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    (Original post by Lord of the Flies)
    So how many kids do you have?
    One kid. Lots of other deformed monsters though.
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    (Original post by Lord of the Flies)
    So how many kids do you have?

    (Original post by ukdragon37)
    One kid. Lots of other deformed monsters though.
    Ermm... WTF?
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    (Original post by bananarama2)
    Ermm... WTF?
    Spoiler:
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    (Original post by bananarama2)
    Ermm... WTF?
    In-joke
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    (Original post by Mladenov)
    I do not understand this. From f(x)(f(-x)-x)=0 it does not follow that f(-x)=x for all x or f(x)=0 for all x. Can I not take f to be zero over the rationals and -x over the irrationals?
    Oops there was a few typos! Especially in having f(x)=0 in the last paragraph but not at the end! Is it fine now?
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    We should allow ciphers on this thread
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    (Original post by ukdragon37)
    I've decided to give you a taste of what I do
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    Problem 224*/**:

    Whenever we say "function" we mean total function.

    Denote for every n \in \mathbb N _0 the set \underline n = \{ m \in \mathbb N _0 | 0 < m \leq n\}.

    The mystery concept of a strange product \underline m \otimes \underline n for two such sets \underline m and \underline n satisfies the following property:



    There exist functions \pi_1 : \underline m \otimes \underline n \to \underline m and \pi_2 : \underline m \otimes \underline n \to \underline n such that:

    For each k \in \mathbb N_0 and each pair of functions f: \underline k \to \underline m and g: \underline k \to \underline n there exists a unique function h : \underline k \to \underline m \otimes \underline n such that for all x\in \underline k we have f(x) = \pi_1(h(x)) and g(x) = \pi_2(h(x))



    Show that regardless of what the exact definition for \otimes is, if it satisfies the above property then for all sets X and (non-zero) natural n, |X \otimes \underline n| = n iff |X| = 1 (in other words X = \underline 1).

    EDIT1: made it clearer, but question is still the same.

    EDIT2: Hmmm I just realised that the elegant solution is far beyond *-level, and the inelegant one may be quite long.

    EDIT3: Actually for the above reason I'm going to withdraw it as a formal problem for this thread, but it's open to anyone who wants to try it.
    Haha, no idea what's going on I'm afraid! :lol:

    Those diagrams look cool though! Your child look awesome! :lol:

    Problem 224*

    'Bananarama 2' is at the top of the Eiffel Tower with twelve coins. All coins are of equal size and appearance though one of them is of a different weight (as it is a counterfeit coin). We do not know whether it is heavier or lighter. He has two bags of equal size and weight labelled A and B and the french government have told him that, if he is going to drop coins off the Eiffel tower, he is obviously going to have to do so using these bags otherwise he would be violating the health and safety regulations. He may place however many coins in bag A and however many coins in bag B and drop them from the tower. 'LOTF' is at the bottom of the tower and, every time the coins are dropped from the tower, he notes which bag hits the ground first or whether they both hit the ground at the same time. Unfortunately, the joke is getting old and so, whilst 'Bananarama 2' would happily drop coins from the Eiffel Tower all day, 'LOTF' had only agreed to stick around for 20 minutes from the time the first pair of bags are dropped. The lift in the Eiffel tower takes 6 minutes to queue for each time 'LOTF' brings the bags and coins back up. We cannot predict exactly how long the lift will take but the time taken to go from the bottom to the top, measured in minutes, can be modelled as having a continuous uniform distribution on the interval [\frac{1}{2} , 2].

    Are they going to be able to find the counterfeit coin? Give reasons for your answer.
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    (Original post by Jkn)
    Problem 224*

    ...
    I'm going to guess no because Galileo says so
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    (Original post by ukdragon37)
    I'm going to guess no because Galileo says so
    **** Galileo! We may assume that 'LOTF' is incredibly good at seeing which (if either) bag hits the ground first.

    Spoiler:
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    You should recognise at least half of this problem, being a compsci


    Problem 225*

    Let \rho_0 be a density such that, if the density of the universe were greater than \rho_0, the gravitational forces would be strong enough to halt the expansion of the universe (i.e. the universe would be closed).

    Assuming that the universe is both isotropic and homogenous, prove that \displaystyle \rho_0 \approx \frac{3H^2_0}{8 \pi G}, where H_0 is Hubble's constant and G is the universal gravitational constant.
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    (Original post by Jkn)
    **** Galileo! We may assume that 'LOTF' is incredibly good at seeing which (if either) bag hits the ground first.

    Spoiler:
    Show
    You should recognise at least half of this problem, being a compsci
    :confused: Yes I recognise the problem but the way you framed it makes the solution impossible. The latter outcome of one hitting the ground before the other is not going to happen, as famously shown by Galileo. You can put one coin in one bag and 100 in the other and they are still going to land at the same time.
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    Solution 224

    The queue to the lift of the Eiffel tower does not take 6 minutes, it takes 2 hours. Thus LotF never gets to the top, people get injured from the falling coins, banarama gets arrested, and LotF goes to a nearby café to enjoy a glass of 1981 Petrus.
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    (Original post by Lord of the Flies)
    Solution 224

    The queue to the lift of the Eiffel tower does not take 6 minutes, it takes 2 hours. Thus LotF never gets to the top, people get injured from the falling coins, banarama gets arrested, and LotF goes to a nearby café to enjoy a glass of 1981 Petrus.
    Cheers for that! :rolleyes: You're just going to roll up at Cambridge and fill your room with vintage wine aren't you?
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    (Original post by ukdragon37)
    :confused: Yes I recognise the problem but the way you framed it makes the solution impossible. The latter outcome of one hitting the ground before the other is not going to happen, as famously shown by Galileo. You can put one coin in one bag and 100 in the other and they are still going to land at the same time.
    I'm guessing the fact that there is air resistance means that the bag that weighs more will indeed hit the ground first.
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    (Original post by und)
    I'm guessing the fact that there is air resistance means that the bag that weighs more will indeed hit the ground first.
    But as Jkn so strongly protested yesterday, that's such an important unstated assumption like assuming I am Obama for "How likely I am to go into the White House?". :laugh:
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    (Original post by ukdragon37)
    :confused: Yes I recognise the problem but the way you framed it makes the solution impossible. The latter outcome of one hitting the ground before the other is not going to happen, as famously shown by Galileo. You can put one coin in one bag and 100 in the other and they are still going to land at the same time.
    ................... are they?
    Spoiler:
    Show

    (Original post by Lord of the Flies)
    The queue to the lift of the Eiffel tower does not take 6 minutes, it takes 2 hours. Thus LotF never gets to the top, people get injured from the falling coins, banarama gets arrested, and LotF goes to a nearby café to enjoy a glass of 1981 Petrus.
    Hahahahahaha PRSOM :lol: Is that the French equivalent of going to the Winchester, pouring a nice cool pint and waiting for everything to blow over? :lol:

    What makes you think LOTF needs to go up in the lift with the coins? If he did, we would not have enough information as to how long it would take him to get down We may assume that they're going to let him chuck the bag of coins in every time the elevator is about to go up and that this time is, in fact, such that he has to wait 6 minutes each time!
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    (Original post by und)
    I'm guessing the fact that there is air resistance means that the bag that weighs more will indeed hit the ground first.
    Given the size, the quadratic drag law would be best to describe the motion and mass cancels out. The shape has an impact and volume but not mass.
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    (Original post by ukdragon37)
    But as Jkn so strongly protested yesterday, that's such an important unstated assumption like assuming I am Obama for "How likely I am to go into the White House?". :laugh:
    Hahahaha :lol:

    Quite the contrary! The nature of assumption is, in this case, such that if the problem does not suggest air resistance is being ruled out, we must assume that it is not! Lack of assumption in Physics forces you towards more sophisticated theories whilst lack of assumption is mathematics often renders a problem un-doable!
    (Original post by und)
    I'm guessing the fact that there is air resistance means that the bag that weighs more will indeed hit the ground first.
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    (Original post by bananarama2)
    Given the size, the quadratic drag law would be best to describe the motion and mass cancels out. The shape has an impact and volume but not mass.
    Note the ratio of drag to weight
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    (Original post by bananarama2)
    Given the size, the quadratic drag law would be best to describe the motion and mass cancels out. The shape has an impact and volume but not mass.
    The force will be the same for any given velocity, but the acceleration will be different since F=ma.
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    (Original post by Jkn)
    Note the ratio of drag to weight
    Drag  F= \frac{1}{2}\frac{m}{V}C_d v^2 A

    Weight  mg

    Ratio is independent of mass...
 
 
 
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