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    (Original post by EnglishMuon)
    Solution 563*
    (Please let me know of mistakes/improvements, Im only a 'learner'!). I also apologise for the lack of LaTeX as I wasnt planning on sharing it.
    Attachment 504271
    Maybe it was a bit of an overkill for the first part, but Im never too sure when it says to prove something which seems obvious.

    Problem 565*
    By considering the results from Problem 563, show that
    \displaystyle\int^1_{-1} \frac{T_{n}(x)T_{m}(x)}{\sqrt{1-x^{2}}}\ dx =

                                       0 for\ n\not=m

\pi for\ n=m=0, 

\frac{\pi}{2}\

                                          for n=m\not=0
    I've just done a matlab project on this stuff *weeps in corner*
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    (Original post by Gome44)
    I've just done a matlab project on this stuff *weeps in corner*
    Sorry to bring it up again! But you have to admit, there are some extremely interesting results about these areas . What was your project on?
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    (Original post by EnglishMuon)
    Sorry to bring it up again! But you have to admit, there are some extremely interesting results about these areas . What was your project on?
    https://www0.maths.ox.ac.uk/system/f...HTManual16.pdf

    Project A (was actually pretty easy, but i spent more time on the first exercise than i did the rest of the project....)
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    Problem 566***

    This is quite a nice one

    Evaluate

    \displaystyle \prod_{k \geqslant 1} \left(\dfrac{\Gamma \left(2^k + \frac{1}{2}\right)}{\Gamma(2^k)}  \right)^{2^{-k}}
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    Problem 567***

    Show that

    \displaystyle  \sum_{n \geqslant 1} \sum_{m \geqslant 1} \dfrac{(n-1)! (m-1)!}{(m+n)!}H_{m+n}  =  3 \zeta(3)
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    Problem 568*

    I'm putting this up as a problem as

    a) there is far too little applied maths here, and this is a nice STEP(ish) level mechanics problem and

    b) it turned out to be far trickier to sort out than I'd anticipated when I started on it yesterday (I think that there is a small twist in it that may not be apparent to a lot of A level mechanics students):

    ----------------------------------------

    A smooth block of mass m is held stationary on the inclined plane of a smooth wedge of mass M. The inclined plane of the wedge makes angle \theta with the horizontal. The wedge is resting on a smooth surface.

    Find the acceleration of the wedge when the block is released.
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    (Original post by atsruser)
    Problem 568*

    I'm putting this up as a problem as

    a) there is far too little applied maths here, and this is a nice STEP(ish) level mechanics problem and

    b) it turned out to be far trickier to sort out than I'd anticipated when I started on it yesterday (I think that there is a small twist in it that may not be apparent to a lot of A level mechanics students):

    ----------------------------------------

    A smooth block of mass m is held stationary on the inclined plane of a smooth wedge of mass M. The inclined plane of the wedge makes angle \theta with the horizontal. The wedge is resting on a smooth surface.

    Find the acceleration of the wedge when the block is released.
    With little justification and less confidence, I shall throw in

    mgsin(2theta)/2M

    I feel I have drastically undercomplicated this.

    Edit: I'm fairly sure this is wrong. This is why I should ignore the lisping priest and not do midnight maths.

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    Attempt 2:

    mgsin(2theta)/2(M+m)

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    (Original post by Krollo)
    With little justification and less confidence, I shall throw in

    mgsin(2theta)/2M

    I feel I have drastically undercomplicated this.
    I think that you have. But it's hard to see where you've gone wrong without working.

    This is why I should ignore the lisping priest and not do midnight maths
    :confused:
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    (Original post by Krollo)
    Attempt 2:

    mgsin(2theta)/2(M+m)
    Nope, but closer, I would say. I'll put up a solution later if no one gets this by the end of the day. But in the meantime:

    Hints:

    Spoiler:
    Show

    1. Since the wedge is accelerating, the normal reaction force on the block is not the same as that in the "block on a slope" scenario since ..

    2. .. there is a component of acceleration perpendicular to the slope.

    3. Applying Newton II correctly gives you 3 equations in 4 variables. Where will the fourth equation come from?
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    (Original post by atsruser)

    :confused:
    A truly dire pun.

    I coincidentally was looking at step ii, 1998 q10 which looks somewhat related; might come back to this later.

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    (Original post by atsruser)
    Problem 568*

    I'm putting this up as a problem as

    a) there is far too little applied maths here, and this is a nice STEP(ish) level mechanics problem and

    b) it turned out to be far trickier to sort out than I'd anticipated when I started on it yesterday (I think that there is a small twist in it that may not be apparent to a lot of A level mechanics students):

    ----------------------------------------

    A smooth block of mass m is held stationary on the inclined plane of a smooth wedge of mass M. The inclined plane of the wedge makes angle \theta with the horizontal. The wedge is resting on a smooth surface.

    Find the acceleration of the wedge when the block is released.
    Spoiler:
    Show

    Here is the solution:

    The top diagram in the attached picture shows the wedge in its original position and after it has moved displacement s to the left.

    We can see from the initial and final positions of the block that it has a displacement d_p perpendicular to the inclined surface of the wedge. Since d_p is actually a function of time, we can differentiate twice to find that the wedge has acceleration \ddot{d_p}=a_p perpendicular to the block.

    Hence by Newton II, there is a component of force on the block perpendicular to the inclined surface, and we cannot simply say that the normal reaction is equal to the component of weight perpendicular to the inclined plane (we *can* do this if the wedge doesn't move, though, as no such component of acceleration then exists)

    In the bottom diagram, we can see the forces on the block, and the horizontal acceleration of the wedge a - you can draw the forces on the wedge yourself (we only need to consider the Newton III reaction to R). Applying Newton II to the block and the wedge, we have:

    mg\cos\theta-R = ma_p
    R\sin \theta = Ma

    Now we are stuck as we have 2 equations in 3 unknowns, R,a,a_p, and Newton's laws are unable to give us any more information.

    However, in this case, we do indeed have some extra information - we require that the block remain on the surface of the wedge at all times - this is called a kinematic constraint - it's extra information that we are adding to the problem to ensure that we solve the problem that we want to solve, and it reduces the degrees of freedom of motion of an object. Mathematically it allows us to eliminate one of the variables in the equations of motion. To write down this constraint algebraically, we note that, if the block remains on the surface, we must have:

    d_p = s \sin \theta

    (a bit of trig is your friend) and by differentiating twice we get:

    a_p = \ddot{s} \sin \theta = a \sin \theta

    Our equations of motion now reduce to

    mg\cos\theta-R = m a \sin \theta
    R\sin \theta = Ma

    which solve to give a=\frac{mg \sin\theta \cos\theta}{M +m\sin^2 \theta} = \frac{mg \sin 2\theta}{2(M +m\sin^2 \theta)}

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    Problem 569

    Let f:\;\mathbb{R}\to 2^{\mathbb{R}} be such that f(x) is finite and x\notin f(x) for all x. Is there a subset X\subseteq \mathbb{R} disjoint from f(X) that has cardinality the continuum?
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    Solution to the painting problem I posted a while ago, which unfortunately went unsolved. I am posting it because the solution is brilliant, and it sounded like people were interested.

    Spoiler:
    Show
    Suppose player 2 has a winning strategy. Let player 1 paint the top right square. Player 2's next move is a winning move. But player 1 could have begun by making that very move. Contradiction.


    (Original post by Renzhi10122)
    ...
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    (Original post by Lord of the Flies)
    Solution to the painting problem I posted a while ago, which unfortunately went unsolved. I am posting it because the solution is brilliant, and it sounded like people were interested.

    Spoiler:
    Show
    Suppose player 2 has a winning strategy. Let player 1 paint the top right square. Player 2's next move is a winning move. But player 1 could have begun by making that very move. Contradiction.
    And this solution indicates either a cop out or that when the question was initially considered am that time ago we went down the wrong route, as in completely misunderstood the question, and you let us just keep walking

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    (Original post by Jammy Duel)
    And this solution indicates either a cop out or that when the question was initially considered am that time ago we went down the wrong route, as in completely misunderstood the question, and you let us just keep walking
    1. no one seemed to have misunderstood the question, which was very clearly stated anyway.
    2. it's a problem: you're supposed to go down quite a few wrong routes before hitting a solution.
    3. I don't check this thread every day to see how people are tackling a question, so that I can nudge them in the right direction if they need it.

    Please appreciate the brilliance of the solution rather than make unhelpful comments.
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    (Original post by Lord of the Flies)
    Solution to the painting problem I posted a while ago, which unfortunately went unsolved. I am posting it because the solution is brilliant, and it sounded like people were interested.
    Spoiler:
    Show
    Suppose player 2 has a winning strategy. Let player 1 paint the top right square. Player 2's next move is a winning move. But player 1 could have begun by making that very move. Contradiction.
    Ah, that's nice. Is it always true in general then, that a move is either a winning or losing move in a game where you can only win or lose? It's a question I've always been meaning to ask someone after encountering it somewhere else.
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    (Original post by Renzhi10122)
    Ah, that's nice. Is it always true in general then, that a move is either a winning or losing move in a game where you can only win or lose? It's a question I've always been meaning to ask someone after encountering it somewhere else.
    Well the concept of winning/losing moves presupposes knowledge of the game's outcome, which some sense answers your question. It requires that the winning player be perfect (i.e. will not make mistakes beyond said move), and that one player plays only winning moves, the other losing moves. That should be quite clear.

    So it is more whether the idea of a winning move makes sense in any game where you can only win or lose, or equivalently, whether there is a "correct" outcome to any such game. The answer is essentially yes under appropriate, almost obvious conditions, and is given in full detail by Zermelo's theorem in game theory:

    For every (finite) game of perfect information (that is, one where all useable information is available to both players at all times; chess, go, or reversi are examples), not involving chance (so not backgammon for instance), between two players who take moves in alternation, there is either a winning strategy for one of the players, or both players can force a draw. If we disregard the possibility of draws (as in our painting game), then we have existence of winning & losing moves/strategies.
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    Follow-up to the painting question: how about with an n\times\omega board?
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    (Original post by Lord of the Flies)
    Follow-up to the painting question: how about with an n\times\omega board?
    Is omega a torus or something?
 
 
 
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