Large Hardon Collider

[Psion]

All I know is that trig is kicking my ass. Fucking variables that need need to be substituted for other variables, which need to be substituted for more variables. Shit makes my head hurt trying to keep track of all the fucking formulas. And the book is terrible at explaining how it derives formulas from Pythagorean theorem, it just starts to go a = b, so b = c, so c = f, and f = g. That's great and all, but where the fuck are d and e? Fucking thing keeps skipping steps in explanation, so I have no clue how it actually derived a certain thing from another. -,-

Oh, and my trig teacher gets half her problems wrong when she solves them for us on the board. That's always fun. <.<;

[Max™]

Well, d and e are getting stoned off to the side...

[Miragemaiden]

Not sure how screwed up your book is, but Pythag shouldn't be too hard. Umm, maybe try this site and see if it explains it any better? http://www.jimloy.com/geometry/pythag.htm

[Max™]

Tattoo this on the neck of whoever is sitting in front of you:
http://schools-wikipedia.org/images/226/22690.png

Then shave their head, and put this on there:
http://blog.ssis.edu.vn/chrischoi/fi...UnitCircle.gif

[Ferion]

I was going to just PM this problem to Woozie, but I figured I would post it here instead in case anyone is interested or enjoys this stuff. If you guys are busy/don't care, feel free to ignore this. It's a workshop question I thought I understood, but after class tonight my professor confused the hell out of me. It goes like this: Suppose you have two functions f and g such that (H1 and H2 stand for hypothesis 1 and 2) [H1] p is an accumulation point of the intersection of Dom(f) and Dom(g). [H2] There is a positive radius r such that for all x in the intersection of Dom(f) and Dom(g) if 0 < |x - p|< r, then f(x) = g(x).

Now the question is: Show that if a limit for f(x) exists at p, then a limit exists for g(x) at p and the two limits are equal.

What I basically did to answer this was I said: Let (xn) be a sequence converging to p, (xn) is an element of the intersection of Dom(f) and Dom(g), and (xn) can not equal p. A limit L exists for f(x) at p if for all e > 0 (e= epsilon), there is a d > 0 (d = delta) such that if 0 < |xn - p| < d, then |f(xn) - L| < e. Since (xn) converges to p, there is an N such that for n(>=)N |xn - p| < d. So in a more formal way I basically said that this shows that f(x) has a limit L at p. Then I go on to say that since we know f(x) converges it must be true that 0 < |x - p| < d. Let d=r, so 0 < |x - p| < d=r, hence f(x) = g(x). Since they are the same function in this radius, they must both converge to the same L.

The problem is I feel like there's some flaws in my logic or I'm kind of misunderstanding the problem. If anyone has some comments or criticisms, feel free to let me know. Sorry if you can't really understand my notation, I'll reread through it tomorrow and try to be more clear.

Edit: Forgot to add the part that confused me. Today in class my professor said she needed to add another hypothesis or the question wouldn't make any sense. She added [H3], let V(p) be a neighborhood of p with radius r. V(p) is a subset of the intersection of Dom(f) and Dom(g). To be honest, I have no idea what the hell that has anything to do with this problem, which is what threw me off.

[Max™]

Looks good as far as I can see, leave it to Woozie to point out some super fuckup that isn't standing out though.

[Anton]

Psion: This is the basics of all calculus. I SERIOUSLY WISH I Could have seen this instead of heard someone try to explain it.

http://i36.photobucket.com/albums/e4...or/Sec2tan.gif

Basically as the two points approach each other, the secant slope gets closer and closer to the Calc answer(in yellow). When the two points are so close together as makes absolutely no difference, you can completely disregard having two points in the first place.

Thus, the slope of a tangent with only one point. That's calc. It's not as scary as people say.

[Max™]

Also: what Anton just said, seriously, look into the visuals, it's SOOOOOOOOOO much nicer after knowing for sure you're picturing the right shit.

[Kaylia]

calculus = (y2-y1)/(x2-x1)

true story.





[edit]
@Ferion
Your proof seem right, but I would need to pull out my analysis textbook to make sure (been a while since in had my class, and i dont remember the definition well enough) I dont understand what your teacher hypothesis add to the problem

[Kaylia]

Spoiler: show

Halp (again)


http://img607.imageshack.us/img607/2203/integral.png

Anyone can give me a hint how to solve this (entropy the normal distribution). I know the answer already since it's pretty common, but none of my last 2 attempt were remotely close to it.

Nvm, I think I found how. I fucking hate solving integral...

[edit]
or not...ffs. I could use a hint.

[BerenTebogo]

Have you tried ripping off your shirt and punching skeletor in the face?

[Kaylia]

I should!

[Ferion]

Thanks Max/Kaylia, at least my proof makes sense to other people. I wish I could help with your integration problem Kaylia, but anything I suggest you've probably tried already since I'm way out of practice when it comes to integration. Also, I don't really know anything about entropy.

My professor also decided to add another part since she gave us an extra week to work on the problem. Figured I'd post it in case anyones interested. I'm actually pretty stuck on it. The additional question goes: Give an example as to why [H1] can not be weakened to say, p is an accumulation point of the Dom(f) and p is an accumulation point of the Dom(g) (instead of the intersection). I basically started off by saying suppose f(x) and g(x) are floor functions where dom(f) = [0,1) and dom(g) = [1,2). So f(x) = 0 for every input in its' domain and g(x) = 1 for every input in its' domain. With this example we know that the intersection of dom(f) and dom(g) is the empty set and the common accumulation point they would have is p = 1. She pretty much said this part is correct, but now I'm completely lost from here. I think I'm supposed to show how this causes a problem with [H2]. The only thing I can think of is since the intersection of dom(f) and dom(g) is the empty set, |x - p| < r will always be true, so f(x) = g(x) which is a contradiction. I'm not really sure if this is right though. I'm also not sure if I'm suppose to change the part in [H2] that says "for all x in the intersection of the dom(f) and dom(g)" to "all x in the dom(f) and the dom(g)" since it was changed for accumulation points in the beginning of the question. I'm starting to see why a lot of students really hate this class haha.

[Max™]

I dunno, those are interesting questions.

Wonder what Woozinator will think of em.

[Woozie]

I'll get to your question when I get home, Ferion (around 4:30ish est). My professor decided to throw a bunch of quantum chemistry stuff into our quantum physics exam as if any of us gives a crap about chemistry. So now I have to learn some chemistry crap before 3:20 or I wont be able to get an A on this exam.

I'm supposed to be in an actual chemistry class right now as we speak. If I wont even do chemistry in my chemistry class, what makes him think I'll do it in my quantum mechanics class? But then again, I rarely show up for quantum mechanics class either. Or any other class this semester for that matter. But Chemistry is the worse. I only show up on exam days or when I need a quit spot to grade papers (chemistry class is more quiet than my office or the library).

True story: I thought I died in chemistry class a few weeks ago. I went there for a nap (like I said, it's more quiet than my office and the library) and no one woke me up when class ended. Everyone just left, shut the door and turned off the lights. So when I woke up everything was black. And due to me being an idiot, my first thought was "death" instead of "blindness" or "the lights are off".

Edit: Wait, Psion, you're a chem major, right? I'll take your calc class for you if you take my chem class. I never realized I could be so bored in a science class.

[Max™]

lol

'Fuck, I'm dead aren't I? Man, I hope that Jesus guy isn't around here... this'll be awkward.'

[Shiroikage]

I'll get to your question when I get home, Ferion (around 4:30ish est). My professor decided to throw a bunch of quantum chemistry stuff into our quantum physics exam as if any of us gives a crap about chemistry. So now I have to learn some chemistry crap before 3:20 or I wont be able to get an A on this exam.

I'm supposed to be in an actual chemistry class right now as we speak. If I wont even do chemistry in my chemistry class, what makes him think I'll do it in my quantum mechanics class? But then again, I rarely show up for quantum mechanics class either. Or any other class this semester for that matter. But Chemistry is the worse. I only show up on exam days or when I need a quit spot to grade papers (chemistry class is more quiet than my office or the library).

True story: I thought I died in chemistry class a few weeks ago. I went there for a nap (like I said, it's more quiet than my office and the library) and no one woke me up when class ended. Everyone just left, shut the door and turned off the lights. So when I woke up everything was black. And due to me being an idiot, my first thought was "death" instead of "blindness" or "the lights are off".

Edit: Wait, Psion, you're a chem major, right? I'll take your calc class for you if you take my chem class. I never realized I could be so bored in a science class.

If you're not sleeping like you're dead in classes, you aren't getting all you can out of your classes.

[Ferion]

I'll get to your question when I get home, Ferion (around 4:30ish est).

Thanks a lot, I appreciate it. This problem isn't due for almost another week so no rush though, especially if you have your own work to worry about.

[Psion]

Lol, I'm only in chem 3, the basic chem course so far, but if you have a question on basic chemistry such as valence electrons or the like I can probably answer it. :D

Though it can't hurt to try asking, if nothing else I can learn ahead of time what horrible things I'm getting myself into heh.

[Woozie]

I was going to just PM this problem to Woozie, but I figured I would post it here instead in case anyone is interested or enjoys this stuff. If you guys are busy/don't care, feel free to ignore this. It's a workshop question I thought I understood, but after class tonight my professor confused the hell out of me. It goes like this: Suppose you have two functions f and g such that (H1 and H2 stand for hypothesis 1 and 2) [H1] p is an accumulation point of the intersection of Dom(f) and Dom(g). [H2] There is a positive radius r such that for all x in the intersection of Dom(f) and Dom(g) if 0 < |x - p|< r, then f(x) = g(x).

Now the question is: Show that if a limit for f(x) exists at p, then a limit exists for g(x) at p and the two limits are equal.

What I basically did to answer this was I said: Let (xn) be a sequence converging to p, (xn) is an element of the intersection of Dom(f) and Dom(g), and (xn) can not equal p. A limit L exists for f(x) at p if for all e > 0 (e= epsilon), there is a d > 0 (d = delta) such that if 0 < |xn - p| < d, then |f(xn) - L| < e. Since (xn) converges to p, there is an N such that for n(>=)N |xn - p| < d. So in a more formal way I basically said that this shows that f(x) has a limit L at p. Then I go on to say that since we know f(x) converges it must be true that 0 < |x - p| < d. Let d=r, so 0 < |x - p| < d=r, hence f(x) = g(x). Since they are the same function in this radius, they must both converge to the same L.

The problem is I feel like there's some flaws in my logic or I'm kind of misunderstanding the problem. If anyone has some comments or criticisms, feel free to let me know. Sorry if you can't really understand my notation, I'll reread through it tomorrow and try to be more clear.

Edit: Forgot to add the part that confused me. Today in class my professor said she needed to add another hypothesis or the question wouldn't make any sense. She added [H3], let V(p) be a neighborhood of p with radius r. V(p) is a subset of the intersection of Dom(f) and Dom(g). To be honest, I have no idea what the hell that has anything to do with this problem, which is what threw me off.

Let's ignore [H3] for a moment. In the problem, it is stated that if |x-p|< r, then f(x) = g(x). But how do we know that both f(x) and g(x) are defined? After all, if x is in dom(f), that doesn't mean x is in dom(g) as well (and likewise, x being in dom(g) doesn't guarantee that it's also in dom(f) ). That's why we need [H3]. It guarantees that if |x-p|< r, then x is in both dom(f) and dom(g).

I'm going to warn you, grading papers has made me really nitpicky when it comes to reading proofs. So here it goes:

Okay, one more warning. I didn't proof read my post. I'll fix any mistakes tomorrow. If you see something that doesn't seem right, then I'm probably wrong.

A limit L exists for f(x) at p if for all e > 0 (e= epsilon), there is a d > 0 (d = delta) such that if 0 < |xn - p| < d, then |f(xn) - L| < e. Since (xn) converges to p, there is an N such that for n(>=)N |xn - p| < d.

So in a more formal way I basically said that this shows that f(x) has a limit L at p.

"So in a more formal way I basically said that this shows that f(x) has a limit L at p"? You're allowed to assume this from the start. It's not something that needs to be proven. It's so confusing when students do that. Sometimes I end up marking off points just because of a lack of clarity. Of course, I know you're just typing something up online and you didn't bother being as precise as possible. Plus I'm not paying full attention so I could just be missing something here. If so, disregard this paragraph.

Then I go on to say that since we know f(x) converges it must be true that 0 < |x - p| < d.

This is also a little confusing. I read this as "f(x) converges implies 0 < |x-p| < d", which makes no sense. I know that's probably not what you meant, but that's how it reads.

Let d=r, so 0 < |x - p| < d=r, hence f(x) = g(x). Since they are the same function in this radius, they must both converge to the same L.

Compare this quote to this one:

A limit L exists for f(x) at p if for all e > 0 (e= epsilon), there is a d > 0 (d = delta) such that if 0 < |xn - p| < d, then |f(xn) - L| < e. Since (xn) converges to p, there is an N such that for n(>=)N |xn - p| < d.

Notice that this last quote gives a bunch of information but then your reasoning and conclusion doesn't use this information. So why is this information there? You never used the fact that "there is an N such that n>=N implies |xn-p|. In fact, the proof doesn't use sequences at all. Having a lot of unused information is so confusing to the grader.

In fact, the second to last quote (i.e. your reasoning and conclusion) stands completely on its own. It's not a complete proof, but the info before it doesn't make it any more or less complete. I feel like it doesn't contribute at all. I'm going to slightly rephrase your proof so that you can see what I mean and why.

"Let lim f(x) = L.

Let |x-p|<r. Then f(x)=g(x). Since they are the same function in this radius, they must both converge to the same L."

I feel like this is basically all you're saying. I could be misinterpreting your proof, but I feel like my rephrased proof contains all the same information that yours does and that the stuff I omitted was extraneous in you proof. This trimmed-down version of the proof should help you see the issues that I'm having with your proof. Of course, that assumes that my trimmed proof and your proof really are the same. I could be misinterpreting, but that could mean that your professor or whoever is grading your paper is also misinterpreting.

Your conclusion and reasoning is correct, but you haven't really "shown" it mathematically. To show that the limit exists, you need to either take an arbitrary sequence in dom(g) and show that it goes to L (so you should have something like "n>N implies |g(xn)-L|<ε") or you somehow produce or show the existance of a δ that satisfies the definition of a limit (so your proof should end with or contain the statement "|x-p|<δ implies |g(x)-L|<ε")

To reach a final statement of this form, you'll probably have to use the information that I quoted and said it wasn't really used.

You're definitely on the right track, and your reasoning in your conclusion was definitely the correct reasoning. You just need to be a bit more rigorous and precise.

Of course, for all I know your proof really is more precise. Maybe what you have on paper is more rigorous than what you typed out because you didn't feel like typing up a whole proof. In that case, if you went along the lines of the reasoning of what's in your post, then your proof is probably fine as long as it contains one of the two statements I said it should have.

I typed up an answer, then realized that since this is homework I probably shouldn't just outright give it to you. So I'm going to spoiler it and trust that you wont look at it unless you have to.

Spoiler: show

The way I would do it would be more like this.

Let ε >0, and let L = lim f(x) as x -> P

Let xn be a sequence in dom(g) converging to p. Then there exists an N1 such that n > N1 implies |xn-p| < r. For such xn, g(xn) = f(xn). Since lim(f) exists at p and the limit is L, then there exists an N2>N1 such that n>N2 implies |f(xn)-L|<ε

Let N =N2 Then n>N implies |f(xn)-L| = |g(xn)-L| < ε. Thus, lim g(x) = L

(Note that it's important that N2>N1 (or equal to). Otherwise f(xn) may not be equal to g(xn) for n>N2 (in fact, f(xn) may not even be defined)).