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3:11 PM
@Givemeabreak very sick in January, and busy after.
 
3:36 PM
@robjohn get well soon
Oh, I'm late by a few months
 
Hi @Dami
 
How's everything going
 
tired, hungry
i literally had 10 hours of sleep, and i just woke up and im still tired as hell
 
Mhm...
I think we're all tired, especially students
 
I was never as tired as a student as I am as a PhD student/ TA :P
 
3:49 PM
Lol, the worst is yet to come
Though I'll say that first quarter this year was my harshest
 
4:12 PM
i think second year is the worst... though differential geometry was alot of work i find most 3rd n 4th years class are less effort then the second year counterparts but perhaps im just a slow learner
 
Well, it depends a lot on what courses or projects you get at what time...
My third year was horrible, but that's because I did a twin-bachelor in maths and physics and had to make 2 bachelor projects at the same time :P
 
lol
physics labs are so nice though...
i did stuff its reasonable im going home now.
9/10
 
@Faust7 Physics labs for me were painful
And I mean, I'm a second year so it's possible that things are going to get much harsher
 
chem labs its go home write a dam report for 10hrs and get 8.5 for me anyway
 
But my first quarter professor basically assigned us 60 problems each week for homework
20-25 of which were linear algebra problems. We didn't cover that in class, so we also had to read the chapter where the problems came from
 
4:17 PM
i was in physics and biochem before switching to math third year is less work if you stick out the second year imo
 
My physics labs in my second year were awful, since I had to go to a different campus, which took 3 hours by bus+train...
 
(50 one time, which was an experience to do in 4 days...)
Then the remaining ones came from Rudin, Sally, and later on in the quarter Buck, which were all somewhat long
 
honestly if you just take as many math class as you can that you need each semester while in the physics program life becomes really easy
 
But that professor had particular tendencies which did not translate over to the other quarters
 
yeah one of my profs in a second year physics class gave us a problem on our final that u had to use stokes theorem 3 times on to integrate where most of the class myself included didnt even know what it was called
 
4:20 PM
@Faust7 That's cruel
 
you could do it with a quad integral if you really stuck to it aswell.
but if you knew the math it was a really simple integral from 0 to r over the volume of a spherical shell (which of course doesnt actually have volume) but w.e
 
4:48 PM
hi chat
why in doing complex integral we can change sin x with e^iz ?
 
@KasmirKhaan You have to take the imaginary part when you do that substitution
 
What do you mean ?
 
$e^{iz} = \cos z + i \sin z$ right?
 
am doing this integral sinx / x
 
Hi @MeowMix
 
4:50 PM
Hi @Balarka
Oh and hey @Alessandro
 
So what they wrote is : sin x /x can be replaced with e^(iz) / ( iz )
 
over what contour?
 
how are you
 
5:16 PM
Hey Zach
 
Hi Nate
 
And how are you today?
 
tired, as always
 
Ah, perhaps an iron deficiency?
notadoctor
 
Idk.
I don't eat that much
 
5:26 PM
Hm.
So what are you working on today?
 
Thinking about one of Ted's problems
 
I need to finish a unit of chemistry today or else I am basically screwed and won't be able to go to the university I want to go to. I'll have to go to a different one for a year, and the courses aren't as good.
 
reading linear algebra
 
linear algebra seems interesting.
 
Morning @MikeMiller
 
5:27 PM
Like I want to take this first year class, it's got something to do with mathematics of computing, and it's proof based.
 
morning
 
but this other school only has something random for computing that you need to take
I think it's called like "media and the world"
or something.
Like screw that.
Hey @BalarkaSen
Hey @MikeMiller
 
How are you guys today?
 
aright
 
5:31 PM
That's good!
 
@Dodsy Do you need another hint on the convex functions one?
Or do you just want a different problem
 
I can't even think about math right now Zach :C
So much chemistry.
And, I decided I'm going to upgrade my english mark, which is sitting at like 75% (I did that back in 2013...)
I figure I can get a mark in the 90's
 
heh "upgrade"
 
Yes :)
But really I'm recompleting the entire course.
So the first assignment is on comparing which is more tragic, hamlet or agommenon
I think that's what it's called.
anyways, I've already read hamlet. So I've got a one up.
I've read hamlet three times, actually.
Long story.
 
i don't know but what's tragic is going through the ordeal of doing that assignment
 
5:35 PM
Haha yes!
The best part is that it isn't even for mark!
marks*
The course set up is very strange, I need to complete 4 units to send the marks, and 2 units are "marked" but only one is worth marks, another is a peer assessment, and the third is a test.
This is helpful because you can't hand in the next assignment until the one before it is marked.
Sorry for flooding the chat with insignificant and off topic details!
 
It's not like there's anyone talking here, so it's fine :)
 
hi everyone
did anyone ping me ?
 
@MeowMix I found this interesting property of squares, but I also found in an elementary number theory book the reason for it being the case, but what I've found is simpler and a bit more interesting.
 
hi @MeowMix
 
So we know that if we have a single square and we add three more squares around it we have 2^2 , and if we add 5 more squares we have 3^2.
This is already widely known.
but then what I've found is that any number n^2 = 1 + 1 + 2 + 2 ... + n
 
5:50 PM
squares??
You mean just adding up?
 
Yeah
So $5^2 = 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5$
 
Do you know why that's the case?
 
I've already said.
 
Uhh
no you didn't
LOL
 
because %n^2 = 1 + 3 + 5 + ...$
 
5:53 PM
wait what
 
I suppose that's true
 
I don't remember the exact formula of course, my book is upstairs.
 
the $n$-th odd number is just $n + n+1$ or $2n+1$
 
2n + 1
that's right.
 
5:54 PM
Nate, do you know induction?
 
Guys, does this mean that we can multiply the terms in a series as often as we want by $x$?
 
Yeah
the radius of convergence will always be the same
because $x$ can't be infinite by definition...
 
like the form of proof?
 
sorry i misread it
 
But zach, if we look at that other way of how squares come together, (we can see it geometrically) then we can see why my way works, it's a redistribution of the integers.
1 + 3 = 1 + 1 + 2
 
5:55 PM
Well, for a series to make sense, there may be conditions on $x$
 
1 + 3 + 5 = 1 + 1 + 2 + 2 + 3
 
Yeah
Just be careful when you use that line of reasoning
 
But these conditions don't change if you multiply or divide a series by a power of $x$
 
why not?
I just don't see it
 
(assuming that you don't end with negative powers, of course, then things get a bit more tricky)
 
5:56 PM
I don't think my idea is revolutionary or important in any way, and the already done way is certainly more intuitive and better.
 
@Dodsy Another reason you can understand why that works is because $1 + 2 + 3 + 4 + \dots = n(n+1)/2$
 
I think I know why
 
pull out the $x^n$ and put it in front of the series.
 
because limsup stays the same
 
5:57 PM
And so your sum equals $n(n+1) + (n+1)$ which is the same as $(n+1)^2$
 
then $|x^n| = |x|^n < R^n$ if $R$ is the radius of convergence
 
ah that makes sense!
 
and the other factor is then the series...
 
$\limsup \vert a_n\vert ^{1/n}$
 
but @MeowMix I was thinking of a way to find the perfect square root given this summation.
 
5:58 PM
Huh?
 
yeah let's say we're given 25, just for ease
we know that 25 = 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5
 
@SteamyRoot what do you actually mean by this? you mean $\sum a_nx^{n+1}=\sum a_n\cdot x^n\cdot x$ ?
 
I guess it doesn
 
$\sum a_n x^{n+m} = x^m \left( \sum a_n x ^n \right)$
 
't work.
you'd have to add (1 + 1) (2 + 2) (3 + 3)
and then sum the roots of these integers.
It wouldn't make sense.
 
6:01 PM
hi Ted
 
Oh, hey it's @Ted
 
Hey ted.
 
hi @ShaV, Zach, Nate, @Steamy
 
Ohi
 
Hi prof @TedShifrin
 
6:01 PM
@SteamyRoot and what did you mean by this? I don't see its meaning (sorry)
 
and hi, skull
 
what kind of meaning does $R^n$ have?
as in, what useful meaning?
 
@Ted Did you already give me a list of suggested problems for section 5?
 
No, probably not.
 
If $|\sum a_n x^{n}| < \infty$ when $|x| < R$, then $| \sum a_n x^{n+m} | < \infty$ when $|x| < R$
 
6:03 PM
Oh, well could you, please?
 
Because $| \sum a_n x^{n+m} | = |x^m \sum a_n x^{n} | = |x|^m | \sum a_n x^{n} | \leq |R|^n | \sum a_n x^{n} |$
 
Um, yes, sir. 3, 9, 10, 13, 15, 17 [very cool], 18, 19, 20. [Do ones with asterisks for practice.]
 
Ted did you see what I found about squares?
 
They're not as round as circles?
4
 
%n^2 = 1 + 1 + 2 + 2...n$
 
6:05 PM
Thank you :] I'll get to work
LOL
 
okay cool, thanks. do any problems occur when $R=\infty$ though?
because then your argument collapse :P @Steamy
 
Say again, @Dodsy?
 
So you know the summation of squares series : %n^2 = 1 + 3 + 5 + 7 ... 2n+1$
 
@Ted $(n+1)^2 = (2\sum_{k=1}^{n}k) + (n+1)$
or, less ugly-ly, what he put above
 
@Steamy I would stick to my argument then
because then we just calculate limsup
 
6:06 PM
Woah meow mix, no!
 
@ShaVuklia You can just work "$x$ by $x$"
 
and it doesn't matter that the exponent is shifted
 
as long as you fix an $x$, then that $x$ is finite
 
Right, the sum of the odd numbers gives perfect squares.
Because $(n+1)^2 = n^2 + (2n+1)$.
 
So, @TedShifrin what I've found is that for example $5^2 = 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5$
 
6:07 PM
ah kay
 
You need $ on both sides, Nate.
 
Oh, I'm putting percentage signs...
And this works for any integer.
 
Oh, this is what Zach said. $1+2+3+\dots+n = n(n+1)/2$, so $2(1+2+\dots+n) =n^2+n$.
What you're writing is $2(1+2+\dots+(n-1))+n = n^2$.
 
Oh yes, that's perfectly reasonable, sorry Zach.
 
It's fine :P
 
6:09 PM
@TedShifrin hi
 
Hi @Liad.
 
Here's a cool equation: $(1 + 2 + \dots + n)^2 = 1^3 + 2^3 + \dots + n^3$
 
Anonymous
Hello people! Does anyone have any idea for this question: math.stackexchange.com/questions/2229616/…
 
Anonymous
0
Q: Prove that either $a=2l$ and $b=m$ or $b+m=al$

blueIf by eliminating $x$ between the equations $x^2+ax+b=0$ and $xy+l(x+y)+m=0$ then a quadratic in $y$ is formed whose roots are the same as the original quadratic in $x$. Then prove that either $a=2l$ and $b=m$ or $b+m=al$. It seems a very lengthy process to eliminate $x$. Does anyone know of any...

 
That is cool Zach!
 
Anonymous
6:10 PM
I think it should have a simple solution.
 
do you remember our discussion about $\Bbb R / A$ where $A = [0,1] ,[0,1)$ ?@TedShifrin
 
Yeah.
 
Hey what's the summation of integers from n - 1 formula.
n to 1
because technically it would be n to 1 + n - 1 to 1.
 
I just wrote that down up there, @Dodsy. ^^^^
 
right?
Oh!
 
6:11 PM
i tried to think about taking $A=(0,1)$
 
n(n+1)/2
 
but isn't this the same as $[0,1)$ ? $\Bbb R/A$ would still not be Hausdorff, correct?
 
so it would be $n(n + 1)/2 + n(n)/2$ ?
 
OK, @Liad. That one's a bit stranger than what we did, but similar.
 
no that doesn't work...
 
6:12 PM
Substitute $n-1$ in the formula for $n$, Nate.
 
oh true.
 
@TedShifrin what do you mean by "stranger" ? does it different from $[0,1)$ ?
 
No, @Liad. To me it looks like $(-\infty,0] \cup \{1/2\} \cup [1,\infty)$, and then you have to decide the topology.
 
so it would be $n(n+1)/2 + n-1(n)/2$
 
No, @Dodsy: $(n-1)(n-1+1)/2 = (n-1)n/2 = n(n-1)/2$.
 
6:14 PM
oh right.
:)
 
@TedShifrin i mean with $A=[0,1)$ , $1',A' $ couldn't be separated, here this is the same doesn't it?
 
It's a pretty cool property of squares.
 
So @Liad, I'm calling $A' = 1/2$ for convenience. But now any neighborhood of $0'$ contains $A'$ and any neighborhood of $1'$ contains $A'$.
 
At first I couldn't find anything online about it, then I started reading an elementary number theory book and it showed the summation of odd numbers thing.
and I was like ohhhhhhhhhhhhhhhh
 
Nate: Can you see why it holds geometrically? Start drawing squares of dots.
 
6:16 PM
yes!
 
@TedShifrin huh , so both $0'$ and $1' $ can't be separated
 
The booklet showed the odd numbers as squares
 
They can from each other, @Liad, but not from $A'$.
 
and them being added to form a larger square
so I can certainly see it geometrically.
 
Right, Nate. That's what I wrote down up there somewhere. Cool.
 
6:16 PM
@TedShifrin yeah that is what i meant. Thanks!
 
It's very cool!
I was pretty excited about it, honestly.
Well, I suppose I'll go do some chemistry.
 
The American Math Monthly sometimes has a page of "proofs without words" that are like this and more.
 
huh, I'll check it out.
I'll talk to you later Ted, Zach, everyone else.
 
Bubye.
 
6:18 PM
Adios
 
Prof @Ted !!
 
Heya @Studentmath :) Happy Pesach :)
 
Thank you :) Did you join someone for a Seder?
 
Nah, I'm not nearly that religious :)
Last seder I participated in was my last year of grad school, and a neighbor and I cooked a totally gourmet feast for it. :)
 
So you were spared the boringness ;)
 
6:19 PM
LOL
 
Haha well at least you enjoyed your food then?
 
Oh yeah, that food was amazing — but that was 1979. :)
 
Nobody even thought of me in 1979, I'm pretty sure
 
I'm pretty sure, too.
You doing OK?
 
So-so, every day is different. I've hit the phase where I'm just waiting for it to be over and be able to go back to studying. At least I found out I truly enjoy Math - trying to start my graduate studies while I'm still at it.
How about you? Enjoying the retirement?
 
6:23 PM
I know the military is far from fun. I'm just glad you're safe and reasonably sound.
Yeah, you weren't much for math when I first encountered you here.
 
Haha reasonably sound indeed, even better than that :)
 
I'm frustrated because I keep sending emails volunteering to tutor/teach and people don't even have the politeness to respond. :(
 
True.. I think I owe these forums/q&a and many people in here, you included, a lot for the progress and my growing like-ness of it.
That's rather impolite indeed.. and foolish of them too, imo
 
I guess I'm going to send some reminder emails soon, but that might be perceived as annoying. Sigh.
 
6:27 PM
I've managed to sneak in couple of hours a month volunteering in tutoring students without means, it's really great..
I read that as 'reindeer emails', all Christmas spirit.
 
LOL @ reindeer.
Good for you, @Studentmath. Kudos.
 
I found out relatively recently that I suck at teaching. LOL
 
Well, you have to think more about where people are and what they're having trouble with, Zach. Many professors aren't so good at that, either.
 
I always find myself making mistakes in my words
 
You're not good at being careful before you speak/type. It takes attention and practice.
 
6:34 PM
@Ted Danke, though no kodus, it is self-rewarding.. any interesting probability questions you encountered on the site lately?
 
LOL, no, not really. Sorry :(
 
still stuck on this question from a few days ago:

A​ hot-air balloon left the ground rising at 3 feet per second. 12 seconds​ later, Victoria threw a ball straight up to her friend Colleen in the balloon. At what speed did she throw the ball if it just made it to​ Colleen?
 
I usually mention you if someone wants help with probability, though.
 
Haha, thanks
 
Show me your work, @WillN.
 
6:36 PM
hey there @ted any good news ?
 
Huh? @Idle001
 
@TedShifrin all I've got so far:

for the balloon: $v(t) = 3t$ $s(t) = \frac {3t^2}{2}$
 
No, no.
 
I cant figure out an equation for the ball
 
The balloon is wrong.
 
6:39 PM
oh dear
 
$v(t)=3$, so $s(t)=3t$.
 
ahh ok
 
That doesn't soundright
 
Does HCN have the extra thing at the end for the structural formula?
 
Because @Ted Last night we had an argument about whether his other question meant meters per second or meters per second squared
 
6:40 PM
I'm going to have to start hanging out at the chem chat
 
the other one was ambiguous about that also
It said an "acceleration of 3 meters per second"
 
No, "rising at" is a rate of change.
No ambiguity, and the units are there.
 
rising at in this case velocity
 
Yup.
 
my mistake
 
6:41 PM
I know, I'm just saying because for the other problem they forgot m/s^2
 
Now, how are we going to take into account the fact that Victoria's time starts 12 seconds later?
 
so I don't know if that's also supposed to be m/s^2
 
Zach, stop muddling.
 
oh yeah we were talking about that last night Zach.
 
6:42 PM
This problem is complicated enough without your making it worse.
 
I swear!
 
I thought that the function should be modelled by $2t^3$
 
Stop, guys.
Seriously.
 
Sorry, Ted.
 
GUYS "rising at" is velocity. Sorry for not making it more clear
"rising at an increasing/decreasing rate of" is acceleration
 
6:43 PM
So, if I want $t$ to be the time after Victoria throws her ball, I need to realize that the height of the air balloon will be $3(t+12)$.
@WillN: Pay attention to me, not to them.
 
combinatorics ?
 
@TedShifrin I was thinking the balloon's position might be given bu: $s(t) = v(t) * t$?
 
@Will: Only because $v$ is constant, so when you integrate you get $vt+\text{constant}$.
Can we move on to the ball?
 
yes please
 
She throws it from ground at speed $v_0$ upwards. What is the position $t$ seconds later?
 
6:47 PM
$v_0t$
 
No, there's gravity pulling it down.
 
oh ok
we've been using -10 instead of -9.8 up till now so thats what ill go with for a(t)
 
We're using feet here, aren't we?
 
$v(t) = -10t$
 
You can't mix feet and meters.
 
6:49 PM
oh yes i didnt notice
 
So use $-32$ ft/sec$^2$ for the gravitational acceleration.
 
$v(t) = -32t$
 
$+v_0$
Now integrate to get the height of the ball at time $t$.
 
$s(t) = -16t^2+\frac{v^2}{2}+36$
wait no thats wrong
 
It's very wrong.
(I didn't mean to shut up everyone else in the room, but the silence is sorta nice. :) )
 
6:54 PM
@MeowMix have you ever studied diophantine equations from Arithmetica? I have Stephen Hawking's "God Created the Integers" but there's so many errors in the text that there's no way you could study it seriously.
 
@TedShifrin i dont know how to get C :(
 
We agreed that she tossed it from ground level. So $s(0)=0$.
 
@TedShifrin Honestly I like just reading what you guys are talking about, to learn. I am not very good at physics, which I why I posed the earlier problem to Zach to see what he thought. I wanted a lesson :P
 
This is actually a bit more complicated problem because of the time difference. Do you see that?
 
yesss
 
6:56 PM
So what's $s(t)$, @WillN?
You had $v(t)=-32t+v_0$, so ...
 
$s(t) = v_0t$
 
Come on, man.
 
$V_0$ is a constant isnt it?
 
Yes.
I just typed $v(t)$ there for you.
 
$v(t) = -16t^2+v_0t+C$
 
6:59 PM
$s(t) = ...$
 

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