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    This is favourite of mine. A lot of you will know it, so please don't give the answer away if you've seen it before

    Problem 43 **/***

    Evaluate \displaystyle \lim_{m \rightarrow \infty} (\lim_{n \rightarrow \infty} \cos^{2n}(m!\pi x))

    (To do it rigorously you obviously need analysis, but don't bother if you've done no Analysis)
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    (Original post by shamika)
    This is favourite of mine. A lot of you will know it, so please don't give the answer away if you've seen it before

    Problem 43 **/***

    Evaluate \displaystyle \lim_{m \rightarrow \infty} (\lim_{n \rightarrow \infty} \cos^{2n}(m!\pi x))

    (To do it rigorously you obviously need analysis, but don't bother if you've done no Analysis)
    My guess is that it's 1 if x is rational and 0 otherwise.
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    (Original post by und)
    My guess is that it's 1 if x is rational and 0 otherwise.
    Indeed! Can you explain why?
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    (Original post by shamika)
    Indeed! Can you explain why?
    The only thing I'm not sure about is the order of the limits. It feels like they should be the other way round, although perhaps it doesn't matter?
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    (Original post by Felix Felicis)
    Problem 42*/**

    Show that \displaystyle\sum_{r=0}^{88} \dfrac{1}{\cos r \cdot \cos (r+1)} = \dfrac{\cos 1}{\sin^{2} 1}

    (angles measured in degrees)
    Solution 42:

     \dfrac{1}{(cosr)(cos(r+1))} = \dfrac{A}{cosr} + \dfrac{B}{cos(r+1)}

     Acos(r+1)+Bcos(r) = 1

    Let r = 90:

     Acos91 = -Asin1=1 \Rightarrow A = -cosec1

    Let r = 89:

     Bcos89 = Bsin1 = 1 \Rightarrow B = cosec1

    As such, we are summing:

     f(r) = \dfrac{cosec1}{cos(r+1)} - \dfrac{cosec1}{cosr}

    In this sum, all terms except for two will cancel out:

     S = \dfrac{1}{cos89sin1}-\dfrac{1}{sin1}

     = \dfrac{1-sin1}{sin^21}

    We know that:

     cosec1cosr-cosec1cos(r+1)=1

    Let r = 0:

     1 - cos1 = sin1

    So cos1 = 1 - sin1, so our sum is \dfrac{cos1}{sin^21} , as required.
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    (Original post by DJMayes)
    Solution 42:

     \dfrac{1}{(cosr)(cos(r+1))} = \dfrac{A}{cosr} + \dfrac{B}{cos(r+1)}

     Acos(r+1)+Bcos(r) = 1

    Let r = 90:

     Acos91 = -Asin1=1 \Rightarrow A = -cosec1

    Let r = 89:

     Bcos89 = Bsin1 = 1 \Rightarrow B = cosec1

    As such, we are summing:

     f(r) = \dfrac{cosec1}{cos(r+1)} - \dfrac{cosec1}{cosr}

    In this sum, all terms except for two will cancel out:

     S = \dfrac{1}{cos89sin1}-\dfrac{1}{sin1}

     = \dfrac{1-sin1}{sin^21}

    We know that:

     cosec1cosr-cosec1cos(r+1)=1

    Let r = 0:

     1 - cos1 = sin1

    So cos1 = 1 - sin1, so our sum is \dfrac{cos1}{sin^21} , as required.
    See that's what I did except.

    cos(1)=0.9998476952
    1-sin(1)=0.9825475936.

    So I got confused an gave up. I checked the first bit out numerically too and that was dodgy too.
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    (Original post by bananarama2)
    See that's what I did except.

    cos(1)=0.9998476952
    1-sin(1)=0.9825475936.

    So I got confused an gave up. I checked the first bit out numerically too and that was dodgy too.
    I know. I don't like it; and wouldn't have submitted the solution if it weren't for the partial fractions bit saying that it's apparently true.
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    (Original post by DJMayes)
    I know. I don't like it; and wouldn't have submitted the solution if it weren't for the partial fractions bit saying that it's apparently true.
    I think the problem is with the partial fractions bit. You're finding A and B only for the cases where r= 90 and 89. It isn't an identity.
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    (Original post by und)
    The only thing I'm not sure about is the order of the limits. It feels like they should be the other way round, although perhaps it doesn't matter?
    It definitely matters which way round the limits are. I just did this in my head and think these are the right way round
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    (Original post by bananarama2)
    I think the problem is with the partial fractions bit. You're finding A and B only for the cases where r= 90 and 89. It isn't an identity.
    I dunno. We'll probably have to find out what Felix was after; as the solution isn't an agreeable one.
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    (Original post by und)
    The only thing I'm not sure about is the order of the limits. It feels like they should be the other way round, although perhaps it doesn't matter?
    Spoiler:
    Show
    Were you thinking the bit in the brackets tended to an integer multiple of pi?
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    Solution 43

    Lemma: If x is not rational then m!x is not an integer for any m. We will prove this by contraposition. If m!x is an integer then m!x=k \Rightarrow x=\frac{k}{m!}, hence x is rational. Therefore if x is not rational then m!x is not an integer.


    We consider the two cases: x is rational; x is not rational.

    If x is rational, then we can write x=\frac{p}{q} for some p,q with hcf(p,q)=1, and hence m!x=\frac{m!p}{q}, which is clearly an integer when m\geq q. Since m \to \infty in the limit, we can make m arbitrarily large, so m!\pi x is an integer multiple of \pi and hence cos^{2}(m!\pi x)=1 \Rightarrow cos^{2n}(m! \pi x)=1 for all n.

    If x is irrational, then from the lemma m! \pi x is not an integer multiple of \pi, and hence \displaystyle |cos(m! \pi x)|<1 \Rightarrow \lim_{n \to \infty}  cos^{2n}(m! \pi x)=0.
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    Solution 41

    For black color we write 1, for white - 0.
    By S we denote the set of all the students and K=\{0.1\}.
    Let S \to K be the set of all maps (colorings) from S to K.
    For each student a from S, we say f \approx g if the set \{a \in S | f(a) \not= g(a)\} is finite. Since \approx is an equivalence relation on S \to K, by AC there exists map T, T: S\to K \to S\to K such that T \approx f. Also f \approx g leads to T(f)=T(g).
    Define G_{a}(f) to be a guessing function for the student a under the coloring f.
    Since each student knows T(f), we define G_{a}(f)=T(f)(a). Consequently, all but finitely many students will guess correctly.

    Let me try something different.
    S - the set of all the students, K - the set of colors, in our case \{0,1\}. We consider again the set of all mappings S\to K. For each student a \in S, define the equivalence relation \approx_{a} on S\to K by f \approx_{a} g if and only if the colorings f and g are not distinguishable for the student a. Let E_{f}(a) denote the equivalence class of the coloring f under the relation \approx_{a}.
    Now, let \preceq be a well-ordering of S\to K (AC). Define \mu to be the guessing function such that \mu(E_{f}(a)) is the \preceq least element of E_{f}(a).
    Let T=\{a\in S| \mu(E_{f}(a))\not= f(a)\}.
    Here, I am left to define a ''good'' binary relation \bigtriangledown over S. I mean, if the set T has no infinite \bigtriangledown -chains, then \bigtriangledown is "good".
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    (Original post by ukdragon37)


    Problem 41***


    A (countably) infinite class of Cambridge students who believe in the Axiom of Choice are taking their finals for the Mathematical Tripos. At the start of the exam, the examiner places either a white or a black hat on each student at random. There is only one question in the exam: "What colour is your hat?" If only a finite number of students answers incorrectly then everyone becomes a Wrangler (very good), otherwise the Wooden Spoon is given to the whole class (very bad). Everyone can see the hats of everyone else besides their own, and since it's exam conditions they are not allowed to communicate. Students may not remove their hats or try to look at it in any way.

    What strategy could the students devise together so to avoid failure and the impending doom of unemployment? Being clever clogs from Cambridge, memorising infinitely large amounts of data is no problem to the students.

    Note: A student suggested that each person just guess at random. However this has already been ruled out by the cleverer students who realised that if they were really unlucky, infinite of them could guess wrong.
    I've got it. It you plot lateral binary graph associated with the symbolic mathematical solutions, you can reduced the problem down to simple SNF's. And by using Dalek's contraction algorithm the answer is: the physical quantity of dimensions m to the ml to the lq to the qt
    to the ttheta to the x has planck quantity root g minus m plus l minus 2q
    plus t hbar to the m plus l minus q plus t minus xe to the minus m minus l minus 7q minus 3t plus 6xk to the minus 2x epsilon 0 to the q.
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    (Original post by bananarama2)
    I've got it. It you plot lateral binary graph associated with the symbolic mathematical solutions, you can reduced the problem down to simple SNF's. And by using Dalek's contraction algorithm the answer is: the physical quantity of dimensions m to the ml to the lq to the qt
    to the ttheta to the x has planck quantity root g minus m plus l minus 2q
    plus t hbar to the m plus l minus q plus t minus xe to the minus m minus l minus 7q minus 3t plus 6xk to the minus 2x epsilon 0 to the q.
    Nice one. :rofl:

    http://youtu.be/lTDcYi_uG08?t=9s

    Also, The proof is trivial! Just view the problem as a Noetherian complexity class whose elements are countable logistic systems.
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    (Original post by und)
    This is what I was thinking so I'm guessing it's wrong!

    Spoiler:
    Show
    Lemma: If x is not rational then m!x is not an integer for any m. We will prove this by contraposition. If m!x is an integer then m!x=k \Rightarrow x=\frac{k}{m!}, hence x is rational. Therefore if x is not rational then m!x is not an integer.


    We consider the two cases: x is rational; x is not rational.

    If x is rational, then we can write x=\frac{p}{q} for some p,q with hcf(p,q)=1, and hence m!x=\frac{m!p}{q}, which is clearly an integer when m\geq q. So in the limit m \to \infty, m!\pi x is an integer multiple of \pi and hence cos^{2}(m!\pi x)=1 \Rightarrow cos^{2n}(m! \pi x)=1 for all n.

    If x is irrational, then from the lemma m! \pi x is not an integer multiple of \pi, and hence \displaystyle |cos(m! \pi x)|<1 \Rightarrow \lim_{n \to \infty}  cos^{2n}(m! \pi x)=0.
    You're almost there. You don't need to consider the case where m < q. Why? This is subtle so don't worry if you're getting confused
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    (Original post by shamika)
    You're almost there. You don't need to consider the case where m < q. Why? This is subtle so don't worry if you're getting confused
    Because q doesn't necessarily divide m! then?
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    (Original post by shamika)
    You're almost there. You don't need to consider the case where m < q. Why? This is subtle so don't worry if you're getting confused
    This is why I said I found the order of the limits confusing. In the case m<q, the first limit is either 0 or 1 depending on the factors of q, but then in the second limit we have m->∞. I would then say that for m<q, the limit is 0, but you told me that it was 1. Hmm...
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    (Original post by und)
    This is why I said I found the order of the limits confusing. In the case m<q, the first limit is 0, but then in the second limit we have m->∞. I would then say that for m<q, the limit is 0, but you told me that it was 1. Hmm...
    Given that in the second limit you're going to take the limit to infinity anyway, how about in the first limit you...

    (Remember that in the first limit, m is arbitrary; the limit is with respect to n.

    Also note that the limits have to be this way around, because otherwise it causes issues when you have irrational x)
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    (Original post by DJMayes)
    I dunno. We'll probably have to find out what Felix was after; as the solution isn't an agreeable one.
    I did all this, but the partial fractions are not equivalent, therein lies the problem.
    Felix's solution is correct, though I'm trying to think of a clever method to show that it is
 
 
 
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