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Take $x^2=x+x+x+\cdots$ ($x$ times). Now differentiating both sides wrt $x$, we get:

$$2x=x.$$

This means $x=0$ or $2=1$.

How? Where did I go wrong?

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    $\begingroup$ More simply, differentiate both sides of $x=1+1+1+\cdots+1$ ($x$ times) and get $1=0.$ $\endgroup$
    – bof
    Mar 10, 2016 at 6:45
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    $\begingroup$ $$\begin{align} \frac{\mathrm d}{\mathrm dx}(\underbrace{x+x+x+\cdots}_{\text{$x$ times}}) &= \underbrace{\frac{\mathrm d}{\mathrm dx}x+\frac{\mathrm d}{\mathrm dx}x+\frac{\mathrm d}{\mathrm dx}x+\cdots}_{\text{$x$ times}} + (\underbrace{x+x+x+\cdots}_{\text{$\frac{\mathrm d}{\mathrm dx}x$ times}}) \\ &= \underbrace{1+1+1+\cdots}_{\text{$x$ times}} + (\underbrace{x+x+x+\cdots}_{\text{$1$ time}}) \\ &= x + x \\ &= 2x. \end{align}$$ :) $\endgroup$
    – user856
    Mar 10, 2016 at 6:49
  • $\begingroup$ @Rahul it's nice to see that chain rule solves the problem even without considering the ill founded notation problem $\endgroup$ Mar 10, 2016 at 7:43

4 Answers 4

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What you are not considering is that while you are taking derivatives of the function on the right hand side with respect to x, the number of components is not fixed. For your example, what you are doing is basically looking at the function $$ g(x,n) = xn $$ and taking the derivative with respect to $x$ and evaluating the partial derivative at $n=x$, i.e. $g_x(x,n)|_{n=x}=n|_{n=x}=x$.

You would need to look at the total derivative of $g$ at $n=x$, i.e. $$ dg = g_{x}|_{n=x} + g_n|_{n=x} = 2x. $$

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The problem, in short, is that the equation $$x^2 = \underbrace{x+\cdots+x}_x$$ only holds when $x$ is a natural number.


To truly see why this is a problem, however, you will need to already understand universal quantification and lambda abstraction. Perhaps look up some YouTube videos about these concepts, or just browse on the internet for awhile until you find good explanations.

Explicitly, what we know is:

$$\left(\mathop{\forall}_{x \in \mathbb{N}}\right)\;x^2 = \underbrace{x+\cdots+x}_x$$

In function notation:

$$\left(\mathop{\lambda}_{x \in \mathbb{N}} x^2\right) = \left(\mathop{\lambda}_{x \in \mathbb{N}} \underbrace{x+\cdots+x}_x\right)$$

So if by $D$ we mean differentiation of functions $\mathbb{R} \rightarrow \mathbb{R}$, we can't use the above equation to deduce

$$D\left(\mathop{\lambda}_{x \in \mathbb{N}} x^2\right) = D\left(\mathop{\lambda}_{x \in \mathbb{N}} \underbrace{x+\cdots+x}_x\right),$$

because neither left- nor right-hand-sides are well-defined.

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My Explanation:

Looking at the definition of a derivative: $$(x^2)'=\lim_{h \to 0} \frac{(x+h)^2-x^2}{h}$$ $x+h\notin\mathbb{N}$, if $x\in\mathbb{N}$.

$x^2 = \underbrace{x+\cdots+x}_x$ is true only when $x\in\mathbb{N}$.

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  • $\begingroup$ Even if we consider $x^2=x+x+x+.... (n \text{ times})=nx$ , solving the derivated equation $2x=n$ gives $x=n/2$ which is not a root of the initial equation. $\endgroup$
    – JJacquelin
    Mar 10, 2016 at 7:09
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The idea to differentiate a function in order to find the zeros of the function is a big mistake because the zeros of the function are different from the zeros of it's derivative.

A elementary example : Solve $x^3-x=0$ for $x$

$f(x)=x^3-x$ . The roots of $f(x)=0$ are $-1\:,\:0\:,\:1$

$f'(x)=3x^2-1$ . The roots of $f'(x)=0$ are $-\frac{1}{\sqrt{3}}\:,\:\frac{1}{\sqrt{3}}$

Solving $f'(x)=0$ for $x$ doesn't give the roots of $f(x)=0$

That is the same in case of an equation made of the equality of two different functions : $f(x)=g(x)$

Of course, the functions $f$ and $g$ are different and we have $f(x)=g(x)$ only for a few values of $x$ that we call the roots of the equation. Except for the roots we have $f(x)\neq g(x)$ and thus $f'(x)\neq g'(x)$.

Now, if we consider the equation $f'(x)=g'(x)$ this equality holds only for a few values and for all other values $f'(x)\neq g'(x)$. There is not raison at all for the roots of the derivated equation be the same as the roots of the original equation.

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  • $\begingroup$ True but if $f(x)=g(x)$ holds for any $x$ than you can say also that $f'(x)=g'(x)$ is true for any $x$. $\endgroup$ Mar 10, 2016 at 7:30
  • $\begingroup$ @goblin can you find pointwise identical function with different derivatives? $\endgroup$ Mar 10, 2016 at 7:40
  • $\begingroup$ @MarcoDice : if $f(x)=g(x)$ holds for any $x$, then $f$ and $g$ are the same function: all values of $x$ are roots of the equation : they are an infinity of roots . One must not confuse with the equation $f(x)=g(x)$ where $f$ and $g$ are different functions : for any $x$ , except a few roots, $f(x)\neq g(x)$ and $f'(x)\neq g'(x)$. I maintain strongly that derivating the functions in order to solve the equation is a big mistake and whose who don't agree don't understand this fundamental fact. $\endgroup$
    – JJacquelin
    Mar 10, 2016 at 8:00
  • $\begingroup$ @JJacquelin Ok, you are right, I was interpreting the expression "$x^2=x+x+x+\cdots$ ($x$ times)" as an identity between two functions, not as a definition of a set of roots. $\endgroup$ Mar 10, 2016 at 8:17

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