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Let's discuss Mathematics

Discussion in 'Science & Technology' started by ParadigmShifter, Mar 16, 2009.

  1. PlutonianEmpire

    PlutonianEmpire Socially Awkward Goofball

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    I've been trying to figure out how to convert to Julian dates manually, for scifi worldbuilding purposes. I found this: ( http://www.aavso.org/computing-jd ), but the value I found on Wikipedia for an existing wide binary star is given in decimal years (1991.605 in this case), so I'm not sure how to convert it directly from that.

    Any ideas?
     
  2. Kyriakos

    Kyriakos Alien spiral maker

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    nm- probably did not help.
     
  3. nc-1701

    nc-1701 bombombedum

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    Well 1991 is obvious I suppose. Then there are approximately 365 days/year so .605*365~=220.8.

    Now count days
    31 (January)
    28
    31
    30
    31
    30
    31 (July)
    add up and get 212 days so far so 8.8 days left...
    So it happened on the 8th of August 1991 at around 7pm.

    Because the months aren't uniform I'm not sure of a clean way to do it with less work than that. Though if you just say months are 30 days you could easily place it within a day or two of being correct without as many calculations.
     
  4. Mise

    Mise isle of lucy

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    in excel you can subtract 1900 from the number, then multiply by 365.2425 days (as per Gregorian calendar), then format it as a date. To convert this to a Julian calendar date, follow the following instructions: http://office.microsoft.com/en-us/excel-help/insert-julian-dates-HP003056114.aspx

    EDIT: wait, I didn't realise that "Julian date" is a different thing to the "Julian calendar"!! In that case you can just do "=ROUNDDOWN(A1,0)&(A1-ROUNDDOWN(A1,0))*365.2425". If you don't want decimals you can wrap the whole thing in "=round( )" or do it in a separate cell.
     
  5. Atticus

    Atticus Deity Retired Moderator

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    I have a question about confidence intervals.

    In some lecture notes this guys says that once you've calculated the CI, say 95% CI, it isn't correct to say that it contains the parameter with 95% chance. It either contains it or not. He goes on with a lengthy reply what you are allowed to say, which could be compressed into "If we produce intervals with this method, 95% of them will contain the right value".

    Now, I accept that the interval either contains or doesn't the right value. But if the value is unknown, isn't it "philosophical nitpicking" to say that it's incorrect to say that there's a 95% chance it does?

    Couldn't you similarly say that Brazil either winds or doesn't win the world cup 2014? So assigning a probability value to it would be incorrect.

    Or to have an example that doesn't involve time: playing cards, either the card on the top of the deck is or is not the card you're wishing to draw. There's no sense in calculating probabilities.

    So, my whole point is: isn't this kind of rigorousness with the terminology conserning CI, loosing all the power that the statistics have? Isn't it similar kind of fatalism that makes people to do bad desicions against odds?
     
  6. Mise

    Mise isle of lucy

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    I suppose it's a nuanced way of looking at it, but I think they're practically equivalent. "There is a 95% chance that any individual CI will contain the value" presents a slightly different perspective to saying that there is a 95% chance that the value is within this interval. It treats the value is that absolute thing, and the CI as the thing that is ephemeral and moveable. I guess people in practice have a tendency to view the model as the real thing and the actual value as a bit of unwanted randomness that is adequately encompassed by sticking CIs on everything. At least I do that, anyway.
     
  7. Samson

    Samson Deity

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    I would say that these are 2 different things. To use your deck of cards example, if you know that the deck is a normal 52 card deck well shuffled you could say that P(red card) = 0.5 with zero confidence interval. However, if I put a deck of cards in front of you, and offer you $1.10 if the next card is red for $1, you may want to check. You could take a few cards and estimate a P(red card) and calculate a confidence interval for this estimate. This would allow you to make a better decision on whether to play or not.
     
  8. azzaman333

    azzaman333 meh

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    Not certain, but I would guess it's because the confidence interval is taken from a random sample of the population at a whole. So depending on the specific sample you get, you could get vastly different means and standard deviations and therefore confidence intervals. You can't strictly say that the specific CI you've calculated has a 95% chance of containing the true mean, as that may not be a property that is strictly true to the calculated interval.

    I'm sure I remember being given a good reason for it at some point during my undergrad, whether or not this is right I can't recall though.
     
  9. Atticus

    Atticus Deity Retired Moderator

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    Thanks for the input! :goodjob:

    I'm not sure we're talking about the same thing here. That's an a priori probability for the card being red, not a result of a statistical investigations.

    The whole card example wasn't meant to be about the CI either, but to make the point about the power of statistics and probabilites: while the card is unknown, it makes sense to assign a probability to it's colour, although strictly speaking it either is or is not red. Similarly, while the paratmeter you're trying to estimate by statistical inference is unknown, it makes sense to say that it has 95% probability on being in the interval.
     
  10. Rashiminos

    Rashiminos Fool Prophet

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    In a sense you could consider the population parameter to be a random variable** with a probability density function, and then construct an interval for which the integral of that function over that interval is not necessarily .95, but the process which generates that interval has at least a 95% chance of generating an interval that contains the value of the variable for this specific population.

    With X as a random variable; constants aci, bci
    Compare:
    P( aci < X < bci), *

    versus

    P{Generate an interval (a,b) containing X}

    _______________________________________________

    * or P (aci < X < bci | sample data)

    **Something more often done in the context of Bayesian inference where the variable represents uncertainty in the knowledge of the parameter. I wouldn't consider myself sufficiently versed in credible intervals to discuss them here.

    If you don't consider a random variable approach like the above, then the parameter itself is not random, so saying the probability of it lying in the observed confidence interval is 95% {P(in the interval)=.95} is problematic. Saying it is in the interval or not seems to imply P(in the interval)=1 XOR P(in the interval)=0 . The 95% probability of a confidence interval has to do with the sample estimate and not the parameter.

    More succinctly, it shouldn't make sense:
    How would you model the probability of Brazil winning the cup? Would it be like flipping a (biased) coin?

    In a typical coin model, the probability of getting heads on a fair coin is 50%. We can simulate flipping a coin a number of times to estimate the probability. Suppose we get a 95% confidence interval from a sample that is between 35% and 43%. What sense does it make to you to say that the true probability of getting heads (assumed to be 50%) has a 95% chance of lying within our interval between 35% and 43%?
     
  11. Mise

    Mise isle of lucy

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    This is cool: http://plus.maths.org/content/non-transitiv-dice

    So there's a set of 3 dice, A, B and C, but the numbers on the dice are set up such that, even though they still average out to 3.5, dice A beats dice B, B beats C, and C beats A.
     
  12. dutchfire

    dutchfire Deity Retired Moderator

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    If you want to play a dice game, but don't like chance, you could do the following:

    Every player gets an empty dice and is allowed to put 21 dots on there. You calculate which die will win according to probability theory, and this player wins the round.

    (I assume for every die A with 21 points, there is a die B with 21 points with B>A.)
     
  13. timtofly

    timtofly One Day

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    How would that work in playing Civ games?
     
  14. PlutonianEmpire

    PlutonianEmpire Socially Awkward Goofball

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    :bump:

    I have a trigonometry question about space. If I know the distance in light years and orbital separation in astronomical units of a binary star, how do I find out the angular separation between the two in arcseconds?
     
  15. Timsup2nothin

    Timsup2nothin Deity

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    You don't have enough information. You need to know the angle between the line connecting the two stars and the line of sight from Earth. If they are lined up they will have no angular separation at all, while if that angle is ninety degrees the angular separation will be maximized.
     
  16. Samson

    Samson Deity

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    WARNING: I am not an expert (or even close).

    I do not think it matters, if you are asking what the maximum seperation a star system will have. I think that whatever the angle of the plane of rotation is, there will be a point that they are seperated by that distance relative to earth. So it becomes simple trignometry, angle = atan(diameter of rotation/distance from earth). For example, from here Eta Cass is 19 light years (1.8e17m) away, and seperated by 70 AU (1.0e13m), diameter 35 AU (5e12). So the angle between them is atan(5e12/1.8e17) = 2.9e-05 radians = 5.7 arcseconds.

    I will re-interate that I could be wrong at any point in this working. I have left more significant figures in my calculation than I have shown.
     
  17. Timsup2nothin

    Timsup2nothin Deity

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    If you are going for a maximum value this seems correct. It does seem like there will be that point no matter what the plane of rotation, though I'd hate to have to prove it.

    Edit: Okay, proving that would not be that bad. So if you are looking for the maximum potential angular separation you can find it.
     
  18. Harv

    Harv Emperor

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    From the same source:

    The short answer to the question is:

    a = 3.26 x / D, where:

    a is the angle between the stars, in arcseconds
    x is the distance between the stars, in AU
    D is the distance from the Earth to the binary system, in Light Years

    So if I take 3.26 times 70, divided by 19, I get 12.0 arcseconds. The reference says 11.9.

    Here is my derivation:

    According to the Wikipedia article on "light-year," a light-year is 63241 AU.

    If I assume the distance to the binary star system is much greater than the separation between them, then the tangent of the angle a is approximately equal to the sine of the angle, which is approximately equal to the angle in radians.

    A ~= sin(a) ~= tan(a) = x / D / 63241, where A is the angle between the stars in radians.

    a = 60 * 60 * 180 * A / pi = 206265 A

    a = 206265x / D / 63241 = 3.26x / D

    Also note that 3.26 is also the number of light-years in a parsec.
     
  19. Robert FIN

    Robert FIN Monty n' Roll

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    I have a maths test tomorrow and I can't figure out these two examples. I don't know math terms in english but I think the task is to divide the thing in 'factors' which basically means that there shall not be any + or - outside ( ) signs. Let's say e=power, for example "3 e2" is "3 power 2" which is 3×3=9.

    1) Divide to factors: (1 - x)e2 + (1 - x)e3

    2) Same task: (&#960; - 4)e2 + &#960; - 4

    Thanks
     
  20. Harv

    Harv Emperor

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    Is this what you mean? It does not directly answer your question, but it might make the question more clear.

    EDIT: I am editing in solutions with the assumption that I correctly understood the question:

    Problem 1: Find the factors for (1-x)e2 + (1-x)e3

    A. I determined f(x) = (1-x)e2 + (1-x)e3 to be equal to -xe3 + 4xe2 - 5x + 2 which we are solving to equal 0. We already know x=1 works, so:

    B. Using long division, divide (-xe3 + 4xe2 - 5x + 2) by (x-1) to get -xe2 + 3x - 2. We can solve this for zero with the quadratic formula.

    C. The quadratic formula gives solutions for x = 1 and x = 2.

    D. There are three roots for this formula. x = 1, x = 1, and x = 2. Note that x = 1 is repeated. This means the root x = 1 has a multiplicity of 2, or that it is used two times.

    E. The function (1-x)e2 + (1-x)e3 can therefore be expressed as: (x-1)(x-1)(x-2)
    EDIT: This should be -(x-1)(x-1)(x-2) or (1-x)(1-x)(2-x)

    F. Check the work. The function above multiplies out to xe3 - 4xe2 + 5x -2.
    EDIT: The edit will work out to -xe3 + 4xe2 -5x +2.

    Problem 2: Same task: (&#960; - 4)e2 + &#960; - 4

    A. This function expands out to ne2 - 7n + 12. We can solve this for zero with the quadratic formula.

    B. The roots of this function are 3 and 4. You can use n = 3 and n = 4 to get 0. Therefore:

    C. We can rewrite the function as (n-3)(n-4).

    D. Check work: This expands out to ne2 - 7n + 12.
     

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