Funny Pictures. Safe For Work, and Safe For The Kids Too.

Discussion in 'Discussions' started by OmniaNigrum, Aug 30, 2012.

  1. Kazeto

    Kazeto Member

    You mean you are a necromancer?

    Cool, I want to see that.


    Also, does the one with the colour puzzle (or whatever) imply that Jesus can sparkle?
     
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  2. OmniaNigrum

    OmniaNigrum Member

    Oh, you mean the FrankenZombieVamp? Yeah. He can sparkle and makes a mean Cappuccino too. :D
     
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  3. mining

    mining Member

    Maths at large:

    x^2 = 1
    Hence, x=±1. Note that this does not restrict x to any one value.

    x±1=±1±1
    1=(±1±1)/(x±1)
    We now take limit as x tends towards 1.
    1=(±1±1)/(1±1)
    1±1=±1±1
    1=±1
    1 is thus equal to minus one.

    Alternatively:
    Let x^2 = 1.

    Notice this does not explicitly show that x=1.

    +-1 = x
    +-2 = x +-1
    +-2-2x=+-1-x
    2(+-1-x)=1(+-1-x)
    Take the limit as x-> 1.
    2=1

    First person to disprove this wins a picture that's sitting on my hard drive.
     
  4. Warlock

    Warlock Member

    [​IMG]
    do I win?

    ...... say, I just got reminded of the time when I left math at school to focus on science and eventually medicine.
     
  5. Kazeto

    Kazeto Member

    Technically speaking, when you operate on values that can be both positive or negative, you are supposed to operate on their absolute values.

    Meaning that it goes like this:
    1 = |x|
    2 = 1 + |x|

    And from there you can only go to something like this:
    2 - (2 * |x|) = 1 - |x|
    Which can be then changed into:
    2 (1 - |x|) = 1 - |x|
    And by subtracting (1 - |x|) from both sides, into this:
    1 - |x| = 0
    Which finally changes into this after adding |x| to it:
    |x| = 1

    The point is, while this is true:
    x = {1; -1}
    This is also true:
    |x| = 1
    Because both |1| and |-1| are equal to 1.

    Basic math, and it really made me feel nostalgic...


    Now I want that picture.

    Edit: And again Omni, you are stalking me. Awesome, I have my own stalker.
     
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  6. OmniaNigrum

    OmniaNigrum Member

    I will have to remind my bank that I never have the capacity for negative funds since negatives are really just another option of positives in mathematics. :D
     
  7. mining

    mining Member

    [​IMG]

    Another physics/maths thing:

    Letf(x) be Bob's position function on an axis, f.
    Let g(x) be Paul's position function on an axis, g.
    Then f(x) * g(x) is the area they trace out as they run.
    They both begin at 0m from the origin of the axis, at non zero velocity. (i.e. f'(x) and g'(x) are strictly increasing)
    Intuitively, the area they trace out is always increasing. Why is this false.
     
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  8. OmniaNigrum

    OmniaNigrum Member

    Because it is an axis? The area is the same from the moment they start so long as they do not change the axis.
     
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  9. Kazeto

    Kazeto Member

    Because neither f'(x) nor g'(x) are equal to 0 on the beginning, but there's nothing said about them not being negative. And if both of these were negative at the beginning, there would be a time when the area derivative would be negative.

    I am not going to create half a page of scribbles just to make a proof that it's possible to prove what I wrote easily using mathematical logic, but that is because it would be the same thing I wrote higher, only with more numbers, and I am simply tired now.

    And it was maths only here. No physics involved.
     
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  10. mining

    mining Member

    Nope, that's not the reason why - the reasoning is the exact same if their initial velocity is negative, zero, or positive.

    While there's no physics involved in the proof, the result implies something physically.
     
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  11. Nicholas

    Nicholas Technology Director Staff Member

    My best guess for the 'tricky' answer here is: because the Earth is round.
     
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  12. Kazeto

    Kazeto Member

    Then do notify.

    Otherwise it's a question that does belong to mathematical logic and I gave you an answer that was appropriate for such.

    But if you want to place it in the real world, then there is a limit on the size of whatever they are running on, and thus after some time the area will stop increasing.
     
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  13. mining

    mining Member

    Well, it's an abstract physical sense - the end resolution is that it indicates that there are no absolute reference points.

    The simple sense of the proof is that the derivative of f(x) * g(x) is f'(x) * g(x) + f(x) * g'(x) - and at x=0, g(0)=0 and f(0)=0. Hence, (f(x)*g(x))'(0)=0, despite the fact that intuitively the area must be increasing, as its two side lengths are.
     
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  14. Kazeto

    Kazeto Member

    That is great, but if, once again, you ventured into the area of mathematical logic, then it's not a rule that my solution has to be the same as yours, for as long as I do have a solution. Which means that my answer is correct too, just different from your answer. You just checked it for f(0) and g(0) being equal to 0, while I checked for f'(0) and g'(0) being negative (and even then, one could argue that my solution is more solid than yours, though I am not going to venture into that area because we are derailing the whole thread).
     
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  15. Warlock

    Warlock Member

    This is totally unrelated to the discussion at hand. Yes, sir, it isn't.

    [​IMG]
     
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  16. mining

    mining Member

    Well, the entire point is that given initial conditions - i.e. the bit where it would intuitively seem that the area should be growing - its not. It's an interesting paradox.
     
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  17. Kazeto

    Kazeto Member

    Actually, the point is to prove that it is not always right, not to prove that it's a paradox. And some people would argue that f'(0) and g'(0) don't really exist because f'(0-) and g'(0-) don't exist as the initial conditions do preclude negative x values from existing (x can only increase and has the initial value of 0).

    Let us stop talking about it here, really. You can create a conversation or a new thread for that, but I really think people are going to be annoyed if we continue talking about things relating to maths in this thread.
     
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  18. OmniaNigrum

    OmniaNigrum Member

    Technically. I could argue that I won that. :D
     
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  19. Mashirafen

    Mashirafen Member

    Being a mathematician is the number one trick for annoying people in the business of money.
     
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  20. Turbo164

    Turbo164 Member