have in mind? Does my solution seem correct, though?

leslie townes

user: everything in that argument is true, although going from the limit involving g at 0 to g having a pole is maybe leveraging at least some knowledge of the classification of isolated singularities of analytic functions.

robjohn

@user193319 you need to verify that this is a pole (not an essential singularity), but other than that, it looks good.

user193319

Okay, thank you two for the input! I'm studying for the complex prelim, so I pretty much have every tool at my disposal (unless the problem explicitly tells me not to use a certain idea/theorem).

AMDG

@Semiclassical In other words, my system describes orientation as a complex vector, $\vec{u} = \langle \alpha, \beta\rangle, \vec{X} = \vec{u} + i\gamma$. Three real parts and one imaginary part instead of three imaginary parts and one real part.

Semiclassical

for a reference on quaternions, this seems pretty good: neil.dantam.name/note/dantam-quaternion.pdf

AMDG

17:16

Supposing that notation makes sense in itself. I'm sure you get the gist, though :)

Semiclassical

they also talk about dual numbers in there

AMDG

Bruh, I hate the 5min comment timeout

leslie townes

user: in that kind of exam context you can leverage the fact that lots of those things have many definitions that the examiner knows to be equivalent, and depending on the definition adopted, it may be clear that you can make moves like that. you might just be applying a definition. :)

AMDG

u should actually be a pair of vectors, $(e^{i\alpha}, e^{i\beta})$

Also forgot a comma. $\vec{X} = \vec{u} + i\gamma$

But in this case, the imaginary term would be more properly written as $i e^{i\gamma}$ rather

Meh, I give up. You get what I'm trying to say. :L

I just don't know what the notation would be.

Semiclassical

one suggestion i'd make in trying to describe your system (along the lines of what you said initially) would be to have your initial point be in the (0,0,1) direction

that's more standard

Ted Shifrin

17:25

@user193319 Use Casorati-Weierstrass.

robjohn

which was mentioned in an answer to the question I linked

Semiclassical

@ted re: our argument a day or so ago, the Wikipedia page here is kind enough to actually be precise about which symbols it's using: en.wikipedia.org/wiki/…

Ted Shifrin

Your proof is circular. Your Laurent expansion assumes a pole at infinity.

AMDG

@Semiclassical Well I think I'd prefer to define it from a more intuitive notion using complex numbers to define orientation. Namely the intuitive notion of orientation as $\vec{x} + i\vec{y}$ where $\vec{x}$ is a real number line oriented in $\Bbb{R}^n$ and $i\vec{y}$ is an imaginary number line perpendicular to $\vec{x}$.

robjohn

$\lim\limits_{|z|\to\infty}|f(z)|=\infty$ and $\lim\limits_{|z|\to0}|g(z)|=\infty$

user193319

17:28

@TedShifrin Which Laurent expansion assumes a pole at infinity? $g$'s or $f$'s?

Semiclassical

still @ ted so $\otimes_{\text{outer}}$ for tensor product of vectors implemented as outer product vs $\otimes_{\text{Kron}}$ for its implementation as Kronecker product

AMDG

wha- oh

Ted Shifrin

Your Laurent expansion of $g$ assumes $f$ has a pole.

leslie townes

the conceptual touchstone here which robjohn has linked is that if a non identically zero analytic function on an open set has an isolated zero, then it has a multiplicity, i.e. there is a positive integer m with the function taking the form (z-a)^m h(z) near the zero a, and h is analytic and nonzero at a.

or, casorati-weierstrass.

Ted Shifrin

Typical abuse of notation, Semiclassic,

Semiclassical

17:30

i mean, gotta have some notation for those operations, and having to write $\otimes_{\text{Kron}}$ is tedious

user193319

Hmm...I see...So, there is no way of fixing my proof?

robjohn

@user193319 cite and use Casorati-Weierstrass. Did you look at the post I linked?

Semiclassical

also i'm not sure physics books ever define Kronecker product in those terms explicitly, but

the way they organize the basis for tensor products amounts to it

Ted Shifrin

I work with tensor product way more than Kronecker product (almost never).

@leslie I agree that we cannot have a zero of infinite order.

user193319

@robjohn Yeah, I'm taking a look at it right now, actually. Thanks!

leslie townes

17:33

so you can identify the order of the pole from 1/f(1/z) which by hypothesis has a removable singularity at 0. this is done in the robjohn link. or something equivalent.

Semiclassical

e.g. if you have basis vectors $e_0,e_1$ for each vector space, then the tensor product gets organized as $e_0\otimes e_0, e_0\otimes e_1, e_1\otimes e_0, e_1\otimes e_1$

leslie townes

we hear a lot about weierstrass, but we never hear more about casorati. what was his deal?

user193319

So, should I redefine $g$ as $g(z) = 1/(f(1/z))$?

Semiclassical

and then representing these as basis vectors $e'_0,e'_1,e_2',e_3'$ gives you the Kronecker product

obv. there's not a unique choice, but it's as good as any if you insist on having everything represented with matrices

Ted Shifrin

You should know the proof of and use C-W. Super important..

leslie townes

17:36

felice casorati was born in pavia and died at age 54. he shares a name with a painter who was born near his death, apparently unrelated, and more english-wikipedia-famous than he is.

Ted Shifrin

@leslietownes Actually dunno!

leslie townes

italian wikipedia is no better. apparently he lived to prove the casorati-weierstrass theorem.

Semiclassical

huh. Casorati matrix : linear difference equations :: Wronskian : linear differential equations

leslie townes

i think it's particularly insulting if you do some good math and then someone comes along with exactly the same name and gets a better reputation in another field.

Semiclassical

there's a suggestive comment in Google Books (link):

"At the ICM in Paris in 1900 , Volterra gave a celebrated lecture on the roles Betti, Brioschi, and Casorati had played in the rebirth of mathematics in Italy after political unification ."

leslie townes

17:38

there is so much stuff about the other felice casorati, even on italian wikipedia.

Semiclassical

books.google.com/…

leslie townes

he had no students, but the same advisor as beltrami.

Semiclassical

for the french readers among us

leslie townes

also the same advisor as some guy named cremona that nobody has ever heard of.

Ted Shifrin

Oh, Cremona is well-known in algebraic geometry circles.

I will read Semiclassic's link a bit later.

leslie townes

17:42

cremona, advisor of veronese, some other person nobody has ever heard of.

until now.

i'm putting people on to all of these forgotten italian mathematicians.

Semiclassical

15 pages

but not especially dense looking

leslie townes

basically the whole algebraic geometry branch of the mathematics genealogy family tree seems to run through casorati's advisor.

i respect him for not going that route

casorati was like, i'm not the type of guy to chase a trend. i'm going to make a trend. like some other guy you may have heard of, i dunno, weierstrass? that's the route that i am taking.

Semiclassical

makes me think a little of the Euler vs Gauss comparison

Euler as the pioneer / explorer, Gauss as the conquerer

leslie townes

gauss was better. i say this not because he is my mathematical ancestor but because he was objectively better.

euler could have been good if he applied himself.

Semiclassical

euler was happy to report what results he had, whereas gauss insisted on knowing a subject to completion first

i'm not sure one is better, but they are different

Euler covered more subjects than Gauss, i think, but Gauss penetrated his subjects more thoroughly

Ted Shifrin

17:47

Veronese is very well-known, @leslie. You're just an ignoramus.

leslie townes

gauss is the beatles, and euler is wings.

Semiclassical

though in terms of personality i appreciate Euler more

leslie townes

ted: i'm pretty sure nobody has ever heard of any of those people.

Ted Shifrin

You know them personally, Semiclassic?

Semiclassical

lol

Ted Shifrin

17:49

Well, no one has heard of Cauchy, either.

leslie townes

that's fine. we don't need to hear about the french. although that doesn't stop them, does it

Semiclassical

i'm basing my judgment on poorly remembered historical references, of course

leslie townes

euler was chiefly known for getting a little bit 'in his cups' and then beating up people who called him "you-ler"

Semiclassical

but see for instance irishtimes.com/news/science/…

leslie townes

why is he wearing that shirt with that scarf? some genius.

he looks like he dressed up to impress his curtains. or perhaps a sofa.

his shirt looks like a kitchen towel.

i'm going to do real work today, i promise.

Semiclassical

17:52

i'm in the "waiting for updates" part of the winter break

what classes will i be assigned to? will it be in-person or is omicron surge going to mess that up? etc

leslie townes

do they give you any notice at all? worst notice i got was hearing on friday evening what i'd be teaching monday morning. when previously i had been told that i was not teaching anything at all.

took a bit of doing to get a copy of the textbook over the weekend.

Semiclassical

it's usually better than that

for the fall there was more lead time b/c of what i was attached to

leslie townes

my wife has her schedule but i think there is still some static around whether any of it might be in-person, or become in-person.

Semiclassical

but if i end up attached to bog-standard intro courses then the notice will be slow

i haven't heard anything yet about that, but given that the omicron surge hasn't fully struck MN yet

i'm not sure what to expect

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$Mathematics$ Mathematics