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    Solution 3

    The condition is equivalent to \displaystyle \int_{0}^{1} (1-x^n)^{\frac{1}{m}}dx=\int_{0}^{1  } (1-x^m)^{\frac{1}{n}}dx. After the substitutions x \mapsto x^m and x \mapsto x^n, we have to prove that \displaystyle \int_{0}^{1} mx^{m-1}(1-x^{mn})^{\frac{1}{m}}dx= \int_{0}^{1} nx^{n-1}(1-x^{mn})^{\frac{1}{n}}dx. Now let \displaystyle x \mapsto (1-x)^{\frac{1}{mn}}. Hence \displaystyle \int_{0}^{1} mx^{m-1}(1-x^{mn})^{\frac{1}{m}}dx= \int_{0}^{1} \frac{1}{n} (1-x)^{\frac{1}{n}-1} x^{\frac{1}{m}}dx, \displaystyle \int_{0}^{1} nx^{n-1}(1-x^{mn})^{\frac{1}{n}}dx= \int_{0}^{1}\frac{1}{m}(1-x)^{\frac{1}{n}}x^{\frac{1}{m}-1}dx. Finally, integrating by parts, for example, \displaystyle \int_{0}^{1}\frac{1}{m}(1-x)^{\frac{1}{n}}x^{\frac{1}{m}-1}dx, we see that the equality holds true.
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    (Original post by ukdragon37)
    As expected, the problems I'm interested in are more quaint than the usual Here's one that's strange but rather easy once you get it.

    Problem 6**/***

    Let \mathbb{N}_\omega = \mathbb{N} \cup \{\omega\} where we define for all n \in \mathbb{N}_\omega, \omega \geq n. For any set S \subseteq \mathbb{N}_\omega we write \sqcup S to denote the smallest member of \mathbb{N}_\omega such that \sqcup S \geq n for all n \in S. (Hence \sqcup\mathbb{N}_\omega = \omega.)

    Let f:\mathbb{N}_\omega \to \mathbb{N}_\omega be a function such that:

    • For all m, n, m \leq n \implies f(m) \leq f (n).
    • For all S = \{a_1, a_2, \dots\} \subseteq \mathbb{N}_\omega, we have f(\sqcup S) = \sqcup\{f(a_1), f(a_2), \dots\}.


    Show that the value F given by F = \sqcup\left\{f^n(1)|n \in \mathbb{N}\cup\{0\}\right\} satisfies f(F) = F and is the smallest member of \mathbb{N}_\omega to do so.
    Just to clarify... \mathbb{N}_\omega is the set of natural numbers up to and including \omega, and \sqcup S is the greatest element of any subset S of \mathbb{N}_\omega or in case of the empty set it is 1?

    I see why that is the case but the problem for me here will be clarity and rigour in how I present the proof.
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    (Original post by und)
    Just to clarify... \mathbb{N}_\omega is the set of natural numbers up to and including \omega, and \sqcup S is the greatest element of any subset S of \mathbb{N}_\omega or in case of the empty set it is 1?

    I see why that is the case but the problem for me here will be clarity and rigour in how I present the proof.
    Yes all you said is correct.

    EDIT: Actually just a slight clarification/correction: suppose S is infinite but does not include \omega, in this case \sqcup S is still clearly \omega so your definition that it is the "the greatest element of any subset S of \mathbb{N}_\omega" is not technically right.
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    (Original post by ukdragon37)
    Actually just a slight clarification/correction: suppose S is infinite but does not include \omega, in this case \sqcup S is still clearly \omega so your definition that it is the "the greatest element of any subset S of \mathbb{N}_\omega" is not technically right.
    But how can S be infinite if S\subseteq \mathbb{N}_{\omega}?
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    One simple number theory problem.

    Problem 7*

    Let \displaystyle p be a prime number. Find all triples of positive integers \displaystyle (a,b,p) such that \displaystyle 2^{a}+p^{b}=19^{a}.

    @ukdragon37 I am still unable to comprehend your definition of \mathbb{N}_\omega. Can we write \mathbb{N}_\omega = \{ \alpha | \alpha\in \{\mathbb{N} \cup \{\omega\}\}, \alpha \le \omega \}? And if this is the case, then is \mathbb{N}_\omega not finite?
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    (Original post by Lord of the Flies)
    But how can S be infinite if S\subseteq \mathbb{N}_{\omega}?
    For example let S be all the even numbers.

    (Original post by Mladenov)
    @ukdragon37 I am still unable to comprehend your definition of \mathbb{N}_\omega. Can we write \mathbb{N}_\omega = \{ \alpha | \alpha\in \{\mathbb{N} \cup \{\omega\}\}, \alpha \le \omega \}? And if this is the case, then is \mathbb{N}_\omega not finite?
    :confused: I'm not sure what you are trying to do. All am saying is \mathbb{N}_\omega is the same as the set of natural numbers except with the addition of an "infinity" \omega element that is defined to be geq all the naturals (and trivially geq itself). \mathbb{N}_\omega is definitely infinite.
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    (Original post by joostan)
    x
    Solution 5
    I can't even begin to comprehend ukD's question so here's joostan's :ninja:

    part i
    e^{i \theta} \cdot e^{i \phi} = e^{i (\theta + \phi)} = \cos (\theta + \phi) + i \sin (\theta + \phi)

    e^{i \theta} = cos \theta + i \sin \theta and e^{i \phi} = \cos \phi + i \sin \phi

    e^{i \theta} \cdot e^{i \phi} = (cos \theta + i \sin \theta ) \cdot (\cos \phi + i \sin \phi ) = \cos \theta \cos \phi - \sin \theta \sin \phi + i( \sin \theta \cos \phi + \cos \theta \sin \phi)

    Equating real and imaginary parts

    \cos (\theta + \phi ) = \cos \theta \cos \phi - \sin \theta \sin \phi and \sin (\theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi as required. Really joostan?


    S_1
    Consider the expansion of (1 + i)^{n} = \displaystyle\binom{n}{0} + \displaystyle\binom{n}{1} i - \displaystyle\binom{n}{2} - \displaystyle\binom{n}{3} i + ... + (i)^{n} \displaystyle\binom{n}{n}

    and (1-i)^{n} =  \displaystyle\binom{n}{0} - \displaystyle\binom{n}{1} i - \displaystyle\binom{n}{2} + \displaystyle\binom{n}{3} i + ... + (-i)^{n} \displaystyle\binom{n}{n}

    The sum in question is \displaystyle\sum_{r=0}^{m} (-1)^{r} \displaystyle\binom{n}{2r} = \frac{1}{2} \left[(1+i)^{n} + (1-i)^{n} \right]

    1 + i = \sqrt{2} \left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) and 1 - i = \sqrt{2} \left(\cos \frac{\pi}{4} - i \sin \frac{\pi}{4} \right)

    \Rightarrow \frac{1}{2} \left[(1+i)^{n} + (1-i)^{n} \right] = \frac{1}{2} \left( 2 (\sqrt{2})^{n} \cos \frac{n \pi}{4} \right) = 2^{n/2} \cos \frac{n \pi}{4} as required


    S_2
    Similarly, \displaystyle\sum_{r=0}^{m} (-1)^{r} \displaystyle\binom{n}{2r+1} = \frac{1}{2} i \left[ (1-i)^{n} - (1+i)^{n} \right]

    \frac{1}{2} i \left[ (1-i)^{n} - (1+i)^{n} \right] = \frac{1}{2} i \left(-2 (\sqrt{2})^{n} i \sin \frac{n \pi}{4} \right) = 2^{n/2} \sin \frac{n \pi}{4}
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    (Original post by Felix Felicis)
    Solution 5
    I can't even begin to comprehend ukD's question so here's joostan's :ninja:

    part i
    e^{i \theta} \cdot e^{i \phi} = e^{i (\theta + \phi)} = \cos (\theta + \phi) + i \sin (\theta + \phi)

    e^{i \theta} = cos \theta + i \sin \theta and e^{i \phi} = \cos \phi + i \sin \phi

    e^{i \theta} \cdot e^{i \phi} = (cos \theta + i \sin \theta ) \cdot (\cos \phi + i \sin \phi ) = \cos \theta \cos \phi - sin \theta sin \phi + i( \sin \theta \cos \phi + \cos \theta \sin \phi)

    Equating real and imaginary parts

    \therefore \cos (\theta + \phi ) = \cos \theta \cos \phi - \sin \theta \sin \phi and \sin (\theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi as required. Really joostan?


    S_1
    Consider the expansion of (1 + i)^{n} = \displaystyle\binom{n}{0} + \displaystyle\binom{n}{1} i - \displaystyle\binom{n}{2} - \displaystyle\binom{n}{3} i + ... + (i)^{n} \displaystyle\binom{n}{n}

    and (1-i)^{n} =  \displaystyle\binom{n}{0} - \displaystyle\binom{n}{1} i - \displaystyle\binom{n}{2} + \displaystyle\binom{n}{3} i + ... + (-i)^{n} \displaystyle\binom{n}{n}

    The sum in question is \displaystyle\sum_{r=0}^{m} (-1)^{r} \displaystyle\binom{n}{2r} = \frac{1}{2} \left[(1+i)^{n} + (1-i)^{n} \right]

    1 + i = \sqrt{2} \left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) and 1 - i = \sqrt{2} \left(\cos \frac{\pi}{4} - i \sin \frac{\pi}{4} \right)

    \Rightarrow \frac{1}{2} \left[(1+i)^{n} + (1-i)^{n} \right] = \frac{1}{2} \left( 2 (\sqrt{2})^{n} cos \frac{n \pi}{4} \right) = 2^{\frac{n}{2}} \cos \frac{n \pi}{4} as required


    S_2
    Similarly, \displaystyle\sum_{r=0}^{m} (-1)^{r} \displaystyle\binom{n}{2r+1} = \frac{1}{2} i \left[ (1-i)^{n} - (1+i)^{n} \right]

    \frac{1}{2} i \left[ (1-i)^{n} - (1+i)^{n} \right] = \frac{1}{2} i \left(-2 (\sqrt{2})^{n} i \sin \frac{n \pi}{4} \right) = 2^{\frac{n}{2}} \sin \frac{n \pi}{4}
    The first part was a warm up, to get you thinking about De Moivre's Theorem, some people need a hint Felix - Lol same about ukD's
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    (Original post by Felix Felicis)
    I can't even begin to comprehend ukD's question so here's joostan's :ninja:
    Objectively the difficulty of actually doing the question I think is lower than a lot of the interesting techniques that I know people here are capable of (and in fact it only relies on A-level knowledge and manipulation on sets). It's just the material seems alien since nobody at high school level, even STEP, emphasises set theory to any depth. :sigh: It's a trend that I think should change, especially if we want to be able to produce the next generation of theoretical computer scientists.
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    Partial Solution 6

    The first assertion is true since f(F) = \sqcup\left\{f^{n+1}(1)|n \in \mathbb{N}\cup\{0\}\right\}, and f(1)\ge 1
    I shall edit, if I manage to prove that F is the smallest element such that f(F)=F.
    By induction, we obtain f^{n}(1)\le f^{n+1}(1). Now suppose that there is F' such that:
    F' < F;
    f(F')=F'.
    Then there is m with F'=f^{m}(1)=f(f^{m}(1)). Therefore, for every n>m, we have f^{n}(1)\le f^{m}(1) - contradiction.
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    (Original post by joostan)
    The first part was a warm up, to get you thinking about De Moivre's Theorem, some people need a hint Felix - Lol same about ukD's
    Ha, fair enough

    (Original post by ukdragon37)
    Objectively the difficulty of actually doing the question I think is lower than a lot of the interesting techniques that I know people here are capable of (and in fact it only relies on A-level knowledge and manipulation on sets). It's just the material seems alien since nobody at high school level, even STEP, emphasises set theory to any depth. :sigh: It's a trend that I think should change, especially if we want to be able to produce the next generation of theoretical computer scientists.
    Sets does look quite interesting, definitely something to look at over the summer
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    (Original post by ukdragon37)
    Objectively the difficulty of actually doing the question I think is lower than a lot of the interesting techniques that I know people here are capable of (and in fact it only relies on A-level knowledge and manipulation on sets). It's just the material seems alien since nobody at high school level, even STEP, emphasises set theory to any depth. :sigh: It's a trend that I think should change, especially if we want to be able to produce the next generation of theoretical computer scientists.
    IMHO it looks like a horrendous mess of notation
    But still, I agree things like groups, sets and rings seem fascinating and applicable.
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    Problem 8*

    A harder version of problem 0. Prove that for any n, it is possible to fine n consecutive integers such that none of them are prime powers.
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    (Original post by joostan)
    IMHO it looks like a horrendous mess of notation
    But still, I agree things like groups, sets and rings seem fascinating and applicable.
    To be fair, I am trying to sneak in a few interesting concepts, which is why I had to expend notation defining them. But this kind of question is very much like what a nasty university examiner (or even interviewer ) would give you - define some new concepts and see what you can make of them.

    Spoiler:
    Show
    \sqcup S is the supremum or least upper bound of S. The conditions that f satisfies are exactly that of a Scott-continuous function. \omega is of course the lowest transfinite ordinal.
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    (Original post by ukdragon37)
    To be fair, I am trying to sneak in a few interesting concepts, which is why I had to expend notation defining them. But this kind of question is very much like what a nasty university examiner (or even interviewer ) would give you - define some new concepts and see what you can make of them.

    Spoiler:
    Show
    \sqcup S is the supremum or least upper bound of S. The conditions that f satisfies are exactly that of a Scott-continuous function. \omega is of course the lowest transfinite ordinal.
    I can see you being a totally nasty supervisor to the undergrads at Cornell
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    (Original post by Felix Felicis)
    I can see you being a totally nasty supervisor to the undergrads at Cornell
    They don't have supervisions there, only recitations to medium-sized groups where the grad student demonstrates example questions. I don't get to actually set the questions for undergrads. :sad:
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    (Original post by ukdragon37)
    They don't have supervisions there, only recitations to medium-sized groups where the grad student demonstrates example questions. I don't get to actually set the questions for undergrads. :sad:
    Aww You'll have your dictatiorial rise to power some day
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    (Original post by Felix Felicis)
    Aww You'll have your dictatiorial rise to power some day
    I look forward to it

    (Original post by Mladenov)
    Partial Solution 6

    The first assertion is true since f(F) = \sqcup\left\{f^{n+1}(1)|n \in \mathbb{N}\cup\{0\}\right\}, and f(1)\ge 1
    I shall edit, if I manage to prove that F is the smallest element such that f(F)=F.
    By induction, we obtain f^{n}(1)\le f^{n+1}(1). Now suppose that there is F' such that:
    F' < F;
    f(F')=F'.
    Then there is m with F'=f^{m}(1)=f(f^{m}(1)). Therefore, for every n>m, we have f^{n}(1)\le f^{m}(1) - contradiction.
    Nearly there, you want to make the last bit clearer. For example it is not a contradiction if f is a constant function. But I'm glad someone is able to look beyond the notation and make quite a bit of headway.
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    (Original post by ukdragon37)
    Nearly there, you want to make the last bit clearer. For example it is not a contradiction if f is a constant function. But I'm glad someone is able to look beyond the notation and make quite a bit of headway.
    Let me try again.

    Solution 6

    Firstly, \displaystyle f(F) = \sqcup\left\{f^{n+1}(1)|n \in \mathbb{N}\cup\{0\}\right\}, and f(1)\ge 1 lead to f(F)=F
    Secondly, by induction, we obtain \displaystyle f^{n}(1)\le f^{n+1}(1). Consequently, \displaystyle \left\{f^n(1)|n \in \mathbb{N}\cup\{0\}\right\} is a directed complete partial order, and its supremum is obviously F.
    If f is a constant, let f(n)= \alpha for some \alpha \in \mathbb{N}_\omega. Then obviously the smallest n \in \mathbb{N}_\omega such that \displaystyle f(n)=n is n=\alpha. Furthermore, F=\alpha.
    Now suppose that f is not a constant function. Suppose also that there is F' such that:
    F' < F;
    f(F')=F'.
    Then there is m with F'=f^{m}(1)=f(f^{m}(1)). Therefore, for every n>m, we have f^{n}(1)\le f^{m}(1), which implies that F is not the supremum of \displaystyle \left\{f^n(1)|n \in \mathbb{N}\cup\{0\}\right\}.
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    Solution 7:

    Spoiler:
    Show

     2^a + p^b = 19^a  \Rightarrow p^b = 19^a - 2^a

\Rightarrow p^b = (19-2)(19^{a-1}+ \cdots + 2^{a-1})

\Rightarrow p=17
    since p is prime.

    Now considering

     2^a + 17^b = (2+17)^a

    If a=1 we have the solution (1,1,17)

    It is easily verified that there is no solution if a=2

    If a \geq 3,
    (2+17)^a = 17(17k+2^{a-1})+2^a

\Rightarrow 17^b = 17(17k + 2^j)

    which clearly has no solutions.

    Hence the only solution is (1,1,17)
 
 
 
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