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The Soc for People of 'GRDCT2008' Mk VI Watch

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    Attempting to claim insomniac victory? :awesome:
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    i was till you came along:p:


    but aren't i/we insane?:eek4: this time..is...just....SOMETHING...t o say the least:jiggy:
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    :teeth:
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    :yep:
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    Nothing better than early morning maths:
    Spoiler:
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    Theorem: For all  n \in \mathbb{N}

     \dfrac{1}{2} < \dfrac{\sigma (n) \varphi (n)}{n^2} \leq 1

    where \sigma (n) is the sum of the positive divisors of n, and  \varphi (n) is the number of positive integers less than n that are coprime to n.

    Proof Let  n = p_1^{u_1} p_2^{u_2} \cdots p_r^{u_r} be the canonical decomposition of n, then

     \sigma (n) = n\displaystyle\prod_{i=1}^r \dfrac{1-p_i^{-u_i-1}}{1-p_i^{-1}}

    and

     \varphi (n) = n\displaystyle\prod_{i=1}^r (1-p_i^{-1}) .

    Hence

     \dfrac{\sigma (n) \varphi (n)}{n^2} = \displaystyle\prod_{i=1}^r (1-p_i^{-u_i-1}) .

    The upper bound follows immediately. And for the lower bound, we see that

     \displaystyle\prod_{i=1}^r (1-p_i^{-u_i-1}) \geq \displaystyle\prod_{i=1}^r (1-p_i^{-2}) \geq \displaystyle\prod_{m=2}^n \left(1 - \dfrac{1}{m^2} \right)  = \dfrac{n+1}{2n} > \dfrac{1}{2}

    as required. :qed:
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    ghosh

    you make me cry with amazement:p:

    :adore:
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    (Original post by Malsy)
    you make me cry with amazement:p:
    Well, none of it was that spectacular, although it took me ages to spot the clever bit, when we decided that

     \displaystyle\prod_{i=1}^r \left(1- \dfrac{1}{p_i^2}\right) \geq \displaystyle\prod_{m=2}^n \left(1 - \dfrac{1}{m^2}\right) = \dfrac{n+1}{2n}

    :yep:
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    ooo nice and simple on the right hand side of the equals sign

    :p:

    :redface:

    morning:w00t:
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    Ooo:
    Spoiler:
    Show
    Theorem: For all n \in \mathbb{N}

     \displaystyle\prod_{m|n} m = n^{\frac{1}{2} d(n)}

    where d(n) is the number of divisors of n.

    Proof Again, let  p_1^{u_1} p_2^{u_2} \cdots p_r^{u_r} be the canonical decomposition of n. Then we find that our product can be rewritten as

     \displaystyle\prod_{m|n} m = \displaystyle\prod_{x_r =0}^{u_r} \cdots \displaystyle\prod_{x_2 = 0}^{u_2} \displaystyle\prod_{x_1=0}^{u_1} \left(p_1^{x_1} p_2^{x_2} \cdots p_r^{x_r}\right) = \displaystyle\prod_{x_r =0}^{u_r} \cdots \displaystyle\prod_{x_2 = 0}^{u_2} (p_1^{\frac{u_1}{2}} p_2^{x_2} p_3^{x_3} \cdots p_r^{x_r})^{u_1+1}

     \iff \displaystyle\prod_{m|n} m = \displaystyle\prod_{x_r =0}^{u_r} \cdots \displaystyle\prod_{x_3 = 0}^{u_3} \left(p_1^{\frac{u_1}{2}} p_2^{\frac{u_2}{2}} p_3^{x_3} \cdots p_r^{x_r} \right)^{(u_1+1)(u_2+1)} = \cdots = \left(p_1^{\frac{u_1}{2}} p_2^{\frac{u_2}{2}} \cdots p_r^{\frac{u_r}{2}}\right)^{k}

    where  k = (u_1+1)(u_2+1)\cdots (u_r+1) . But this is d(n), and the result follows. :qed:


    Sidenote We have an interesting abstract way to lay out all the divisors of n: consider an r-dimensional lattice/matrix, which is of size (u_1+1) \times (u_2+1) \times \cdots \times (u_r+1) , and the entry at  (x_1, x_2, \cdots x_r) , with  1 \leq x_i \leq u_i +1 represents the divisor of n, namely,  p_1^{x_1-1} p_2^{x_2-1} \cdots p_r^{x_r-1} . Then this lattice/matrix contains every divisor of n once and only once.
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    Sometimes, I wish I had taken maths to A level just so I could humour Ghosh's little mathmo moments :o:
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    (Original post by Melting Sugar.)
    Sometimes, I wish I had taken maths to A level just so I could humour Ghosh's little mathmo moments :o:
    Umm, you'll find that A-level maths won't help you understand anything I've written above, not even the notation, really, however, any part you don't understand, I could probably explain to you.
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    (Original post by Melting Sugar.)
    Sometimes, I wish I had taken maths to A level just so I could humour Ghosh's little mathmo moments :o:
    lol are you joking mals, Ive got less of a clue than you
    all I recognise are the greek symbols and then.... its all greek to me :awesome:
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    University Challenege
    ftw
    age difference compared to last times is obvious
    lolled at the 30, 000 seconds
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    (Original post by clad in armour)
    University Challenege
    ftw
    age difference compared to last times is obvious
    lolled at the 30, 000 seconds
    Same. Jeremy Paxman's face: :lolwut:
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    (Original post by natty_d)
    Same. Jeremy Paxman's face: :lolwut:
    yay :woo: I got the superconductor question right and she said semiconductor
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    (Original post by clad in armour)
    yay :woo: I got the superconductor question right and she said semiconductor
    What you upto today?
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    :hello:
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    (Original post by natty_d)
    What you upto today?
    not much, read the Mail to some service users at the care home
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    (Original post by GHOSH-5)
    Umm, you'll find that A-level maths won't help you understand anything I've written above, not even the notation, really, however, any part you don't understand, I could probably explain to you.
    are you joking???
    have you even slept?!
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    me and ghosh were up so late as you can tell by the post times:redface: :ninja: :p:
 
 
 
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