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.
Noah Myers
You are not me.
Charles Foster
Hah, you retard. I was meta meta meta baiting by pretending I was meta meta baiting by posing to be you but what I was actually trying to accomplish was to bait you into replying to me, which you just did.
Levi Allen
So how will this be applied to electrical engineering?
Benjamin Kelly
It is not precisely clear but my theory is that all the current problems you may have in electrical engineering, stuff that has no practical solution according to what we know, maybe because of the fact that we are basing everything on a wrong assumption, that calculus is consistent. It isn't.
After my result travels around the globe then my hope is that people will dust off all those old analysis textbooks and find where the initial contradiction lies and then we will reform all of calculus correctly. After that, we may end up with a better calculus that could solve every problem in electrical engineering.
Adam Scott
You know that all you've shown is that the derivative of a constant is 0 right?
Jose Hall
>So we have proven 1=0 The absolute state of Veeky Forums
Joseph Cruz
No. I am differentiating x with respect to x.
Benjamin Butler
Which gives you a constant. Can be 0, can 1 or pi.
Jackson Anderson
Do you want to ask me how I know you are failing calculus?
Adam Russell
>P(A) is true >P(B) is also true >Therefore A = B
Isaac Hughes
Holy fuck
Jaxon Morales
this is like saying that any function satisfying [math]f \circ f = f[/math] (which is basically any retraction on a subspace) means that [math]f[/math] is actually two functions, one of them the zero function.
First off, according to your own site Wolfram Alpha, [math]x\circ x=x(x)=\mathbb{R}[/math]. Therefore, [math]x\circ x \ne x[/math], and [math]\frac{d}{dx}(x \circ x) = \frac{d}{dx}(x(x))=x'(x)\times x'[/math]. Can you fuck off now?
Jackson Smith
how many levels are you on right now
Sebastian Anderson
nice
Eli Rodriguez
For those wondering, where OP's proof fails is that it shows that d/dx(x) solves the polynomial equation u^2-u=0, not that it is equal to every solution. It can be shown (by other means) that the derivative is unique (where it exists), and the derivative of a variable with respect to itself is 1.
Daniel Green
>be OP x = 1; x = x^2, because 1 = 1^2 x^2 - x = 0 x = 0 or 1 1 = 0
Carson Garcia
Okay, I had my fun. Yeah, this thread was bait. Anyone who is not retarded can spot where the error is. But what amuses me is that so many people answered the most retarded shit:
I know I came to troll but the fact that there are so many actual stupid people on Veeky Forums makes me sad.
Luis Perry
I suppose a map D that takes every element in a ring to 0 is a derivation in the sense of