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Apoapsis

In here, we all will write anything in white font so that we highlight it. Please put (parentheses) around writing so we know where to go.

(I'm amazed at how grandmasters play. They make moves that seem ridiculous and win with a new move. How do they do it? Thoughts?)Cool

hptchess
(I am figuring that out myself.)
Apoapsis
(Yay! This thread could gonon forever.)
itaibn

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

!!!!live os ma I

 

 

 

 

 

 

 

 

 

 

fruitallergy

 

(I like chocolate)

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((what up my chess dog im chillin like a spy yo  dont be haten be appreiceating)
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Apoapsis
(White font, JFercan!)
calvinhobbesliker

(the quadratic formula is as follows: negative b plus or minus the square root of b squared minus 4ac all over 2a)


BasicLvrCH8r

(One can use the quadratic formula to conclude that the cube root of one is negative .5 +- radical three over two times i.

x^3 = 1

X^3 - 1 = 0

(x^2 + x + 1)(x-1) = 0

x^2 + x + 1 = 0

x = (-1+-(1-4)^.5)/2

x= (-1 +- 3^.5 * i)/2

x = -.5 +- (3^.5*i)/2


calvinhobbesliker
(the cube root of 1 is 1)
Apoapsis
(Duh)
savy_swede
(white font is the greatest idea ever!!!!)
calvinhobbesliker
BasicLvrCH8r wrote:

(One can use the quadratic formula to conclude that the cube root of one is negative .5 +- radical three over two times i.

x^3 = 1

X^3 - 1 = 0

(x^2 + x + 1)(x-1) = 0

x^2 + x + 1 = 0

x = (-1+-(1-4)^.5)/2

x= (-1 +- 3^.5 * i)/2

x = .5 +- (3^.5*i)/2) 


(you divided each side by (x-1), but that is impossible beacuse since x cubed is one, x is one, and 1-1=0. you can't divide by 0


x-5058622868
(If i told you, then i'd have to kill you!!)Tongue out
BasicLvrCH8r
calvinhobbesliker wrote: BasicLvrCH8r wrote:

(One can use the quadratic formula to conclude that the cube root of one is negative .5 +- radical three over two times i.

x^3 = 1

X^3 - 1 = 0

(x^2 + x + 1)(x-1) = 0

x^2 + x + 1 = 0

x = (-1+-(1-4)^.5)/2

x= (-1 +- 3^.5 * i)/2

x = -.5 +- (3^.5*i)/2) 


(you divided each side by (x-1), but that is impossible beacuse since x cubed is one, x is one, and 1-1=0. you can't divide by 0


(x does not have to be one, however.

I will show that my expression cubed is equal to one. I will only show the positive solution (if you want to verify the negative, do it on your own). 

The expression for (a+b)^3 is a^3 + 3a^2*b + 3ab^2 + b^3

a = -.5

b=  3^.5*i/2

x = ((-.5)^3) + (3 * (-.5)^2 * 3^ .5 *i/2) + (3 * -.5 * (3^.5*i/2)^2) + ((3^.5*i/2)^3)

x = -.125 + .6495190528i + 1.125 + -.6495190528i

(The ugly decimal is (-3*3^.5*i)/8)

x = (-.125 + 1.125) + (.6495190528-.6495190528)

x = 1+0

x = 1


Apoapsis
(What????)
Apoapsis
(A month and a half lost! Are any James Bonds still there?)
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