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    (Original post by bananarama2)
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    Yup, it is about the square brackets.
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    I think, I have done too much number theory .


    As we speak, let me suggest something in this direction.

    Problem 68**

    Evaluate \displaystyle \int_{0}^{1}\int_{0}^{1}...\int_  {0}^{1} [x_{1}+x_{2}+...+x_{n}]dx_{1}dx_{2}...dx_{n}.
    Of course, [x] is the largest integer not greater than x.
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    (Original post by dknt)
    Another physicsy maths one :awesome:

    Problem 65 **/***

    A cone of base radius R, height H, mass M and homogeneous mass density ϱ is orientated as shown below.

    Attachment 209111

    The angle the cone makes to the xy plane is α. The cone is then attached to an axis along its outer mantle in the direction of  \hat{\mathbf{u}}.

    Calculate the moment of inertia of the cone, with respect to this axis.
    I don't *know* how to do this. Is there anywhere you can point me to to read about it. (Inertia tensor? )
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    (Original post by bananarama2)
    I don't *know* how to do this. Is there anywhere you can point me to to read about it. (Inertia tensor? )
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    That's definitely the easiest way and if you can do it that way you should, however it's not necessary. Consider an arbitrary mass element dm at point r=(x,y,z). Decompose the vector r into its parallel and perpendicular components (relative to the axis). \mathbf{r}_{||}=(\mathbf{r} \cdot \hat{\mathbf{u}})\hat{\mathbf{u}  }. Then you can find the perpendicular component and then just use the integral definition for the moment of inertia. Strictly speaking this could be shorter than the tensor method, depending how you go about using the tensor.
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    Problem 69***

    For x  \neq 0

    evaluate

    \displaystyle \sum_{r=1}^{\infty} \dfrac{1}{r^2 + x^2}
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    Solution 69

    Consider the function \displaystyle f(z)=\frac{(-1)^{z}}{z^{2}+x^{2}}, which has poles at ix and -ix

    We have \displaystyle Res(\pi \csc (\pi z)f(z),ix)= \frac{(-1)^{ix}\pi \csc (\pi x i)}{2ix} and \displaystyle Res(\pi \csc( \pi z)f(z),-ix)= \frac{(-1)^{-ix} \pi \csc (\pi x i)}{2ix}

    Notice \displaystyle \csc (\pi x i)= -i csch (\pi x) and \displaystyle csch (\pi x)= \frac{2e^{\pi x}}{e^{2\pi x}-1}

    So,

    \begin{aligned}\displaystyle Res(\pi \csc( \pi z)f(z),ix)&= \frac{(-1)^{ix}\pi \csc (\pi x i)}{2ix} \\ &= -\frac{e^{-\pi x}\pi e^{\pi x}}{x(e^{2\pi x}-1)}\\ &= -\frac{\pi}{x}\frac{1}{e^{2\pi x}-1} \end{aligned}

    \begin{aligned}\displaystyle Res(\pi\csc( \pi z)f(z),-ix) = -\frac{\pi}{x}\left(1+\frac{1}{e^  {2\pi x}-1}\right) \end{aligned}


    Therefore \displaystyle -\sum_{a_{k}} Res(\pi \csc( \pi z)f(z),a_{k}) = \frac{\pi}{x}\left(1+\frac{2}{e^  {2\pi x}-1}\right) = \frac{\pi}{x} \coth (\pi x)

    Now, since \displaystyle \sum_{r= -\infty}^{\infty} (-1)^{r}f(r) = -\sum_{a_{k}} Res(\pi\csc (\pi z) f(z)), where a_{k} are the poles of f, we obtain \displaystyle \sum_{r=-\infty}^{\infty} \frac{1}{r^{2}+x^{2}}= \frac{\pi}{x}\coth( \pi x)
    \displaystyle \sum_{r= -\infty}^{-1} \frac{1}{r^{2}+x^{2}} + \sum_{r=1}^{\infty} \frac{1}{r^{2}+x^{2}} = \frac{1}{x^{2}}(\pi x\coth(\pi x)-1)

    Hence \displaystyle \sum_{r=1}^{\infty} \frac{1}{r^{2}+x^{2}} = \frac{1}{2x^{2}}(\pi x\coth(\pi x)-1)
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    (Original post by Mladenov)
    Solution 69

    Consider the function \displaystyle f(z)=\frac{(-1)^{z}}{z^{2}+x^{2}}, which has poles at ix and -ix

    We have \displaystyle Res(\pi \csc (\pi z)f(z),ix)= \frac{(-1)^{ix}\pi \csc (\pi x i)}{2ix} and \displaystyle Res(\pi \csc( \pi z)f(z),-ix)= \frac{(-1)^{-ix} \pi \csc (\pi x i)}{2ix}

    Notice \displaystyle \csc (\pi x i)= -i csch (\pi x) and \displaystyle csch (\pi x)= \frac{2e^{\pi x}}{e^{2\pi x}-1}

    So,

    \begin{aligned}\displaystyle Res(\pi \csc( \pi z)f(z),ix)&= \frac{(-1)^{ix}\pi \csc (\pi x i)}{2ix} \\ &= -\frac{e^{-\pi x}\pi e^{\pi x}}{x(e^{2\pi x}-1)}\\ &= -\frac{\pi}{x}\frac{1}{e^{2\pi x}-1} \end{aligned}

    \begin{aligned}\displaystyle Res(\pi\csc( \pi z)f(z),-ix) = -\frac{\pi}{x}\left(1+\frac{1}{e^  {2\pi x}-1}\right) \end{aligned}


    Therefore \displaystyle -\sum_{a_{k}} Res(\pi \csc( \pi z)f(z),a_{k}) = \frac{\pi}{x}\left(1+\frac{2}{e^  {2\pi x}-1}\right) = \frac{\pi}{x} \coth (\pi x)

    Now, since \displaystyle \sum_{r= -\infty}^{\infty} (-1)^{r}f(r) = -\sum_{a_{k}} Res(\pi\csc (\pi z) f(z)), where a_{k} are the poles of f, we obtain \displaystyle \sum_{r=-\infty}^{\infty} \frac{1}{r^{2}+x^{2}}= \frac{\pi}{x}\coth( \pi x)
    \displaystyle \sum_{r= -\infty}^{-1} \frac{1}{r^{2}+x^{2}} + \sum_{r=1}^{\infty} \frac{1}{r^{2}+x^{2}} = \frac{1}{x^{2}}(\pi x\coth(\pi x)-1)

    Hence \displaystyle \sum_{r=1}^{\infty} \frac{1}{r^{2}+x^{2}} = \frac{1}{2x^{2}}(\pi x\coth(\pi x)-1)
    Don't you just love elementary solutions?
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    (Original post by ben-smith)
    Don't you just love elementary solutions?
    As Nikolai Nikolov, the leader of our mathematics team, says: Why simple, when it could be complicated?!
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    Methinks that I have digested quite well his advice.
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    (Original post by Mladenov)
    As Nikolai Nikolov, the leader of our mathematics team, says: Why simple, when it could be complicated?!
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    Methinks that I have digested quite well his advice.
    Out of curiosity, how old are you and are you planning on studying at university here? (Take to PM if you'd like)
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    This one is a bit easy compared to some of what's been posted but it's quite a nice question;

    Problem 70 */**

    Consider the set \{1, 2, 3,..., N\}. Two sets A and B are chosen independently and equally likely among all the subsets of \{1, 2, 3,..., N\}.

    Find P(A \subseteq B).

    I hope it's not already been done.
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    (Original post by Mladenov)
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    Which IMO team are you on then?
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    (Original post by shamika)
    Out of curiosity, how old are you and are you planning on studying at university here? (Take to PM if you'd like)
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    I am 19 years old, and I am in my last year of high school.
    Yes, I am planning to study at university either in UK or in France.


    (Original post by bananarama2)
    Which IMO team are you on then?
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    I have not participated in the IMO yet.
    In tenth grade, I finished 11th (out of 12) on our team selection test. Last year I was second reserve (finished 8th). I hope I will make it this year.
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    Only if there was no geometry..
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    (Original post by Mladenov)
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    I have not participated in the IMO yet.
    In tenth grade, I finished 11th (out of 12) on our team selection test. Last year I was second reserve (finished 8th). I hope I will make it this year.
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    Only if there was no geometry..
    That's literally amazing. Good luck!
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    (Original post by bananarama2)
    ...
    Thanks.


    Solution 70

    Note that there are 2^{N} subsets of \{1,2,..,N\}.
    Let card(B)=i,  i \in \{0,1,..,N\}. Then \displaystyle P(card(B)=i)= \frac{\dbinom{N}{i}}{2^{N}}. So, since there are 2^{i} subsets of B, we have \displaystyle P(A \subseteq B)= \sum_{i=0}^{N} \frac{\dbinom{N}{i}}{2^{N}}\frac  {1}{2^{N-i}}= \frac{1}{2^{2N}}\sum_{i=0}^{N} \dbinom{N}{i} 2^{i}=\frac{3^{N}}{2^{2N}} .
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    (Original post by Mladenov)
    **Complex analysis awesomeness**
    My way:
    \dfrac{sinx}{x}= \Pi (1-\dfrac{x^2}{(\pi r)^2})

\Rightarrow \dfrac{sin(i \pi x)}{i \pi x}= \Pi (1+\dfrac{x^2}{ r^2})

\Rightarrow ln(\dfrac{sin(i \pi x)}{i \pi x})= \displaystyle \Sigma ln(1+\dfrac{x^2}{r^2})
    Differentiating both sides wrt x:
    \displaystyle \Sigma \dfrac{2x}{r^2+x^2}= \dfrac{d}{dx}[ln(\dfrac{sin(i \pi x)}{i \pi x})]=\dfrac{d}{dx}[ln(\dfrac{sinh(\pi x)}{\pi x})]=1/x(\pi x coth(\pi x)-1)

\Rightarrow \displaystyle \sum_{r=1}^{\infty} \frac{1}{r^{2}+x^{2}} = \frac{1}{2x^{2}}(\pi x\coth(\pi x)-1)
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    (Original post by Mladenov)
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    I am 19 years old, and I am in my last year of high school.
    Yes, I am planning to study at university either in UK or in France.




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    I have not participated in the IMO yet.
    In tenth grade, I finished 11th (out of 12) on our team selection test. Last year I was second reserve (finished 8th). I hope I will make it this year.
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    Only if there was no geometry..


    That really is incredible. Make sure you have a look at STEP and MAT, assuming you apply to Oxford / Cambridge / Imperial / Warwick.

    Also, you'll be pleased to know that there is no Euclidean plane geometry of the kind that's useful for IMO at university!
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    (Original post by Mladenov)
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    Only if there was no geometry..
    (Original post by shamika)

    Also, you'll be pleased to know that there is no Euclidean plane geometry of the kind that's useful for IMO at university!
    This may be a silly question but does anybody like geometry?? Even Mladenov who appears to be a maths genius dislikes it
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    Problem 71**

    Find all x,y \in \mathbb{C} such that x^y = y^x.

    EDIT: The question is asking for a way to generate solutions of that kind. To be precise, define the pair of parametric functions x,y : S \to \mathbb{C} such that x(a)^{y(a)} = y(a)^{x(a)} for all a of some set S (for example, \mathbb{R} or \mathbb{C}), and the functions must be able to generate all the possible solutions in \mathbb{C}. You shouldn't need to use any function definitions beyond A-level.
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    (Original post by joostan)
    This may be a silly question but does anybody like geometry?? Even Mladenov who appears to be a maths genius dislikes it
    At least in the UK, the problem is that it's not taught properly. I remember at GCSE I had to randomly learn the circle theorems a day before the exam because no one had bother teaching them at school. Similarly with vectors; in retrospect my teacher had really confused the hell out of us, despite spending weeks trying to explain. The fact that a vector could be 'free' just made no sense to me until several months into my A-Levels. Needless to say, my mind was blown
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    (Original post by joostan)
    This may be a silly question but does anybody like geometry?? Even Mladenov who appears to be a maths genius dislikes it
    I don't Dunno if they like it or not but the Chinese have some really freaky Euclidean geometry in their syllabus - someone mus like it
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    (Original post by joostan)
    This may be a silly question but does anybody like geometry?? Even Mladenov who appears to be a maths genius dislikes it
    I quite like it The only real disadvantage is trying to get a compass that doesn't screw up your circle.
 
 
 
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