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    (Original post by Flauta)

    Solution 425

    Consider a trapezium with base 1, and two parallel sides of lengths 14 and 21. Its area is given by A= \frac{1}{2}(14+21) = 17.5
    How did you solve it so quickly??! :eek:
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    (Original post by Lord of the Flies)
    How did you solve it so quickly??! :eek:
    My little secret.
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    (Original post by Lord of the Flies)
    How did you solve it so quickly??! :eek:

    I would've used the Hardy-Littlewood inequality to solve it. I'm surprised at his geometric approach though.
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    (Original post by Zakee)
    I would've used the Hardy-Littlewood inequality to solve it. I'm surprised at his geometric approach though.
    Why surprised?

    There can be more than one proof, why don't you post yours?
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    (Original post by Lord of the Flies)
    Problem 425*/**/***

    Evaluate \displaystyle\int_2^3 7x\,dx
    Let B be a topological algebra with continuous trace  \phi , then, it's clear that:

     \phi(\omega_f^p) = p \displaystyle \sum_{j=1}^{nCp} \sum_{\pi \in s(p,j)} sgn(\pi) \phi (W_{ij} W_{\pi}) dz^j .

    (Consider wedge products and that r is a cyclic action over the symmetry group mapped out by  \pi .)

    Substitute p=7, and evaluate. Voila.
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    Problem426***

    Evaluate \displaystyle\int^3_0 145020202781202010213347791x^2\ dx

    and deduce what was going through my mind when I thought of this question.
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    (Original post by MW24595)
    Let B be a topological algebra with continuous trace  \phi , then, it's clear that:

     \phi(\omega_f^p) = p \displaystyle \sum_{j=1}^{nCp} \sum_{\pi \in s(p,j)} sgn(\pi) \phi (W_{ij} W_{\pi}) dz^j .

    (Consider wedge products and that r is a cyclic action over the symmetry group mapped out by  \pi .)

    Substitute p=7, and evaluate. Voila.
    Did you just string together a load of random mathy things, or is that actually a genuine formula?
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    (Original post by james22)
    Did you just string together a load of random mathy things, or is that actually a genuine formula?
    I was just typing out a proof. Give me a moment, my dear friend!
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    (Original post by james22)
    Problem426***

    Evaluate \displaystyle\int^3_0 145020202781202010213347791x^2\ dx

    and deduce what was going through my mind when I thought of this question.
    Damn, that question's too hard for me but I'm guessing it's the number of presents you're hoping for Christmas?
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    (Original post by Zakee)
    I'm currently in year 13, and I think it may be University level, I'm not sure. I heard from someone that it comes up in FP3 (A-level), but I haven't studied FP3 yet. I'm just extremely interested in Pure Mathematics and have a knack for problem-solving.
    Almost finished FP3- it definitely does not come up.
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    (Original post by Flauta)
    Damn, that question's too hard for me but I'm guessing it's the number of presents you're hoping for Christmas?
    Nope.

    The number of presents I am hoping for it actually A(112,63).
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    (Original post by nahomyemane778)
    Almost finished FP3- it definitely does not come up.
    Are you on the mickey mouse exam board or something?
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    (Original post by james22)
    Nope.

    The number of presents I am hoping for it actually A(112,63).
    Last year I got 37 so I'm hoping for 38 this time around
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    (Original post by Lord of the Flies)
    Problem 425*/**/***

    Evaluate \displaystyle\int_2^3 7x\,dx
    The solutions posted so far are good, but I will use a different approach.

    Firstly, it's obvious by the finiteness of \displaystyle\lim_{n \to \infty}\inf(p_{n+m}-p_n ) (Zhang, 2013) that the integral is well defined.

    It is left to calculate its value. Consider an algebra A over an 4k-dimensional Hyperkahler manifold as k \to \infty. Then it is possible to construct a Banach space X such that every operator on the space is close to an operator of A. (Gowers, Maurey, 1994).

    Thence is follows that \displaystyle\int_2^3 7x\,dx = 17.5 \Box
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    (Original post by nahomyemane778)
    Almost finished FP3- it definitely does not come up.
    Wait what?! You don't have the Bertrand-Chebyshev?! O.o What about the Hardy-Littlewood?
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    (Original post by Flauta)
    Last year I got 37 so I'm hoping for 38 this time around
    Glad to hear it Dudley
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    (Original post by Lord of the Flies)
    Are you on the mickey mouse exam board or something?

    Im on edexcel. Am I in a dream world? Why the hell would Bertrand's Postulate appear in FP3?
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    (Original post by MW24595)
    Let B be a topological algebra with continuous trace  \phi , then, it's clear that:

     \phi(\omega_f^p) = p \displaystyle \sum_{j=1}^{nCp} \sum_{\pi \in s(p,j)} sgn(\pi) \phi (W_{ij} W_{\pi}) dz^j .

    (Consider wedge products and that r is a cyclic action over the symmetry group mapped out by  \pi .)

    Substitute p=7, and evaluate. Voila.
    (Original post by james22)
    Did you just string together a load of random mathy things, or is that actually a genuine formula?
    I realized it wasn't all that obvious. So, I decided to elaborate a little. The proof is nothing too extraordinary, but it does make some delightfully intricate use of symmetry.

    Consider that:

     \phi(\omega_f^p) = \phi(\omega \land \omega \land .... \omega) 



\Rightarrow \phi(\omega_f^p) = \displaystyle \sum_{j=1}^{nCp} \sum_{\pi \in s(p,j)} sgn(\pi) \phi (W_{\pi(j_1)} W_{\pi(j_2)}....W_{\pi(j_p)}) dz^j

    Now, recall that r is the cyclic action on the permutation group Sp defined by
     r(j) = (j+1), 1 ≤ j ≤ p−1 and  r(p) = (1). Clearly  rSp = S_p , and  sgn(r^{-1} \pi) = (−1)^{p−1} sgn(\pi) . Since p is odd,  sgn(r^{−1} \pi) = sgn(\pi) . And so we have our trace. Moving  W_{j_i} to the first in every product, we are done.

    From here on, the proof is trivial.
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    (Original post by und)
    The solutions posted so far are good, but I will use a different approach.

    Firstly, it's obvious by the finiteness of \displaystyle\lim_{n \to \infty}\inf(p_{n+m}-p_n ) (Zhang, 2013) that the integral is well defined.

    It is left to calculate its value. Consider an algebra A over an 4k-dimensional Hyperkahler manifold as k \to \infty. Then it is possible to construct a Banach space X such that every operator on the space is close to an operator of A. (Gowers, Maurey, 1994).

    Thence is follows that \displaystyle\int_2^3 7x\,dx = 17.5 \Box
    I don't quite understand the statement "4k-dimensional Hyperkahler manifold", since all Hyperkahler manifolds have dimension 4k. It's like me saying "consider an associative group".
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    (Original post by james22)
    I don't quite understand the statement "4k-dimensional Hyperkahler manifold", since all Hyperkahler manifolds have dimension 4k. It's like me saying "consider an associative group".
    That's just to help increase clarity. Otherwise  k \to \infty could refer to absolutely anything.
 
 
 
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