Revolutionary calculus proof

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?

Other urls found in this thread:

en.wikipedia.org/wiki/Generalizations_of_the_derivative#Derivatives_in_algebra
en.wikipedia.org/wiki/Zero_ring
twitter.com/SFWRedditGifs

XD

what the fuck is x composed with x

x is identity element for composition so x composed with x is just x.

>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

>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.

I can't believe there would be even dumber responses to such a retarded bait.

It's metabait

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.

You are not me.

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.

So how will this be applied to electrical engineering?

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.

You know that all you've shown is that the derivative of a constant is 0 right?

>So we have proven 1=0
The absolute state of Veeky Forums

No. I am differentiating x with respect to x.

Which gives you a constant. Can be 0, can 1 or pi.

Do you want to ask me how I know you are failing calculus?

>P(A) is true
>P(B) is also true
>Therefore A = B

Holy fuck

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.

[math]\frac{d}{dx}(x\circ x)=(\frac{d}{dx}x \circ x)\frac{d}{dx}x[/math]
[math](1\circ x)\times 1[/math]
[math]1 \times 1[/math]
[math]1[/math]
???

[math] e^{2\pi i}=1[/math]
[math] 2\pi i=\log(1)[/math]
[math] 2\pi i=0[/math]

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?

how many levels are you on right now

nice

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.

>be OP
x = 1;
x = x^2, because 1 = 1^2
x^2 - x = 0
x = 0 or 1
1 = 0

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.

I suppose a map D that takes every element in a ring to 0 is a derivation in the sense of

en.wikipedia.org/wiki/Generalizations_of_the_derivative#Derivatives_in_algebra

I find no way of making sense of 1=0, though, unless you even reduce the function space to one over

en.wikipedia.org/wiki/Zero_ring

>Yeah, this thread was bait
>I came to troll

HAHA XD DUDE!!!

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