yeah, but we can generalize this though to general setting.

Hello @TedShifrin :))

Heya @Ali :)

@TedShifrin I was reading about alternating algebras and that grading the tensor algebra and quotienting it can have their order swapped. Is this the most natural construction for it?

I am not familiar with alternating algebras. Do you mean exterior algebra?

Well let m be an ideal of k dim tensor algebra over E

where elements of m are of the form x1 x x2 ... xn

but have repeated terms

then quotient out m from k dim tensor algebra over E

then direct sum for all k

I guess the exterior derivative is a natrual cohmology thingy T^k(E)/m --> T^(k+1)/m

Sorry for not typing in Latex I can't see my keyboard too good

No, I'm not talking about exterior derivative. I'm talking about exterior algebra. That seems to be what you're doing.

Oh ok I am confusing things

Why was I calling it an alternating algebra

sorry for that yes Exterior algebra

Basically you skew-symmetrize tensors. You can do it either by quotient or by skew-symmetrizing (summing $\sum (-1)^\sigma \sigma$ applied to the tensor), probably with a $1/k!$ in there.

Which one do you prefer?

I generally do the latter, but I have certainly used the former too

Generally, a sub-thing is easier to work with than a quotient-thing.

hmm I can see the former being more difficult to actually work with

It depends what you want to do.

Anyhow, not much point worrying about it until you confront a practical instance.

Certainly for undergraduates I've done the former. Let me look back at notes for my grad geometry course and see what I did there.

Ok thanks

Yeah, I used skew-symmetrization there.

nice I will read up in 2 weeks

after exams

I can see in some algebraic cases the latter might be more useful.

Also @Ted have you ever seen something like this before?

I'm not going to think about that. I'm trying to work on a diff geo question on main.

@MithleshUpadhyay Huh? I'm saying it is text copied from a pdf due to the character style. I don't know where it's from. I'm just telling you what kind of file it was copy/pasted from.

ok @Ted I think I will go to bed now anyways, thank you for the discussion :)

Night, @Ali.

@TheGreatDuck , I got it, thanks.

hi chat

Anybody seen Steamy Root?

not recently

spectral-categories.blogspot.ca/2017/01/…

rehi DogAteMy

rehi Duck

(waits to see if anyone responds)

just as I thought ... invisible.

sorry

one handed

LOL, I won't ask. ... I know you were too busy ducking.

drinking tea

hows it going?

doing just fine ... winding down ... how're you? ready for school to start up again?

yup

tomorrow it starts

mixed emotions, but in the end you're glad

nah. I'm very happy.

what math are you taking this term?

abstract math

some kind of proofing class

What year are you?

junior

OK ... I used to try to get my advisees to take that sophomore year.

to be fair, I am primarily a CS major. I added the math major last spring.

Ohh, so it's half a repeat of the discrete math course you took for CS. But more proofs.

also, I got a D+ in calc 1 the first time so that probably threw things off lsightly

@TedShifrin no I actually skip discrete math for math. The CS one counted for both.

hello guys

I will also recommend to you the book I mentioned to mick. Houston's How To Think Like a Mathematician.

I'll think about it.

LOL ... you can always ignore me :)

Professor Elmendorf actually has a book suggested for the class if we need extra material. I don't remember the name off the top of my head.

hi @Kushal

There are a lot of books for that course. I wasn't very happy with the one I used the last time I taught it.

Apparently the class doesn't actually have required books. He just recommended it in the syllabus to anyone interested in extra reading (as in beyond the course content).

@TedShifrin How are you doing?

Duck, I was tempted to teach the course without a book, to break students of the habit of trying to copy things out of the textbook. But I never quite got there.

Doing well, @Kushal, and you?

@TedShifrin to be fair, I hear the professor is a bit smarter than most professors. Just from what other students have said. Rumors are rumors. :p

I certainly always tried to be a dumb professor :D

@TedShifrin fine. Ok I am looking for a introductory probability book with a good exercise. Do have something in mind? Besides Sheldon Ross, and William Feller.

Well I've heard he's made a few important advancements in group theory.

And he also has a tendency to teach complex integration in calc 2 from what other students have said.

Those are the two totally standard books, @Kushal. There's also one by two electrical engineers at MIT which is quite good. Hold on a sec.

Duck, that is cool, but in the intro to proofs course it's part of the task not to blow away most of the students.

it's not intro to proofs. It's abstract math. Discrete math was technically the intro to proofs class.

unless I misunderstand?

I think you do.

ah

@Kushal: Bertesekas & Tsitsiklis. Google it.

@TedShifrin Ok me see. Thanks.

I heard the main difference between the two was that after predicate logic the math version did a bunch of proofs whereas we learned haskell.

I just kind of assumed that was where the intro to proof came in.

Duck, as Associate Head, I had to examine a lot of course transfers. I'm just saying that the math version of "abstract math" is more proof oriented than "discrete math," but there are similar contents.

most of the students I know who have said they took the class (with him) said he wasn't too difficult to understand. If anything, he's just very strict about writing in complete sentences, which makes sense.

If you haven't written proofs a fair amount, this is the right course for you, and maybe this guy will push you hard, which is good for you. I'm just saying, based on having taught it a few times, that most students slow the course down and need coddling.

I'm totally supportive of complete sentences. Totally.

@TedShifrin well I'm actually not a course transferer per se. I'm currently a double major.

unless you count honors as a third major but I tend to not mention that. Sounds.. self-inflating to mention it. :p

I didn't say you were. I just said I dealt with hundreds of transfers and examined course syllabi from lots of places.

ah I see

no, honors is not a major.

(well at my university it's actually an honors college)

you're at one of the branches of Purdue?

what gave it away?

I looked up the prof's name, silly.

:p

Some schools have honors colleges, some have honors programs ...

I think your anonymity is safe with Duck.

of course

I dont care really

I was just curious how you knew. XD

I thought maybe I mentioned it at one point.

I hope you kick butt in the class.

Nope.

I try not to pry.

of course. I just mean I probably brought it up at some point when talking to other people so you might have just remembered.

:p

To be fair, I actually have a pretty decent list of things I've wrote down that at some point i would want to know how to prove (if only to avoid more hand-waving as people put it).

BTW, your university website absolutely sucks.

granted, obviously irrelevant to the class. I just mean I am very much interested in learning how to prove things formally rather than just algebraic identities and whatnot.

No, no, you won't be doing algebraic identities.

@TedShifrin I know

I mean that my knowledge of proofing is mostly algebraic type things.

You'll be doing different proof methods, induction, understanding functions and relations, some modular arithmetic or introductory analysis stuff.

The sort of things I would prefer to prove (that I believe are true) are on the more abstract level.

I just mean it's something I am definitely interested in learning. That's all.

:p

I mean I suppose I'm interested in math in general but I just mean it's a class that interests me in particular.

I hope that made sense.

Stay enthusiastic, and be prepared to bug him in office hours. You may have questions and concerns that aren't appropriate for derailing the class :)

trust me. I'm used to that by now. Part of the honors thing is doing stacked courses which are basically extra project things. So I'm used to meeting professors during office hours. :p

:)

thanks for the advice

Well, I don't mean to butt in.

you're not at all

But I had some students who were very disruptive (even though I encouraged disruption to a large extent). I repeatedly had to tell them to stop and come see me.

I'm not that type. I'm more the sort that doesn't ask a lot of questions because I usually understand fairly well.

and ask maybe 2 or 3 questions to clarify precise things

At some point, if you want, I can send you my exams from that course, Duck.

maybe

I'll think about it. :)

No obligation :)

bleh, it's just cold enough inside to be annoying

heya @Semiclassic :)

hey

@Semiclassical If you don't go outside at night you will not be annoyed by the cold and you won't even notice it.

I escaped Atlanta from the icemageddon that never happened and had 70º+ in San Diego today. So I know your pain.

um. I am inside.

then turn your heat up.

XD

I deduced that from the "inside," @Semiclassic.

Duck, that costs lots of money.

then grab a blanket

I should find a thicker pair of socks.

XD

indeed. Thick socks are a must this time of year.

OK, I'm disappearing. Time for dinner. Take care, Semiclassic and Duck.

later

@TedShifrin cya

@TedShifrin btw.I know you sometimes acted annoyed when I discussed the weird piecewise constant integral stuff.Mostly my reason for looking at it is because it makes solving differential equations easier most of the time.So I just look at it as recreation.I figure a healthy understanding of valid faster methods is a good thing.Laplace transforms are good for solving equations and work but one has to admit any opportunity to bypass them and use versions of the easier methods is a good one.

So I just figure it's something nice and relaxing to do. If I improve my knowledge, great. If I merely spent time in the evening practicing math, also great. It's basically just for fun. I ask about it on here cause I figure other people might enjoy it as an interesting 'different' way to try doing things. :-)

@Semiclassical what are you up to on this fine evening?

nothing interesting.

you could be watching the mgs2 speedrun

@MikeMiller the what?

'the' mgs2 speedrun?

idk what that is

i know of speedruns

but what is mgs2

good evening

Maybe nobody answered them because not many people know the topics

I decided to go all in

@IwillnotexistIdonotexist Interesting username, by the way.

@AkivaWeinberger I have been thinking about having the questions moved to MO to try for better luck

And yes, my name is my way of being self-contradictory

TheGreatDuck what are you trying to do

@GFauxPas hm?

you mean my question?

yes

Yeah maybe that's a good idea @IwillnotexistIdonotexist

look at alternate versions of the derivative/antiderivate/differential equation rigorously

Good luck on deciphering existence

you want to have a system where a non-constant function can have a zero derivative?

yes

:/

A nonexistent existence is valid if you think carefully about it

just because you're curious?

here's a good example why

try solving the differential equation y'' + 2floor(x)y' + floor(x)^2y = 0

let me guess. You'd use the laplace transform?

I'd question whether that has a solution at all

I solve the auxilliary equation r^2 + 2r*floor(x) + floor(x)^2 = 0

when I see "floor" and "derivatives" in the same place I get suspicious

which has a double solution -floor(x)

(different equation I was thinking of)

therefore

the solution is of the form

How would you use Laplace transforms, by introducing an auxilliary variable and treating it as a partial de?

$x = t$ or something?

I didn't say I was using a laplace transform

you suggested I would use the LT

I'm thinking if that's possible

I'm saying you would. x is the dependent variable.

that's not the point

y = C(x)e^{-floor(x)x} + D(x)xe^{-floor(x)x} is the solution where D and C are piece-wise constants and y is continuous

consider $y'' =xy$

how would you solve that using LT?

use dollar signs for mathjax

I don't actually know how via laplace transform

Laplace transform is generally for constant coefficients

oh

more generally, if $y$ is a function of $t$, the constants can be constant WRT $t$, I think. my diff eq is rusty

y is dependent on x

then it's not a Laplace transform situation afaik

anyway then

so what equation are we dealing with

$y'' + 2floor(x)y' + floor(x)^2y = 0$ has the solution $y = ce^{-floor(x)x + \frac {floor(x)^2}{2} + \frac {floor(x)}{2}} + dxe^{-floor(x)x + \frac {floor(x)^2}{2} + \frac {floor(x)}{2}}$

put it in dollar signs please

I guess I'll manage

How are you defining $\dfrac {\mathrm d \operatorname{floor}}{\mathrm dx}$

huh?

what's the derivative of $e^{-\operatorname{floor}(x)x}$?

it's a weak solution

what does that mean

it is undefined at integer points

then how are you differentiating it

I'm not. That's the solution to the differential equation.

If it's a solution to a differential equation, it means it satisfies the differential equation

thats what a solution is

you're telling me there exists a $y$ that satisfies $y'' + 2 \lfloor x \rfloor y' + \lfloor x \rfloor ^2 y = 0$

what is the integral of floor?

hold up

I'm making a point

are you telling me that there is a function that satisfies that equation?

I didn't finish making mine

y' = floor(x) has the solution of the indefinite integral of floor, right?

let's say yes

well no

indefinite integrals arent solutions

the integral of floor is $x*floor(x) + floor(x)^2/2 + floor(x)/2$

let's make it $\displaystyle \int_0^t \lfloor x \rfloor \, \mathrm dx$

@Semiclassical agdq is on

@GFauxPasby your reasoning how can the heavyside function be in the solution to a differential equation?

the heaviside function has a distributional derivative at its point of discontinuity

...it's point of discontinuity is a jump discontinuity

which is why you need to change the definition of the derivative if you want to say something like

nothing about it's values make it any more special than the jump discontinuities of floor

$\dfrac {\mathrm du}{\mathrm dt} = \delta (t) $

are you referring to that?

but the dirac delta function is an object that exists by proof, not by definition.

yes

isn't it something one derived as the derivative of heavyside?

well you define it, and prove that it makes sense and that it does what you want it to do

eh there's more than one way to define it, but to define it that way seems very handwavy to me

no I mean I thought given nothing about the derivative but the limit definition one can actually derive the dirac function. Is that wrong?

Well if you're okay with veering off topci

are we?

no not really

technically the whole solution existing was off topic

Then yes, it is wrong

"look at alternate versions of the derivative/antiderivate/differential equation rigorously"

^what we were discussing at first. :p

true

but anyway, the dirac delta cannot be defined using epsilon-delta

I was saying it bypasses the laplace transform sometimes.

there are ways to make it work

carefully

take this differential equation instead: y'' + 2y' + y = u(x) where u is heavyside

you either have to change the way you think about functions, or the way you think about limits, or the way you think about derivatives, or something

i dont know for sure but I believe laplace transform might solve it

ah fair enough

that you can solve using laplace transforms, sure

what if you used the method of undetermined coefficients, instead?

but if you want to consider $x \in \mathbb R$ instead of just $x > 0$

I've never tried undetermined coefficients with dirac delta, I guess it might work

well let me back up a second. I'm getting ahead of myself

Let's solve the differential equation in the system where this is true: "if and only if a function is piecewise-constant does it have a derivative of 0 for all real numbers"

im pretty sure based on undetermined cooeficients with just a constant on the RHS the coefficient is that constant

okay, let's say I believe such a system exists

...why would you claim it doesn't?

you'd have to win me over though

because you're saying that it has derivatives at points where it's not continuous

and that those derivatives are zero

you just won't let me finish will you? XD

I said "okay, let's say I believe such a system exists" :P

go on

the roots of the auxilliary equation would be -1 and -1

no I think you use a first degree polynomial in that case

at least

therefore the solution would be $y = C(x)e^{-x} + D(x)xe^{-x} + h(x)$

no

y'' + 2y' + y = 0

would be the homogenous equation

correct

that's a second degree auxilliary equation

this isn't the way I'm familiar with

those arent constant coefficients

y'' + 2y' + y = 0 has constant coefficients

...

so it will be $y_h = C_1 e^x + C_2 e^{-x}$, no?

yes

except the one flaw you missed

okay, I agree

which is?

piecewise constants have derivative 0 in this system

so wouldn't the coefficients be piecewise constant?

:-)

no I don't think so

um yeah

$C_1$ and $C_2$ are being pulled from a scalar field

not from a space of functions

sigh

no?

I don't know, I've never done this

we are saying $y$ can be expressed in terms of the basis $\{e^x, e^{-x}\}$

the c's come from the logic of antidifferentiation

"all these functions fulfill the antiderivative because c has derivation 0 and addition rule"

I prefer to look at it from a linear algebra perspective, I guess that's why we're not seeing eye to eye

piecewise constants have derivative of 0 so c gets an upgrade

constant coefficients means

in that system, it's all piecewise linear combinations

we are solving an equation of the form

$P(D_x)y = 0$

that's what constant coefficients menas

agree or disagree?

Huh. Any degree $d$ map $S^2 \to S^2$ with $|d| > 1$ which is furthermore $C^1$ has infinitely many periodic points. That's kind of interesting.

$P$ is a polynomial in $D_x$

that's an an ODE with constant coefficients means

$P(D_x)y =\text{functions in the space}$

$y = C(x)e^{-x} + D(x)xe^{-x} + u(x)$ is the solution to the equation $y'' + 2y' + y = u(x)$ in the alternate system where u is heavyside. This is a fact. I know it to be so. :-)

anyway. wanna know the important bit?

sure

do piecewise-constants form a scalar field?

that's important

the "weak solution" en.wikipedia.org/wiki/Weak_solution to the equation is the solution in that alternate system such that y is continuous

if they don't, it's not a useful system

not familiar with this topic, reading

@BalarkaSen weird

i don't know it much either. Basically, it's the solution that best works and we just ignore the issue at the point of discontinuity. I.e. what you said about dirac delta.

better still

solutions to differential equations are always continuous

Wikipedia says to use weak solutions means youre looking for distrbutions

I'm okay with that

shrugs heard it in an answer to a question once.

anyway

@MikeMiller The reference seems to be Shub-Sullivan

solutions to a differential equation are always continuous?!?

except when diverging or undefined

let me think about that

:p

it's a fact

feel free to think about while I carry on

except when undefined, boo

@GFauxPas Depends what you mean by "differential equation" and "solution".

this means $y = C(x)e^{-x} + u(x)$ and $y = D(x)xe^{-x} + u(x)$ are each individual continuous solutions for some undetermined C and D

if you're dealing with distributions and not just functions

I'm not sure what a continuous distribution is

small changes in $f$ results in small changes in $L[f]$?

i dont even know what a distribution is

we're solving the differential equation y'' + 2y' + y = u(x)

I'm about to pretty much finish solving

a distribution is like a function, but is never evaluated at points. it only works by affecting other functions, such as through multiplication

Like let's say $L$ is a distribution on the space of all functions

@GFauxPas That's assumed in the definition of distribution (though you should know what you mean by "small changes in $f$"). A continuous distributional solution just means... a distribution that's representable by a continuous function.

you couldn't ask me what $L(3)$ is because that's now how we use distributions

But you can ask what $L\sin(3)$ is

therefore $ye^{x} = C(x) + u(x)e^{x}$ and $ye^{x} = D(x)x + u(x)e^{x}$ which means $f = C(x) + u(x)e^{x}$ and $f = D(x)x + u(x)e^{x}$ where f is continuous

Mike I was being imprecise because I wasn't sure what I meant

which means C(x) = -u(x)c where c is an arbitrary constant based on simple comparison of the limit at 0

Like you never evalute $\delta(x)$ at any $x$, it doesn't make sense

@GFauxPas but the dirac function isn't in the final solution

but you can multiply it by a function and then integrate it, like $\int_0^{2\pi} \delta(x-\pi/2)\sin x \, \mathrm dx$

I'm just saying what a distributional solution would be

it's just a normal impulse differential equation

it's used all the time in engineering from what i hear and is solvable

these impulses and distributions were used by physicists before mathematicians were able to find a way for them to make sense

when engineers used them they used them as an approximation

and it gave solutions that seemed to work

but mathematicians don't like approximate solutions being treated as exact

do you argue that the laplace transform method gives inaccurate results?

for several reasons

It should be true that as long as $u$ is continuous, the solution to that ODE is $C^1$.

all I am stating is that I can give the same solutions it would give without using it.

I'm saying you have to know what you're saying

you're saying a constant is a special type of constant, and a function is a special type of function, to which I say "okay, I'll hear you out. but be careful"