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    (Original post by Renzhi10122)
    Is omega a torus or something?
    Most probably the ordinal number, i.e. the board is infinite to one direction.
    More exactly, the positive y direction.
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    (Original post by EricPiphany)
    Most probably the ordinal number, i.e. the board is infinite to one direction.
    More exactly, the positive y direction.
    Ah, makes sense.
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    (Original post by Renzhi10122)
    Ah, makes sense.
    Yeah, more precisely countably infinite. ``\mathbb{N}" if you will.
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    (Original post by Lord of the Flies)
    Yeah, more precisely countably infinite. ``\mathbb{N}" if you will.
    Then the game is infinite I guess?
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    (Original post by Renzhi10122)
    Then the game is infinite I guess?
    I shouldn't really get involved with this, because I don't really understand what's going on, but as far as I can see, while the game can be made arbitrarily long in terms of moves, a game will always be finite. Notice that infinitely many squares will be painted on the first go, and then there is only a finite number of moves before that has to happen again etc.
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    (Original post by EricPiphany)
    I shouldn't really get involved with this, because I don't really understand what's going on, but as far as I can see, while the game can be made arbitrarily long in terms of moves, a game will always be finite. Notice that infinitely many squares will be painted on the first go, and then there is only a finite number of moves before that has to happen again etc.
    Lol, I don't really fully understand this either. So my reasoning was that on each move, there is an infinite number of unpainted squares, and an arbitrarily large (as opposed to infinite) number of new squares are painted, still leaving an infinite number of squares unpainted. Hence, the game never ends.
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    (Original post by Renzhi10122)
    Lol, I don't really fully understand this either. So my reasoning was that on each move, there is an infinite number of unpainted squares, and an arbitrarily large (as opposed to infinite) number of new squares are painted, still leaving an infinite number of squares unpainted. Hence, the game never ends.
    The board has finite bounds in the x coordinate direction, but is unbounded in the positive y coordinate direction. If on the first turn the player paints all the y coordinates greater or equal to his chosen coordinate with the same x coordinate or greater, he will be painting infinite boxes.
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    (Original post by EricPiphany)
    The board has finite bounds in the x coordinate direction, but unbounded in the positive y coordinate direction. If on the first turn the player paints all the y coordinates greater and equal to his chosen coordinate, he will be painting infinite boxes.
    Ah, positive y...
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    In this network, each switch is closed with probability p, independent of any other switch. What is the probability that current can flow from A to B?
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    (Original post by Gregorius)
    In this network, each switch is closed with probability p, independent of any other switch. What is the probability that current can flow from A to B?
    (I am really bad at prob. and stats. but l tried).
    p^2(2p^3-5p^2+2p+2).
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    (Original post by EricPiphany)
    (I am really bad at prob. and stats. but l tried).
    p^2(2p^3-5p^2+2p+2).
    Haha ur first answer was a number 😂 i wudve got something like 2/1 😂


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    (Original post by physicsmaths)
    Haha ur first answer was a number 😂 i wudve got something like 2/1 😂


    Posted from TSR Mobile
    I jumped in without reading the question correctly. I was thinking equal probability i.e. p = 0.5, and I counted cases, missing one out by error. so I got 17/32. :p:
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    (Original post by EricPiphany)
    (I am really bad at prob. and stats. but l tried).
    p^2(2p^3-5p^2+2p+2).
    Well done! But only half marks so far as you haven't shown your working!
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    (Original post by Gregorius)
    Well done! But only half marks so far as you haven't shown your working!
    I'm terrified.
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    I haven't explained the counting properly, but l would find it hard to explain.
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    (Original post by EricPiphany)
    I'm terrified.

    I haven't explained the counting properly, but l would find it hard to explain.
    PRSOM! I must admit, if I saw your first method as an answer, I would want more explanation, as it's not transparently clear why the two routes that go through the middle switch give you independent probabilities. The second method, however, is completely above board, conditioning on the event that the middle switch is closed and then using the law of total probability.
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    (Original post by EricPiphany)
    I'm terrified.
    I haven't explained the counting properly, but l would find it hard to explain.
    This is lovely!
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    The harder questions aren't gathering very much attention it seems, so here's a nice dinner-table problem!

    Problem 570

    If a rectangle can be subdivided into smaller rectangles, each having at least one side of integer length, must the initial rectangle also have at least one side of integer length?
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    (Original post by Gregorius)
    PRSOM! I must admit, if I saw your first method as an answer, I would want more explanation, as it's not transparently clear why the two routes that go through the middle switch give you independent probabilities. The second method, however, is completely above board, conditioning on the event that the middle switch is closed and then using the law of total probability.
    (Original post by Zacken)
    This is lovely!
    Thanks guys!
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    Problem 571

    Find the abscissa(e) of the turning points of

    f(x) = x\ln x +x^x + e^{x^x} + e^{e^{x^x}} + \cdots + e^{e^{.^{.^{e^{x^x}}}}}

    where the final term is the n'th tetration of e tetrated by x^x
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    (Original post by atsruser)
    Problem 571

    Find the ordinates of the turning points of

    f(x) = x\ln x +x^x + e^{x^x} + e^{e^{x^x}} + \cdots + e^{e^{.^{.^{e^{x^x}}}}}

    where the final term is the n'th tetration of e tetrated by x^x
    How pretty are you expecting the answer to be?
    Spoiler:
    Show
    At present I have the rather unpleasant expression you have, x\ln(x) \mapsto -\frac{1}{e}, and with the x^x=e^{x\ln(x)}=e^{-\frac{1}{e}}, and I don't much fancy seeing if it simplifies.
 
 
 
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