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7:00 PM
@Faust, where are you?
 
@TedShifrin it doesnt seem to define it the onyl thing i can find in the real analysis book is everything regarding the compleness axiom
 
It may be in an earlier section.
Where they did absolute value, maybe.
 
the best that i have is if S contains a largest element $s_0$ then we call it the MAX S
 
OK, so let's do the induction argument.
Induction on what?
 
thats actually the part that im having trouble figuring out
we have a set
with a countable number of things in it
since R is an ordered field id liek tothink S has some order on it
 
7:03 PM
No. What did the problem say again?
 
so perhaps we have a smallest or largest element
Prove by mathematical induction that every non-empty finite subset $ S \subset \mathbb {R}$ of real numbers has a maximum element.
 
FINITE.
What does that mean?
 
that we can find some integer to represent the number of elements we have?
 
Right.
That's what we're going to do induction on.
So, start with a set with $1$ element.
Hi, DogAteMy. And keep quiet :P
 
well the that element is themax
 
7:05 PM
reacts
 
Well then that element*
 
Right, Faust. What do we do next with induction proofs?
 
assume it true for all sets with n elements prove it for n+1
 
Great. So you start with a set $S$ with $n+1$ elements. What do you do with it?
 
(this is where my deleted comment from earlier was pointing)
 
7:07 PM
Sure, @Semiclassic. Most of us wouldn't bother belaboring this, but at the beginnings of learning proofs ...
 
Right.
 
i wanna break it down
 
Be specific?
 
well i would like to know if the biggest element is in the n elements or the extra one
like make a subset or something
 
I need the proof of sine summation series. Research effort: Seen my notes and the first answer here: math.stackexchange.com/questions/17966/… but it's incomplete and I don't know how to proceed.
 
7:09 PM
Well you could- remembers vow of silence (re Faust)
 
So how do we make a set with $n$ elements when we started with $n+1$ elements?
 
we could intersect it with a set on n elements
of*
 
Nope, @Faust. I don't know how to do that.
Be as simple and direct as possible.
@Abcd: Do you know about $e^{i\theta}$? (I'm guessing not.)
 
@TedShifrin No.
 
Could we just take one specfic element out of the set?
 
7:10 PM
If it helps to make things concrete, take $n=2$.
 
Perfect, @Faust.
@Abcd: So you need to figure out how to use the addition formulas for sin and cos to get $\sin x + \sin y$.
 
Would a proof through compactness be overpowered
 
So you write $S = S' \cup \{a\}$, where you just picked $a\in S$. @Faust
 
@TedShifrin I understood the steps shown in the linked answer. How do I proceed?
 
then we can say $S^{'}$ has n elements so true by IH
 
7:12 PM
Give the maximum element of $S'$ a name, @Faust.
 
oh
b
 
82
Q: How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

QuixoticHow can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? For example here is the sum of $\cos$ series: $$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$ There ...

 
then the max (a,b) = max S
 
Hm, how would you write a computer program that finds the maximum of an unordered list
 
Right, @Faust.
 
7:13 PM
Thank you!
 
It looks complete to me, @Abcd. What is your specific question?
 
@TedShifrin How to continue from where he left off
 
i had no idea where that was going or coming from but that makes alot sense the way you explained it thanks =)
 
@Faust: And if you haven't defined $\max(a,b)$, just say the answer is $a$ if $a\ge b$ and $b$ if $b>a$. :)
 
@TedShifrin How do I continue after the last step he has shown? He says, "Then by doing the same thing you will have some terms cancelled out. You can easily see which terms are going to get Cancelled. Proceed and you should be able to get the formula."
 
7:14 PM
I really should of taken analysis before now
most notation for the it is $max \{a,b ]" $
 
I feel restless without proofs of the formulas I come across. (and am supposed to learn)
 
@Abcd: You have to write out the $n+1$ terms $\sin (a+kd)\cdot\sin (d/2)$ and add them up as $k$ goes from $0$ to $n$.
 
bleh
 
You should get a series where lots of cancellation occurs.
It's called "telescoping."
 
@TedShifrin I would be highly obliged if you show these steps with very few terms ....continuing till n
 
7:17 PM
@Faust: Good advice when you see "proof by induction" — induction on what? for starters.
I'm not going to type all that out, @Abcd, and I have to leave. Write out his formula for $k=0$, for $k=1$, and for $k=2$. Write them above one another.
Look at the three lines carefully.
 
@TedShifrin okay
 
Bye, all. DogAteMy: I apologize for the muzzle :P
 
See you @Ted!
 
have a good one ted
 
I mean ffs. If you actually know the residue theorem then that's a one-liner.
 
7:20 PM
yeah for residue theorem
@Semiclassical being able to read the question can be hard though
 
Sometimes.
 
i find understanding the question to always be harder than the math ;P
also do you have the summer off or something? every time i come here your always here
 
i blame bad habits for that.
 
7:36 PM
$Ax^4 +Bx^3 +Cx^2 +Dx +E=0 $ assuming A,B,C,D,E are integers can i use the rational zeros theorem to prove that this equation does or does not have any rational roots?
for some reason i thought none of the rules applied to roots once we broke the $x^3$ barrier
 
[Random] The point opposite to $z$ on the Riemann sphere is $-1/\bar z$
 
Oh that's nifty
 
@Faust7 You can
Nothing changes for higher polynomials with regards to the rational root theorem
 
Sankyuu
 
@Daminark So in other words you invert it and reflect it across the origin
 
7:42 PM
@AkivaWeinberger opposite as in, antipodal if you take the inverse stereographic projection?
 
@SteamyRoot Yeah
 
Heh... makes perfect sense, but I never really though about it...
 
Also, if $z=a+bi$ and $-1/\bar z=c+di$, I think $(a,b,1)$ and $(c,d,1)$ are perpendicular
which seems to make sense geometrically
The thing I was reading said something that I'm 99% sure is equivalent to that ^
 
Heh. Pretty nifty.
You never really think about that when you actually work with the Riemann Sphere, because anything involving $\bar{z}$ is heresy
 
But the antipodal map has negative degree, so…
^Nice gif I found
It's that combing of the sphere that only has one singular point
It's also what a translation $z\mapsto z+c$ looks like on the Riemann sphere (only fixes $\infty$)
 
8:10 PM
@PVAL you got your original photo back, huh?
 
did I?
 
At least the photo I'm seeing right now is back to the blue one
 
It never changed on this computer for me.
I always saw the blue one.
 
Weird
 
I see a purple one
 
8:14 PM
I see it as blue though
I mean I'm also probably a bit colorblind so...
 
It has like four diamonds in the center
 
Yeah exactly
triple plot twist: they're not diamonds and we're both shapeblind
4
 
My attempts: $cos x = sin^{1/2}(x)$. Then I substituted this in the given equation and kept solving and solving and solving but couldn't reach any answer.
 
Hm
This probably sounds like it's coming out of nowhere, but
@Daminark Subtract sin^2 and take roots
 
$\cos^2 x=\sin x$...hrm.
 
8:19 PM
Yeah no I misread his statment
 
I agree that that seems like the natural starting point.
 
@AkivaWeinberger There's a condition in the statement which semi has stated above
 
The expression on the right is even with respect to $x$
 
You can keep reducing the highest power
 
So maybe even-ize the condition:
$(\sin x+\sin^2x-1)(-\sin x+\sin^2x-1)=0$
And then we can convert it all to cosines
 
8:20 PM
@Semiclassical hrm?
 
Oh derp
 
just thinking out loud
 
That's awful, Semi's thing will lead to the same thing quicker:
 
Or, actually, if you want the fast way
factorise the equation :^)
 
or use $y=\sin x$ and solve the resulting quadratic.
 
8:20 PM
So we have $\sin^6(x) + 3\sin^5(x) + 3\sin^4(x) + \sin^3(x) - \sin^2(x) - \sin(x)$
 
$\cos^2x=\sin x$. Square both sides:
$\cos^4x=1-\cos^2x$
 
@Daminark Yes, I got that
 
Oh that's clever
 
@Semiclassical hrm = thinking out loud?
 
8:21 PM
$y^6 + 3y^5 + 3y^4 + y^3 - 1 = (y^2 + y - 1)(y^4 + 2y^3 + 2y^2 + y + 1) = 0$
 
@SteamyRoot oooo
 
That is, if $u=\cos x$, then $u^4+u^2-1=0$
so try to factor it out
or something
 
That's what I just did :P
 
Wait secret your $y$ is my $u^2$?
Oh so it's 0 then QED
 
yeah
 
8:22 PM
Good job team
 
I had $y = \sin x$
 
\('-')/
 
@SteamyRoot What no you didn't
 
@SteamyRoot Factorising isn't easy for a 6 degree equation :/
 
What's funny is that the solutions are $\sin x=-1.61803,0.61803.$
 
8:23 PM
I had one of the solutions as $\approx0.666\dots$ @Semiclassical
 
$y^2 + y - 1$
 
The first one isn't valid, but the second is 1/golden ratio=golden ratio -1
 
How do I solve?
 
Has $\varphi, -1/\varphi$ as roots
 
Oh whoops misread @Semiclassical
@Abcd OK so
 
8:24 PM
Am I off by a minus sign?
 
Ijsut did polynomial division to get $(y^2 + y - 1)(... ) + (...)$
 
$\sin x+\sin^2x=1$ @Abcd
 
turned out the remainder was zero
 
$\sin x=1-\sin^2x$
 
@AkivaWeinberger yes
 
8:24 PM
@SteamyRoot I think thats off by a minus sign.
 
$\sin x=\cos^2x$
Square both sides
$\sin^2x=\cos^4x$
$1-\cos^2x=\cos^4x$
Let $u=\cos x$ to make writing easier
$1-u^2=u^4$ or $u^4+u^2-1=0$
 
@Semiclassical Oh, right, indeed.
 
You want to find $u^{12}+3u^{10}+3u^8+u^3-1$ @Abcd
 
@AkivaWeinberger Yes
 
Do polynomial division with $u^4+u^2-1$
 
8:27 PM
@AkivaWeinberger This is a complex method. Any other way to solve this problem?
 
lolno
 
It doesn't end up with any remainder
 
this is about as simple as you'll get with this problem.
 
@Semiclassical Is it possible to solve using my approach?
 
so $u^{12}+3u^{10}+3u^8+u^6-1$ is a multiple of $u^4+u^2-1$, which equals $0$
so it equals $0$
 
8:28 PM
What exactly was your method?
 
@Abcd The other solution is noting that $\sin^2(x) = 1 - \sin(x)$ lowers the degree by one, and repeatedly subsituting.
 
@AkivaWeinberger Woah!
 
Which, well, is going to take forever, is painful, and you'll likely make a tiny mistake that invalidates your answer at some point.
 
@SteamyRoot Okay..
 
Polynomial division isn't much harder than long division, really. And it's a good tool to have, so embrace it!
 
8:30 PM
...
They have the answer listed as C?
 
yes because there's a -1 in the end
 
so it is C
 
The thing with the -1 was a multiple of the other thing
 
No. Akiva included the -1 in his factorization.
 
8:31 PM
so it should still be (A) 0
 
No. I am really sorry. It's A
I misread.
Sorrrrry
 
OK good
 
@AkivaWeinberger Thanks for your innovative approach!
 
You're welcome!
 
Gr8 joke: Why did the blind man fall into the well?
 
8:33 PM
@Daminark ROFL
 
@Daminark He couldn't see that well
 
Dammit
Also I was cracking up at "He c"
 
No he blind he don't c
Dat da whole point
Chau
 
0
Q: Generalization of $n$-th dimensional vector space isomorphic to $n$-th product of a field

ALannisterThis question is related to one asked by somebody else here; however, unlike them, I do not want to show that a vector space over a field $F$ and of dimension $n$ is isomorphic to the $n$-th direct power $F^{n}$ of $F$. Instead, I want to generalize this to the following case: If $B$ is a bas...

Yello
 
can someone tell me if that part is unnecessary for the proof? :S
btw there's a typo, it's $f$ instaed of $f_j$, I saw it on the errata's book
you can skip case I
WAIT I already saw it's not unncessary, thanks anyway...
 
8:52 PM
It's always good when you can answer your own questions. :)
 
@ALannister Is there a reason why you want to map to the direct sum rather than the direct product?
 
That's the way it's stated in the problem. Isn't it usually easier, too? Because then all but finitely many coordinates must be zero.
 
You're imposing that on the codomain
But why should it be true on the domain?
 
I guess it doesn't.
Not sure where I said that it had to be for the domain, though.
 
Well, you didn't.
But if you want an isomorphism between the two...
 
8:58 PM
You can do it for when there are only $n$ elements in the basis. I've linked to a problem like that.
Maybe I chose the wrong function?
 
Well, sure, for finite dimensions the direct sum and direct product are the same
 
Okay, so I probably did not define the mapping correctly. Part of what I'm asking is whether my mapping is correct and if not how do I fix it.
 
The idea behind your map is fine.
 
The full text of the problem says "Using representation of elements of a direct sum $\oplus_{\alpha \in I}F$ as functions $f: I \to F$, prove the second statement of Proposition 15:18: If $B$ is a basis for a vector space $V$ over $F$, then $V$ is isomorphic to the direct sum $\oplus _{\alpha \in B}F$ of copies of $F$ indexed by $B$."
Proposition 15:18 states the following "A vector space over a field $F$ and of dimension $n$ is isomorphic to an $n$-th direct power $F^{n}$ of $F$. More generally, if $B$ is a basis for a vector space $V$ over $F$, then $V$ is isomorphic to the direct sum $\oplus_{\alpha \in B}F$ of copies of $F$ indexed by $B$".
@SteamyRoot Maybe the very first part helps?
 
How does the text you're using define vector space and direct product?
 
9:07 PM
It defines a vector space as a triple $(V, +, F-\text{multiplication})$
A direct product it doesn't define - it's not actually a text. We have no text. It's typed up lecture notes.
I think the semester before we defined direct product (it's a 2 semester course)
lemme check.
 
Well, as far as I can tell, the statement is wrong.
And it should have direct products instead of direct sums
 
Which statement is wrong?
 
For example, $\mathbb{R}^\mathbb{R}$, i.e. the functions from $\mathbb{R}$ to $\mathbb{R}$, are a vector space over the field $\mathbb{R}$
 
We defined direct products of groups last semester: The direct product of a family of groups $\{ G_{\alpha}\}_{\alpha \in I}$ is the group $\prod_{\alpha \in I}G_{\alpha}$ with componentwise multiplication as the group operation.
 
But unless a function is zero on all but finitely many $x \in \mathbb{R}$, it won't map into $\oplus_{x \in \mathbb{R}} \mathbb{R}$
 
9:12 PM
Well, you're supposed to use a basis for $\mathbb{R}$ as an indexing set.
I'll ask my prof.
So, specifically, I need finite support for the vector space?
 
Either that, or you want to map to direct products instead of direct sums
 
It seems like doing it for $n$ is a very common exercise. Doing it for an arbitrary indexing set is not.
Let's say you do want to map to direct products, then what?
 
Then... your map works?
 
But the specifics of carrying out the additions...part of what I asked is how the mechanics works in showing it's a homomorphism.
 
Well, if you denote an element of the vector space as $\sum_{b \in B}f_b b$ where $B$ is the basis
 
9:23 PM
Another element would be say $\sum_{b \in B}g_{b}b$?
No overline like I put in my question on the b's. Same basis elements
 
hi
 
Then for $v = \sum_{b \in B}f_b b$, $v' = \sum_{b \in B}f'_b b$, you have $v + v' = v = \sum_{b \in B}(f_b + f'_b) b$
 
Hey Tyrion ;)
 
0
Q: Suppose that $x^2 = b$ is an irrational positive real number. Show that there exists a nontrivial unit in Z[x].

TyphonSuppose that $Z[x]$ is composed of all numbers of the form $c + dx$ where $c$ and d are integers and $x$ is an irrational positive solution to the polynomial $x^2 + b = 0$ where $b$ is an integer. I seek to show that there exist any elements that are units aside from $1$ and $-1$. I know that...

@ALannister that's not my name! >:(
 
And, inside $\times_{b \in B} F$, you have pretty much the same thing.
 
9:24 PM
LOLZ! I know!
All right, @SteamyRoot thank you.
 
@ALannister I know you know.
 
I'll try to fix up what I have and in the meantime, I have sent an email to my prof.
 
Hello Anna.
 
Since the operations and such are pretty much element-wise, they all behave like for finite-dimensional spaces.
 
Who's Anna?
 
9:31 PM
you are
 
10:06 PM
Chicken!
 
10:19 PM
https://repl.it/Jfcr/18
@AkivaWeinberger Well, I made it bigger. This one goes up to $\omega^{\omega^\omega}$ in the fast growing hierarchy
And a little bit past that
Shoot, made a mistake D:
 
wth
 
Don't worry, I'm fixing it lol
 
ok
 
The last one terminates too quickly ;)
There we go
Yeup
Nope, this one is better:
This now goes, erm, beyond $\omega^{\omega^\omega}+1$ at least
@Faust7 :D Better now?
 
Dude i clicked on all the links and quite honestly im 100% certain i do not speak the same dialect as you =)
whats your doing could be a random number generator for all i can tell.
 
10:30 PM
:D You can read Ruby?
 
rehi @Faust. Howdy, SBA.
 
no i was just pushing the play button n reading all the funny numbers
 
Nah, if you look carefully on the printed stuff on the side window, it should be pretty clear that its doing something logical
Howdy Ted
 
mine kept erroring our say f element not found for pages
 
10:31 PM
l):1:in `f'
(repl):1:in `block in f'
(repl):1:in `times'
(repl):1:in `f'
(repl):1:in `block in f'
(repl):1:in `times'
(repl):1:in `f'
(repl):1:in `block in f'
(repl):1:in `times'
(repl):1:in `f'
(repl):1:in `block in f'
(repl):1:in `times'
(repl):1:in `f'
(repl):1:in `block in f'
(repl):1:in `times'
(repl):1:in `f'
(repl):1:in `block in f'
(repl):1:in `times'
(repl):1:in `f'
(repl):1:in `block in f'
(repl):1:in `times'
(repl):1:in `f'
(repl):1:in `block in f'
(repl):1:in `times'
(repl):1:in `f'
 
Oh, yeah, that just means stack overflow
 
enough spamming
 
Morning @ ted
 
Its set to "give up" after so many steps
 
morning? where are you, Faust?
 
10:32 PM
Victoria, bc , canada
 
It ain't morning!
 
but i just woke up so its morning =)
 
No, you didn't just wake up — we were talking hours ago.
 
hey there, @chat
 
10:32 PM
hi @Lucas
 
hey lucas
 
i went to beafter you left i wasnt feeling well. feeling a bit better now so back at it
 
Oh, glad to know I make you ill, Faust :D
 
@Ted I think you missed this
:P
 
haha i wish it was that simple then id know how to feel better
 
10:33 PM
I feel sorry for you, Lucas. Which book?
 
HAHAHAHAH
It was "Linear Algebra: A Geometric Approach"
 
Oh, that one. Somehow I thought you were working on abstract algebra stuff.
 
you got a book on aa?
 
Oh, Lucas, the linear algebra one is the one that's translated into Portuguese!
yeah, Faust, first one I wrote.
 
@TedShifrin I passed the algebra exam !!
 
10:41 PM
Congrats, @Maks.
Did I know about it?
 
@TedShifrin i got asked to jion the honours program at my uni this summer and finally caved and said yes. im alittle worried about having to do an honours project and thesis, which is why i have refused until now. (mostly because anything in math thats unsolved seems incredibly difficult to solve.) what exactly will i be doing a project on?
 
Don't ask me!!
 
rofl
 
And that's a long way away.
 
thought u were a teacher for a long time
and not really! im suppose to start it in my second semester of this year
 
10:43 PM
It's not uniform across all programs
 
Although you could do some investigative ("baby research") work, most Honors theses people do are just a book report or exposition of some material you've learned in some depth.
 
@BalarkaSen That problem I was thinking about related to the non separating curve graphs, someone posted a solution on the arxiv today.
 
@TedShifrin well thats less scary thanks ^^
 
Well, there are things you could probably do based on a solid diff geo course, for example. But I didn't think you'd gotten far enough in math yet.
 
@AkivaWeinberger We have $f(100)\gg g\left(10^{19729}\right)$, where $f(n)=f(n,-n,[n,n],[n,n])$ is my function and $g$ is the Goodstein sequence.
 
10:44 PM
i havent done any analysis but i only got 3 semesters left
i was orgionally a physics and biochem student who got spirited away to the math dept so a good number of my credits come from there
 
You should definitely pick something you are really interested in, Faust ... have some background and want to learn more depth.
If you're strong with computation, you could possibly try to make it more experimental :)
 
abstract algerbra has been my favorite thus far i also really enjoyed some stuff i did on chaotic systems some bifurcations of systems in 3d specifically homoclinical loop bifurcations and im rather interested in topology as well
i dont really like discreete math very much except that graph theory is cool
also analysis is way too hard lol
 
All of math is hard. :)
 
i took analysis as part of an advanced calculus class at third year it was the first 2 weeks of the class even at 30hrs per week of hw in it i still barely understand the crap.
on the bright side my university split it up into 3 classes so no one else has to suffer through it
 
Have you studied multivariable analysis (like derivative as a linear map, inverse function theorem)?
 
10:53 PM
yes
in that calculus class
 
Ah ... so you might find Guillemin & Pollack a good thing to read/work on.
 
the first 2 weeks of which is now an entire second year course
 
Lie groups are where algebra and manifolds meet.
 
ill check it out
 
Most of my lectures on YouTube are (rigorous) multivariable stuff, with linear algebra.
 
10:55 PM
ive been meaning to put all the pieces together to understand C* algebras properly
 
That's some infinite-dimensional analysis.
 
i grabbed your linear algerbra textbook off amazon mine sucks and yours looks pretty good
 
I'm really not here to solicit interest in my books. :P Which was the one you used?
 
Hey @TedShifrin
 
heya @Danu
 
10:59 PM
I've got a Riemannian geometry question
 
i have 2, introduction to linear algebra for science and engineering and someone gave me linear algebra by Gilbert strange which is somehow alot worse then the first one
 

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