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    (Original post by shamika)
    Given that in the second limit you're going to take the limit to infinity anyway, how about in the first limit you...

    (The limits have to be this way around, because otherwise it causes issues when you have irrational x)
    I mentioned that I took m->∞ when I was finding the first limit; is that not enough? Or why can't we just write m=n?
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    (Original post by und)
    I mentioned that I took m->∞ when I was finding the first limit; is that not enough? Or why can't we just write m=n?
    Ordinarily, you would care what m is doing in the first (inner) limit, because as you've pointed out, the result you've stated only holds for m \geq q . However, this isn't a problem overall, since when you look at the second limit...
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    Problem 44***

    Let P(n) be a polynomial with coefficients in \mathbb{Z}. Suppose that \deg(P)=p, where p is a prime number. Suppose also that P(n) is irreducible over \mathbb{Z}. Then there exists a prime number q such that q does not divide P(n) for any integer n.


    Problem 45*

    Let p be a prime number, p> 3. Given that the equation p^{k}+p^{l}+p^{m}=n^{2} has an integer solution, then p \equiv -1 \pmod 8.
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    (Original post by shamika)
    Ordinarily, you would care what m is doing in the first (inner) limit, because as you've pointed out, the result you've stated only holds for m \geq q . However, this isn't a problem overall, since when you look at the second limit...
    Can we make m arbitrarily large in the first place?
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    (Original post by und)
    Can we make m arbitrarily large in the first place?
    Yes. I'm trying to hint strongly as to why but clearly I'm being useless at hinting
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    (Original post by DJMayes)
    ...
    (Original post by bananarama2)
    ...
    (Original post by joostan)
    ...
    Yeah, as has been said, the problem with that method is the partial fraction decomposition, it's not an identity
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    (Original post by shamika)
    Yes. I'm trying to hint strongly as to why but clearly I'm being useless at hinting
    The only thing that I see in that limit is m->∞, but I know nothing about repeated limits so I don't know what the conditions are for using this fact in the first limit.
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    (Original post by und)
    The only thing that I see in that limit is m->∞, but I know nothing about repeated limits so I don't know what the conditions are for using this fact in the first limit.
    In this case, assume you can use that fact

    I think the thing that's confusing you is that you might be thinking that one limit is applied first and then the other. They're both being applied at the same time, that is, the following doesn't happen:

    1) fix an m
    2) calculate the inner limit
    3) then calculate the outer limit.

    Think of it as both occurring simultaneously. To be honest, I was envisaging the difficulty of trying to swap the limits around, but in the case x is rational, it's ok to do so (at least, the result is the same either way)

    However, I think you've sufficiently understood to be able to credit yourself with the answer. Let me know if you want me to write something up, but it'll essentially be what I've already tried to explain

    EDIT: this happened to be on a tripos paper - and Gowers also has a crack at explaining it (see the comments in: http://gowers.wordpress.com/2012/04/...s-ii/#comments
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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

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

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

     = \dfrac{1}{\sin(1)} \displaystyle\sum_{r=1}^{89} \tan(r) - \dfrac{1}{\sin(1)} \displaystyle\sum_{r=0}^{88} \tan(r)

     = \dfrac{\tan(89)}{\sin(1)} = \dfrac{\cot(1)}{\sin(1)} = \dfrac{\cos(1)}{\sin^2(1)}
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    (Original post by Noble.)
    ...
    Awesome
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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.
    You know the answer to a set theory question is wrong when it contains the word "planck" :pierre:
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    (Original post by ukdragon37)
    You know the answer to a set theory question is wrong when it contains the word "planck" :pierre:
    :teehee:

    tehee

    \displaystyle  \langle \psi | \hat{G}| \psi | \rangle = \int_{- \infty}^{\infty} \psi^* \hat{G} \psi dx
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    Solution 43:

    Firstly, we have to apply the limit as n becomes arbitrarily large. As the power increases, there are one of three possibilities:

     -1<cosx<1 , in which case it will approach zero.

     cosx = 1 , in which case it will remain at one.

     cosx = -1 , in which case it will oscillate between -1 and 1.

    Next, look at the different possibilities of m with regards to x. If x is rational, then sufficiently large m will contain the denominator of x and a multiple of 2, making x a multiple of  2\pi so that the overall thing will reach a limit of 1. However, if x is irrational then it is impossible for it to become an integer multiple of  \pi . Because of this, cosx will approach zero as n tends towards infinity as the magnitude of cosx will be less than one.
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    My solution to #42 is absolutely horrendous :lol: even the latex code hates it.
    (Original post by bananarama2)
    ..And by using Dalek's contraction algorithm the answer is:
    :nope::nope::nope:
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    (Original post by DJMayes)
    Solution 43:

    Firstly, we have to apply the limit as n becomes arbitrarily large. As the power increases, there are one of three possibilities:

     -1<cosx<1 , in which case it will approach zero.

     cosx = 1 , in which case it will remain at one.

     cosx = -1 , in which case it will oscillate between -1 and 1.

    Next, look at the different possibilities of m with regards to x. If x is rational, then sufficiently large m will contain the denominator of x and a multiple of 2, making x a multiple of  2\pi so that the overall thing will reach a limit of 1. However, if x is irrational then it is impossible for it to become an integer multiple of  \pi . Because of this, cosx will approach zero as n tends towards infinity as the magnitude of cosx will be less than one.
    You don't need the third case because you're looking at the limit of  \cos^2 x (and hence you can simplify the rest of the proof) But yes, that's essentially it
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    (Original post by shamika)
    EDIT: this happened to be on a tripos paper - and Gowers also has a crack at explaining it (see the comments in: http://gowers.wordpress.com/2012/04/...s-ii/#comments
    Thanks for this. I guess it's the sort of thing where introducing some rigour can clear up any confusion.
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    (Original post by Mladenov)
    Problem 45*

    Let p be a prime number, p\ge 3. Given that the equation p^{k}+p^{l}+p^{m}=n^{2} has an integer solution, then p \equiv -1 \pmod 8.
    k = l = m = 1,\; p = n = 3 satisfies that equation. Is there something missing from the question?
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    (Original post by Lord of the Flies)
    k = l = m = 1,\; p = n = 3 satisfies that equation. Is there something missing from the question?
    What's even more worrying is p isn't congruent to -1mod8. I just spent ages on this question!
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    (Original post by Lord of the Flies)
    That was my point.
    Yeah I know Anyway, let me type up where I got to. I think p must either be congruent to 3mod8 or 7mod8.
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    (Original post by ukdragon37)
    You know the answer to a set theory question is wrong when it contains the word "planck" :pierre:
    :rofl:

    This feels like a challenge to come up with something. Now if only I knew any physics...
 
 
 
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