+ Reply to Thread
Page 56 of 284 FirstFirst ... 6 46 54 55 56 57 58 66 106 ... LastLast
Results 1101 to 1120 of 5661

Thread: Large Hardon Collider     submit to reddit submit to twitter

  1. #1101
    Banned.

    Join Date
    Jul 2005
    Posts
    17,472
    BG Level
    9
    FFXI Server
    Ifrit
    WoW Realm
    Area 52

    Anyone here is good with labview? It was my first time using the program the other day, and I kept getting some retarded error. Basically, I was gathering data through a card and sending it to a graph.

    DAQ -> convert to double -> graphic

    However, instead of updating the graphic in real time, it was waiting for the experiment to end to send the data and show the result. Problem is, it can only save up to 10sec of data before crashing (overflow after 10sec), and I needed a good 60 seconds of data.


    I'm about to program the whole shit in C, debugging labview is fucking retarded.

  2. #1102
    Title: "HUBBLE GOTCHU!" (without the quotes, of course [and without "(without the quotes, of course)", of course], etc)
    Join Date
    Jul 2006
    Posts
    3,141
    BG Level
    7

    I've never used labview. They had a labview demo at the APS meeting and gave me some sort of coupon or something for a discount or free trial or something like that, but I never gave it a try.

  3. #1103
    Banned.

    Join Date
    Oct 2005
    Posts
    3,442
    BG Level
    7

    woozie what is strange matter and why does it have such a bland name

  4. #1104

    Quote Originally Posted by Woozie View Post
    Also, if your library has a mathematical methods book or advanced engineering math book, that would probably help you a lot in terms of understanding fourier transforms.
    My Circuit Analysis book has a chapter on Fourier Analysis (I also looked at the syllabus for this semester, and that is material covered in my Circuits class this semester), so I'll give that a look through for tomorrow. I think I understand it better now that I know the complex transformations, which makes the integral much easier to solve.

  5. #1105
    Title: "HUBBLE GOTCHU!" (without the quotes, of course [and without "(without the quotes, of course)", of course], etc)
    Join Date
    Jul 2006
    Posts
    3,141
    BG Level
    7

    Quote Originally Posted by Zalius View Post
    woozie what is strange matter and why does it have such a bland name
    There are 6 types of quarks: Up, Down, Top, Bottom, Strange, and Charm. Protons and Neutrons are made of Up and Down quarks. Electrons belong to a separate family, called Leptons. So basically, all the matter we see around us are made of Up, Down, and Electrons.

    Particles will always decay to its most stable configuration. In particle physics, "more stable" usually means "less mass". Of course, a particle can't just decay into any other lighter particle. A particle can only decay in a way so that no conservation laws are violated (charge/mass/energy/spin/isospin/color/etc). So if there exist a lighter particle that a given particle can decay into (without violating conservation laws), then the particle usually will do so.

    The reason we only see Up and Down quarks is because these are the two lightest quarks. Any heavier quark produced will keep decaying until it becomes Up or Down.

    Strange matter is what we call matter that is made of Strange quarks instead of Up and Down quarks (actually it would be a combination of Up, Down, and Strange). There was a this idea floating around a while ago that under certain conditions, Strange matter would be more stable than regular matter (i.e. Up and Down only matter). We call these stable configurations of strange matter "strangelets" Now let's suppose this were the case, that strange matter was more stable than normal matter. Then a strangelet interacting with normal matter could actually turn the normal matter into strange matter as well. Then this new strange matter would convert the matter it touches into strange matter, which would convert the matter it touches into strange matter. So one strangelet on the earth could start a chain reaction that would eventually turn the entire earth into strange matter. It's one of those Ice-nine or Grey Goo type of scenarios that you may have heard about in fiction.

    Of course, if this were true it would have happened by now in Neutron stars. There's about a billion other reasons this can't happen as well.

  6. #1106
    Renegade Philosopher
    Join Date
    Jul 2004
    Posts
    4,551
    BG Level
    7

    Quote Originally Posted by Eliseos View Post
    So i just had a "duh" moment with myself. Modern physics exam on Monday (I lied before about my exam schedule and I can't read a calendar, so I have this weekend to study for it), going over the first homework assignment which involved complex numbers and adding waves. I completely missed the identities my teacher gave us for complex numbers and sinusoidal functions, and as a result had no idea how to convert from either to the other lol. Now how you went through those transformations makes even more sense. Just thought I'd share my "Hey I got it but I'm an idiot" moment, lol

    Now I get to move on to studying Fourier Transforms, Interference, and Michelson Interferometers.
    De Moivre's formula is one of the most useful things I ever learned from complex variables. Being able to simplify cos(nx) into a polynomial in cos(x) without having to memorize trig identities is insanely useful at times.

  7. #1107
    Banned.

    Join Date
    Jul 2005
    Posts
    17,472
    BG Level
    9
    FFXI Server
    Ifrit
    WoW Realm
    Area 52

    We call these stable configurations of strange matter "strangelets" Now let's suppose this were the case, that strange matter was more stable than normal matter. Then this new strange matter would convert the matter it touches into strange matter, which would convert the matter it touches into strange matter
    If everything is turned into strange matter, shouldnt we call it normal matter.

  8. #1108

    Quote Originally Posted by Kaylia View Post
    If everything is turned into strange matter, shouldnt we call it normal matter.


    Oh, so you know what normal is?

  9. #1109
    Banned.

    Join Date
    Jul 2005
    Posts
    17,472
    BG Level
    9
    FFXI Server
    Ifrit
    WoW Realm
    Area 52

    Quote Originally Posted by Kryssan View Post
    Oh, so you know what normal is?
    The standard?

  10. #1110
    Salvage Bans
    Join Date
    Jul 2009
    Posts
    902
    BG Level
    5
    FFXI Server
    Ifrit

    I know this is the science thread and not the math thread but I didn't want to make a thread that doesn't really need more than one reply to. So I had this type of question asked in a test the other day and I didn't know how to answer it:

    1. If point (-3, 6) is on graph y= f(x) where is it on the graph y= 1/f(x)

    2. If point (-3, 6) is on graph y=f(x) where is it on graph y= |f(x)|

  11. #1111

    Quote Originally Posted by Kaylia View Post
    If everything is turned into strange matter, shouldnt we call it normal matter.
    Maybe it's just way too early in the morning, but Woozie already said that doesn't happen?

  12. #1112
    Title: "HUBBLE GOTCHU!" (without the quotes, of course [and without "(without the quotes, of course)", of course], etc)
    Join Date
    Jul 2006
    Posts
    3,141
    BG Level
    7

    Quote Originally Posted by Stilzkin View Post
    I know this is the science thread and not the math thread but I didn't want to make a thread that doesn't really need more than one reply to. So I had this type of question asked in a test the other day and I didn't know how to answer it:

    1. If point (-3, 6) is on graph y= f(x) where is it on the graph y= 1/f(x)

    2. If point (-3, 6) is on graph y=f(x) where is it on graph y= |f(x)|
    If I'm understanding the question right, they want you to find y(-3) on the new graph, right? I have a strong feeling I'm misinterpreting this. The wording doesn't sound right. A straight interpretation of this is confusing. If (-3,6) is on y=f(x), there's no garuntee that it's also on y=1/f(x) so it makes no sense to ask where it is on the graph. And if it does happen to be on the graph, it's going to be at (-3,6), obviously (there's no way the point (-3,6) could be at any point other than point (-3,6). Otherwise it wouldn't be (-3,6), it would be something else. For example, (-3,6) can't be at (5,-4) at the same time in the same coordinate system. It's impossible).

    If your teacher meant what I think he meant, then every point (x,f(x)) goes to (x,1/f(x)), so (-3,6) -> (-3,1/6) and in the second, (x,y) goes to (x,|f(x)|), so (-3,6) -> (-3,6)

    Quote Originally Posted by Eliseos View Post
    Maybe it's just way too early in the morning, but Woozie already said that doesn't happen?
    He was joking. If everything in the world was Strange matter, then strange matter would be "normal" (because it would be the most common, which is usually how we define something to be normal) and anything else would be considered "odd" or "abnormal" forms of matter.

  13. #1113
    assburgers
    Join Date
    Mar 2007
    Posts
    10,999
    BG Level
    9

    Wouldn't x = -3 where f(x) was the reciprocal give y = 1/6?
    Similarly, wouldn't (-3, |f(x)|) put the answer in the positive planes, since absolute value can only be positive?

  14. #1114
    Banned.

    Join Date
    Jul 2005
    Posts
    17,472
    BG Level
    9
    FFXI Server
    Ifrit
    WoW Realm
    Area 52

    Quote Originally Posted by Eliseos View Post
    Maybe it's just way too early in the morning, but Woozie already said that doesn't happen?
    That was just a bad joke.

  15. #1115
    Relic Shield
    Join Date
    Oct 2006
    Posts
    1,696
    BG Level
    6

    Here's a real math question:

    I'm trying to make an epsilon delta prove for lim x > c, x^2 = c^2 (limit as x tends to c of x squared is c squared). I'm just going computationally, so far...

    0 < |x-c| < δ

    <=> c-δ < x < c+δ Let δ ≤ c

    => (c-δ)^2 < x^2 < (c+δ)^2
    = c^2 - 2cδ + δ^2 < x^2 < c^2 + 2cδ + δ^2
    <=> -2cδ + δ^2 < x^2 - c^2 < 2cδ + δ^2

    Here I'm getting close to the form |f(x) - L| < ε, but I'm not sure how to finish it.

  16. #1116
    The Mizzle Fizzle of Nikkei's Haremizzle

    Join Date
    Feb 2006
    Posts
    22,050
    BG Level
    10
    FFXI Server
    Bismarck

    Unless someone opens up a paypal or something I shalt not say!

  17. #1117
    Salvage Bans
    Join Date
    Jul 2009
    Posts
    902
    BG Level
    5
    FFXI Server
    Ifrit

    ugh now it looks so easy, don't know why it looked harder in the test

  18. #1118

    What did the photon say to the neutron?

    Nothing, it just waved!

  19. #1119

    Quote Originally Posted by Woozie View Post
    He was joking. If everything in the world was Strange matter, then strange matter would be "normal" (because it would be the most common, which is usually how we define something to be normal) and anything else would be considered "odd" or "abnormal" forms of matter.
    Quote Originally Posted by Kaylia View Post
    That was just a bad joke.
    Yeah, brain wasn't functioning this morning lol, sorry. It especially wasn't working for my Physics exam

  20. #1120
    Banned.

    Join Date
    Jul 2005
    Posts
    17,472
    BG Level
    9
    FFXI Server
    Ifrit
    WoW Realm
    Area 52

    Quote Originally Posted by Cadsuane View Post
    Here's a real math question:

    I'm trying to make an epsilon delta prove for lim x > c, x^2 = c^2 (limit as x tends to c of x squared is c squared). I'm just going computationally, so far...

    0 < |x-c| < δ

    <=> c-δ < x < c+δ Let δ ≤ c

    => (c-δ)^2 < x^2 < (c+δ)^2
    = c^2 - 2cδ + δ^2 < x^2 < c^2 + 2cδ + δ^2
    <=> -2cδ + δ^2 < x^2 - c^2 < 2cδ + δ^2

    Here I'm getting close to the form |f(x) - L| < ε, but I'm not sure how to finish it.
    Hmmm, let me try


    Definition

    Lim(x) = C ⇔ ∀ε> 0 ∃δ>0, for all x ∈ f domain and 0 < | x - x0 | < δ => |f(x) - L | < ε
    x->x0

    Proof
    For f(x) = x²
    We are trying to show that

    ∀ε> 0 ∃δ>0, 0 < | x - C| < δ => |x² - C²| < ε
    x->C

    Since we can write
    |x² - C²| = |x - C|*|x + C| ≤ (|x|-|C|) * (|x -C|)

    Let's assume
    (|x -C|) ≤ 1, then we have
    |x|-|C| ≤ ||x|-|C|| ≤ |x -C| ≤ 1 => |x| ≤ C +1

    Going back to our original statement, we can say that
    |x² - C²| ≤ (2*|C| +1) (|x -C|) < ε
    => (|x -C|) < ε/(2*|C| +1)

    As long δ < Minimum[ ε/(2*|C| +1) , 1 ], the statement ∃δ>0, making the lim x² = C² true


    [edit]

    Might want to wait for woozie seal of approval before writting this in your homeworks. There is probably easier way to do it too.

Similar Threads

  1. Two Nuclear Submarines Collide in Atlantic Ocean
    By Firedemon in forum General Discussion
    Replies: 33
    Last Post: 2009-02-18, 05:38
  2. The Large Hadron Collider goes online tonight...
    By alt in forum General Discussion
    Replies: 8
    Last Post: 2008-09-10, 00:50
  3. Large Hadron Collider...
    By Jotaru in forum General Discussion
    Replies: 71
    Last Post: 2007-11-05, 21:42