Large Hardon Collider

[Kaylia]

This document is the one I wrote. I simply used the identity to go from (1) to (3) because I know it works. Right now, I'm just trying to demonstrate that we can use this identity for complex number, because I never used geometric progression on complex before, and I don't understand very well why it's working.<


[edit]

If it was real number, it would be easy to do the proof (especially since there is like 5 on wiki)

http://upload.wikimedia.org/math/a/7...c851b39cde.png

[Woozie]

Why does it matter if your magnitude is shorter than one? That's only important if you're summing over an infinite amount of terms. But in the picture you posted it only goes to N_a - 1.

Edit: In your demonstration, did you assume anywhere that the number you were summing was real? From the looks of it, you didn't have to assume that, so your answer is already valid for the complex case. Just rewrite the proof and say "let q be any complex number" instead of real. Maybe I'm missing something. Like I said...sleeping pills lol

Edit 2: And then of course you're using the fact that any complex number is just r*e^(i*t) where r and t are real numbers. So you substitute that expression in for q. But you've already done that, so from the looks of it, you've already given a perfectly legitimate proof for what you're asking for.

[Eliseos]

Ugh, my homework in theory of computation is due Thursday, and he just started a chapter today, which this homework has problems in. I found solutions online to the three homework problems in the new chapter, and my professor is cool with using those as long as we cite sources. I don't really have the time to spend on these three problems anyway, and two days isn't enough time to learn it properly anyways. My brain doesn't think the way this class requires to be successful, I can never understand the proofs until I see the solutions. Apparently most of my class is in the same boat as I am, as the class average is ~68%. End rant.

[Woozie]

http://upload.wikimedia.org/math/a/7...c851b39cde.png

This derivation doesn't assume q to be real, does it? If you were to go through their exact same steps assuming q was complex, where would your reasoning break down?

[Kaylia]

It's not important in this case since it's finite.


[edit] And it's working fine using the recurrence. I was getting lost in trigonometric identity using other methods.

[Woozie]

It's not important in this case since it's finite.

Now I'm really confused. Are you agreeing with me or are you telling me that I'm still missing something? I don't think you understand how these sleeping pills affect me. I'm a total idiot for the next 5 hours or so, so right now so you're going to have to be really clear in what you say lol.

[Kaylia]

Now I'm really confused. Are you agreeing with me or are you telling me that I'm still missing something?

I'm lost in the unordered edit of our post lol.

I agree with you, I thought you were asking if it was finite or not. To that, I answered that for exp(ix), it shouldn't matter since it will always be smaller or equal to 1, but that's unimportant since we are dealing with a finite number anyway.

[Woozie]

Okay, so you agree that your derivation is perfectly fine (because the sum is finite and therefore it doesn't matter of q is real or not), right?

So are you asking me how to take this to infinity? If that's what you're asking, the sum should be divergent (unless the argument in the exponential function happens to be a purely real negative number). If the argument is some imaginary number a*i, then the sum cannot converge because the sequence of terms itself doesn't go to zero (for any convergent series, the corresponding sequence goes to zero). If you want to see where your sum goes as you take your sum to infinity, just look at what happens in equation (3) as N_a goes to infinity. The limit doesn't exist.

If that wasn't what you were asking, then I still have no clue what you were originally asking x_x

Edit: Yeah, I think the ninja editing in our posts is why this got so confusing lol

[Max™]

He's gonna be dreaming of infinite ordinals now, you meanie.


(because of transfinite induction, wherein you can be certain that a sequence will end if it is well ordered and every member has a "lesser" step in it's subset, Cantor was a crazy motherfucker)

[Kaylia]

Okay, so you agree that your derivation is perfectly fine because the sum is finite and therefore it doesn't matter of q is real or not), right?

I simply wanted to demonstrate geometric progression for complex number in general, before using it on a complex series. The only proof I had were for real number.


The derivation I posted should be fine, but I simply used the result of the geometric progression, a result that I could not demonstrate for the complex domain.

[Max™]

Whoa, yeah, I just tried to see a geometric progression for a complex domain... ow.

It kersplode for you too?

[Kaylia]

Whoa, yeah, I just tried to see a geometric progression for a complex domain... ow.

It kersplode for you too?

Convert it to R² and imagine it there instead, it's pretty much the same.

[Woozie]

Edit: I'm somehow using the letters q and r interchangeably in this post for some reason. My bad @_@

The derivation I posted should be fine, but I simply used the result of the geometric progression, a result that I could not demonstrate for the complex domain.

This is the part that confuses me. You DID demonstrate it for the complex domain. Is there anywhere in your derivation where you said "let r be a real number"? Your derivation doesn't need to assume q is real as long as you're not summing to infinity.

http://upload.wikimedia.org/math/5/f...dc6e9a8d26.png

We can find a simpler formula for this sum by multiplying both sides of the above equation by 1 − r, and we'll see that

http://upload.wikimedia.org/math/3/0...abb2dd761f.png

since all the other terms cancel. Rearranging (for r ≠ 1) gives the convenient formula for a geometric series:

http://upload.wikimedia.org/math/b/0...e10ba6d29b.png

I'm assuming that's the derivation you used, right? Nowhere in the derivation do you need to assume r is real. Your reasoning wont change at all if r does happen to be complex. So to show this for the complex case, you do the exact same thing you did for the real case, except instead of saying "let r be some real number", you say "let r be some complex number". The rest is exactly the same. You've already proven what you're asking about. As long as you didn't say "let r be a real number" in your proof, I don't see why you're saying you haven't demonstrated it for the complex domain. It looks to me like you have (you just didn't explicitly state it).

[Max™]

Convert it to R² and imagine it there instead, it's pretty much the same.

Yeah, but I started out looking at it as a slice of R^3, so when I tried to progress from each step to the next one I kept wanting to adjust the axis so it spirals backwards, and inwards, while growing larger outwards, and taller... which hurt my brain.


I think I've got Ordinalitis.

[Kaylia]

This is the part that confuses me. You DID demonstrate it for the complex domain. Is there anywhere in your derivation where you said "let r be a real number"? Your derivation doesn't need to assume q is real as long as you're not summing to infinity.

In my very first post, I simply used the result, it wasn't verified (at the time) for complex number.


Right now, there is no issue, I understand both demonstration (the one I posted after) and the one you just posted.


The reason why I wasnt able to demonstrate it earlier is that I was writting my r as "a+ ib", and when you do r^n, you get stuck with huge term. Why was I doing it like this, instead of using the same demonstration? I don't know, I thought it would works...

[Kaylia]

I'm off to bed too, I'm also running out sleep lately. Thanks for the help, and sorry for holding you up.

[Max™]

The reason why I wasnt able to demonstrate it earlier is that I was writting my r as "a+ ib", and when you do r^n, you get stuck with huge term. Why was I doing it like this, instead of using the same demonstration? I don't know, I thought it would works...

This is why it hurt my brain, setting it as a + ib = bkoosh, holy shit my head asploded.


Incidentally, what do you guys think about the Fine Structure Constant?

[Woozie]

In my very first post, I simply used the result, it wasn't verified (at the time) for complex number.


Right now, there is no issue, I understand both demonstration (the one I posted after) and the one you just posted.


The reason why I wasnt able to demonstrate it earlier is that I was writting my r as "a+ ib", and when you do r^n, you get stuck with huge term. Why was I doing it like this, instead of using the same demonstration? I don't know, I thought it would works...

My bad, I knew I was misunderstanding something x_x

[Woozie]

Okay, I get it now. I thought you were asking "How do I get from equation (1) to equation (3)?". So my response was "Uh...you use equation (2), but you've already done that, so you've answered your own question..."

What you were really asking was "How do I know equation (2) is valid for complex numbers?". I assumed you had derived equation 2 on paper, in which case I was thinking "Look back at your derivation and note that it works for any r (or q) not just for real r (or q).

[Mizango]

Interesting article on the recurring issue of Mars' methane production.

http://www.sciencedaily.com/releases...1208132349.htm

Life on Mars Theory Boosted by New Methane Study

ScienceDaily (Dec. 8, 2009) — Scientists have ruled out the possibility that methane is delivered to Mars by meteorites, raising fresh hopes that the gas might be generated by life on the red planet, in research published in Earth and Planetary Science Letters.

Methane has a short lifetime of just a few hundred years on Mars because it is constantly being depleted by a chemical reaction in the planet's atmosphere, caused by sunlight. Scientists analysing data from telescopic observations and unmanned space missions have discovered that methane on Mars is being constantly replenished by an unknown source and they are keen to uncover how the levels of methane are being topped up.

Researchers had thought that meteorites might be responsible for Martian methane levels because when the rocks enter the planet's atmosphere they are subjected to intense heat, causing a chemical reaction that releases methane and other gases into the atmosphere.

However, the new study, by researchers from Imperial College London, shows that the volumes of methane that could be released by the meteorites entering Mars's atmosphere are too low to maintain the current atmospheric levels of methane. Previous studies have also ruled out the possibility that the methane is delivered through volcanic activity.

This leaves only two plausible theories to explain the gas's presence, according to the researchers behind the latest findings. Either there are microorganisms living in the Martian soil that are producing methane gas as a by-product of their metabolic processes, or methane is being produced as a by-product of reactions between volcanic rock and water.

Co-author of the study, Dr Richard Court, Department of Earth Science and Engineering at Imperial College London, says: "Our experiments are helping to solve the mystery of methane on Mars. Meteorites vaporising in the atmosphere are a proposed methane source but when we recreate their fiery entry in the laboratory we get only small amounts of the gas. For Mars, meteorites fail the methane test."

The team say their study will help NASA and ESA scientists who are planning a joint mission to the red planet in 2018 to search for the source of methane. The researchers say now that they have discovered that meteorites are not a source of Methane on Mars, ESA and NASA scientists can focus their attention on the two last remaining options.

Co-author, Professor Mark Sephton, Department of Earth Science and Engineering at Imperial College London, adds: "This work is a big step forward. As Sherlock Holmes said, eliminate all other factors and the one that remains must be the truth. The list of possible sources of methane gas is getting smaller and excitingly, extraterrestrial life still remains an option. Ultimately the final test may have to be on Mars."

The team used a technique called Quantitive Pyrolysis-Fourier Transform Infrared Spectroscopy to reproduce the same searing conditions experienced by meteorites as they enter the Martian atmosphere. The team heated the meteorite fragments to 1000 degrees Celsius and measured the gases that were released using an infrared beam.

When quantities of gas released by the laboratory experiments were combined with published calculations of meteorite in-fall rates on Mars, the scientists calculated that only 10 kilograms of meteorite methane was produced each year, far below the 100 to 300 tonnes required to replenish methane levels in the Martian atmosphere.