Theorem: [math] \frac{d}{dx} x = 0 [/math] Proof: [math] \frac{d}{dx} (x \circ x) = \frac{d}{dx} x \times \frac{d}{dx}x = \left( \frac{d}{dx} x \right)^2 [/math] according to the chain rule. But obviously [math] x \circ x = x [/math] so [math] \frac{d}{dx}(x \circ x) = \frac{d}{dx} x [/math]
If we combine these two equalities we get that [math] \left( \frac{d}{dx} x \right)^2 = \frac{d}{dx} x [/math]. This is actually a polynomial in the derivative of x. The polynomial is [math] u^2 - u = 0 [/math]. If you put this polynomial into wolfram alpha you get two solutions. Therefore x has two derivatives, 1 (the one you know) and 0. It is interesting that all this time no one has picked this up. Calculus is actually wrong because if we assume calculus is right when we have that [math] 1 = \frac{d}{dx} x = 0 [/math]. So we have proven [math] 1 = 0 [/math].
I wonder, what does this mean? All of modern math stands on calculus so now that the house of cards has fallen what will we do?
x is identity element for composition so x composed with x is just x.
Juan Diaz
>derivative of x is zero >x is just a placeholder for a constant >derivative of any constant is zero Gee wizz you're like Isaac Newton wow I can't believe you Simon Cosgrove the derivative of a constant is 0 I'm fucking floored call the press
Nathaniel Russell
>x is just a placeholder for a constant >[math] dx [(math]
What did he mean by this? I mean, what was not clear? I am differentiating x, with respect to x.
Chase Cook
I can't believe there would be even dumber responses to such a retarded bait.
Jace Turner
It's metabait
Chase Barnes
No, you retarded. I am doing meta meta bait by preteding I did not know that you were baiting when you were meta baiting so that way I was baiting you to reply to me by baiting you into thinking I was actually not aware you were meta baiting.
