1. #1
    Bitter Old Fart Dribble Joy's Avatar
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    Default [OT] Since we're in the mood....

    Show that: e^(i*pi) = -1.

  2. #2
    Spermy made my hands messy :'( Asurmen Spec Op's Avatar
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    Quote Originally Posted by Dribble Joy
    Show that: e^(i*pi) = -1.
    e^x = Sum(n=0 -> infinity) x^n/n!
    x=i*pi we get:
    1 + (i*pi) + ((i*pi)^2)/2 + ((i*pi)^3)/6 + ((i*pi)^4)/24 ...
    since i^2 = -1 we get:
    1 + i*pi - (pi^2)/2 + -1(i*pi^3)/6 + ((pi)^4)/24...

    split this into (edit: by split I mean simply move shit)
    (1-(pi^2)/2 + (pi^4)/24...) + ((i*pi) - (i*pi^3)/6) + (i*pi^5)/120 ...)
    we can factor i from the right, getting
    (1-(pi^2)/2 + (pi^4)/24...) + i((pi) - (pi^3)/6) + (pi^5)/120 ...)
    these are the power series for cos and sin, so we get
    cos(pi) + i * sin(pi)
    cos(pi) = -1, sin(pi) = 0
    -1.

    Edit:
    the power series for sin is
    sum(n=0->infinity) x-(x^3)/3! + (x^5)/5! -(x^7)/7!...
    the power series for cos is
    sum(n=0->infinity)1-(x^2)/2! + (x^4)/4! - (x^6)/6!...

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  3. #3

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    Well, that was a quick thread. Assman is just showing off.
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    Registered User Neallys's Avatar
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    Now draw us a graphic using ASCII showing the function x = (3x/57) * pi

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  5. #5
    Andy is a ginger minger. ashley watts's Avatar
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    Don't even think computers would ever do those calculations
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    Banned User Glok's Avatar
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    Quote Originally Posted by Neallys
    x = (3x/57) * pi
    5 dimensional equation using split differential calculus?

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    Final Boss of the Internet Kanedax's Avatar
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    Ugh.....I hate math.

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    [uneX] - Developer Mighty Max's Avatar
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    Quote Originally Posted by Glok
    5 dimensional equation using split differential calculus?
    Well i'd just say ____________________ the flatline

    But if you want such a challenge in that direction *shrug*:
    Be y' := dy/dx, y'' := dy'/dx ...

    Find y(x) so that
    y'(x) = y^4(x) * cos(x) + y(x)*tan(x)

    and

    y(0) = 0.5


    (Did i tell i had exams on that last saturday? *g*)

    :edit: but i can't even read my notes anymore. That exercise was not the forth differntiation but power. Corrected above
    Last edited by Mighty Max; 10-02-09 at 20:48.
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  9. #9
    Banned User Glok's Avatar
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    MM

    f(x')=0.(S)^1/0 x^2 - y'^2

    Did I got my no math symbol shorthand right?
    Someone should dig up that old math symbol usage post (or unhallowed? pixel demons )

  10. #10
    [uneX] - Developer Mighty Max's Avatar
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    You can use latex syntax if that's better for you. Your answer in that form doesn't make any sense to me
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  11. #11
    Banned User Glok's Avatar
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    Bloody Hell. Fine then.


    x/y' = (x^2)/0 + xy' - 0/(y'^2)



    (Unhallowed 3 plane complex roots?)

  12. #12
    Banned User Glok's Avatar
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    Quote Originally Posted by Mighty Max
    Find y(x) so that
    y'(x) = y^4(x) * cos(x) + y(x)*tan(x)
    y(0)=0.5 -> solved.

    As for the quote (MM 3 part dang equation? Fruitcakes! Quantum Gravity possibly? Newton, Fermat, and Fermi were such nitwits trying to use euclidian geometry for this one. )

    y=(|(4x^3 * y'^2)| - (16y*(x'^2/y^2)))/0


    Absolutes? Nah. Why are 3space/2time gravity equations even remotely connected to the square of the volume of the spheres of equal size with the 2 gravitational bodies at the mathematical centers of each?
    Last edited by Glok; 11-02-09 at 03:33.

  13. #13
    Registered User Neallys's Avatar
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    Irony not even good in maths yeh? :/
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  14. #14
    [uneX] - Developer Mighty Max's Avatar
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    Quote Originally Posted by Glok
    y(0)=0.5 -> solved.
    Uhm No.
    The trival solution does not work here.
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  15. #15
    Banned User Glok's Avatar
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    MM why do you bedevil me? Ok let's be simple retarded stupid fucked up and expand the thing:

    Quote Originally Posted by Glok
    y=(|(4x^3 * y'^2)| - (16y*(x'^2/y^2)))/0
    The pure acceleration due to gravity between the centers of gravity of 2 separated massive bodies including an assumption of a zero separation scenario = (The absolute value of (the second integral of the cube of their summed mass multiplied by the integral of the complex root of the speed of light squared squared)) - (The linear sum of the energy inherent in the tidal forces experienced on both bodies)

    Better? Probably not, but FU neways. :P

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