Mathematics - 2015-10-08 (page 2 of 2)

15:01

Go the way with the system of equations for a really nice solution (and very easy at the same time).

UserX

I dont get how that would help. Havent solved them but I guess I+J would end up in something easy to calculate, I-J too and then I'd do I+J-(I-J) right?

Chris's sis the artist

@UserX you get a system of equations from which you get $J$, the integral you need.

UserX

That sounds lengthier than calculating J elementarily though. Haven't tried the system way yet.

Chris's sis the artist

@UserX Using the system of equations makes things easier.

For instance $I+J =\int_0^1 \frac{1 + 1/x^2}{(x - 1/x)^2 + 2} \ dx$ is immediately done after making the variable change. The same thing for $I-J$.

UserX

What variable change? I'm using partial fractions and it's HUUUUGE

Chris's sis the artist

15:09

@UserX What is the derivative of $x-1/x$? :-)

UserX

1/X^2+1

Wait

Lol

Chris's sis the artist

@UserX :-) You have that in numerator. Thus the integral becomes extremely easy. You make the variable change $x-1/x=y$.

Then $(1+1/x^2) dx = dy$. Fit the new integration limits, perform the elementary integration and you're done extremely fast.

@UserX Think that both $I+J$ and $I-J$ can be computed in one line each one. You really have a very fast way at hand.

:D

UserX

15:25

That was awesome

I knew you would be the right person

Chris's sis the artist

@UserX :D

The stuff I was playing with

$$\int_0^1 \int_0^1 \int_0^1\frac{\ dx \ dy \ dz}{(1+x+y+x y ) (3
+x+y+z)(1-x-y-z+x y+x z+y z+7 x y z) }$$

UserX

That feels like it doesn't have a nice closed form

Chris's sis the artist

@UserX It has a very nice one.

@UserX $\displaystyle \frac{\pi^2}{36}$

UserX

Where do you find these?

Chris's sis the artist

@UserX I don't find them, I create them daily. :-)

UserX

15:39

I mean you can't come up with that integral and magically find that it has that of a perfect solution

Chris's sis the artist

@UserX They usually come from research, often I'm also surprised by the closed forms of some integrals.

Mike Akers

Hello all, I have a question that's probably not appropriate for a MO question

What's the most mathematically technical and pedantic way to say "Tying a knot in a rope?"

J. M. is back.

@MikeAkers Whatever for?

Mike Akers

Mainly to poke fun at a mathy friend

J. M. is back.

"knot" is sufficiently technical as it is.

Mike Akers

15:44

well, a knot is closed isn't it?

so an overhand knot in an unclosed length of rope would not be a "Math Knot"

J. M. is back.

Ah, right; mathematical knots are closed.

anon

@J.M.isback. nice avatar

J. M. is back.

@anon Thanks, it's an old elliptic function plot I had lying around. :)

anon

@J.M.isback. I see the triangular lattice, but the origin seems special

@MikeAkers I think you mean tying a rope in a knot

Mike Akers

ok fine :)

I was hoping for something like "deforming a high aspect ratio manifold embedded in R^3" or something like that

but i'll go with "tying a knot"

anon

15:51

what do you mean by "aspect ratio"?

J. M. is back.

@anon Yes, the origin is a multiple zero. That's why the phase coloring varies quite a lot there.

Mike Akers

well, i don't know how to say it technically

but i'm trying to say that a rope is basically a really long thin cylinder

that's super bendy

anon

if you just have a strand with free-hanging ends then there's no topological way to distinguish the rope being knotted or not. you'd have to fix the endpoints. also, do you want to refer to the act of knotting a rope, or the end result where the rope is knotted (by colloqial standards)?

Mike Akers

the act of knotting

J. M. is back.

@MikeAkers That's not entirely accurate; it's a bundle of intertwined fibers. ;)

anon

15:53

@J.M.isback. if the origin is a multiple zero, then so are its translates. numerical artifact?

mathematically, a rope would be a function from [0,1] or S^1 into a space...

Mike Akers

@J.M.isback. true

anon

or maybe the picture doesn't show the full fundamental parallelogram...

J. M. is back.

@anon Well, it's actually more akin to the theta functions than the elliptic functions, so I think "quasi-periodic" is the accurate description.

anon

gotcha

J. M. is back.

i.e., dividing different translations of that function yield genuinely doubly-periodic functions.

anon

15:57

say a rope is a path [0,1]->Dx[0,1] (where D is the unit disk in R^2) that is smooth isotopic to a line segment from (0,0) to (0,1). the act of knotting it would be an smooth isotopy to another path with the same endpoints which fixes the first endpoint (0,0) the entire time.

that probably works

J. M. is back.

Mike Akers

that sounds pretty good

J. M. is back.

Now, that is a genuine elliptic function.

anon

hexagons are bestgons

J. M. is back.

@anon Yes, that's part of the reason I'm obsessed with this particular family of elliptic functions. :)

anon

16:02

what'd you use to make the picture btw? mathematica with tweaks of standard options?

J. M. is back.

@anon Mathematica, yes. But I had to write a bunch of image-processing stuff myself.

There is of course an analytic proof that the function is invariant under the transformation $z \to z \exp(2 i\pi/3)$, but I found the picture way more convincing.

bentham

16:20

how do I obtain the joint pdf of 3 random variables?

I read that using the jacobian

bentham

16:48

please help

Mary Star

17:57

What dot products? Do you mean the following?

the unit vector a is perpendicular to n : $a \cdot n =0$

the angle between a and t is theta : $a \cdot t =|a| |t| \cos \theta=|t| \cos \theta$

J. M. is back.

...and since $\cos(\pi/2)=0$, point 1 follows from point 2.

Huy

18:13

wtf is going on here

new people

J. M. is back.

Like who?

Huy

idk

where's mike

evinda

@DanielFischer @robjohn @Huy
Show that $r(t)=\left (\cos^2 t-\frac{1}{2}, \sin t\cos t, \sin t\right )$ is a parametrization of the curve
of intersection of the circular cylinder of radius $\frac{1}{2}$ and axis the $z$-axis with the sphere of radius $1$ and centre $\left (-\frac{1}{2}, 0, 0\right )$.

Do we have to show that r(t) satisfies the conditions $\left( x+ \frac{1}{2}\right)^2+y^2+z^2=1$ and $x^2+y^2=\frac{1}{4}$ or do we have to find the intersection and show that r(t) satisfies the latter?

Huy

whichever you prefer

so probably the former

J. M. is back.

The former is easier to do.

Huy

18:18

no offense but those are not very manifold exercises you've been working on these days

@J.M. What does JM stand for?

evinda

@Huy @J.M.isback.
Substituting $x= \cos^2 t-\frac{1}{2}$ into the equation $x^2+y^2=\frac{1}{4}$ we get $y^2=\frac{1}{4}- \left( \cos^2 t-\frac{1}{2}\right)^2=\frac{1}{4}- \cos^4 t+ \cos^2 t-\frac{1}{4}= \sin^2 t \cos^2 t \Rightarrow y= \pm \cos t \sin t $. Why is the latter equal to $y= \cos t \sin t$ ? Or isn't it?

@Huy manifold?

J. M. is back.

@Huy They are my initials.

Huy

@J.M. I assumed that, which is why I asked "what do they stand for?" - but if you don't want to tell me that's fine of course

J. M. is back.

@evinda The idea is to replace all instances of x,y,z with the corresponding components.

Huy

@evinda: Why are you taking the square root?

J. M. is back.

18:21

@Huy As you say. :)

Huy

Hm. I'm trying to remember someone who might be back.

J. M. is back.

@evinda You should end up with an expression entirely in terms of the parameter.

Huy

was your previous username your full name or just initials too, @J.M.?

evinda

Now I want to show that any point on that intersection is of the form $(\cos^2 t-\frac{1}{2}, \sin t \cos t, \sin t)$. @J.M.isback. @Huy

J. M. is back.

@Huy Just initials, yes.

Huy

18:23

Ok.

Hippalectryon

@Huy Not yet :P

Huy

@Hippalectryon: I'll start doing it tomorrow, so you better do too.

evinda

@Huy Shouldn't I take the square root?

Hippalectryon

@Huy Okay, I'll see this weekend. I still have Prasalov's book to read too :D

Huy

@Hippalectryon: I want to study it a bit more in detail to then apply it to curved spaces too, because my prof told me it could be interesting for me with my background

do you have any knowledge of functional analysis already or none at all, @Hippa?

Mary Star

18:33

@TobiasKildetoft So we can write $a$ in the form $a=A t+B n+C b$ but we cannot find the coefficients $A$ and $C$, but since $a$ is in the plane spanned by $t$ and $b$, we know that $B=0$. Is this correct?

Tobias Kildetoft

@MaryStar yes

Chris's sis the artist

@Hippalectryon $$\int_0^1 \int_0^1 \int_0^1\frac{\ dx \ dy \ dz}{(1+x+y+x y ) (3 +x+y+z)(1-x-y-z+x y+x z+y z+7 x y z) }$$

Huy

how's it going @Khallil

Hippalectryon

@Huy a bit. not much.

@Chris'ssistheartist :o new forms everyday

Chris's sis the artist

@Hippalectryon because I work every day, without exception.

;)

Hippalectryon

18:39

:D

Chris's sis the artist

@Hippalectryon $\displaystyle \frac{\pi^2}{36}$

Hippalectryon

That's unexpectedly nice

Chris's sis the artist

@Hippalectryon :D

Huy

@Hippalectryon: which of Prasalov's books are you referring to btw ?

Hippalectryon

@Huy Problems in Linear Algebra

Huy

18:43

ok

anything interesting to share?

Hippalectryon

Well, it has some cool theorems I didn't know of. The proof are a bit too short sometimes (like a "it is straightforward" that takes you 5 minutes to understand), but overall I really like it

Huy

like which?

J. M. is back.

@Hippalectryon "straightforward" is always a relative term.

Hippalectryon

ugh Readcube is downloading updates ಠ_ಠ

Huy

wat is readcube

Hippalectryon

18:51

readcube.com

Huy

yea

ELI5

why is it useful

Hippalectryon

It auto adds references for your pdfs

Huy

wat

ELI5

Hippalectryon

Basically it makes it easier to organize research papers and find new ones. Unlike having 50 pdfs in one folder with weird names.

Huy

ic

so you're like one of those researchers

very organized

reading many papers

respect

Hippalectryon

18:55

I'm just a student :/

With so many pdfs, I don't even know who is what anymore :P

Huy

ah ok

respect withdrawn

Hippalectryon

:(

Huy

aren't there like many versions of this readcube? I mean alternatives

I heard about this stuff before but never bothered trying it out because I don't feel like I'd use it remotely to its full extent

Hippalectryon

No idea. I found readcube somehow and I liked it, so I didn't look any further.

Huy

and you use so many pdfs simultaneously?

takes me weeks to get through a short article

references? years

Hippalectryon

18:57

Well I download quite many pdfs. Too many.

Huy

ok

Lepidopterist

19:17

Can someone explain to me what trace property is being used in this answer? math.stackexchange.com/questions/853696/…

To get the last inequality

Robert Israel has high rep, so I assume it's true

Hippalectryon

@Lepidopterist I don't see any answer by R.Israel there, which one are you talking about ?

Lepidopterist

sorry, it was the wrong link @Hippalectryon

1

A: Convergence of a product under the trace

Let $A = U \Sigma V^*$ be the singular value decomposition of $A$. Then $$|\text{tr}(A (B_k - I))| = |\text{tr}(\Sigma V^* (B_k - I) U) | \le \text{tr}(\Sigma) \|B_k - I\|$$ where $\text{tr}(\Sigma)$ is the sum of the singular values of $A$. Moreover, this is best possible (it's an equality if...

that's it there @Hippalectryon

Hippalectryon

Hm I don't know much about spectral norms :/

Lepidopterist

he doesn't explain it as if it should be an obvious property

Huy

@Lepidopterist: $\|AB\| \leq \|A\| \|B\|$

oh wait is it a different norm?

Lepidopterist

19:28

i'm not sure about your question. the norm is the spectral norm

that is how i defined it in the question and he didn't specify another one

Huy

isn't it Cauchy Schwarz?

Lepidopterist

does that apply to the trace?

can you explain your thinking?

Ramanewbie

@bentham what is 'pdf' ?

Lepidopterist

probability distribution function

Ramanewbie

@Lepidopterist Ok

evinda

19:37

If $(x,y,z)$ lies on the sphere $\left( x+\frac{1}{2} \right)^2+y^2+z^2=1$ then $z^2=1-\left( x+\frac{1}{2} \right)^2-y^2 \leq 1$. Therefore $|z| \leq 1$. It follows that $z$ must be of the form $z= \sin t$ for some $t$.

Substituting $z= \sin t$ into the equation $z^2+x=\frac{1}{2}$ we get $\sin^2 t+x=\frac{1}{2} \Rightarrow x= \cos^2 t-1$.

Substituting $x= \cos^2 t-\frac{1}{2}$ into the equation $x^2+y^2=\frac{1}{4}$ we get $y^2=\frac{1}{4}- \left( \cos^2 t-\frac{1}{2}\right)^2=\frac{1}{4}- \cos^4 t+ \cos^2 t-\frac{1}{4}= \sin^2 t \cos^2 t \Rightarrow y= \pm \cos t \sin t $.

Lepidopterist

@Huy, why do you think it's Cauchy-Schwarz?

Huy

@Lepidopterist: $|\operatorname{Tr}(\Sigma V^*(B_k-I)U)| = |\langle (B_k-I)U, V \Sigma \rangle| \leq \|(B_k -I)U\| \|V \Sigma\| = \|B_k-I\| \|\Sigma\| = \operatorname{Tr}(\Sigma) \|B_k - I\|$

Lepidopterist

@Huy i don't quite follow. the first equality expresses the trace as an inner product. what kind?

Huy

@Lepidopterist: $\langle A, B \rangle = \operatorname{Tr}(B^*A)$

Lepidopterist

isn't that the frobenius norm?

Huy

19:41

idk what you call it, we called it Hilbert-Schmidt

and it is an inner product

the norm induced by it could be called Frobenius norm

Lepidopterist

i see. so the norm you are using is not the spectral norm

Huy

yea, because you got a trace

Lepidopterist

his answer was deceptive, because i asked for the spectral norm

thanks, though

Huy

you said or any other matrix norm

norms are equivalent anyways since you're finite dimensional

Lepidopterist

well maybe he should have specified...

Huy

19:43

idk, maybe he did a different thing in his head with your spectral norm

Lepidopterist

equivalent, but he presented it as a tight inequality

Huy

but it looks a lot like Cauchy Schwarz and so does his comment "this is the best possible inequality"

Lepidopterist

right

you are probably right. his answer should be edited, though

thanks @Huy

Huy

actually, according to wiki you can write the spectral norm in a very convenient way as an inner product, @Lepidopterist

using that inner product and that $U, V$ are unitary, you should get the same with Cauchy Schwarz immediately

Lepidopterist

i don't see how, since you can't express the spectral norm as a trace, @Huy

Huy

19:47

wait a second, maybe I've been too hasty

Razieh Noori

@Lepidopterist see (D.49) theorem of abstract harmonic analysis hewitt 2, page 704 , so $$|\text{tr}(A (B_k - I))| \leq \Vert A\Vert_{\infty} \|B_k - I\|_1 $$

Huy

trace and inner product are linear though

so it should be ok, no?

Lepidopterist

@RaziehNoori how does that relate to the spectral norm and the trace?

@Huy maybe i'm missing something, let me see

@Huy I guess I'm not sure what you mean unless you want to use some other inequality between the trace and spectral norm

@Huy What you did works because the H-S norm IS the trace of a certain matrix

Huy

well yeah because $\Sigma$ is as in SVD

is that not given here?

I just assumed by notation

Razieh Noori

The spectral norm of a matrix A is the largest singular value of A . trace norm is sum of singular values

so spectral norm\leq trace norm

Lepidopterist

19:58

yes, @Huy. As @RaziehNoori says the spectral norm is the largest singular value

Huy

I don't see the problem

is my argument wrong?

Lepidopterist

@Huy i don't quite understand what your argument is

Huy

the one before with the Hilbert Schmidt norm

Lepidopterist

that made sense. but you implied you could relate it to the spectral norm because the trace and inner product were linear. i don't get what you mean

Huy

yeah my idea was wrong

I was too hasty

sorry about that

Lepidopterist

20:03

thanks for thinking through it with me. i would have prefered the spectral norm, but the frobenius norm is probably enough for my purposes

@RaziehNoori i don't think the infinity norm is the spectral norm. it's the maximum column norm as i recall

the maximum column 1-norm, that is

Razieh Noori

Matrix norm

In mathematics, a matrix norm is a natural extension of the notion of a vector norm to matrices. == DefinitionEdit == In what follows, will denote the field of real or complex numbers. Let denote the vector space containing all matrices with rows and columns with entries in . Throughout the article denotes the conjugate transpose of matrix . A matrix norm is a vector norm on . That is, if denotes the norm of the matrix , then, iff for all in and all matrices in for all matrices and in Additionally, in the case of square matrices (thus, m = n), some (but not all) matrix n...

Lepidopterist

yes @RaziehNoori look at wikipedia's definition of the infinity norm. it's the maximum column 1-norm

Razieh Noori

hewii uses my symbol

Lepidopterist

@RaziehNoori on what page does he define this?

Razieh Noori

see page 703

my phd thesis is about $\frak{E}_p(I)$

i know these spaces

Lepidopterist

20:17

i don't see it

earlier he defines it as $\|\cdot\|_{sp}$

Razieh Noori

‎$$\|E_i\|_{\varphi_p}=\Big(\sum_{j=1}^{n}{|s_{j}^{i}|}^p\Big)^{1/p}$$ and‎
‎$$\|E_i\|_{\varphi_{\infty}}=sup\lbrace{s_{1}^{i},s_{2}^{i},...,s_{d_{i}}^{i}}\rbrace,$$‎

Lepidopterist

isn't that just the $l_\infty$ norm?

it even says these are the $l_p$ norms. that is not the spectral norm

@RaziehNoori the spectral norm is defined on page 397

Razieh Noori

The spectral norm of a matrix A is the largest singular value of A

Lepidopterist

i understand that

Razieh Noori

now see $\|E_i\|_{\varphi_{\infty}}$

Ramanewbie

20:24

Do you agree that the diameter of a hexgon equals twice its side.

Razieh Noori

yes

Lepidopterist

@RaziehNoori what is $E_i$?

Razieh Noori

i copied it, you can use $E$

$$\|E\|_{\varphi_{\infty}}=sup\lbrace{s_{1},s_{2},...,s_n\rbrace,$$‎

Lepidopterist

this text is unreadable. you are citing D49 to prove the inequality?

it would be nice if you made an argument instead of citing a text that is impossible to read

unless you have spent tons of time with it. he's not using any standard notation

Razieh Noori

excuse me d.39

in your symboles |trac XY|\leq \|X\|_{spectral norm}\|Y\|_{trace norm}

Lepidopterist

20:44

Ok, Razieh. I will upvote your answer because I think you are probably right. I have to read it carefully but I trust you. However, if you want it to be generally understood you should use the standard notation. It's an interesting theorem, but I'm not sure I get the proof

Lucas

20:59

Hey guys, given a binary operation, does the non-existence of an inverse imply that there is no identity?

Lepidopterist

@Lucas no

try Z as a monoid under multiplication

Learner

21:23

Hello someone can help me with statistics?

to get the joint pdf I have to calculate the determinant of the jacobian?

Karim Mansour

@TedShifrin !!!

my prof gave me 90 % in my first assignment in topology

he is very picky !!!

Learner

ok better ask in the site

skill patrol

Try here @Learner

:-)

Ten fold

CrossValidated's general room for gossip, grumbles, and idle c...

robjohn

21:41

@J.M.isback. Nice to see you around again :-)

Chris's sis the artist

22:45

Artwork is everywhere here in my work today.

(some sleep is needed though)

Agawa001

does anyone hear about zalgo or zolgo text

b̢̢̯̺̜̺̹̊͑̒͂͛̑ͨ̓ͣͣ͊ͮ̆̋͗͂̾̍ȯ̴͍͙͈͕̤̫̥ͭ̈̅̍̍̆̕͟͠o̊̓̋̑̇ͪͣͭ͊͑̑̌̿̔̾ͥ͐ͩ̍͏̴̰̯͙̹̠͔̠̜͈͈͔̖̱‌̦̟̱̱͓o̗̥͉͓͚͈̊ͣͪ̔̅ͩ͋̆̀ͩ͘͢͢o̯̦̖̭̞͊ͤͭ̆͒͒ͣ̊̌̌͌͗̓͆ͫ̄̓́ö̼͈̹̤͍̭͚̼̹͈̟̗̰̯͓̲̺̤͙̒̽ͧ̔̿̍͟͠͡o͆‌̴̷͇͇̱͉͚̠̼̼̱ͩ͆̍̌̈́͌͌͋̓͌͜

Learner

23:04

thanks @skillpatrol

Mary Star

Could someone of you take a look at my question:

0

Q: Signed curvatures

Let $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ be two plane curves. Show that, if $\textbf{$\tilde{\gamma}$}$ is obtained from $\gamma$ by applying an isometry $M$ of $\mathbb{R}^2$, the signed curvatures $κ_s$ and $\tilde{κ}_s$ of $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ ar...

?

J. M. is back.

@robjohn Just got my computer back the other day, so... :) great to be back!

JMoravitz

Sometimes I hate being on an active college campus. Music blasting in the big greenspace while I'm trying to get work done. sigh

robjohn

@J.M.isback. How long were you without it?

Ted Shifrin

heya @JMoravitz and @robjohn

robjohn

23:19

@TedShifrin How are things today?

Ted Shifrin

Yucky ... it's getting hot and my cold is not going away. So I'm an unhappy smurf.

J. M. is back.

@robjohn More than a year, I think.

Ted Shifrin

I don't even recognize J.M.

although I do know a mathematician with that name

heya @PVAL

PVAL

hi

im done with my slave labor for the week

Ted Shifrin

congrats

I'm trying to give away volunteer math tutoring to a community college, and they won't even answer my email :(

robjohn

23:22

@J.M.isback. Wow... why so long?

J. M. is back.

@robjohn I had to pawn it, you see.

Ted Shifrin

ah ...

morphic

@TedShifrin Wow, they are missing out..

Ted Shifrin

thanks, mr eyeglasses ... You wanna give me a testimonial? :P

PVAL

I bet it read like a phishing attempt.

Ted Shifrin

23:34

so what math should we talk about tonight, mr eyeglasses?

Or it went to spam because of attachments ... so I re-emailed her this afternoon with no attachments.

I don't think what I wrote looks like any phishing I ever got

PVAL

Well you wouldn't believe how many mothers moved from Sweden to Texas and were looking for tutoring for their sons or daughters by trusted professionals like me.

Ted Shifrin

LOL

I got a few of those when I was on the faculty at UGA.

I suppose I could just drive up there and try to walk in and talk to the person, but I think that's rude. I could try snail mail, too. It's ridiculous.

PVAL

Maybe the guy teaching the class with his masters from polynomial institute of Arkansas would feel threatened by you.

morphic

lol

They want to rip off grad students with cheap pay but won't accept a free established mathematician

JMoravitz

@PVAL between the "please tutor my jailbait daughter" emails and the "your mailbox needs to be upgraded to the newest version or it will be locked" emails, I think I hate the ones from HonorSociety.org and Qollaboration the most.

morphic

23:42

omg I keep getting e-mails from those scam honor society things

they are coming from my school too, so I can't block them or anything

PVAL

@JMoravitz I hate the ones written by my university president far more than anything else.

Ted Shifrin

@PVAL, I didn't contact the math department, although I would if I wanted to adjunct. I contacted the academic support center.

PVAL

@morphic I doubt there are any funded grad students at a community college.

Ted Shifrin

No, there aren't grad students at all.

morphic

@PVAL Oh, I guess I was just referring to my own school system. Our PhD candidates teach at our community colleges

Ted Shifrin

23:45

as adjuncts, yes

PVAL

@TedShifrin This is SDCC?

Ted Shifrin

San Diego Mesa College, @PVAL, part of the same system.

morphic

@TedShifrin I thought all the teaching grad students were considered adjuncts

Ted Shifrin

not ones teaching in their own university, @morphic

@PVAL: It's just a two-year college.

morphic

Well I mean our graduate school doesn't have any teaching positions available for grad students, so...

Ted Shifrin

23:48

at most research universities, grad students TA large calculus lectures and often teach their own courses.

PVAL

I had to teach people about springs...

JMoravitz

proof of that here

PVAL

I don't know anything about springs...

morphic

@PVAL Did you get to play with Slinkys in class

Ted Shifrin

You'd rather teach about falls and winters?

morphic

23:49

lol

PVAL

I'd rather teach mathematics

morphic

ಠ_ಠ

Ted Shifrin

don't be a snoot about applications, @PVAL

JMoravitz

I've never heard it as a noun before...

I've only heard the adjective, "snooty"

Ted Shifrin

Ironically, two of my worst evaluations about 20 years ago in a multivariable calculus class (regular) came from two chemistry majors, who both complained that I did too much physics in the class.

@JMoravitz: I endeavor to keep you on your toes :)

JMoravitz

23:51

Your unicity is palpable @Ted

Ted Shifrin

ponders what the hell that means

PVAL

@TedShifrin If I never need to hang a weight from a metal spring and throw it straight downward, I'll know what 2nd order ODE to solve.

Ted Shifrin

You are teaching engineers and physics majors, not just pure math students.

My first quarter at Berkeley I was assigned to TA the sophomore DE class. I was upset that they didn't give me multivariable, but then, ultimately, I had a blast teaching the class and had some wonderful students.

PVAL

At least we avoided the chapter (with the pretty insane flow chart) about what first-order ODE I need to solve if I want to detect painting forgeries.

Ted Shifrin

Radioactive decay? :)

morphic

23:58

Radioactive Spider-man

Ted Shifrin

ugh, the chiropractor bruised my ribs, and it still hurts a week later :(

Ten fold

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$Mathematics$ Mathematics