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    (Original post by jack.hadamard)
    Problem 257 */**

    How many subsets of \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} are there such that the sum of the smallest and largest element is 11?
    Solution # 257

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    2^8+2^6...2^2 +2^0= 1(4^5-1)/3= 341
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    (Original post by Blutooth)
    ...
    Do you want to put "Solution #" on top.
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    (Original post by jack.hadamard)
    Do you want to put "Solution #" on top.
    no problem
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    (Original post by theterminator)
    http://theproofistrivial.com/ - What? :P
    LOL :lol: Blatantly gonna start sending people that link hahahahahaha :lol:

    (Original post by und)
    x
    What did you call it "The Proof is Trivial!" btw?
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    (Original post by Jkn)
    What did you call it "The Proof is Trivial!" btw?
    The name of this thread comes from that link (it used to be in my signature)
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    (Original post by Lord of the Flies)
    Here is a truly sensuous result.

    Problem 192*

    f is a twice-differentiable function with continuous derivatives, and satisfies the following conditions over (a,b):

    (\text{i})\;f(x)>0\qquad (\text{ii})\; f''(x)+f(x)>0

    Additionally,

    (\text{iii})\; f(a)=f(b)=0

    Show that b-a>\pi
    Has anybody got a solution they'd like to post?
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    (Original post by bananarama2)
    Why does the first one seem so familiar? :pierre:
    I remember it, too. I think it was the first problem I solved which you guys (current 6th formers) were interested in.
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    (Original post by Lord of the Flies)
    The name of this thread comes from that link (it used to be in my signature)
    You've got a good taste in music BTW
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    (Original post by Felix Felicis)
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    Problem from ages ago :L

    Solution 168 (2)


    This was a slog but the Beta and Gamma functions are sexy :sexface:

    \displaystyle \int_{0}^{ \frac{\pi}{2}} \ln \sin x \ln \cos x dx

    Let t = \sin^{2} x :

    \displaystyle \int_{0}^{ \frac{\pi}{2}} \ln \sin x \ln \cos x dx = \frac{1}{8} \int_{0}^{1} \frac{ \ln t \ln ( 1 - t)}{\sqrt{t} \sqrt{1 - t}} dt

    Now, consider I = \displaystyle\int_{0}^{1} t^{ a - 1/2} (1-t)^{b - 1/2} dt = \text{B} (a + 1/2, b + 1/2) = \dfrac{ \Gamma(a + 1/2) \Gamma ( b + 1/2)}{ \Gamma ( a + b + 1) }

    Differentiating under the integral sign, we get:

    \displaystyle \frac{ \partial^{2} I}{ \partial a \partial b} = \int_{0}^{1} \ln t \ln (1-t) t^{a - 1/2} (1- t)^{b - 1/2} dt

    which is x8 of the desired integral for a, b = 0

    Furthermore, we have:

    \displaystyle\begin{aligned} \frac{ \partial^{2} I}{ \partial a \partial b} &= \frac{ \partial^{2}}{ \partial a \partial b} \text{B} ( a + 1/2, b + 1/2) \\ \\ &= \frac{ \partial^{2}}{ \partial a \partial b} \frac{ \Gamma (a + 1/2) \Gamma (b + 1/2) }{ \Gamma ( a + b + 1)} \bigg|_{a = 0, b = 0} & (*) \\ \\ & = \text{B} \left( 1/2 , 1/2 \right ) \cdot \left( \left[ \psi (1/2) - \psi (1) \right]^{2} - \psi ' (1) \right) \\ & = 4 \pi \ln^{2} 2 - \frac{ \pi^{3}}{6} \\ & \Rightarrow \int_{0}^{ \frac{ \pi}{2}} \ln \sin x \ln \cos x dx = \frac{ \pi}{2} \ln^{2} 2 - \frac{ \pi^{3}}{48} \end{aligned}

    stuff

    (*) & \displaystyle \frac{ \partial^{2} }{\partial a \partial b} \frac{ \Gamma (a + 1/2) \Gamma ( b + 1/2)}{\Gamma ( a + b+ 1)} =

    \text{B} ( a + 1/2 , b + 1/2 ) \cdot \left\{ \left( \psi ( a + 1/2) - \psi (a + b + 1) \right) \cdot \left( \psi ( b + 1/2 ) - \psi ( a + b + 1) \right) - \psi ' (a + b + 1) \right\}

    Using the result that \displaystyle \psi (z) = - \gamma - \frac{1}{z} + \sum_{r=1}^{\infty} \frac{z}{r(r+z)} we get that:

    \psi (1) = - \gamma where \gamma is the Euler-Mascheroni constant

    and that \psi (1/2) = - \gamma - 2 \ln 2

    Using the definition that \psi ' (z) = \displaystyle\sum_{r=0}^{\infty} \frac{1}{(z+r)^{2}}

    we get \psi ' (1) = \dfrac{\pi^{2}}{6} (sum reduces to Basel problem)
    Gorgeous.
    Shamelessly so.
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    (Original post by theterminator)
    Has anybody got a solution they'd like to post?
    There's a link to the solution in the OP
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    (Original post by Felix Felicis)
    There's a link to the solution in the OP
    I see. Thank you. I only just realised that there was a second post.. :P
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    Is there a function which takes prime numbers as values on all positive integers?

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    I am looking for a general comment.
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    (Original post by jack.hadamard)
    Is there a function which takes prime numbers as values on all positive integers?

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    I am looking for a general comment.
    Could you elaborate a bit more on the conditions? I can think of periodic functions where every integer input of x gives prime numbers but only a specific subset. However, I'm not sure if this is quite what you're asking.

    (E.g.  f(x) = 5+2cos( \pi x ) gives a prime output for every integer input, but only 2 specific ones.)
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    (Original post by DJMayes)
    However, I'm not sure if this is quite what you're asking.
    What you gave is a good example. I am looking for a function that assumes infinitely many primes as values.
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    (Original post by jack.hadamard)
    Is there a function which takes prime numbers as values on all positive integers?

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    I am looking for a general comment.
    Surely there is no known function which gives different primes for each integer. If I find it do I get a millions pounds (and a job which GCHQ)?
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    (Original post by bananarama2)
    Surely there is no known function which gives different primes for each integer. If I find it do I get a millions pounds (and a job which GCHQ)?
    I claim there exists a constant \mathfrak{m} such that \lfloor \mathfrak{m}^{3^n} \rfloor is a prime for all n \in \mathbb{N}.
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    (Original post by jack.hadamard)
    Is there a function which takes prime numbers as values on all positive integers?

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    I am looking for a general comment.
    5(n2n) + 1
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    (Original post by jack.hadamard)
    I claim there exists a constant \mathfrak{m} such that \lfloor \mathfrak{m}^{3^n} \rfloor is a prime for all n \in \mathbb{N}.

    (Original post by MAyman12)
    5(n2n) + 1
    Interesting, it appears I'm very much mistaken. Pfft all this maths nonsense
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    (Original post by jack.hadamard)
    Is there a function which takes prime numbers as values on all positive integers?

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    I am looking for a general comment.
    Yes,

    f(x)= x if x is prime, 0 otherwise

    will work.
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    (Original post by bananarama2)
    Interesting, it appears I'm very much mistaken. Pfft all this maths nonsense
    I just think Mills' constant is under-appreciated and decided to promote it. Actually, there are uncountably many such \mathfrak{m}.
 
 
 
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