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    (Original post by CherGalois)
    i understand that my mathematical ability may well look so astonishing to you that it requires an unnatural means to be as good as me, but i can assure you this is not so. Atleast i waited till morning to reply so you could have one last dream, one last hope...

    unfortunately, my mathematical abilities are intrinsically natural and have only been developed thus far due to my own natural abilities.

    This is the same for all oxford mathmos....maybe one day with all your hard work you may bathe in a glory similar to mine.


    (Original post by souktik)
    Umm, hey, does anyone have some nice problems on combinatorics? Or on elementary number theory? I feel like solving problems, but not scary integrals.
    What sort of level of number theory do you know?
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    (Original post by Lord of the Flies)
    This is merely an exercise if one knows a bit of complex analysis, and is a massive slog otherwise...<br />
    <br />
    Here is something for the new year!<br />
    <br />
    <b>Problem 420**</b><br />
    <br />
    If \displaystyle \sum_{k\geq 0}x_k converges, does \displaystyle \sum_{k\geq 0}x_k^{2014} converge? What about \displaystyle \sum_{k\geq 0}x_k^{2013}?<br />
    <br />
    <b>Problem 421***</b><br />
    <br />
    \displaystyle \int_{-\infty}^{\infty} \left[1+\left(\frac{1}{x+1}+\frac{2}{x  +2}+\cdots +\frac{2013}{x+2013}-x\right)^{2014}\right]^{-1}\,dx
    <br />
    <br />
    What method might one use without complex analysis that turns out to be a slog ?
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    (Original post by DJMayes)
    What sort of level of number theory do you know?
    Err, just normal math olympiad stuff - I'm still in school. Flt, Euler's phi, quadratic reciprocity, etc.

    Posted from TSR Mobile
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    Problem 422*

    Easy number theory one.

    Find all prime numbers p &lt; 1000 such that for some integer n, the expression n^2-pn is a prime power.
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    (Original post by IceKidd)
    What method might one use without complex analysis that turns out to be a slog ?
    Laplace transforms. My point was that the question is either very straightforward, very painful, or completely inaccessible, depending on who is looking at it - so I don't really see the point.
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    (Original post by Tarquin Digby)
    Problem 422*

    Easy number theory one.

    Find all prime numbers p &lt; 1000 such that for some integer n, the expression n^2-pn is a prime power.
    n^2 - pn = q^m , where p and q are primes.

     n^2 - pn = n(n - p) .

    So n = q^a and n - p = q^b, where a &gt; b.

    By substituting n = q^a into  n - p, we get that q^a - q^b = q^b(q^{a - b} - 1)= p.

    Since p and q are both primes q^{a - b} - 1 = 1 \Rightarrow q^{a - b} = 2

    \therefore p = 2.

    Also consider the case n - p = 1.

    q^a - 1 = p \Rightarrow (q - 1)(...) = p.

    Then either p = q - 1, which gives us the result  p = 2, as 2 and 3 are the only consecutive integers that are both primes.

    Or q - 1 = 1 \Rightarrow q = 2. Now 2^a - 1 = 2^{a - 1} + ... + 1 = p. The only possible values that the Mersenne primes less than 1000.
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    (Original post by 0x2a)
    n^2 - pn = q^m , where p and q are primes.

     n^2 - pn = n(n - p) .

    So n = q^a and n - p = q^b, where a &gt; b.

    By substituting n = q^a into  n - p, we get that q^a - q^b = q^b(q^{a - b} - 1)= p.

    Since p and q are both primes q^{a - b} - 1 = 1 \Rightarrow q^{a - b} = 2

    \therefore p = 2.
    No - There are many more primes actually.

    Consider p = 3 and n = 4.
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    (Original post by Tarquin Digby)
    No - There are many more primes actually.

    Consider p = 3 and n = 4.
    Woops did not consider the case n - p = 1. My bad
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    (Original post by Tarquin Digby)
    Problem 422*

    Easy number theory one.

    Find all prime numbers p &lt; 1000 such that for some integer n, the expression n^2-pn is a prime power.
    Is this correct?

    Solution 422

    Spoiler:
    Show


    Write  n^2-np = n(n-p) = q^m for some prime q. From this, we can deduce that either:

     n \equiv 0 (q) or  n - p \equiv 0 (q)

    Assume both are true. Then, we must have  p \equiv 0 (q) , so p=q. Equivalently, n=kq, from which we obtain:

     k(k-1)q^2 = q^m

     k(k-1) = q^{m-2}

    As q is prime, this is clearly only possible if k=2, q=2, m=3. Then, p=2.

    Next, assume that only one is true. Clearly we cannot have the RHS true without the LHS. Then, we must have n - p = 1, n=p+1, giving us:

     (p+1)^2-p(p+1) = q^m

     p^2+2p+1-p^2-p = q^m

     p = q^m - 1 = (q-1)(q^{m-1}+...)

    So, q=2, and all our other primes must be of the form 2^m - 1, i.e. all possible primes are:

    2, 3, 7, 31, 127

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    (Original post by Lord of the Flies)
    If you try proving it, you will find that it is more interesting than it initially looks.
    You're right, it isn't so simple if it isn't absolutely convergent. The Cauchy cirterian does not give it trivially as I thought it would. I think I'm onto something though.
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    (Original post by DJMayes)
    Is this correct?

    Solution 422

    Spoiler:
    Show


    Write  n^2-np = n(n-p) = q^m for some prime q. From this, we can deduce that either:

     n \equiv 0 (q) or  n - p \equiv 0 (q)

    Assume both are true. Then, we must have  p \equiv 0 (q) , so p=q. Equivalently, n=kq, from which we obtain:

     k(k-1)q^2 = q^m

     k(k-1) = q^{m-2}

    As q is prime, this is clearly only possible if k=2, q=2, m=3. Then, p=2.

    Next, assume that only one is true. Clearly we cannot have the RHS true without the LHS. Then, we must have n - p = 1, n=p+1, giving us:

     (p+1)^2-p(p+1) = q^m

     p^2+2p+1-p^2-p = q^m

     p = q^m - 1 = (q-1)(q^{m-1}+...)

    So, q=2, and all our other primes must be of the form 2^m - 1, i.e. all possible primes are:

    2, 3, 7, 31, 127

    Yes that's right.
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    (Original post by CherGalois)
    u=-x, and im not suggesting its an odd function
    There is a much cleaner way to explain this but a solution is a solution I guess.

    Problem 423***
    Prove that
    \displaystyle \lim_{x\to+\infty} \sum_{n=1}^{\infty} \frac{nx}{(n^2+x)^2} = \frac{1}{2}

    Extension:
    Then show that there exists a positive constant k for every x \in [1,\infty) such that:
    \displaystyle \lvert \sum_{n=1}^{\infty} \frac{nx}{(n^2+x)^2} - \frac{1}{2} \rvert \leq \frac{k}{x}
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    (Original post by Lord of the Flies)
    This is merely an exercise if one knows a bit of complex analysis, and is a massive slog otherwise...
    Was interested in the massive "slog" alternatives, or I wanted to attempt to plug the gap in the previous absence of *** as there had been none posted in a while. Apologies
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    (Original post by Llewellyn)
    There is a much cleaner way to explain this but a solution is a solution I guess.<br />
    <br />
    <b>Problem 423***</b><br />
    Prove that<br />
    \displaystyle \lim_{x\to+\infty} \sum_{n=1}^{\infty} \frac{nx}{(n^2+x)^2} = \frac{1}{2}<br />
    <br />
    Extension:<br />
    Then show that there exists a positive constant k for every x \in [1,\infty) such that:<br />
    \displaystyle \lvert \sum_{n=1}^{\infty} \frac{nx}{(n^2+x)^2} - \frac{1}{2} \rvert \leq \frac{k}{x}
    <br />
    <br />

    What would you say the cleaner way to explain it is?
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    (Original post by Tarquin Digby)
    I thought you're not allowed to integrate over a discontinuity? (denominator is 0 when x=0)
    Sorry I missed this reply.
    If you consider the limit on either side then the integral is still possible; although thinking critically like this is good. I can't remember if that is covered at A-Level so if you are an A-Level student don't worry about it too much.
    (For instance, think about the graph of sin(x).(x^(-1))
    (Original post by CherGalois)
    What would you say the cleaner way to explain it is?
    Well most coherent replies would qualify as a better explanation.

    For instance, what DJM did was nice to show it can be reduced down to sin(nx)/sinx. Alternatively you can split the limits and apply the substitution to just one integral. Then you can mention a reduction formula can be shown via basic trig identities. You don't have to latex it, you can briefly go over your method, but I think "u=-x" is insufficient to claim a 'solution'. If you are going to take the time to reply, it's probably best to give a meaningful reply.
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    My u=-x substitutiom clesrly led to sinnx/sinx djmayes just copied my solution.. which is fine as he pereservered with it which is always a nice quality to see in mathematicians.
    (Original post by Llewellyn)
    Sorry I missed this reply.
    If you consider the limit on either side then the integral is still possible; although thinking critically like this is good. I can't remember if that is covered at A-Level so if you are an A-Level student don't worry about it too much.
    (For instance, think about the graph of sin(x).(x^(-1))
    Well most coherent replies would qualify as a better explanation.

    For instance, what DJM did was nice to show it can be reduced down to sin(nx)/sinx. Alternatively you can split the limits and apply the substitution to just one integral. Then you can mention a reduction formula can be shown via basic trig identities. You don't have to latex it, you can briefly go over your method, but I think "u=-x" is insufficient to claim a 'solution'. If you are going to take the time to reply, it's probably best to give a meaningful reply.

    (Original post by Llewellyn)
    Sorry I missed this reply.
    If you consider the limit on either side then the integral is still possible; although thinking critically like this is good. I can't remember if that is covered at A-Level so if you are an A-Level student don't worry about it too much.
    (For instance, think about the graph of sin(x).(x^(-1))
    Well most coherent replies would qualify as a better explanation.

    For instance, what DJM did was nice to show it can be reduced down to sin(nx)/sinx. Alternatively you can split the limits and apply the substitution to just one integral. Then you can mention a reduction formula can be shown via basic trig identities. You don't have to latex it, you can briefly go over your method, but I think "u=-x" is insufficient to claim a 'solution'. If you are going to take the time to reply, it's probably best to give a meaningful reply.
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    (Original post by Tarquin Digby)
    It's not covered very well at A level at all. We're just told: If the denominator is 0 in the open interval then 'you can't do it'. It would be nice to know exactly why.
    Isnt it a bit obbious...if it is undefined at a point in the interval then there is an asymptote there and the graph never finitrly finishes as it infinitely tends to this point so the area is infinite. It can then be evaluated in scenarios where a limit at this value exists on both sides (if integratint across the discontinuity)
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    (Original post by Tarquin Digby)
    It's not covered very well at A level at all. We're just told: If the denominator is 0 in the open interval then 'you can't do it'. It would be nice to know exactly why.
    ..Well actually, you can. Over (countably) infinitely many discontinuities, if you so wish. Advanced integration theory uses the notion of a 'measure', which would make this less seemingly paradoxical; but it's still possible to show that (Riemann!) integration doesn't really care even without measures: http://math.stackexchange.com/questi...iemann-integra
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    (Original post by CherGalois)
    Isnt it a bit obbious...if it is undefined at a point in the interval then there is an asymptote there and the graph never finitrly finishes as it infinitely tends to this point so the area is infinite. It can then be evaluated in scenarios where a limit at this value exists on both sides (if integratint across the discontinuity)
    I'm not quite sure what you mean, but

    \displaystyle \int_{-1}^1 \frac{dx}{\sqrt{x}}

    has finite area, despite having an asymptote.

    Edit: Please be clearer, I don't see how the above is not a counterexample to your claim.
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    (Original post by henpen)
    I'm not quite sure what you mean, but

    \displaystyle \int_{-1}^1 \frac{dx}{\sqrt{x}}

    has finite area, despite having an asymptote.
    You have just provided an example to aid my explanation, thankyou
 
 
 
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