Large Hardon Collider

[Max™]

So, a friend just turned 24 while we were chatting, I said "I would say that I'm 30 years old, but that's far too vague", so I pulled up a google page, calculator, and worked out that I'm a bit over 8,696,965,064,961,600,000 vibrations of a cesium atom old.


Also:
http://www.sciencedaily.com/releases...1102185722.htm

ScienceDaily (Nov. 3, 2010) — The results of a high-profile Fermilab physics experiment appear to confirm strange 20-year-old findings that poke holes in the standard model, suggesting the existence of a new elementary particle: a fourth flavor of neutrino.

[Max™]

Hadn't seen this on here yet.

http://www.nasa.gov/topics/technolog...r-preview.html

http://www.jwst.nasa.gov/videos_science.html

[Woozie]

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.

Actually, since we have H3, we can toss out H1 all together. That's the answer I'd tell the professor.

Edit: I'll add an explanation in a moment (or later today).

Okay, here's the reason. H1 follows from H3. Hence, if we have H3, we don't need to also have H1. So your teacher's question doesn't make sense (or I'm missing something here. If I am missing something I hope someone points it out).

Anyways, suppose H3 is true. So v(p) is a neighborhood around p and v(p) is in the intersection of dom(f) and dom(g).

Let s1 be a point in v(p). Let s2 be a point in v(p) such that 0<|s2-p| < min {1/2,|s1-p|}. I know such a point exist because if there were no points in v(p) closer to p than min {1/2,|s1-p|}, then v(p) is not a neighborhood of p. [if I wanted to be rigorous I'd use induction at this point. But I'm not going to bother]

Likewise, I can find an s3 such that 0<|s3-p|< min{1/3,|s2-p|}. In general, I can make a sequence such that 0<|sk-p|<min{1/3,|s(k-1)-p|}. Since each 0 < |sk-p| < 1/k, then sk converges to p. And since sk is an element of v(p) for all k, and v(p) is a subset of the intersection of dom(f) and dom(g), then it follows that p is an accumulation point of the intersection of dom(f) and dom(g). Hence, [H1].

[Max™]

Like a boss...

[Ferion]

Thanks for the feedback Woozie I really appreciate it. In regards to your first post, don't worry about being harsh when it comes my work. Being blunt about what I'm doing wrong is the only way that I'll learn it better. To be honest my post was typed from bed out of memory before I was about to go to sleep, so it was worded very awkwardly and poorly. But you made some very good points. The proof you posted was pretty much what I was attempting to do, except I made the incredibly dumb mistake of proving the "if" part of an "if,then" statement.

Then there exists an N1 such that n > N1 implies |xn-p| < r.

The only question I have about you proof is this part. I'm assuming you're setting epsilon equal to r, or am I completely confusing what you're doing here?

Edit: There's possibly a quicker way, but I'm not too sure about the definition of neighborhood. As far as I know, all it means is that v(p) is some set in which p is interior. If that is the case, then p is in the intersection of both sets and thus is automatically an accumulation point (the sequence p,p,p,p,p,.... converges to p). So in that case, your proof would be a one line or one sentence proof. But if neighborhood is defined in such a way that p doesn't have to be in v(p) (e.g. v(p) is some sort of punctured disk or punctured ball, whichever word you class uses), then this one line proof wont work and you'll have to use the one above.

I see what you're doing with your proof and I really like your reasoning, although I only quickly read through it. I think you may have slightly ignored the question though, unless I'm really missing something. I think my professor may be looking for some kind of contradiction from when you weaken [H1] to "and" instead of "intersection". However, I'm sure that once I have some more time to got through your proof that I will find some kind of serious contradiction if we assume p is an accumulation point of the dom(f) and the dom(g). I could be completely wrong about this though as I need to spend a lot of time on proofs to really understand them and I only quickly read through yours.

[Max™]

It struck me that there might be issues if it was a Hausdorff topology, where the neighborhood might not be the same, but I doubted it would be such without it being implicitly stated.

[Woozie]

Thanks for the feedback Woozie I really appreciate it. In regards to your first post, don't worry about being harsh when it comes my work. Being blunt about what I'm doing wrong is the only way that I'll learn it better. To be honest my post was typed from bed out of memory before I was about to go to sleep, so it was worded very awkwardly and poorly. But you made some very good points. The proof you posted was pretty much what I was attempting to do, except I made the incredibly dumb mistake of proving the "if" part of an "if,then" statement.



The only question I have about you proof is this part. I'm assuming you're setting epsilon equal to r, or am I completely confusing what you're doing here?



I see what you're doing with your proof and I really like your reasoning, although I only quickly read through it. I think you may have slightly ignored the question though, unless I'm really missing something. I think my professor may be looking for some kind of contradiction from when you weaken [H1] to "and" instead of "intersection". However, I'm sure that once I have some more time to got through your proof that I will find some kind of serious contradiction if we assume p is an accumulation point of the dom(f) and the dom(g). I could be completely wrong about this though as I need to spend a lot of time on proofs to really understand them and I only quickly read through yours.

r is playing the role that an epsilon would normally play. I could have said

"for any ε1, there exist an N1 such that n>N1 implies |xn-p| < ε1 (it's ε1 instead of just ε because I've already used ε for something else). Let ε1=r. Then |xn-p|< r"

but it's shorter to just bypass that part all together. Afterall, limits work with any positive number, not just epsilon. So since r is already known to be positive, I can go straight to making |xn-p|< r instead of introducing an epsilon and then setting it equal to r.

The problem with the second part is that even if [H1] had been modified to say something different, the original form of [H1] would also still be true as long as [H3] is true. Hence, nothing will have really changed.

It would be like if I said

[H1] r > 5
[H3] r=13.

Now if I were to change [H1] to anything else, the original form of [H1] is still true because it follows from [H3]. If I weaken [H3] to say "r >= 5", then the original stronger statement "r>5" is still true because of [H3]. If I eliminate [H1] altogether, the original [H1] "r>5" is still true. As long as [H3] is there, it doesn't matter what I do to [H1], the original [H1] will still hold true and hence nothing will change. There's absolutely nothing you could do with the new set of rules that you couldn't do with the old set and vice versa.

It's kind of like when you had systems of linear equations in linear algebra. If one was linearly dependent on the others, it didn't matter if it was there or not. It didn't change anything about the system at all. It didn't add anything new or take anything away. H1 is playing the role of that one equation that's linearly dependent on the others. If you take it away, nothing changes.

If you change [H1] to say "The sky is blue", it's still going to be true that "p is an accumulation point of the intersection of Dom(f) and Dom(g)" even if there's no [H#] saying so (because H3 makes this statement true).

[Kaylia]

Anyone can answer this statistics questions?

Why the entropy of a normal distribution (http://upload.wikimedia.org/math/8/8...77b3816494.png) is negative when the standard deviation is small? Unless I'm completely wrong, shouldn't the entropy be equals to 0 for a standard deviation of 0.

[Eliseos]

If e means the charge of an electron, and sigma squared isn't units of coulombs, I'm pretty sure that isn't right (which is how I'm assuming you are getting a negative answer). ln has to be unitless for it to mean anything.

[Kaylia]

nah, e is euler number lol

[edit]
entropy is listed on the right side, and is equal to what i obtain after doing the integral of S= integral(f*lnf) where f is the normal distribution function
http://en.wikipedia.org/wiki/Normal_distribution

[Eliseos]

Oh, well how are you getting a negative value for that then?

[Kaylia]

Entropy graph for a varying sigma
http://img641.imageshack.us/img641/6084/entropy.png


Ln of number that are located between 0 and 1 gives negative values. Because sigma (standard deviation) can be smaller than sqrt[e/2*Pi], you can get negative entropy.


I'm having hard time understanding this...entropy cannot be negative, it shouldnt even be allowed to go under 1.

[Woozie]

Because a logarithm is negative for small arguments (arguments less than one, that is). So if 2pi*e*sigma^2 is less than one, the logarithm is negative.

[Eliseos]

Because a logarithm is negative for small arguments (arguments less than one, that is). So if 2pi*e*sigma^2 is less than one, the logarithm is negative.

Oh, this then. I thought you meant you were getting negative values on the inside of the natural log. And if sigma is 0 you'd get -infinity for that.

[Kaylia]

Well, I know this much. The issue is that entropy cannot be negative in theory since the definition is ln(# of possible state). Is there any reason why sigma wouldn't be able to go under sqrt(e/2*Pi) in this case?

[Eliseos]

Aren't you calculating the change in entropy? Meaning you can have a negative change in entropy. I was always under the impression that you cannot calculate entropy directly, just that it can't be zero. Same reason why you can't calculate a speed of something directly, only relative to something else.

[Woozie]

I have no clue. I suck at thermal physics. I don't even know what that's the standard deviation of.

Edit: My bad, I didn't read your post thoroughly. I thought you were asking why that expression became negative, not why it's physically possible.

[Kaylia]

Aren't you calculating the change in entropy? Meaning you can have a negative change in entropy. I was always under the impression that you cannot calculate entropy directly, just that it can't be zero. Same reason why you can't calculate a speed of something directly, only relative to something else.

From my understanding, entropy is a mathematical concept that come from statistics and information theory. If you add Boltzman's constant to the mix, you get thermo/phys stats definition, but in this case, it's purely mathematics.

Anyway, I guess you're right about it being the change of entropy, and not the entropy....I just don't understand why we call it entropy if it's a change. Hell, why am I not dealing with differential equation?


The definition I'm using come from
http://upload.wikimedia.org/math/a/a...898cdc48f7.png
where omega is the number of microstates, (shouldnt be able to go under ln(1) in this scenario).

I have no clue. I suck at thermal physics. I don't even know what that's the standard deviation of.

Edit: My bad, I didn't read your post thoroughly. I thought you were asking why that expression became negative, not why it's physically possible.

There is no physics here, only statistics!

Standard deviation is just a parameter of the normal distribution. In this case, it would be the standard deviation of "x", a random variable.

[Kaylia]

http://en.wikipedia.org/wiki/Differential_entropy

I guess that's what I was dealing with, and it can be negative...It doesnt really help me understand what happen at sigma = sqrt(e/2pi) thought, but I will figure it out eventually (probably some sort of inflection point in gaussian distribution).

[Max™]

Entropy graph for a varying sigma
http://img641.imageshack.us/img641/6084/entropy.png

http://imgs.xkcd.com/comics/mu.png
"As the CoKF approaches 0, productivity goes negative, as you pull OTHER people into chairspinning contests."