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    (Original post by bananarama2)
    Directly below?
    Well yes? What are you wanting us to account for, the rotation of the earth whilst it's being dropped? :lol:
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    (Original post by Jkn)
    Well yes?
    Well no
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    (Original post by bananarama2)
    Well no
    Well yes! You have phrased the problem as a physics problem... not a mathematics problem or even a Newtonian Mechanics problem. Unless the effect is measurable and consistent under the parameters you have set out, then we must assume it will land directly below

    ...wait so what you have in mind isn't the rotation of the earth... ?
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    (Original post by Jkn)
    Well yes! You have phrased the problem as a physics problem... not a mathematics problem or even a Newtonian Mechanics problem. Unless the effect is measurable and consistent under the parameters you have set out, then we must assume it will land directly below

    ...wait so what you have in mind isn't the rotation of the earth... ?
    You assume nothing. It most certainly doesn't land below and of course it's a Newtonian mechanics problem.

    Well that's for you do decide
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    (Original post by Jkn)
    Edit: Where's this problem from btw? It was rather neat
    Well, I saw it on this webpage. Otherwise, it is USAMO 1994.
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    (Original post by Lord of the Flies)
    Accccch why did I have to miss the integrals? Well done on 159 Mlad, that's a nice solution.
    (Original post by ben-smith)
    Very nice Mlad! I'm too slow again
    Thanks.

    (Original post by Jkn)
    And I don't see what's going on in 159. Where does the first line come from? And how does it link to the second and third? The differentiation thing doesn't seem to make sense (though I'm sure it does), could you clarify it for me My solution looks so different to yours, it's in you form of two symmetrical series expression involving all the square numbers (and no factorials!) :lol: (I checked it though and it does work!)
    The first line is just integration by parts. Then, we represent \sin^{n}x as a sum of sines or cosines, depending on the value of n \pmod 2.

    Problem 163**

    Let f \in C^{1}. Find \displaystyle \lim_{n\to \infty} n\left(\int_{0}^{1}f(x)dx - \frac{1}{n}\sum_{i=1}^{n} f(\frac{i}{n}) \right).

    Problem 164**

    Evaluate \displaystyle I = \int_{0}^{\infty} \frac{\arctan x}{x^{\alpha}}dx, where \alpha \in \left(1,2 \right).

    Problem 165***

    Evaluate \displaystyle \int_{0}^{2} \frac{x^{4}}{(x^{2}+1)(x(2-x)^{3})^{\frac{1}{4}}}dx.

    Problem 166***

    Evaluate \displaystyle \int_{0}^{\infty} \frac{x\ln x}{(1+x^{2})(1+x^{3})^{2}}dx.

    Problem 167*** (one that is interesting, and not trivial)

    Evaluate \displaystyle \sum_{n} \frac{1}{n^{2}}, over the set of all biquadrate free positive integers.
    If you are interested, find \displaystyle \sum_{n} \frac{(-1)^{\frac{n-1}{2}}}{n}, over the set of all odd cube free positive integers.
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    (Original post by jack.hadamard)
    Well, I saw it on this webpage. Otherwise, it is USAMO 1994.
    Awesome cheers! Question 4 looks fun! Btw, are there any UK university maths competitions? Or anything between Oxford and Cambridge colleges and stuff like that?
    (Original post by bananarama2)
    You assume nothing. It most certainly doesn't land below and of course it's a Newtonian mechanics problem.

    Well that's for you do decide
    :pierre:
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    (Original post by Jkn)
    :pierre:
    The best problems are the simplest
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    (Original post by Mladenov)
    Problem 167*** (one that is interesting, and not trivial)
    I like this. Below is a hint; a question that I had posted in the inappropriate place at the inappropriate time.

    Spoiler:
    Show

    The Riemann zeta function \zeta(s) is defined as

    \displaystyle \zeta(s)\ :=\ \sum_{n \geq 1} n^{-s}\ =\ \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots
    and has a finite value for all real numbers s with s > 1. It is given that \displaystyle \zeta(2) = \frac{\pi^2}{6}


    i) Let \displaystyle f(m, n) = \frac{1}{mn^3} + \frac{1}{2m^2n^2} + \frac{1}{m^3n} for positive integers m and n.

    Show that \displaystyle f(m, n) - f(m + n, n) - f(m, m + n) = \frac{1}{m^2n^2}.

    Hence, by summing over all m and n, show that

    \displaystyle \zeta(2)^2 = \left(\sum_{m,n \geq 1} - \sum_{m > n \geq 1} - \sum_{n > m \geq 1} \right) f(m, n) = \frac{5}{2}\zeta(4).


    ii) It is further given that \displaystyle \sum_{1 \leq n < k/2} \zeta(2n)\zeta(k - 2n) = \frac{k + 1}{2} \zeta(k) for all even k \geq 4.

    Show that \displaystyle \zeta(6) = \frac{\pi^6}{945} and express \zeta(8) in terms of \zeta(2), \zeta(4) and \zeta(6).


    iii)* Prove by induction that \zeta(k)\ =\ r \pi^k for all even positive integers k, where r is a rational number.

    EDIT: Nevermind, read it the right way.
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    (Original post by Jkn)
    Btw, are there any UK university maths competitions?
    I don't know of any maths competitions here that can be compared to Putnam.
    However, the IMC is an international competition for university students run by UCL.
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    (Original post by jack.hadamard)
    I don't know of any maths competitions here that can be compared to Putnam.
    However, the IMC is an international competition for university students run by UCL.
    Going to Bulgaria this year I see :lol:
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    (Original post by bananarama2)
    The best problems are the simplest
    Lets assume you drop it off the second platform of height "h" which is given to be 115m

    We take the period of earth's rotation "t" to be 23 hours, 56 minutes and 4 seconds and the radius of the earth to be "r".

    Assuming the immediate curvature of the each in negligible, we can assume that the tangential velocity is equal to the horizontal velocity.

    For the earth, V_e=k\frac{2\pi r}{t},

    For the ball, V_b=k\frac{2\pi (r+h)}{t}.

    Let v denote the speed of the ball with respect to the ground in a direction tangential to the curvature of the earth.

    \Rightarrow v=k\frac{2\pi h}{t} \approx 0.00839k m/s.

    The latitude of Paris is approximately 48.87 N. Hence k=\cos(48.87).

    Let the distance the ball lands (in a westerly direction, i.e. under the Eiffel Tower) be denoted by d.

    We have that d=vt and that 115=\frac{1}{2} g t^2

    \Rightarrow d=\cos(48.87) \times 0.00839 \sqrt{\frac{330}{9.81}} \approx 3.20 cm, to 3 significant figures. \square
    Spoiler:
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    :pierre:
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    (Original post by bananarama2)
    Problem 162 *

    If I drop a ball off the Eiffel tower facing east. Where does it land? (Neglect air resistance)
    What do you want us to do? If we assume nothing we have to take into account that we're in a rotating frame of reference and that certainly is not "*". Not for any A-Level mechanics course that I've seen.
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    (Original post by Jkn)
    Solution 162

    Lets assume you drop it off the second platform of height "h" which is given to be 115m

    We take the period of earth's rotation "t" to be 23 hours, 56 minutes and 4 seconds and the radius of the earth to be "r".

    Assuming the immediate curvature of the each in negligible, we can assume that the tangential velocity is equal to the horizontal velocity.

    For the earth, V_e=k\frac{2\pi r}{t},

    For the ball, V_b=k\frac{2\pi (r+h)}{t}.

    Let v denote the speed of the ball with respect to the ground in a direction tangential to the curvature of the earth.

    \Rightarrow v=k\frac{2\pi h}{t} \approx 0.00839k m/s.

    The latitude of Paris is approximately 48.87 N. Hence k=\cos(48.87).

    Let the distance the ball lands (in a westerly direction, i.e. under the Eiffel Tower) be denoted by d.

    We have that d=vt and that 115=\frac{1}{2} g t^2

    \Rightarrow d=\cos(48.87) \times 0.00839 \sqrt{\frac{330}{9.81}} \approx 3.20 cm, to 3 significant figures. \square
    Spoiler:
    Show
    :pierre:
    But the base of the Eiffel tower is much wider than the middle or at the top. If you dropped it vertically off the second platform, and it only moved 3.20 cm, how did it reach the bottom without hitting the side or the first platform? :pierre:
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    (Original post by SParm)
    What do you want us to do? If we assume nothing we have to take into account that we're in a rotating frame of reference and that certainly is not "*". Not for any A-Level mechanics course that I've seen.
    For this instance the rotating reference frame whilst playing a part isn't anything beyond a-level. There is a way of doing it that involves the Coriolis and centrifugal force, but there is a way which doesn't use it.
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    (Original post by ukdragon37)
    But the base of the Eiffel tower is much wider than the middle or at the top. If you dropped it vertically off the second platform, and it only moved 3.20 cm, how did it reach the bottom without hitting the side or the first platform? :pierre:
    Since when have the compscis become such realists?
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    (Original post by SParm)
    What do you want us to do? If we assume nothing we have to take into account that we're in a rotating frame of reference and that certainly is not "*". Not for any A-Level mechanics course that I've seen.
    To be honest I'd be happy with a "below", "in front" or "behind".
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    (Original post by bananarama2)
    For this instance the rotating reference frame whilst playing a part isn't anything beyond a-level. There is a way of doing it that involves the Coriolis and centrifugal force, but there is a way which doesn't use it.
    Ah I see, fair enough. The centrifugal force (and indeed any forces \mathcal{O}(\omega^{2}) are negligible) but Coriolis was what I had in mind.
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    (Original post by bananarama2)
    To be honest I'd be happy with a "below", "in front" or "behind".
    So.... was I right or not? :ninja:
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    (Original post by Jkn)
    So.... was I right or not? :ninja:
    Nope.
 
 
 
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