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4:00 PM
@Nicholas The new distance and the initial distance are proportional to each other by some factor $k$. If the slope is constant, then that means the new point $(x_2^{'},\,y_2^{'})$ is also proportional to $(x_2,\,y_2)$ by the same factor $k$. Can you take it from there and derive an expression for the new point?
 
@I'mGettingThere \cdot: $\lVert\cdot\rVert$
 
user129943
Thanks @DanielFischer
 
@DanielFischer If we want to compare two recurrence relations of the form $T(n)=aT\left( \frac{n}{b} \right)+f(n)$, do the two relations have to have the same b?
 
Err
:)
Isn't there a simple formula for this
?
 
That's it
 
4:02 PM
@evinda No, you can - assuming a nice enough $f$ - find the $\Theta$ of each individually, and then compare.
 
Looks kinda weird to me
 
I phrased that slightly incorrectly, but thinking geometrically is the key to finding the answer @Nicholas
 
got it
!
assuming curr distance is 50
and x2,y2 = 100,100
new distance is 100
x2,y2 = 200
200,200
correct?
 
@DanielFischer Here at page 11 : http://www.scribd.com/doc/55552407/Master-s-Theorem#scribd
I haven't undersood why since f(n) is the same in both recurrence relations , we try the first case.. Do you have an idea?
 
Let $d_1=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$, and proportionality of the line segments implies $\frac{d}{d_1}=\frac{x_2^{'}-x_1}{x_2-x_1}$
Wait, that's totally wrong...
Hmm...
 
4:08 PM
wait whats wrong with my scenario?
if for each distance point I want to extend my line i add 1point to x and y respectively the line will get extended right?
the current distance is expressed easily like this - so I can get the current distance just fine
var distance = (Math.sqrt((Math.pow((x2-x1), 2))+(Math.pow((y2-y1), 2)))).toFixed(0);
 
I wasn't saying yours was wrong, I'm thinking mine was
 
user129943
Is there a nice way to do -2 in LaTeX as well? For example $-2$ - the minus is too big! (Last Q I promise)
 
I need to stop questioning myself, lol... My approach is correct, not sure about yours since $(x_1,\,y_1)$ wasn't given
 
I already have x1,y1
My line is comprised of x1,y1 - x2,y2 the 2 endpoints
 
Yes, and my formula above gives you $(x_2^{'},\,y_2^{'})$ in terms of $d,\,x_1,\,y_1,\,x_2$, and $y_2$
 
4:13 PM
Hi @AlexWertheim
 
Hello @Balarka :)
How goes Hatcher?
 
Not bad, but homology is a little tough. It's taking a little time.
What about you? Have you been thinking about anything interesting lately?
Interesting as in math, 'course.
 
That's ok, learning hard math well takes a lot of time.
Hard to say whether I think about anything interesting, math or otherwise. :)
 
Well, what are you studying right now?
 
@Nicholas You didn't take into account the initial coordinates $(x_1,\,y_1)$, I can't verify your result above with the information given
 
4:16 PM
Well, I've just begun a somewhat aggressive self-study program.
 
Um, give me 2 secs
 
Aggressive?
 
Aggressive for me, anyway. It'll be a lot of information for me to pick up in a relatively short period of time.
 
What are you planning to study?
 
I'm reading Atiyah and Macdonald for commutative algebra, and Serre's "Linear Representations of Finite Groups" for representation theory. Hoping to make it through all of the first and the first 8 chapters of the second.
 
4:18 PM
I found Atiyah-MacDonald a bit dry. I mean, it dumps a bunch of problems on you way before talking about much of the ideas. Rudin-type, I'd say.
 
I'm also reading Willard's "Topology", since I don't know very much point set topology. I'd like to be able to work through Hatcher as well.
Perhaps. It's got excellent exercises though.
 
Yes, good exercises, but I'd prefer to pick up the ideas from somewhere else before doing them.
Which chapter are you in right now?
 
Haha, perhaps. I think I learn best by struggling through ideas a bit first. I read through Mendelson's "Topology" some time ago and did every exercise. Things were presented so thoroughly and clearly, I don't think I ever really internalized things for myself.
 
PS : If it fits you, you could try Simmons for topology. I found it excellent. Never read about Williams though.
Mendelson is not a good book.
Again, of the Rudin-type. And it wades through the stuff from an analysts point of view.
 
Haha, I liked it. It was just what I needed at the time. I think I need something a bit more comprehensive now.
 
4:22 PM
Try Simmons.
 
Thanks @Bal
Gonna try out Simmons
 
Yes, you should.
It has excellent problems too.
 
What's the exact title
 
"Modern analysis", I think.
 
4:24 PM
It takes an intuitions-first point of view of things. You'd find it fun : it begins with metric spaces.
 
Perhaps if I don't like Willard, @Balarka. I am enjoying it so far, though, and it seems to be a very strong treatment.
 
As you wish. Each has his different tastes.
 
To answer your other question, well, I've had some exposure to A&M before. But I'm reading chapter 1 right now.
 
Have fun! I've only done as much as chapter 2, but then dropped it since I wanted to study algebraic topology thoroughly.
Have you come to the series of spectrum-of-rings exercises?
 
I will. I'm just glad to be doing math again.
Yes. :)
 
4:28 PM
I suggest you should give Spec R a thought.
Have you drawn the pictures?
 
No, not yet. I'd like to, so I can understand Mumford's famous drawing.
 
Mumfords treasure maps are my favorite! =D
 
Out of curiosity, if you don't particularly care for A&M, what commutative algebra books do you like? Eisenbud? Matsumura?
 
Probably Eisenbud, but I haven't looked at it. Heard praises about it being a lot more thicker than A-M though :P
 
Eisenbud is very thick, very extensive. I considered using it.
But I think I'm going to use it to supplement A&M. Many good books, only so much time to read each of them. :)
 
4:34 PM
Well, you have a lot of time. Which year are you in now?
You're a grad student, right?
 
Haha, in some ways. I graduated from college (university) last year. I'm working this year. I'll be a graduate student next year.
Soon. :)
 
Oh then you definitely will have a lot of time. Quals are years away, so no pressure ;)
Pick up some book, skim through a few chapters, try to do some exercises, skim through the chapters again -- that's how I learn.
 
Maybe so. But I'd like to be well prepared going in, and really make the most of my graduate years. I've only seriously studied math for about two years now, and many know a great deal more than me. I'd like to bridge that gap.
It doesn't have to happen over night. But sooner rather than later would be good. :)
 
Don't worry, you'll be fine =)
So, how much topology have you studied so far, @Alex?
 
Thanks. :) Well, through Mendelson, I was exposed to the basic theory of topological spaces. Things like interior, closure, continuity from a topological point of view. Homeomorphism, weak and strong topologies. Basic notions of connectedness and compactness.
 
4:48 PM
morning
 
morning, @Mike
 
As far as Willard goes, just started. Learning some set theory, since I never did a lot of the basic problems, and would like to just have seen them before. Now onto metric spaces.
Hello, @Mike. I watched Adaptation last night.
 
Hello!! Does someone have an idea for:
 
metric spaces can be fun.
 
0
Q: Do I have to apply an algorithm or just compare them?

Mary StarI have to describe the operation of the processure "RADIX SORT" at the following list of words: COW, DOG, SEA, RUG, ROW, MOB, BOX, TAB, BAR, EAR, TAR, DIG, BIG, TEA, NOW, FOX What does it mean to describe the operation of RADIXSORT?? To write how the list will look like at each step?? O...

??
 
4:49 PM
Enjoy it, @AlexWertheim?
 
@AlexWertheim What is Adaptation about
 
I did, @Mike. I'm still digesting the last 20 minutes though.
 
It was Fargo last night for me. The Coen brothers movie, not the TV show.
 
I really, really enjoyed the first hour or so. I'm not sure what to make of the rest.
Oh? Haha, how did you like it? That movie is polarizing...
 
@AlexW: Note that the movie turned into a standard Hollywood action movie the moment Donald started to help writing... and when Donald was no longer helping with the writing...
(I love that last shot over the freeway.)
 
4:51 PM
@Mike: that's what I figured. A kind of meta-satirization. Funny you should say that, since I was thinking "this is what the movie would look like if Donald wrote it..."
Still, it was so incongruous with Charlie's vision of the movie, which I was really enjoying. Disappointment, loneliness, self-loathing.
 
@DanielFischer Do you have an idea for: math.stackexchange.com/questions/1179564/… ??
 
movies are alright
 
Well, two reasons. It's not really just a difference in vision, but also a difference in life philosophy; and with the exposure Charlie came out in the end to be a happier person and a better writer. And two: it would really be hard to write a movie where nothing truly happens, haha
The only good piece of comprehensible media I've aeen where nothing happens is Waiting for Godot.
Also yeah, I thought Fargo was hilarious. I was cracking up at the woodchipper scene, and I learned that people from Minnesota are inherently funny.
 
Right. I did appreciate the shift, since Charlie's anemic mannerisms are almost sickening after a point. The scene with the waitress...
Frances McDormand is tremendous in Fargo.
 
I have a pretty high tolerance for that sort of thing, but the waitress scene made me cringe.
 
4:59 PM
Right? I also found "The 3" hilarious, both for the script and how everyone loves it...
 
Hahaha, yeah.
Frances McDormand played Marge, right? She was great.
 
Yeah. The absurd but just relentless pursuit.
 
Whoever played Jerry did a great job. What an absolutely pathetic character.
 
The whole sequence to the end is great. Like you said, the woodchipper scene is hilarious. And then when he's in the car with Marge, and she's saying she just doesn't understand... man.
William H. Macy. Agreed, just perfectly done.
Supposedly he insisted on it to the Coen brothers. I'm glad he did, he nailed it.
 
Hi there! :-)
I need hint about a number theory problem.
 
5:06 PM
So what were you guys up to before movie chat?
 
@Balarka and I were just chatting. He had asked what I was studying at the moment, so we were talking a bit about that.
 
@DanielFischer If we have two algorithms A, A' and they have the smae speed, would you say that A' is asymptotically faster than A?
 
Ah, fun stuff.
 
I'm trying to take your advice, and be as prepared for the quals as possible, whatever the outcome may be. :)
 
Hi, can someone lend me a hand on a geometry problem?
 
5:11 PM
Great, @AlexW :)
 
@MikeMiller OK, let's get this straight : I don't understand why Hatcher talks about linear simplices in the proof of excision at all. All what we need is a chain htpy equivalence between $C_n(X)$ and $C_n^\mathfrak{U}(X)$. The chain map $C_n^\mathfrak{U}(X) \to C_n(X)$ is clear. The other map $C_n(X) \to C_n^\mathfrak{U}(X)$ can be obtained by taking a singular simplex, subdividing like mad until I get every piece to be inside the cover by Lebesgue.
Don't see how linear chains interact.
[I'm actually a bit embarrased to ask this question : this is really basic stuff and I should have made this clear to myself long ago.]
 
@DanielFischer Your solution is exactly that I read in the book where the question appeared. I wish my mind were as sharp as yours ;)
 
Wait a tick. $X$ might or might not be a metric space. How can we apply Lebesgue if it's not?
 
@BalarkaSen: He has to do the calculations he did there at some point, whether it's in that section or later. The point of doing it in a convex open subset of Euclidean space is just notational convenience, I think.
you're not applying the lebesgue lemma to X...
 
Oh sorry. hehe. we are subdividing the domain simplex.
@MikeMiller So linear chains don't interact in the proof at all...?
that's confusing.
 
5:20 PM
@AlexWertheim Hi!!! If we have two algorithms A, A' and they have the same speed, would you say that A' is asymptotically faster than A?
 
@evinda common sense commands a clear "no"
 
Doesn't seem that way to me from glancing over it. It seems just like a pedagogical tool: let's do it in a simpler setting first.
 
@LeGrandDODOM http://math.stackexchange.com/questions/1045173/which-is-the-greatest-integer-value-of-a-for-which-a-is-asymptotically-fas/1104806?noredirect=1#comment2402348_1104806
So would you say that the greatest value of a is 48?
 
bah. that just confuses readers.
 
@evinda: I'm afraid I don't know the answer, I'm sorry!
 
5:22 PM
A ok.. no problem :) @AlexWertheim
 
Yeah, I don't really understand it. I thought some of the calculations in that section mattered, but they don't seem to.
 
Weird.
 
Your statement about how the chain homotopy is defined above is how I think of the proof.
 
@DanielFischer What about this problem? Given a regular polyhedron such that two consecutive vertices have integer coordinates and a third vertice has integer coordinates, prove the polygon is a square
 
I'm going to read this thoroughly to see if there's really something I am missing and mail Hatcher if this rubbish is irrelevant, I think.
 
5:28 PM
Oh, yes it does matter. $\sigma_\sharp$ is the pushforward $C_k(\Delta^n) \to C_k(X)$, sending $f$ to $\sigma \circ f$. The formulas he's using to show that things are chain maps or chain homotopies should depend on the calculations above - because $\Delta^n$ is homeomorphic to a convex subset of Euclidean space.
I'm done reading it, though.
 
Hmm, that should be it.
Let me read it first.
 
I think he's only even using the linear chain version on the identity map $\Delta^n \to \Delta^n$, so it remains that the generality was unnecessary; but it gives an idea of what's going on in the proof if he does it like he did there. I no longer object to it.
 
5:43 PM
I like triangles
 
@ʙᴀᴅᴀᴛᴍᴀᴛʜ o/
 
Hi 
I feel kinda bad for making my username this one
 
Kinda negative
 
Meh. I get it, @Mike.
 
5:56 PM
@ʙᴀᴅᴀᴛᴍᴀᴛʜ Then just change it to @-ʙᴀᴅᴀᴛᴍᴀᴛʜ
 
lol, @ɧɿρρԹʅȝՇԵՐՎԾՌ
 
@ɧɿρρԹʅȝՇԵՐՎԾՌ I have to wait a month
 
@ʙᴀᴅᴀᴛᴍᴀᴛʜ You too ? :D
 
@ʙᴀᴅᴀᴛᴍᴀᴛʜ Open an SE account on a new site. You can change your name there. When doing so, apply the change network-wide.
After that, I think you can delete your account on the new site, so that nothing changed but your name.
 
Loopholes, brought to you by @MikeMiller
Let's see if this works
 
5:58 PM
Nah, thank the artist currently known as Woodface
 
Can someone help me at:
-1
Q: Dividing the interval in subintervals of equal length

Mary StarI am asked to describe the operation of the processure BUCKET SORT at the array $$A=\langle 0.75, 0.13, 0.16, 0.64, 0.39, 0.20, 0.89, 0.53, 0.71, 0.42, 0.19 \rangle $$ dividing the interval $[0, 1)$ in $10$ subintervals of equal length. In my book I found the following algorithm: BUCKETSORT...

 
Great guy, him.
 
??
 
I wonder if someone has made a list of his usernames.
 
Hello.
@ʙᴀᴅᴀᴛᴍᴀᴛʜ What are you doing now?
 
6:07 PM
I wonder how I got so much nitpicky all in a little while.
 
@JasperLoy Trying to learn topology
 
Go prove that a metric space is Hausdorff, @ʙᴀᴅᴀᴛᴍᴀᴛʜ
 
@Bal I already know how by contradiction
 
@BalarkaSen Don't give him random exercises.
 
By contradiction?
@ABeautifulMind He is studying metric spaces.
 
6:09 PM
@Bal Yea using the triangle inequality if you assume there is a common element in the neighborhoods of two distinct points you get a contradiction
 
seems the contradiction there is sort of artificial
 
False statement (problem with quantifications : there is nothing called "the neighborhood")
Take two points x, y in your metric space.
I take the nbhd to be complements of these two points in X :P
 
he said the neighborhoods :D
by "in the neighborhoods" he means, i assume, "in every pair of neighborhoods"
 
I just didn't feel like writing it out because I have to learn easier stuff first before Hausdorff spaces and such
 
Hausdorff spaces is not hard.
 
6:12 PM
I think you should not tell people that the things they're working on are not hard.
 
@ABeautifulMind Get your mind out of the gutter
 
@MikeMiller Oh, I didn't mean that. I meant that definition of Hausdorff spaces isn't hard, i.e., it doesn't take a lot of background to think about them.
 
Ok. Using more words is good, as it avoids misunderstanding.
 
Sorry @ʙᴀᴅᴀᴛᴍᴀᴛʜ, just in case you thought I meant something else.
@Mike Well, English is not my native language, and I am neither quite adept at it. I'll try improve it, though.
 
Sure. Advice, not criticism.
 
6:18 PM
Sorry again if it sounded like something else.
 
I'm sort of used to it anyway - most people say some things are obvious and it takes me days/weeks to figure out why
 
I didn't mean that, @ʙᴀᴅᴀᴛᴍᴀᴛʜ. I always get stuck at most obvious stuff. It's nothing to worry about.
 
Everyone gets stuck on something every now and then, it's part of being human
 
I feel like I get stuck on everything all the time, not just something every now and then
 
Best way to learn mathematics is to get stuck.
 
6:23 PM
^
Finding the solution is all the more satisfying after being stuck on something for a while.
 
Except when you never find the solution :(
 
If you do a lot of exercise, you'll see that the amount of getting stuck gets reduced.
 
Mathematicians don't get anywhere with intelligence alone, it's a field that requires lots and lots of practice
 
Experience is also a (big) part of mathematics, not just thinking through something. You don't get striked by brilliant ideas often.
And in the cases where you don't, use your experience =)
 
But when you are struck by brilliant ideas, be sure to show your work. Otherwise, you'll be looking back and scratching your head about the result's validity (that's what's happened to me with this work on polylogs)
 
6:32 PM
That happens to me so often.
 
@evinda That depends on whether "faster" translates to $<$ or to $\leqslant$ in the context.
 
Handwaving can get you only so far, @Balarka :P
 
Yea, I was reading my homework from last semester and it seemed to make sense then, but when I read it now I don't really understand
 
I usually sometimes handwave without writing down the proof rigorously, which sometimes gives devastating results.
 
I guess only Gauss can do it
 
6:36 PM
1
Q: Let i do. Let i do. Let i do.

mike2007Let i do. Let i do. Let i do. Let i do. Let i do. Let i do. Let i do. Let i do.

 
That edit....
 
What is this lol
 
Someone rollbacked that edit.
 
Question hijack or is that actually related to this "Let i do" business
 
I guess he edited the whole question after getting his answer on the comment.
 
6:38 PM
lol...
 
HI @DanielFischer
 
@LeGrandDODOM No immediate idea for that one.
@JohnJack Hi.
 
@ᴇʏᴇs I got suspended.
 
@JasperLoy Oh..
 
Sad huh, people flagged and people agreed.
 
6:46 PM
@DanielFischer Could I ask something. If you are given a function defined by $g(x,y): = \{ 4 \text{ if }x= 0\text{ or } y = 0 \text{ and } x^{2} + y^{2}~~\text{ otherwise} \}$ then does it follow that $g_{x}(0,0) = 0$ by simply noting that $g_{x}(x,y) = \{ 0 \text{ if }x = 0 \text{ or }y = 0 ~~\text{ and }2x ~~\text{ otherwise} \}$. I don't really see how we know that $g_{x}$ exists?
 
@DanielFischer A ok, we haven't said anything about it in class.
We have the recurrence relation $T(n)=7T\left( \frac{n}{2}\right)+n^2$ and the solution is $T(n)=\Theta(n^{\log_{4}{49}})$.
We also have the recurrence relation $T'(n)=\alpha T'\left( \frac{n}{4} \right)+n^2$.

$a=\alpha, b=4, f(n)=n^2$

$n^{\log_b a}=n^{\log_4 \alpha}$
For $\alpha>16$, $f(n)=O(n^{\log_b{\alpha}}- \epsilon), \epsilon>0$

So for $\alpha>16$: $T'(n)=\Theta(n^{\log_4{\alpha}})$

So that $A'$ is asymptotically faster than $A$, it has to hold $n^{\log_4{\alpha}}< n^{\log_4{49}} \Rightarrow \log_4{\alpha}<\log_4{49
@DanielFischer Is it right like that?
 
@JohnJack $g_x(0,y)$ exists only for $y = 0$ and $y^2 = 4$. You need to take the definition of the partial derivatives. Fix the $(x_0,y_0)$ you're interested in, and look at $$\frac{g(x,y_0) - g(x_0,y_0)}{x-x_0}.$$ If the limit exists, that's your partial derivative, otherwise $g_x$ doesn't exist at $(x_0,y_0)$.
 
@JohnJack I'm confused-how is it that $g_x(0,y) = 0$ for say, $(0,3)$?
 
@DavidWheeler Hi, I just got suspended from chat, lol.
 
@ABeautifulMind for which reason?
 
6:53 PM
@DavidWheeler For mentioning boobs, lol.
Anyway, it was just 30 min.
 
lol
 
Seems like a pointless display of power-like police stopping you for acting suspicious, grilling you for half an hour, and then releasing you.
 
@evinda If $\alpha$ is restricted to integer values, and "faster" means "strictly faster", it's correct. If $\alpha$ can be any (positive) real value, then there is no largest $\alpha$ such that $A'$ is asymptotically strictly faster. If "faster" is to be interpreted in the weak sense, then $\alpha = 49$ is the largest value.
 
@DanielFischer Do you mean I should look at the limit $$\lim\limits_{x \rightarrow x_{0}}\frac{g(x,y_0) - g(x_0,y_0)}{x-x_0}.$$?
 

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