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Integrate 1/x^2

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    Within the boundaries of 1 and -1.
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    1/x^2=x^(-2)
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    Why is the answer negative?
    Spoiler:
    Show

    Think about the graph.
    What happens when x=0?
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    @Square: Yes, now integrate it within the boundaries of 1 and -1 to give me a number.


    @TheTallOne: Dammit, its a trick question! A kinda lame one, I know.
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    >_>
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    I thought it can't be done when the domain of integration includes 0?
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    (Original post by Vivisteiner)
    ^Yes, now integrate it within the boundaries of 1 and -1 to give me a number.
    It doesn't converge. Happy?
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    ans = math error :yep:
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    (Original post by Chaoslord)
    ans = math error :yep:
    Not quite sure, but isn't it over 9000?
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    ^lol, CORRECT!

    Ok fine, I'll give you another;

    calculate e^i*pi
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    (Original post by Vivisteiner)
    ^lol, CORRECT!

    Ok fine, I'll give you another;

    calculate e^i*pi
    That's in a simple enough form.
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    (Original post by Vivisteiner)
    ^lol, CORRECT!

    Ok fine, I'll give you another;

    calculate e^i*pi
    Why exactly?

    e^{i \pi} = -1
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    (Original post by Vivisteiner)
    ^lol, CORRECT!

    Ok fine, I'll give you another;

    calculate e^i*pi
    Presumably you mean ei*pi, rather than ei*pi.

    The answer is -1. Back to you - prove it
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    (Original post by Agrippa)
    Presumably you mean ei*pi, rather than ei*pi.

    The answer is -1. Back to you - prove it
    I'd be surprised if someone at interview was asked to expand e^(ix) then note the taylor series for the trig functions (although it is quite a pleasing result).
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    Lol, it was a trick. e^(i*pi) = -1 but e^i*pi is already in its simplest form.


    And I don't know how to ruddy well prove it. xD
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    A hack way of doing it is to differentiate

    f(x) = e^(ix)/(cosx + isinx) and hence work out what f(x) is.
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    (Original post by Vivisteiner)
    Lol, it was a trick. e^(i*pi) = -1 but e^i*pi is already in its simplest form.
    http://xkcd.com/169/

    All there is to say on things like that
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    (Original post by Vivisteiner)
    Lol, it was a trick. e^(i*pi) = -1 but e^i*pi is already in its simplest form.


    And I don't know how to ruddy well prove it. xD
    A strange maths troll indeed. An interview with your kind of questions actually would worry me.
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    Lmao, I'm just kinda bored so I asked inane questions and sounded stupid even though I'm obviously incredibly smart and can even differentiate x^x^x^x^x^x in under ten seconds with respect to x.
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    (Original post by Vivisteiner)
    Lmao, I'm just kinda bored so I asked inane questions and sounded stupid even though I'm obviously incredibly smart and can even differentiate x^x^x^x^x^x in under ten seconds with respect to x.
    Sodding LaTeX

    x^x^x^x^x^x (x^(-1+x^x^x^x)+x^x^x^x^x Log[x] (x^(-1+x^x^x)+x^x^x^x Log[x] (x^(-1+x^x)+x^x^x Log[x] (x^(-1+x)+x^x Log[x] (1+Log[x])))))

    Done

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Updated: December 8, 2008
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