The root of music?

Yes, Akiva is the root of all music

Right now, I have like one or two lines from "A Lovely Night" playing on repeat in my head

Copyright policy: No $\frac{d}{dx}$ works.

(no derivative works)

:P

@Akiva do you want to talk about the QQ problems when admission is over?

I didn't look at them yet, but sure

To be honest, right now I need to study for a test I have tomorrow

Then do so

so I should stop procrastinating at some point

Ok, I'll stop mentioning you for now then

So that you're not compelled to talk

How does one solve for $k$ given $\mathbf{G}, \mathbf{x}, \mathbf{y}$ and knowing that $\mathbf{G}^k \mathbf{x} = \mathbf{y}$?

$k\in \mathbb{Z}$ and $\mathbf{G}$ a square matrix

@DHMO Hola, que tal?

hola

A spherical ballon is inflating with helium at a rate of 200pi ft^3/min. How fast is the balloon's radius increasing at the instant the radius is 5ft?

did this in high school, completely forgot how to go about this, although i remember its fairly simple...

@WillNjundong first let's find a function of the radius

i did that

@WillNjundong dV/dt = 200pi, find dr/dt

Then take the derivative of it

and evaluate at r = 5

i did that too

so... you got the answer then

mymathlab didnt agree with my answer :(

@ZachHauk I misunderstood this at first, I'd call that a function for the radius, or a function of the input

@WillNjundong post your radius function here

so we were at if there's an old lady there we'll have k + 6 cats

no

if there's $k+1$ ladies, there are $k+6$ cats

but remember

there was an assumption we made

what assumption is that?

that n = k

all im given is the rate of change of the balloon's volume

no..

@WillNjundong ok so the sphere's volume is $(4/3)\pi r^3$

that for any k we give we will get the output we expect

@WillNjundong chain rule: dV/dt = (dV/dr) (dr/dt)

i took the volume formula, got its derivative and multiplied with 200pi

eh ill just let @DHMO take this

@CausingUnderflowsEverywhere yes!

So we're assuming that our statement is true for $k$

And we proved that it will work for $k+1$

Right?

@WillNjundong Write a function $f$ giving the radius in terms of the volume. Find the volume that gives the desired radius (find $x$ where $f(x) = 5$). Evaluate $f'(x)$.

@Akiva over here calling me stupid /s

Anyways, go study

@ZachHauk Noo

I'm not going to dignify that quote with a response besides that little quip I just gave

Also,

@ZachHauk Noo

I mean, fine

@ZachHauk Unfortunately I read mathisfun.com while you were gone

Derp, I totally blurred that

@CausingUnderflowsEverywhere idc

I suck at teaching

I'm going to go play some games

It's rather boring this time of day

Tell me if Ted comes on

Well, ping me

@WillNjundong Sorry I said "find the volume", but I meant "find the time"

@ZachHauk well Im sure you're teaching it the way it makes sense to you, but everyone's way of making sense of things is $k$ permutations different than the next, so maybe to improve your teaching skills you need to try to understand the way the person calculates things, and use that information to help them fill the gap preventing them from understanding the topic at hand

@CausingUnderflowsEverywhere Are you still on induction?

I want to play games too but Im so behind in understanding things

@MickLH got it right now! thank you :D

isn't it embarassing, I was on 0.9... is 1 for 3 days I think

If induction isn't making sense then you're just overthinking it :P

It's simple but powerful, no tricks required

I dont understand the algebra used in Zach's first example. if I did I'd probably understand induction, let me go find it

Oh look it's @Ted

I think I have an idea about that problem

@ZachHauk you noticed lol

Fake Ted.

So am I right in saying that the matrices with $a=d$ are $I_2$ and $-I_2$?

Hi fake ted

Yes, because $a=d$ forces $b=c=0$. You need to figure out why.

Well, yeah...

Because then it's of the form $nI_2$

No, no, you're going backwards. Why does $A^2=I$ force that?

Because we have

$b = -b$ and $c = -c$

Only if $\det(A)=1$? How do we know it can't be $-1$?

Because $a = d$ is only if $det(A) = 1$

otherwise it'd be $a = -d$

Why?

Oh, right.

OK.

@ZachHauk confused by the algebra used here

Because we have that $a = \frac{1}{\det{A}} d$

OK, @Zach. I yield.

Okay so here was my insight

I don't only think that (when $a = -d$) shearing is the non-distance preserving

No, there are others, in fact.

I think it's any matrix with determinant 1, reflected over $x$-axis

No, that's definitely not right.

I mean

any matrix reflected over $x$-axis

then you apply the inverse

Why should the square of that be the identity?

You have

3 * (a + 1)! yields 3^a+1 ?

Huh? @Causing

well, let $Y$ be the reflection over $x$-axis

and $X$ be our matrix with determinant 1

You mean $Y=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$?

yep

We say that $A = XYX^{-1}$

Oh, OK. Sure.

So $A^2 = (XYX^{-1})(XYX^{-1})$

and then you use associative

Yes, yes.

OK, so is that all?

That's a general reflection when $X$ is a rotation.

I was wondering if someone could show me a full step process of the algebra used here chat.stackexchange.com/transcript/message/35888201#35888201

Yeah :P

$X$ just has to be any matrix with determinant $1$

starting there and below

But I guess it's going to be a non-orthogonal reflection in general.

@TedShifrin (when you get a moment) Over what timespan is Chapter 7 usually delivered?

No, $X$ needn't have any restriction on its determinant. Just nonsingular.

Oh, that's right

:P

I guess you could scale by the square root of determinant when the determinant is $>0$, but not if the determinant of $X$ is negative.

@CausingUnderflowsEverywhere I (very sloppily and loosely) went over it: chat.stackexchange.com/transcript/message/35888581#35888581

This is conjugacy in group theory, and a general change of coordinates in linear algebra (really the same concept). This is the change-of-basis formula you'll learn later (or relearn).

Ok, now to tackle this funky looking $X^2 = O$

Anyhow, @Zach, you can certainly reduce to $\det X = \pm 1$. I don't know what that gets us.

Oh thank you MickLH I'm sorry I missed it

@CausingUnderflowsEverywhere hope it helps? I realize now I should have said $2 \cdot 2^k \leq 2\cdot k! \leq (k+1)\cdot k!$ to make it more clear

@Daminark: No, no projective geometry in my multivariable course. There's a section in my linear algebra book and a whole chapter in my algebra book. (That's what Zach was working through until he got derailed. :P )

Really cool cheap book: Pedoe's Geometry: A Comprehensive Course.

@Ted Hey! I did it for my own good :P

(This is the Pedoe of Hodge-Pedoe's multi-volume algebraic geometry text. He was a student of the very famous Hodge.)

Hodge Theaters?

Who's that?

It's this weird math thing

in Interuniversal something theory

By that guy Mochizuki. I have no idea what any of it means

Who does?

I'm talking about the Hodge of the Hodge Theorem in geometry/analysis and the Hodge conjecture in complex algebraic geometry, Hodge-Riemann bilinear relations, etc.

I have no clue to what you refer .. :P

Interuniversal Teichmuller theory is probably what you've got on the brain.

kurims.kyoto-u.ac.jp/~motizuki/…

@TedShifrin Am I a horrible student if I don't finish Chapter 7 today?

LOL, @MickLH. I just sent you that stuff for you to play with. It's hard stuff and you probably need to play around with eigenvectors/eigenvalues too.

I was very proud of writing the exercise on how Mathematica actually draws its pictures using projective geometry. :)

Hmm, I know of the Riemann bilinear relations but I hadn't seen Hodge tossed in there.

@Zach Whatever structure he's defined has a name that comes from Ted'a Hodge.

@Semiclassic: Riemann is for Riemann surfaces. Hodge's name goes on for higher-dimensional abelian varieties.

Ah.

Say what? @MikeM

Yeah I knew it :P

@Ted go read that paper. You have homework

Just kidding :P

lol

Here's a question I've been thinking about. Let $X$ be a subset of $D^3$, whose intersection with the ball's boundary is the equator $S^1$. Now, $D^3\setminus X$ may or may not be connected.

Oh, @Zach, I have no clue what's in that paper, but it's the same Hodge.

In addition, there may or may not be a continuous map from $X$ to $S^1$ that fixes $S^1$.

DogAteMy: Do you mean $D^3$?

I do, thank you.

Whew. I was confuzled.

Change the second line too.

So, for example, if $X$ is simply a disk, $D^3\setminus X$ is disconnected, and there's no retract onto its boundary.

Right?

No retract of whom onto whose boundary?

So I'm wondering if those are equivalent, that is, if $D^3\setminus X$ is connected if and only if there's a map from $X\to X\cap S^2=S^1$ that fixes $S^1$.

@Ted I think he was just looking to identify he names, yeah.

@TedShifrin No retract of $X$ onto $\partial X=X\cap S^2=S^1$.

If $X$ is a disk in the ball whose boundary is the equator.

I'm not trying to make excuses but my brain keeps stalling about the complex numbers. I know them pretty well, but I haven't touched anything smooth since at least 2016 and it feels like I have to "decompress" all that knowledge again before using it.

So it's connected only when $X$ is a closed hemisphere?

What? $X$ is a subset of $D^3$, not $S^2$

@MickLH: You can skip the complex stuff. I put it in there because in principle you could have complex eigenvalues and so you'd need them for the JCF.

Oh, I forgot to mention that I specifically want $X$ to be closed

Right, DogAteMy. I can't see how to stick a disk in there that won't disconnect the ball if the boundary is the equator ... unless I stick the disk along the upper or lower hemisphere. Give me an example that I'm missing.

Take, for example, the disk through the equator, minus a smaller open disk in the middle. Its complement is connected, and you can retract it onto its boundary radially.

Oh, I wanted $X$ homeomorphic to a disk.

@TedShifrin $X$ isn't necessarily a disk.

@Ted OK I found that either $a = -d$ or $b = c = 0$

Yeah.

You're allowing it to have boundary components.

For $A^2 = O$

Huh? @Zach ... something seems wrong.

Yeah. It could also have higher genus (and no other boundary components), if you want.

I see. OK.

Did I matrix multiply wrong? I got that $\begin{bmatrix} a & b \\ c & d \end{bmatrix}^2 = \begin{bmatrix} a^2+bc & c(a+d) \\ b(a+d) & d^2+bc \end{bmatrix}$

So this is probably a good fancy duality exercise. Too topological for me at the moment :P

Besides, I need to leave in 5 minutes.

Anecdote: After working exclusively in finite fields for several months, the infinite beauty in every facet of the real domain is shining with bright sparkling color.

It could even be disconnected and whatever. But disconnected components that don't intersect the sphere can just be mapped to an arbitrary point on the equator.

Ah, yeah, I should probably learn duality at some point @TedShifrin

I think you could wait until college :D

Counts as "at some point". College is approaching fast!

@Zach: So, if $a=-d$, you have to consider more.

In the case $a = -d$, we have something of the form $nI_2$ because $a^2 + bc = d^2 + bc$ in that case

But they have to equal $0$ !!

And I don't know why you keep saying $nI_2$.

Off-diagonals equal zero, on-diagonals are equal to each other? @TedShifrin

Oh, wait

OH, nevermind

Let me think about this a bit more

I dun goofed.

Wait, off-diagonals aren't zero!

OK. Uh huh. I am eating dinner and then playing bridge for the evening, so bye until sometime tomorrow, although tomorrow is crazy busy.

Right, DogAteMy.

If the diagonals are $0$ is one case ... But there are more interesting options.

I have a quick question, $$\left|\frac{2+z}{2-z}\right|=\left|\frac{2+z}{2-z}\frac{2+z}{2+z}\right|\leq \left|\frac{(2+a)^2-b}{(2+a)^2+b}\right|$$where $z=a+ib$

Nevermind

What's the question?

Night, all.

night

thanks again

Do we get something like $\begin{bmatrix}xy&x^2\\-y^2&-xy\end{bmatrix}$?

I am not familiar with complex number, am I on the right track?

@ZachHauk The above should satisfy your equation

Stop spoiling and study /s

Anyways

yeah that's correct

Cuz $a \neq -d$ makes the matrix the zero matrix

I suppose you could change variables to something like $\begin{bmatrix}a&b\\-a^2/b&-a\end{bmatrix}$

And $a = -d$ requires that $bc$ = $-(xy)^2$

Anything that makes it so we have $bc = -(a)^2$

Next exercise time, and this tiem don't spoil unless I ask you :P

@AkivaWeinberger feel free to spoil it for me though! I seem to learn best from being given the answer and time to reverse engineer it myself

@MickLH That doesn't sound 100% wise, though I guess it depends on the details of what you're doing while "reverse engineering" it

reverse engineering is illegal in 18 states and 3 provinces

Yeah it's way too easy to mislead yourself that way, but I have a natural defense because I pretty much don't believe anything including my own thoughts.

Hm, why did it only get banned in those eighteen states? starts reverse-engineering

I'm a reverse engineer asm hacker

OOH REALLY

yes

x64 in particular?

I hacked ocarina of time, for example

@ZachHauk brother!

No, x64 is just x86 64 bit

so, I guess yes but

x86 I program in

oh I always dreamed of disassembling that game

Also, MIPS and 6502

I wrote a hack that generates a bijection of the get item table

based on filename

And so, it gives you random items when you open a chest

@MickLH Another thing I like is trying to unravel lemmas, so that I end up with a (messier but lower-level) proof of the thing that doesn't rely on any other results

all those instructions and data you can fit in a 5mb image..