Nice proof (1 Viewer)

no_arg · Feb 5, 2020

Let $x>0.$ .
$x^2=x+x+\dots+x$ $\qquad$ where the sum on the right has $x$ terms.
For example if $x=4$ we have $4^2=4+4+4+4.$
Differentiating $\qquad$ $x^2=x+x+\dots+x$ $\qquad$ we have
$2x=1+1+\dots 1$ $\qquad$ where the sum on the right still has $x$ terms.
Thus $2x=x$ and hence $2=1$ .

Drdusk · Feb 5, 2020

no_arg said:
Let $x>0.$ .
$x^2=x+x+\dots+x$ $\qquad$ where the sum on the right has $x$ terms.
For example if $x=4$ we have $4^2=4+4+4+4.$
Differentiating $\qquad$ $x^2=x+x+\dots+x$ $\qquad$ we have
$2x=1+1+\dots 1$ $\qquad$ where the sum on the right still has $x$ terms.
Thus $2x=x$ and hence $2=1$ .

$\text{I believe it's because you can't take the derivative of}\hspace{2mm}x+x+...+x \\ \text{with respect to x because it's summed 'x' times?}$

blyatman · Feb 5, 2020

Every step up to and including $2x=x$ seems fine. However, you can't simply divide by $x$ much like how you can't divide both sides of $x^2=x$ by $x$ . Doing so is equivalent to dividing by 0, as $x=0$ is a solution. The correct way to solve it would be to subtract $x$ from both sides, which leaves the solution $x=0$ . Thus, the only way for a solution to exist is if $x\geq0$ (i.e. to include 0 in the domain).

In regards to why this is the case, I'm still trying to get my head around it. One would think that the derivative of both sides would be the same for all $x\geq0$ since $x^2=x+x+\cdots+x$ . However, this definition would start to get a bit iffy for non-integer values of $x$ . But then that just raises another question: why would this only be true for $x=0$ , and not all integer $x$ values? Regardless, you can't get $2=1$ since the step you took (i.e. dividing by 0) would result in the end of the universe as we know it.

@Drdusk I'm also thinking about your point of having $x$ terms. On the surface, this seemed OK to me, but now I'm questioning myself whether this is indeed allowed, since the result of $x=0$ seems rather nonsensical.

Carl10101 · Feb 5, 2020

no_arg said:
hence

.

oh ok

Trebla · Feb 5, 2020

The main flaw is that the number of terms is also a variable whereas the additivity of differentiation applies when the number of terms is fixed. A similar example is writing:

x = 1+1+1+....+1 (x-times)

Leading to the 1 = 0 fallacy.

Another flaw is that this decomposition assumes x is an integer so obviously there are complications with limits/continuity etc.

Drdusk · Feb 5, 2020

Trebla said:
The main flaw is that the number of terms is also a variable whereas the additivity of differentiation applies when the number of terms is fixed. A similar example is writing:

x = 1+1+1+....+1 (x-times)

Leading to the 1 = 0 fallacy.

Another flaw is that this decomposition assumes x is an integer so obviously there are complications with limits/continuity etc.

Yeah but if you just look at the equation x+x+....+x (x times) then how can you mathematically tell that you cannot do that. Sure you can factorize it and what not to show it but that equation alone doesn't show the flaw which is weird.

ultra908 · Feb 5, 2020

Drdusk said:
Yeah but if you just look at the equation x+x+....+x (x times) then how can you mathematically tell that you cannot do that. Sure you can factorize it and what not to show it but that equation alone doesn't show the flaw which is weird.

I guess its like trying to do d/dx (x^x) as x*x^(x-1). Bcos "x times" is also varying, you can't just diff it normally.

Trebla · Feb 5, 2020

Drdusk said:
Yeah but if you just look at the equation x+x+....+x (x times) then how can you mathematically tell that you cannot do that. Sure you can factorize it and what not to show it but that equation alone doesn't show the flaw which is weird.

The decomposition step is perfectly valid for integer values of x.

The invalid step is the differentiation step because it is not recognising that the number of terms is also a variable.

stupid_girl · Feb 6, 2020

Trebla said:
The decomposition step is perfectly valid for integer values of x.

The invalid step is the differentiation step because it is not recognising that the number of terms is also a variable.

If the decomposition step is only valid for positive integer values of x, then it's simply not continuous and therefore not differentiatable.

InteGrand · Feb 8, 2020

$\noindent It also works for non-integer but \textbf{constant} multiples of $x$, e.g.$

$\begin{align*} \frac{d}{dx}(\pi x) &= \frac{d}{dx}(\underbrace{x + x + x + \cdots + x}_{\pi \text{ times}}) \\ &= \underbrace{ \frac{d}{dx}(x) + \frac{d}{dx}(x) + \frac{d}{dx}(x) + \cdots + \frac{d}{dx}(x)}_{\pi \text{ times}}\\ &= \underbrace{1 + 1 + 1 + \cdots + 1}_{\pi \text{ times}} \\ &= \pi. \,\checkmark \end{align*}$

$\noindent You can replace $\pi$ with any real constant $a$.$

$\noindent It only fails when the number of terms is variable.$

no_arg · Feb 8, 2020

InteGrand said:
$\noindent It also works for non-integer but \textbf{constant} multiples of $x$, e.g.$

$\begin{align*} \frac{d}{dx}(\pi x) &= \frac{d}{dx}(\underbrace{x + x + x + \cdots + x}_{\pi \text{ times}}) \\ &= \underbrace{ \frac{d}{dx}(x) + \frac{d}{dx}(x) + \frac{d}{dx}(x) + \cdots + \frac{d}{dx}(x)}_{\pi \text{ times}}\\ &= \underbrace{1 + 1 + 1 + \cdots + 1}_{\pi \text{ times}} \\ &= \pi. \,\checkmark \end{align*}$

$\noindent You can replace $\pi$ with any real constant $a$.$

$\noindent It only fails when the number of terms is variable.$

I think you can have x pies but you can't have pi x's!

ultra908 · Feb 8, 2020

no_arg said:
I think you can have x pies but you can't have pi x's!

aren't they the same?

Drdusk · Feb 8, 2020

ultra908 said:
aren't they the same?

How?

ultra908 · Feb 9, 2020

Drdusk said:
How?

oh lol pies yum

dan964 · Mar 4, 2020

Try differentiating the RHS by first principles:
$\begin{align*}\frac{dy}{dx} &= \lim_{h \rightarrow 0}{\frac{1}{h}\left(\underbrace{(x+h)+(x+h)+\dots+(x+h)}_{(x+h) \text{ times}} - \underbrace{x+x+\dots+x}_{x \text{ times}} \right)} \\ &= \lim_{h \rightarrow 0}{\frac{1}{h}\left(\underbrace{h+h+\dots+h}_{x+h\text{ times}}+ \underbrace{x+x+\dots+x}_{h \text{ times}} \right)} \\ &= \lim_{h \rightarrow 0}{\left(\underbrace{1+1+\dots+1}_{x+h \text{ times}}+ x \right)} \\ &= \lim_{h \rightarrow 0}{2x+h} \\ &= 2x \end{align*}$

jyu · Apr 24, 2020

False statement (the sum on the right has x terms), because x is not necessarily a natural number.

If x are natural numbers, x^2 is not differentiable.

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