[OT] Since we're in the mood....

[Dribble Joy]

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

[Asurmen Spec Op]

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!...

[Apocalypsox]

Well, that was a quick thread. Assman is just showing off.

[Neallys]

Now draw us a graphic using ASCII showing the function x = (3x/57) * pi

:angel:

[ashley watts]

Don't even think computers would ever do those calculations :rolleyes:

[Glok]

x = (3x/57) * pi5 dimensional equation using split differential calculus?

[Kanedax]

Ugh.....I hate math.

[Mighty Max]

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

[Glok]

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 :()

[Mighty Max]

You can use latex syntax if that's better for you. Your answer in that form doesn't make any sense to me :D

[Glok]

Bloody Hell. Fine then. :)


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



(Unhallowed 3 plane complex roots?)

[Glok]

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. :D)

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?

[Neallys]

Irony not even good in maths yeh? :/

[Mighty Max]

y(0)=0.5 -> solved. ;)

Uhm No. :)
The trival solution does not work here.

[Glok]

MM why do you bedevil me? Ok let's be simple retarded stupid fucked up and expand the thing:


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

[Tantta]

I didnt understand a fucking thing