Official TSR Mathematical Society

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  1. ben-smith's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by Tobedotty)
    the silver casio one, I think its called the fx-991es. Its basically just an upgraded version of the fx83 and has a couple of handy buttons that can work out definite integrals and the gradient of a line at a particular point, since only does definite things its not banned but its probably the most sophisticated one you'll find that isnt banned.
    It also does stuff like complex numbers and matrices (I think) which is a godsend.
  2. Llewellyn's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by Tobedotty)
    the silver casio one, I think its called the fx-991es. Its basically just an upgraded version of the fx83 and has a couple of handy buttons that can work out definite integrals and the gradient of a line at a particular point, since only does definite things its not banned but its probably the most sophisticated one you'll find that isnt banned.
    Thanks for the input, I've looked it up and it looks really good, problem solved>>
  3. gff's Avatar
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    Re: Official TSR Mathematical Society
    It's very quiet here.


    [*] Let \phi(\alpha) be the coefficient of x^{2012} in the power series expansion of (1 + x)^{\alpha} about the point x = 0.


    Evaluate \displaystyle \int_{0}^{1} \phi(-y-1)\left(\frac{1}{y + 1} + \frac{1}{y + 2} + \frac{1}{y + 3} + \cdots + \frac{1}{y + 2012} \right)\ dy


    Hint:
    Spoiler:
    Show

    Due to Newton, we're fairly confident in writing \displaystyle (1 + x)^{\alpha} = \sum^{\infty}_{n=0} \binom{\alpha}{n} x^n, which later became the Taylor series of (1 + x)^{\alpha} near x = 0.
    Last edited by gff; 16-03-2012 at 00:26. Reason: Added a hint, since nobody is interested by far. :p
  4. Blutooth's Avatar
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    Re: Official TSR Mathematical Society
    What does (y or ¬ x) ^x imply? Surely it implies Y.
    Last edited by Blutooth; 26-03-2012 at 22:34.
  5. ben-smith's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by Blutooth)
    as (¬Y)^Y is a contradiction (X OR ¬Y) ^Y must = X^Y
    I'm by no means an expert but, surely, (¬Y)^Y is the empty set. Where is the contradiction?
  6. TheMagicMan's Avatar
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    Re: Official TSR Mathematical Society
    Try combining the first and third expressions. the combine the simplified first expression with the second
    Last edited by TheMagicMan; 26-03-2012 at 22:28.
  7. TheMagicMan's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by ben-smith)
    I'm by no means an expert but, surely, (¬Y)^Y is the empty set. Where is the contradiction?
    As they are propositions it is a contradiction...e.g. let Y be the proposition that I am a cat...I cannot be both a cat and not a cat simultaneously...

    Spoiler:
    Show
    ...barring some strange quantum effect
  8. Blutooth's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by ben-smith)
    I'm by no means an expert but, surely, (¬Y)^Y is the empty set. Where is the contradiction?
    Why do you think y^(¬Y) would be the empty set?


    Let us assume that y is the empty set, are you telling me that (¬Y) is also the empty set? Surely not. (¬Y) in this case would be something other than the empty set? not the empty set and the empty set is a contradiction (and not the empty set).
    Last edited by Blutooth; 26-03-2012 at 22:25.
  9. ben-smith's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by TheMagicMan)
    As they are propositions it is a contradiction...e.g. let Y be the proposition that I am a cat...I cannot be both a cat and not a cat simultaneously...

    Spoiler:
    Show
    ...barring some strange quantum effect
    errr, well don't view them as propositions
    I guess my point is that if it were a contradiction then the whole initial statement would be a contradiction in which case we aren't really doing anything. Anyway, it was a minor point (if one at all).
    No idea why someone negged you. Let me try and balance it out.
  10. TheMagicMan's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by ben-smith)
    errr, well don't view them as propositions
    I guess my point is that if it were a contradiction then the whole initial statement would be a contradiction in which case we aren't really doing anything. Anyway, it was a minor point (if one at all).
    No idea why someone negged you. Let me try and balance it out.
    If we look at it from a set point of view, aren't we just going to use the fact that taking the union of a set A with the empty set is equivalent to A
  11. multiplexing-gamer's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by ben-smith)
    errr, well don't view them as propositions
    I guess my point is that if it were a contradiction then the whole initial statement would be a contradiction in which case we aren't really doing anything. Anyway, it was a minor point (if one at all).
    No idea why someone negged you. Let me try and balance it out.
    Sorry, I negged him by accident, but I dont have any rep points so it wont affect your score!
  12. ben-smith's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by Blutooth)
    Why do you think y^(¬Y) would be the empty set?


    Let us assume that y is the empty set, are you telling me that (¬Y) is also the empty set? Surely not. (¬Y) in this case would be something other than the empty set? not the empty set and the empty set is a contradiction (and not the empty set).
    Well, my thinking is that ¬Y is the complement of Y and so they are disjoint and therefore, by definition, their Y^(¬Y)= ∅.
    Maybe we have different interpretations of what this means, I mean it in the way this does.
  13. ben-smith's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by TheMagicMan)
    If we look at it from a set point of view, aren't we just going to use the fact that taking the union of a set A with the empty set is equivalent to A
    That's what I was trying to say!
  14. Blutooth's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by ben-smith)
    Well, my thinking is that ¬Y is the complement of Y and so they are disjoint and therefore, by definition, their Y^(¬Y)= ∅.
    Maybe we have different interpretations of what this means, I mean it in the way this does.
    There are different ways of interpreting the problem, I was just teasing you. But seriously, if you can tell me what ∅^(¬∅) equals, I will be impressed as I haven't the foggiest. :cool:
    Last edited by Blutooth; 26-03-2012 at 23:11.
  15. gff's Avatar
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    Re: Official TSR Mathematical Society
    Here we go, a fun question.



    [*] The curves A, B, C and D are defined in the plane as follows.

    \displaystyle A = \left\{ (x,y) : x^2 - y^2 = \frac{x}{x^2 + y^2}\right\}

    \displaystyle B = \left\{ (x, y) : 2xy + \frac{y}{x^2 + y^2} = 3 \right\}

    \displaystyle C = \{ \ (x, y) : x^3 - 3xy^2 + 3y = 1 \}

    \displaystyle D = \{ \ (x,y) : 3x^2y - 3x - y^3 = 0 \}


    Prove that A \cap B = C \cap D.
  16. Oh I Really Don't Care's Avatar
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    Re: Official TSR Mathematical Society
    hhmmmm

    Nobody really knows what is happening as notation is far from standard.

    ¬ denotes a negation. ^c (for instance let A be a set, write A^c) denotes the complement of the set A. etc.

    Do you have a reference for your notation or an explanation?
  17. Unbounded's Avatar
    • TSR Demigod
    Re: Official TSR Mathematical Society
    (Original post by gff)
    Here we go, a fun question.



    [*] The curves A, B, C and D are defined in the plane as follows.

    \displaystyle A = \left\{ (x,y) : x^2 - y^2 = \frac{x}{x^2 + y^2}\right\}

    \displaystyle B = \left\{ (x, y) : 2xy + \frac{y}{x^2 + y^2} = 3 \right\}

    \displaystyle C = \{ \ (x, y) : x^3 - 3xy^2 + 3y = 1 \}

    \displaystyle D = \{ \ (x,y) : 3x^2y - 3x - y^3 = 0 \}


    Prove that A \cap B = C \cap D.
    [Assuming these are curves in the real plane], writing  z = x+iy \in \mathbb{C} gives new sets:

     \bar{A} = \{ z \in \mathbb{C} - \{ 0 \} : \mathrm{Re} (z^2) = \mathrm{Re} (1/z) \}

     \bar{B} = \{ z \in \mathbb{C} -\{ 0 \} : \mathrm{Im} (z^2) - \mathrm{Im} (1/z) = 3 \}

     \bar{C} = \{ z \in \mathbb{C} : \mathrm{Re} (z^3) - 3\mathrm{Re} (iz) = 1 \}

     \bar{D} = \{ z \in \mathbb{C} : \mathrm{Im} (z^3) - 3\mathrm{Im} (iz) = 0 \}

    This gives that  \bar{A} \cap \bar{B} and  \bar{C} \cap \bar{D} are comprised of the zeros of  z^3 - 3iz - 1 . Since we have a nice correspondence between  \mathbb{R}^2 and  \mathbb{C} , we have  A \cap B = C \cap D .
  18. gff's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by Unbounded)
    ...
    I was hoping it's gonna stay for a while... +rep
  19. TheMagicMan's Avatar
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    Re: Official TSR Mathematical Society
    (Original post by gff)
    I was hoping it's gonna stay for a while... +rep
    Here's a nice question:

    i)Take 4 quantities a_i, 1\leq i \leq 4.

    We play a game according to these rules: Player A adds a total of 1 to the quantities, apportioning it however he likes. Player B then reduces any two consecutive quantities, or the first and the last, to 0. They then alternate 'turns'. Assume Player B plays optimally. Player A wins if he makes any one of the quantities greater than a certain value x. Player B's goal is to stop Player A from winning. The game continues indefinitely. What is the maximum value of x such that Player A can win?

    ii) Using the same rules, what is the maximum value of x if there are 5 quantities?

    iii) And if there are n quantities?

    I have a very ugly and lengthy solution for iii) so bonus points for elegance on that one
    Last edited by TheMagicMan; 28-03-2012 at 14:22.
  20. Llewellyn's Avatar
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    • Astray anyway
    Re: Official TSR Mathematical Society
    (Original post by TheMagicMan)
    ~Quantity question^
    Before I potentially waste a lot of time doing it all wrong, is the answer to i)
    Spoiler:
    Show

    16/3 (i.e. this is the maximum value for 4 quantities)?
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