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    (Original post by gff)
    This week seems to be devoted to integrals.

    [*] Determine all continuous functions f : [0, 1] \to \mathbb{R} which satisfy

    \displaystyle \int_{0}^{1} f(x)(x - f(x))\ dx = \frac{1}{12}
    Spoiler:
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    Something along the lines of noting that f(x)(x-f(x)) \equiv \dfrac{x^2}{4} - \left(f(x)-\dfrac{x}{2}\right)^2 so it suffices to find f such that:

    \displaystyle\int_0^1 \left(f(x)-\dfrac{x}{2}\right)^2 dx = 0 and since the integrand is non negative, this holds iff f(x) = \dfrac{x}{2}
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    (Original post by Farhan.Hanif93)
    ...
    Yup, it is a nice little question. I better go back to sleep.
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    A rather amusingly unusual approach works for this one.


    [*] Compute

    \displaystyle \int \sqrt{\frac{e^x - 1}{e^x + 1}}\ dx, \ \ \ \ \ x > 0
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    [Balkan 1997]
    Let m an n be integers greater than 1. Let S be a set with n elements, and let  A_1, A_2,...A_m be subsets of S. Assume that for any two elements x and y in S, there is a set  A_i such that either x is in  A_i and y is not in  A_i or x is not in  A_i and y is in  A_i . Prove that  n \leq2^m
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    (Original post by gff)
    A rather amusingly unusual approach works for this one.


    [*] Compute

    \displaystyle \int \sqrt{\frac{e^x - 1}{e^x + 1}}\ dx, \ \ \ \ \ x > 0
    Hypothesis, since i'm about to sleep.
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    Guessing this has something to do with tanh(1/2x). If you multiply num and denom by e^-1/2x you get root( sinh(x/2)/ cosh(x/2)). Then maybe make the sub t^2=tanh(x/2). Looks like some unfriendly algebra after that though...
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    (Original post by gff)
    A rather amusingly unusual approach works for this one.


    [*] Compute

    \displaystyle \int \sqrt{\frac{e^x - 1}{e^x + 1}}\ dx, \ \ \ \ \ x > 0
    I've just had a go at this one. I don't think I've found your "amusingly unusual approach" though, and I'm not entirely sure that my answer is right.

    I just used one basic substitution to change the form a bit, and then made some informed guesses about what certain things would integrate to.

    Spoiler:
    Show


    Let t = e^x, so \frac{dt}{dx} = e^x,dx = \frac{dt}{e^x}. Then

    \int \sqrt{\frac{e^x-1}{e^x+1}}dx = \int \frac{e^x-1}{\sqrt{e^{2x}-1}}dx = \int \frac{e^x}{\sqrt{e^{2x}-1}}dx - \int \frac{dx}{\sqrt{e^{2x}-1}}dx =

    \int \frac{dt}{\sqrt{t^2-1}} - \int \frac{dt}{t\sqrt{t^2-1}} = \log (2(\sqrt{t^2-1}+t))+\arctan \left( \frac{1}{\sqrt{t^2-1}}\right)+C =

    \log (2(\sqrt{e^{2x}-1}+e^x))+\arctan \left( \frac{1}{\sqrt{e^{2x}-1}}\right)+C

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    (Original post by nohomo)
    I've just had a go at this one. I don't think I've found your "amusingly unusual approach" though, and I'm not entirely sure that my answer is right.
    That's my "good" English - bear with me.
    If you remove the factor of 2 in the log the answer is correct.

    The amusing approach.
    Spoiler:
    Show

    Let \displaystyle u = \sqrt{\frac{e^x - 1}{e^x + 1}}, \ \ 0 < u < 1.


    Complete solution.
    Spoiler:
    Show

    Then, we have \displaystyle x = \ln\left(\frac{1 + u^2}{1 - u^2}\right) = \ln(1 + u^2) - \ln(1 - u^2) and \displaystyle dx = \left(\frac{2u}{1 + u^2} + \frac{2u}{1 - u^2} \right)\ du.

    Hence,

    \displaystyle \int \sqrt{\frac{e^x - 1}{e^x + 1}}\ dx = \int u\left(\frac{2u}{1 + u^2} + \frac{2u}{1 - u^2} \right)\ du


    \displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int \left[\left(2 - \frac{2}{1 + u^2}\right) - \left(2 - \frac{2}{1 - u^2} \right) \right]\ du


    \displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2(\text{arctanh}(u) - \text{arctan}(u)) + c



    (Original post by Blutooth)
    [Balkan 1997]
    Let m an n be integers greater than 1. Let S be a set with n elements, and let  A_1, A_2,...A_m be subsets of S. Assume that for any two elements x and y in S, there is a set  A_i such that either x is in  A_i and y is not in  A_i or x is not in  A_i and y is in  A_i . Prove that  n \leq2^m
    Solution.
    Spoiler:
    Show

    Denote the set containing all subsets as \mathbb{A} = \{ A_1, A_2, ..., A_m \} and define 1_{\mathbb{A}} : \mathbb{S} \to \{0, 1\} by the following.

    1_{A}(x) = \begin{cases}

  1 & \text{if } x \in A \\

  0 & \text{otherwise }

\end{cases}

    Then, let the element x \in \mathbb{S} be represented over \mathbb{A} as the set \rho(x) = \{\ 1_A(x) \ | \ A \in \mathbb{A}\ \}, and note that |\rho(x)| = |\mathbb{A}|.

    By hypothesis, the representations of two elements x,y \in \mathbb{S}, namely the sets \rho(x) and \rho(y), are not equal and ordered identically.
    For if it was true, then 1_A(x) = 1_A(y) for every subset A; e.g. there is no subset that contains x, but not y, and vice-versa.

    Now, we would like to know what is the maximum number of representations over a given \mathbb{A} which satisfy the above conditions.
    As already noted, each \rho(x) contains |\mathbb{A}| elements that are either  0 or 1; all possible permutations are then 2^{|\mathbb{A}|}, and let be contained in the set \Omega(\mathbb{A}).

    Finally, since \rho(x) \not= \rho(y), it follows that the map \rho : \mathbb{S} \to \Omega(\mathbb{A}) is injective.

    Hence, we conclude that |\mathbb{S}| \leq |\Omega(\mathbb{A})|.
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    (Original post by gff)
    That's my "good" English - bear with me.
    If you remove the factor of 2 in the log the answer is correct.

    The amusing approach.
    Spoiler:
    Show

    Let \displaystyle u = \sqrt{\frac{e^x - 1}{e^x + 1}}, \ \ 0 < u < 1.


    Complete solution.
    Spoiler:
    Show

    Then, we have \displaystyle x = \ln\left(\frac{1 + u^2}{1 - u^2}\right) = \ln(1 + u^2) - \ln(1 - u^2) and \displaystyle dx = \left(\frac{2u}{1 + u^2} + \frac{2u}{1 - u^2} \right)\ du.

    Hence,

    \displaystyle \int \sqrt{\frac{e^x - 1}{e^x + 1}}\ dx = \int u\left(\frac{2u}{1 + u^2} + \frac{2u}{1 - u^2} \right)\ du


    \displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int \left[\left(2 - \frac{2}{1 + u^2}\right) - \left(2 - \frac{2}{1 - u^2} \right) \right]\ du


    \displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2(\text{arctanh}(u) - \text{arctan}(u)) + c





    Solution.
    Spoiler:
    Show

    Denote the set containing all subsets as \mathbb{A} = \{ A_1, A_2, ..., A_m \} and define 1_{\mathbb{A}} : \mathbb{S} \to \{0, 1\} by the following.

    1_{A}(x) = \begin{cases}

  1 & \text{if } x \in A \\

  0 & \text{otherwise }

\end{cases}

    Then, let the element x \in \mathbb{S} be represented over \mathbb{A} as the set \rho(x) = \{\ 1_A(x) \ | \ A \in \mathbb{A}\ \}, and note that |\rho(x)| = |\mathbb{A}|.

    By hypothesis, the representations of two elements x,y \in \mathbb{S}, namely the sets \rho(x) and \rho(y), are not equal and ordered identically.
    For if it was true, then 1_A(x) = 1_A(y) for every subset A; e.g. there is no subset that contains x, but not y, and vice-versa.

    Now, we would like to know what is the maximum number of representations over a given \mathbb{A} which satisfy the above conditions.
    As already noted, each \rho(x) contains |\mathbb{A}| elements that are either  0 or 1; all possible permutations are then 2^{|\mathbb{A}|}, and let be contained in the set \Omega(\mathbb{A}).

    Finally, since \rho(x) \not= \rho(y), it follows that the map \rho : \mathbb{S} \to \Omega(\mathbb{A}) is injective.

    Hence, we conclude that |\mathbb{S}| \leq |\Omega(\mathbb{A})|.
    Very good, the above is a nice, rigorous soln. However, I think I would phrase my answer to the above question slightly differently- partly because I'm not too conversant with standard set theory notation, and also because there are others in this forum as ignorant as myself.

    Spoiler:
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    Let us associate with each element x in S a number of m binary digits-eg for a particular x, a(x)=(1011101010).

    How we construct a(x)...
    If the element x belongs to the set A_m set the mth digit from the left in a(x) to 1. If the element x does not belong to to set A_m set the mth digit from the left in a(x) to 0.

    Now clearly given m, the total possible number of different numbers a(x) is 2^m. This is because for each digits column (of which there are m) we can set the value to either 1 or 0==> 2 choices.

    Thus if we have n> 2^m at least 2 of the elements of S have the same number a(x) (pidgin-hole principle). So we have an x_1 and an x_2 where a(x_1)=a(x_2). What does this mean?

    If x_1 belongs to a subset a_m so does x_2. If x_1 does not belong to a subset x_1 neither does x_2. Thus the assumption that we can find a set where there is an x_1 but not an x_2 does not hold.
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    Stupid spammers! Extracted from spammer's thread.

    (Original post by oh_1993)
    I'm not very experienced with this but if an Abelian set is one where the order of applying operations between elements within a set doesn't matter, then surely you need to know what the operation is that is being applied to the set e.g. multiplication modulo 3, addition...

    So g = +/- 1 and prove that 1 * -1 = -1 * 1 ??

    Does this even make sense lol
    I don't see a reason why not to answer your question here, since the OP's request was answered.
    Spoiler:
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    You would probably call it an Abelian group, rather than simply a set, since it has to have an operation associated with it.
    However, the good thing about abstract algebra is that you do not need to know all details - you want to be as general as possible given a few assumptions.

    Another perhaps confusing thing is that 1 is used not as a number, but as a symbol representing the multiplicative identity.
    However, it is the case that the number is the multiplicative identity of familiar groups.

    Also, the notation g^2 is symbolic. It is a short hand for g \cdot g where the dot is the operation of the group, not necessarily ordinary multiplication.


    *****
    Here is a good question related to this thread.
    It comes from OCR MEI's FP3 Additional Further Maths book.

    Spoiler:
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    [*] If A and B are sets, the symmetric difference A \bigtriangleup B is defined by

    A \bigtriangleup B = (A \cap B') \cup (A' \cap B)

    as shown below in the Venn diagram of the set \xi containing them.

    Click image for larger version. 

Name:	200px-Venn0110.svg.png 
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ID:	133946


    (i) Prove that the set of all subsets of the set \xi forms an Abelian group under the operation \bigtriangleup.


    (ii) Given that \xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}, A = \{1, 2, 3\} and B = \{2, 4, 6, 8\},

    solve the equation A \bigtriangleup X = B


    (iii) Prove that if a^2 = e (the identity) for every element a of a group, then the group is Abelian. Given an example of such a group.
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    I like questions which can teach you new ideas - I hope you do as well.


    [*] Let \mathbb{A} be a non-empty set and let f : \mathcal{P}(\mathbb{A}) \to \mathcal{P}(\mathbb{A}) be an increasing function on the set of all subsets of \mathbb{A}, meaning that

    \ \ \  f(X) \subset f(Y), \ \ \ \ \text{if} \ X \subset Y.


    Prove that there exists T, a subset of \mathbb{A}, such that f(T) = T.
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    (Original post by gff)
    I like questions which can teach you new ideas - I hope you do as well.


    [*] Let \mathbb{A} be a non-empty set and let f : \mathcal{P}(\mathbb{A}) \to \mathcal{P}(\mathbb{A}) be an increasing function on the set of all subsets of \mathbb{A}, meaning that

    \ \ \  f(X) \subset f(Y), \ \ \ \ \text{if} \ X \subset Y.


    Prove that there exists T, a subset of \mathbb{A}, such that f(T) = T.

    I use  x \subset y to mean x is a subset of y, and can even be equal to y.

    Spoiler:
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    Define A_{n}=f(A_{n-1}) Where A_0=\mathbb{A}
    Statement to be proved  f(A_{n}) \subset A_{n}
    Assuming  f(A_{n-1}) \subset A_{n-1}


    Proof
     f(A_{n-1}) \subset A_{n-1} \Rightarrow A_n=f(A_{n-1}) \subset A_{n-1}  \Rightarrow  A_{n} \subset A_{n-1}
    Taking f to both sides this  \Rightarrow f(A_{n}) \subset f(A_{n-1})=A_{n}
    This  \Rightarrow f(A_{n}) \subset A_{n} .

    Since statement  f(A_{n}) \subset A_{n} is true for n=0 because A_0=\mathbb{A} and f(\mathbb{A})maps onto itself, the statement is true for for all n\geq 1




    Note if  f(A_n)=A_n we have succeeded in tackling the question. If not we can find a set |A_{n+1}| where |A_{n+1}|<|A_n|. Proof

    f(A_n)\subset A_n and also   f(A_n)\neq A_n \Rightarrow |f(A_n)|<|A_n|   \Rightarrow |A_{n+1}|<|A_{n}|.

    Thus for each n either we succeed in finding a set A_n that satisfies f(T)=T or we generate a set A_{n+1} which has fewer elements than A_n.We keep on repeating this procedure(eg is A_0 the set, If yes stop, if no is A_1 the set with the quality f(T)=T... ) until we have found a set satisfying f(T)=T or until we have a set of one element. This set must be equal to its image, else we have a sets whose image is the empty-set. But this contradicts the original assumptions.


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    Here is an interesting question to solve. An even number of people are sitting down at a table for breakfast. When they come back in the evening for dinner, they are not necessarily seated in the same order. Whatever the new seating arrangement prove that there are at least 2 people who are sitting with the same number of people in between them at both dinner and breakfast.
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    You're assuming the set is finite.
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    (Original post by gff)
    That's my "good" English - bear with me.
    If you remove the factor of 2 in the log the answer is correct.

    The amusing approach.
    Spoiler:
    Show

    Let \displaystyle u = \sqrt{\frac{e^x - 1}{e^x + 1}}, \ \ 0 < u < 1.


    Complete solution.
    Spoiler:
    Show

    Then, we have \displaystyle x = \ln\left(\frac{1 + u^2}{1 - u^2}\right) = \ln(1 + u^2) - \ln(1 - u^2) and \displaystyle dx = \left(\frac{2u}{1 + u^2} + \frac{2u}{1 - u^2} \right)\ du.

    Hence,

    \displaystyle \int \sqrt{\frac{e^x - 1}{e^x + 1}}\ dx = \int u\left(\frac{2u}{1 + u^2} + \frac{2u}{1 - u^2} \right)\ du


    \displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \int \left[\left(2 - \frac{2}{1 + u^2}\right) - \left(2 - \frac{2}{1 - u^2} \right) \right]\ du


    \displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2(\text{arctanh}(u) - \text{arctan}(u)) + c





    Solution.
    Spoiler:
    Show

    Denote the set containing all subsets as \mathbb{A} = \{ A_1, A_2, ..., A_m \} and define 1_{\mathbb{A}} : \mathbb{S} \to \{0, 1\} by the following.

    1_{A}(x) = \begin{cases}

  1 & \text{if } x \in A \\

  0 & \text{otherwise }

\end{cases}

    Then, let the element x \in \mathbb{S} be represented over \mathbb{A} as the set \rho(x) = \{\ 1_A(x) \ | \ A \in \mathbb{A}\ \}, and note that |\rho(x)| = |\mathbb{A}|.

    By hypothesis, the representations of two elements x,y \in \mathbb{S}, namely the sets \rho(x) and \rho(y), are not equal and ordered identically.
    For if it was true, then 1_A(x) = 1_A(y) for every subset A; e.g. there is no subset that contains x, but not y, and vice-versa.

    Now, we would like to know what is the maximum number of representations over a given \mathbb{A} which satisfy the above conditions.
    As already noted, each \rho(x) contains |\mathbb{A}| elements that are either  0 or 1; all possible permutations are then 2^{|\mathbb{A}|}, and let be contained in the set \Omega(\mathbb{A}).

    Finally, since \rho(x) \not= \rho(y), it follows that the map \rho : \mathbb{S} \to \Omega(\mathbb{A}) is injective.

    Hence, we conclude that |\mathbb{S}| \leq |\Omega(\mathbb{A})|.
    I think I just got lucky with that integration question you posted. Just thought you know what the heck let's try that substitution - didn't realise it would actually work.

    Sorry I haven't been posting in a while - been a bit ill .
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    (Original post by SimonM)
    You're assuming the set is finite.
    Good point. I suppose the result doesn't hold for infinite sets.
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    (Original post by Blutooth)
    Good point. I suppose the result doesn't hold for infinite sets.
    PRSOM for the post.

    Can't resist sharing the nice solution.
    Spoiler:
    Show

    Consider the family of sets \mathcal{F} = \{\ K \in \mathcal{P}(\mathbb{A}) \ | \ f(K) \subset K \ \}.

    We observe that \mathbb{A} \in \mathcal{F} and conclude that the family \mathcal{F} is non-empty.
    Hence, let T be the intersection of all sets in \mathcal{F}, and our aim is to show that f(T) = T.

    If K \in \mathcal{F}, then f(T) \subset f(K) \subset K, and by considering the intersection over all K, we deduce that f(T) \subset T and that T \in \mathcal{F}.
    Next, because f is increasing by hypothesis, it follows that f(f(T)) \subset f(T), and hence f(T) \in \mathcal{F}.

    Finally, since T is included in every element in \mathcal{F}, we also conclude that T \subset f(T).

    The double inclusion proves that f(T) = T, as desired.
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    ^^ awesome soln. Exposed me to a newish idea in maths.
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    Check out this documentary about IMO candidates. To sum it up in 3 words: interesting, saddening and inspiring.

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    (Original post by Blutooth)
    Check out this documentary about IMO candidates. To sum it up in 3 words: interesting, saddening and inspiring.

    In my first term (perhaps second) I met Jos. Seemed socially acceptable (well, to the mathmos at least ...).

    Good documentary though - it introduced me to "upper" level mathematics, perhaps even had a role in me studying it currently.
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    (Original post by Oh I Really Don't Care)
    In my first term (perhaps second) I met Jos. Seemed socially acceptable (well, to the mathmos at least ...).

    Good documentary though - it introduced me to "upper" level mathematics, perhaps even had a role in me studying it currently.
    I like Jos's take on the world. He seems like quite an interesting and amusing character, regardless of the asperges.

    I was just wondering- quite a few of those talented mathmos had asperges. Have you noticed any people with asperges while studying for your maths degree?

    I have a friend who is very good at maths, and sometimes I wonder if he has an autistic spectrum disorder-though it might be that he's just a bit socially awkward (as some mathmos often are :P).

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