Mathematics - 2017-09-08 (page 3 of 8)

Balarka Sen

14:00

@Liad Compose with a rotation of $\Bbb C$ so that the line $\{\alpha z_0\}$ gets mapped to the real axis $\{\alpha\}$ probably.

Liad

i thought about it but i want to understand something else. if i use Cauchy–Riemann equations how do i differentiate $\alpha(z) x_0$ with respect to $x$ (or $y$) ? $\alpha$ is a function from $\Bbb R$ to $\Bbb R$ @BalarkaSen

Abcd

Hey!

Balarka Sen

I don't understand the question but I have to leave now

Mike Miller

bye

Abcd

I just wanted to ask whether we can treat functions like algebraic variables (adding subtracting and stuff like we do with variables)? Actually, I used functions like algebraic variables in a physics problem and got the right answer but I am unsure about my solution.

@LeakyNun Mind answering^ ?(Please let me know if more details required)

Leaky Nun

14:07

@Abcd yes, as long as the subtraction is well-defined

Abcd

@LeakyNun well defined?

Leaky Nun

Well-defined

In mathematics, an expression is called well-defined or unambiguous if its definition assigns it a unique interpretation or value. Otherwise, the expression is said to be not well-defined or ambiguous. A function is well-defined if it gives the same result when the representation of the input is changed without changing the value of the input. For instance if f takes real numbers as input, and if f(0.5) does not equal f(1/2) then f is not well-defined (and thus: not a function). The term well-defined is also used to indicate whether a logical statement is unambiguous. A function that is not well...

anon

uh oh

Dodsy

class is awesome :o

learning about sets atm

Leaky Nun

@Dodsy nice

@Abcd in particular, what does $f+g$ mean where $f$ and $g$ are real functions?

Dodsy

14:09

I was so nervous to go haha

Abcd

@LeakyNun I am not sure. Haven't had much exposure to function algebra.

Leaky Nun

@Abcd how would you define $f+g$?

Abcd

@LeakyNun I seriously don't know..

Leaky Nun

have you seen this symbol? $f \circ g$

Abcd

@LeakyNun no

Dodsy

14:10

that's the composition of f and g, correct?

Leaky Nun

@Dodsy yes

@Abcd alright, $f+g$ is defined as a function whose value at $x$ is $f(x)+g(x)$.

i.e. pointwise addition of functions

Dodsy

would be $g(f(x))$ if $f \circ g$

Abcd

@LeakyNun Will I get to know the reason for this ever? PS: There's a big chapter functions in class 12, will I get to know why f(x)+g(x) = f+g?

Leaky Nun

@Dodsy not really

Dodsy

sorry.

Leaky Nun

14:12

@Abcd let's look at $f(x)=x$ and $g(x)=x^2$.

What do you say $f(x)+g(x)$ is?

@Abcd I never said that.

Semiclassical

g sends x to g(x), and then f sends g(x) to f(g(x))

Abcd

@LeakyNun $x+x^2$?

Leaky Nun

@Abcd exactly

Semiclassical

f(x)+g(x)=(f+g)(x) not f+g

Leaky Nun

so it would make sense if $f+g$ is a function that sends $x$ to $x+x^2$

Abcd

14:13

Yes, right @LeakyNun

Dodsy

@semi

I wish I would stop doing that

I always think that by pressing enter, your name will just enter into my text bar

Leaky Nun

@Abcd so what would $f-g$ mean if $f$ and $g$ are real functions?

Semiclassical

@Dodsy use the tab key instead

Abcd

$f(x) + g(x)= value of f(x)+value of g(X)$. Please verify.

Leaky Nun

@Abcd ?

Abcd

14:16

@LeakyNun I can't understand the concept clearly.

Leaky Nun

@Abcd you said "yes, right" above

let's do it again

@Abcd that makes no sense whatsoever

now, $f$ is a function that sends $x$ to $x^3+1$

$g$ is a function that sends $x$ to $x^2+1$

what should $f+g$ as a function send $x$ to?

Abcd

@LeakyNun what do you mean by "sends x"? x is input right?

Leaky Nun

@Abcd yes

a function sends a value to another value

Abcd

@LeakyNun What does $f+g$ mean mathematically, and non- intuitively? What is the definition of $f+g$?

Leaky Nun

7 mins ago, by Leaky Nun

@Abcd alright, $f+g$ is defined as a function whose value at $x$ is $f(x)+g(x)$.

Kasmir Khaan

14:19

all righty

Semiclassical

notationally, that means (f+g)(x)=f(x)+g(x)

Abcd

@LeakyNun Edit makes the post come down now?

Kasmir Khaan

@LeakyNun Hi again >< i know its been more than 30 mins =p

Leaky Nun

@Abcd ?

Kasmir Khaan

@Semiclassical Hello semi

Semiclassical

14:19

so at least notationally it's as though one 'distributes' the x into f and g

Abcd

@LeakyNun Understood.

Semiclassical

one shouldn't take that literally, of course.

Abcd

@LeakyNun x^3-x^2

Leaky Nun

5 mins ago, by Leaky Nun

@Abcd so what would $f-g$ mean if $f$ and $g$ are real functions?

@Abcd right, now answer ^

@KasmirKhaan what is your question?

Semiclassical

@KasmirKhaan heyo

Leaky Nun

14:21

@Abcd no, a function is not a value.

Kasmir Khaan

@LeakyNun well its better to reread the chapters so I know what to ask =p

@LeakyNun Ted and Semi helped me yesterday on that question, but still strugle to make a good proof

Abcd

@LeakyNun $f(x) + g(-x)$?

Leaky Nun

@Abcd same problem.

Kasmir Khaan

@LeakyNun the question was how to show z/2 x z/2 and z/4 are the only groups with 4 elements

Leaky Nun

@KasmirKhaan I said I already scrolled through the conversation

Kasmir Khaan

14:23

Okay

Ehm what I have trouble with is

Leaky Nun

and I will refrain from commenting (on the conversation). Do not get me started.

Kasmir Khaan

What do you mean ?

Semiclassical

I think him explaining his lack of commentary would count as commentary :)

Leaky Nun

@KasmirKhaan never mind. go on.

@Abcd do you know what the difference between a function and a value is?

Kasmir Khaan

the problem I got is , when i have 2 groups , and i know all their elements

also found a 1-1 correspondance between each element

Abcd

14:24

@LeakyNun Function = Takes inputs and gives outputs; Value = Abstract quantity

Kasmir Khaan

in this case am talking about GL_2( Z/2 ) and S_3

Leaky Nun

@Abcd hmm, no.

Kasmir Khaan

both have 6 elements

Leaky Nun

$x$ is a variable, it isn't constant, but it isn't a function.

the dependent variable is also not a function.

and the value can also be dependent.

Kasmir Khaan

How to prove they are isomorphic, without listing all possible products

f(xy) = f(x) f(y) this holds for all, but should I write it out for all 36 possible products

Leaky Nun

14:26

you just need to find what the generator maps to, @KasmirKhaan

@Abcd that's better

Kasmir Khaan

@LeakyNun Why is that enough?

Leaky Nun

@KasmirKhaan because everything else is generated by the generators

@Abcd so $f(x)+g(x)$ is not a function, but a value.

Kasmir Khaan

@LeakyNun all righty , let me try to think for a while why that is true :D

Leaky Nun

@KasmirKhaan they're called the generators for a reason

Abcd

@LeakyNun ok

El'endia Starman

14:27

Leaky Nun

@Abcd so what does $f-g$ mean?

El'endia Starman

I posted this to Facebook a couple years ago. ^^

> Been playing around with Fibonacci-style sequences on groups!

For both, I started with the Cayley Table. Then I traced out the paths that sequences take. For example, the second one is the integers modulo 4, or in other words, {0,1,2,3} (and 3+3=2 in this group). You can start with 0 and 1, and by repeatedly adding them Fibonacci style, you end up with the sequence 0, 1, 1, 2, 3, 1, 0, 1. For Z_4, there are four such sequences:

0, 0, ...
0, 1, 1, 2, 3, 1, 0, 1, ...
0, 2, 2, 0, 2, ...
0, 3, 3, 2, 1, 3, 0, 3, ...

Leaky Nun

you're right

Abcd

@LeakyNun what are f and g?

Leaky Nun

@Abcd real functions

Abcd

14:29

@LeakyNun You tell please.

Leaky Nun

@Abcd $f-g$ is the function that sends $x$ to $f(x)-g(x)$

Abcd

@LeakyNun I said that earlier :( and I meant the same thing.

Leaky Nun

@Abcd no, you said $f-g$ is $f(x)-g(x)$

Abcd

@LeakyNun I meant $f-g$ is a new function that returns output as $f(x)-g(x)$ for an input $x$. Is it correct now?

Leaky Nun

@Abcd yes it is

and I can hardly tell you mean that by simply "$f(x)-g(x)$".

Abcd

14:32

@LeakyNun It's the same thing for the same $x$ inputs IMO

Semiclassical

$f-g$ maps $x\mapsto f(x)-g(x)$

Leaky Nun

@Abcd that's a common misconception

Abcd

@LeakyNun Please recheck.

Leaky Nun

which grade are you in?

Semiclassical

If Leaky had said "what is $(f-g)(x)$?" then $f(x)-g(x)$ would be appropriate

Leaky Nun

14:34

@Abcd ok

1 min ago, by Abcd

@LeakyNun It's the same thing for the same $x$ inputs IMO

Semiclassical

but he didn't. he asked what $f-g$ was. hence the answer should be a function, not the output of that function.

Leaky Nun

@LeakyNun No, a function is not the same as its value evaluated at $x$

Abcd

@LeakyNun Grade 11, why?

Leaky Nun

@Abcd alright, then you probably won't need that distinction

Kasmir Khaan

@LeakyNun we dont have any generators on S_3

Semiclassical

14:35

bull.

Abcd

@LeakyNun When will I need it?

Leaky Nun

@KasmirKhaan of course we have.

Kasmir Khaan

the order of elements are 1 ,2 and 3

Leaky Nun

@Abcd when you get into university perhaps

Semiclassical

@LeakyNun was referring to S3 not having generators.

Abcd

14:36

@LeakyNun :0

Leaky Nun

@KasmirKhaan alright, I meant $S_3$ has generators.

@Semiclassical wrong ping

Semiclassical

oops

Kasmir Khaan

Well but generator by definiton must generate the group

S_3 is not cyclic

Leaky Nun

cyclic is when you have one generator that generates the whole group

Semiclassical

this is what I was getting at earlier. generator (singular) versus generators (plural)

there's not one generator, no. but there is a set of generators.

El'endia Starman

14:37

IIRC there's a theorem that every finite (simple?) group can be generated by exactly two elements.

Semiclassical

definitely not.

Abcd

@LeakyNun But now I am eager to know. Can you explain the difference sometime later (because I am busy with Physics right now)?

El'endia Starman

@Semiclassical Is that in response to my statement? What about this? mathoverflow.net/a/59216/57164

Leaky Nun

15 mins ago, by Abcd

@LeakyNun Function = Takes inputs and gives outputs; Value = Abstract quantity

Semiclassical

I may have been too hasty.

Leaky Nun

14:40

you already explained it

Semiclassical

i'm a bit shocked if it's true

Abcd

@LeakyNun okay.

Semiclassical

@El'endiaStarman yeah, looks like I was wrong. huh.

El'endia Starman

@Semiclassical To be fair, it's certainly a shocking result in general.

Kasmir Khaan

<123> = {e , (1,2,3), (132) }

but how does this help us?

Leaky Nun

14:43

@KasmirKhaan, your group needs to be generated by 2 elements.

Kasmir Khaan

I allready found the 1-1 correspondance between each element of Z/2 X Z/2 and S_3

Semiclassical

as we've been saying, one generator isn't going to cut it here.

Kasmir Khaan

oh

Semiclassical

you need two.

Leaky Nun

@KasmirKhaan a bijection is not an isomorphism.

Kasmir Khaan

14:44

two elements

Semiclassical

(i'd have said at least two, but it turns out we don't need more than that.)

Kasmir Khaan

@LeakyNun okay >< I was thiniking limited to one element =p

this idea opens the door to more glorious ideas then :D

let me think =p

Semiclassical

okay, so what two elements generate the group?

Kasmir Khaan

(123) and (23)

Secret

Semiclassical, let me show you a cute identity of sequences I derived and proofed:

Let $A_n=\sum_{m=1}^na_m,B_n=\sum_{m=1}^nb_m$. Then:

$$\sum_{m=1}^na_m\sum_{k=1}^mb_k = \sum_{m=1}^n(A_n-A_{m-1})b_m$$

SteamyRoot

14:49

Yup, it's indeed possible to generate any finite simple group with two elements...

Semiclassical

$A_n-A_{m-1}=\sum_{k=m}^n a_k$.

SteamyRoot

Though, as far as I'm aware, every proof of this relies quite heavily on the classification :/

Semiclassical

so this is just a special case of swapping the order of summation of a double sum.

(along with swapping the summation labels)

Secret

Ah I see, I guess that's why I saw this weird triangular pattern:
\begin{align}
\sum_{m=1}^na_mB_m & = a_1 (b_1)\\
& + a_2 (b_1+b_2)\\
& + a_3 (b_1+b_2+b_3)\\
& + a_4 (b_1+b_2+b_3+b_4)\\
& ...\\
& + a_n (b_1+b_2+b_3+b_4+\cdots +b_n)\\
\\
& = (a_1+a_2+a_3+a_4+\cdots +a_n) b_1\\
& + (0+a_2+a_3+a_4+\cdots +a_n) b_2\\
& + (0+0+a_3+a_4+\cdots +a_n) b_3\\
& + (0+0+0+a_4+\cdots +a_n) b_4\\
& ...\\
& + (0+0+0+0+\cdots +a_n) b_n\\
& = \sum_{m=1}^n(A_n-A_{m-1})b_m
\end{align}

Kasmir Khaan

@LeakyNun so by that logic , i should just show the homomorphism , 1-1 and onto on the generator elements?

Leaky Nun

14:53

@KasmirKhaan yes

Kasmir Khaan

neet :d

but how to argue formally that all the groups structures are "inside" the generators?

SteamyRoot

Do you mean "neat"? :P

Kasmir Khaan

yes ><

I learned that word by hearing it

Semiclassical

NEET actually is web slang, amusingly enough

SteamyRoot

Not sure if we have many neets here

Leaky Nun

14:55

@KasmirKhaan no idea

Kasmir Khaan

okay I guess its time to keep reading the book

I wish I could find some good lectures on this

anyone know good lectures on abstract algebra?

on youtube or something?

Leaky Nun

brb

Abcd

15:15

If $x+y=2$, then how do I find the minimum value of $x^2+y^2$?

Is it even possible to find it? I don't think so...

Leaky Nun

@Abcd I have never seen anyone giving up in a shorter time

Abcd

@LeakyNun Hey! I tried it for 10 minutes already. Using AM, GM. Possible inequalities but in vain

Leaky Nun

what are your tools?

what topic is it under?

Abcd

@LeakyNun Part of a physics problem's solution. :p

Leaky Nun

@Abcd any further constraint on $x$ and $y$?

SteamyRoot

15:19

Have you tried using that $x + y = 2$ ...?

Leaky Nun

have you tried interpreting the problem geometrically?

(it becomes trivial)

Akiva Weinberger

Hint: $y=2-x$.

Leaky Nun

@AkivaWeinberger I thought $x+y-2=0$ is also one of the standard forms

Akiva Weinberger

Er, sure, I guess, that's also a true equation

I was thinking of writing $x^2+y^2$ as an equation in terms of $x$, using substitution

Mike Miller

also follows from (x-y)^2 >=0 and (x+y)^2=4

Leaky Nun

15:25

I mean, I can recognize $x+y=2$ by itself.

oh, @AkivaWeinberger you weren't following my approach

I really like my approach :P

SteamyRoot

I think at Abcd's level, the standard approach is reducing to a single variable and differentiating :P

Mike Miller

All of the approaches we've all said are more or less equivalent

@SteamyRoot I doubt there's calculus.

SteamyRoot

I mean, technically you're intersecting a paraboloid with a plane to get a parabola, but, yeah

Akiva Weinberger

AM-GM theoretically works as well

Leaky Nun

@SteamyRoot nah, you don't need differentiation to find the minimum value of a quadratic polynomial

you only need to express it in the vertex form

SteamyRoot

15:27

You don't need it, no

Akiva Weinberger

I mean I dunno, the thing that you feel should be the answer turns out to be correct, so

SteamyRoot

But it's the standard approach for any "minimise this stuff" question

Balarka Sen

use lagrange maximization and nuke the whole thing off the galaxy

Leaky Nun

@BalarkaSen right

Akiva Weinberger

Incidentally, given that this is a quadratic in $x$ and symmetric about $x=1$, you don't even have to calculate anything, you just know it's either the endpoints or the middle

('Cause there can be only one extremum)

Leaky Nun

15:28

we have already like tons of approaches

and here goes Abcd "I don't think it's possible"

Balarka Sen

@LeakyNun "except mine is the best"

Leaky Nun

@BalarkaSen come on, isn't mine beautiful?

SteamyRoot

I definitely don't think my solution is the best or most beautiful :P

Balarka Sen

@Steamy Is that your bragging line? Being a normie?

SteamyRoot

Haha, no.

It's just that I've been TA'ing a crapload of first year bachelor courses

So for easy question I default to whatever solution requires the least tools and/or insight

However sad that may be

Abcd

15:31

@LeakyNun No

@LeakyNun -_-. I said that after tryinggg.

Leaky Nun

@Abcd read the conversation ensuing

try out every approach suggested

(except Balarka's)

SteamyRoot

rip

Abcd

@SteamyRoot who?

SteamyRoot

Balarka's approach

Mike Miller

I have been working on a problem for a year and a half now

You can't give up after 10 minutes

Akiva Weinberger

15:36

If we want to continue with the theme of ridiculous ways to solve this problem

Here's another one

Abcd

@AkivaWeinberger Are you sure?

Akiva Weinberger

@Abcd Like 80%

Mike Miller

Haha Balarka got banned

I wonder who reported him

Akiva Weinberger

Another way is to say $(x-1)^2+(y-1)^2=x^2+y^2-2(x+y)+2$${}=x^2+y^2-2(2)+2=x^2+y^2-2$

SteamyRoot

Wait seriously?

Mike Miller

15:37

It's you for sure

Akiva Weinberger

so $x^2+y^2=(x-1)^2+(y-1)^2+2$

SteamyRoot

In that case, RIP Balarka

El'endia Starman

@MikeMiller It's Not Nice to discuss suspensions (however short they may be).

2

Akiva Weinberger

so if we want to minimize it we need both those squares to be as small as possible (i.e. zero)

Leaky Nun

@MikeMiller me?

Mike Miller

15:38

@El'endiaStarman Sorry, who is this not nice to?

SteamyRoot

How the hell is whatever he said even offensive, anyway?

Mike Miller

I'm asking him what it was lol

SteamyRoot

Some joke about normies or so

Leaky Nun

@MikeMiller https://chat.stackexchange.com/transcript/message/39892033#39892033

Mike Miller

Maybe people don't like the F-bomb. Curiously enough, though, search it in this chat and you'll find a lot of use

gian

15:40

What is the the function $f: R \times M \to M$ that accompanies a module usually called? Would it just be a ring action?

Secret

[Waiting's harmonic reciprocal series]

Mike Miller

wp @Leaky

El'endia Starman

@MikeMiller The user that was suspended. They can't say anything about it while they're suspended.

Secret

Reduction formula:
\begin{align}
\sum_{n=1}^m\left(\frac{H_n}{n}\right)^r = \sum_{n=1}^m\frac{H_n^{r-1}}{n}\sum_{s=1}^n\frac{1}{s} = \sum_{n=1}^m\left(\sum_{s=1}^m\frac{H_s^{r-1}}{s}-\sum_{s=1}^{n-1}\frac{H_s^{r-1}}{s}\right)\frac{1}{n}
\end{align}

SteamyRoot

Pffffft

Dodsy

15:41

@Semiclassical I was suspended

SteamyRoot

Totalitarian chatroom incoming

Secret

where the individual sums can be computed in terms of harmonic numbers (to be done later)

Leaky Nun

@MikeMiller :)

Abcd

@LeakyNun Please suggest a method.

Mike Miller

@El'endiaStarman Nah, they can say plenty. I'm having a lovely discussion with him on gChat about normies. I also posted the "offending" message up there. That way we all know the context!

Dodsy

15:42

just a second ago

Mike Miller

@Abcd We gave like 10 brah

Leaky Nun

@Abcd mine :P

Abcd

@MikeMiller I am confused because of so many messages.

Leaky Nun

well, the substitution is probably the most straightforward

Abcd

@LeakyNun Permalink please

Dodsy

15:43

@MikeMiller I was suspending for saying a variation of "oh my god"

SteamyRoot

@MikeMiller Wait what's gChat?

Leaky Nun

@Abcd you just need to substitute the first one into the second

Dodsy

but with just the letters

Mike Miller

google

Dodsy

and not actually written out.

SteamyRoot

15:43

oh

Mike Miller

he's the only reason i have one of those

everyone else in the world just uses facebook

Abcd

@LeakyNun ? What do you mean?

SteamyRoot

I didn't know it still existed

Secret

Also, the fact that Ted gets extremely annoyed of my (accidental??) flooding of the chat room means posting frequency will decrease in maths chat and then increase again in h bar

Leaky Nun

@Abcd $x+y=2$, express $y$ in terms of $x$, substitute to $x^2+y^2$, find minimum.

Abcd

15:44

Ok

SteamyRoot

@Secret It doesn't feel very accidental tbh

Have you ever, like, considered keeping a blog?

Dodsy

Nobody wishes to discuss my suspension sniff

SteamyRoot

That would probably also get rid of your problem of finding your old messages

Leaky Nun

@Secret you might want to just open a room

Mike Miller

@Dodsy Tell me about it

Leaky Nun

15:45

@MikeMiller he wrote "omfg" and got suspended

El'endia Starman

@MikeMiller Not in this chat. The Not-Nice part is a group of people talking about person A in a public space where A cannot say anything. They can't defend themselves, make an appeal, etc.

Leaky Nun

without the quotes

Mike Miller

Oh my god, you were suspended for something non-vulgar? :o

Dodsy

yes.

yes

Mike Miller

@El'endiaStarman Great! Unban Balarka so he can. :)

2

Secret

15:45

[redacted]

Steamy: I have started the blog, but things are still slow...

Dodsy

@El'endiaStarman please unban him

SteamyRoot

saying "omfg" gets you banned?

Dodsy

I was also banned today for no reason

Mike Miller

Find a single person who could possibly have found that message hurtful, lol

El'endia Starman

@MikeMiller The system will unban him soon enough.

Dodsy

15:46

that's right

well it's not fair

my suspension was also not fair.

Secret

My temporary blog before moving to wordpress
http://secretlabsultraviolet.blogspot.com.au/

Mike Miller

Well, it seems mean not to let him defend himself.

SteamyRoot

If the "system" bans people for those messages, I'm losing faith in the system.

Secret

NB No maths yet

Dodsy

Well somebody reported my message.

Abcd

15:46

Is the minimum 0 @LeakyNun?

Mike Miller

We're not allowed to talk about bans, but neither are the banned! It makes good sense.

Leaky Nun

@Abcd why?

Dodsy

interesting development

Mike Miller

Ooh, which one?

El'endia Starman

@MikeMiller Yeah, exactly. That's why suspensions should not be discussed while they're in effect.

Dodsy

15:47

my lin alg class was terrible

Mike Miller

I can't see on mobile.

Dodsy

@El'endiaStarman okay but bans should be discussed.

Mike Miller

@El'endiaStarman Hahaha. Do you not understand how nonsensical that is?

Abcd

@LeakyNun Minimum of "$(2-y)^2+y^2$" can be minimum 4 as y can be minimum zero.

Dodsy

What's the point of continuing to come to a place where I can't even say "omfg" without being banned, and mods join the chat and just let me be banned.

SteamyRoot

15:48

So if a ban was nonsensical, you have to just sit it out anyway?

Dodsy

That's not fair.

Mike Miller

I'm pretty sure, in any case, there was a long series of mods coming in here for a while telling people to stop reporting.

Dodsy

yes there were.

SteamyRoot

Yup...

Mike Miller

I guess now we're down to mods coming in here to have us stop talking about the reporting.

3

Secret

15:48

@LeakyNun The problem of rooms is that they are too many to keep them from frozen. Seriously, if the chat rooms don't get auto deleted or frozen when not in use, I am ok to open as many rooms as I want

Mike Miller

If we don't talk about a problem, there's decent odds we don't have one, right?

Secret

It's that maintainence bit that annoys me about making new rooms

El'endia Starman

@Dodsy For what it's worth, I probably would have dismissed (or not taken action) on that flag if I had been around to see it.

Dodsy

I understand.

Hey daminark.

:D

Mike Miller

Ooh, what about the one from 10 minutes ago? Can I get your thoughts on that in 20?

I understand that for the next 20 you did sign a sworn affidavit not to talk.

Dodsy

15:50

first day of class was cool, already learning about sets. Then in lin alg teacher decided to "motivate us" and ended up teaching vectors very poorly and calling magnitudes velocities and confusing people. But it was very boring and grade 9 level content.

Mike Miller

eh this is boring now

math time zoom zoom

Akiva Weinberger

@Abcd How would it be zero? For what value of $y$?

Abcd

@AkivaWeinberger It would be 4 for y =0

Akiva Weinberger

Oh I see

You can get it to be smaller

Expand out $(2-y)^2$

Leaky Nun

@AkivaWeinberger come on

let him think.

Secret

15:53

8 mins ago, by SteamyRoot

That would probably also get rid of your problem of finding your old messages

Dodsy

@MikeMiller what are you working on Mikey?

Secret

(To be thinking)

anyway, heading to sleep

Dodsy

night secret

Mike Miller

I'm just reading this morning. I don't want to do actual research. I have to do a bunch of lame symbol manipulation

Abcd

@AkivaWeinberger I got $y^2+2y+2\ge0$. But that doesn't take me to a minimum value.

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$Mathematics$ Mathematics