8
$\begingroup$

What's wrong in this equation?

$$\underbrace{x+x+x+x+\cdots+x}_{x \textrm{ times}}=x^2$$ now differentiate w.r.t. 'x' both sides $$\underbrace{1+1+1+1+\cdots+1}_{x \textrm{ times}}=2x$$ So, $$x=2x$$ but how?

My friend gave me this and I know there is some problem with this, but what?

Any help will be appreciated.

$\endgroup$
5
  • $\begingroup$ Differentiate it again, you get $ 0 =2$. That's more question. $\endgroup$
    – Inceptio
    Mar 27, 2013 at 13:17
  • 1
    $\begingroup$ This is a duplicate $\endgroup$
    – Pedro ♦
    Mar 27, 2013 at 13:18
  • $\begingroup$ If you wanted to just go with it, you'd better write $1+\cdots+1_{x\text{ times}}+x+\cdots+x_{1\text{ times}} $ (so you don't forget about the other $x$, and you get a "right" answer $x+x=2x$. $\endgroup$
    – Pedro ♦
    Mar 27, 2013 at 13:23
  • $\begingroup$ Many other fake proofs here $\endgroup$
    – MathGeek
    Mar 27, 2013 at 13:38
  • 1
    $\begingroup$ I was going to try to write an answer that shows you how to define $x+x+\dots(x \rm{times})$ so that it always equals $x^2$, but it just turns into a tautology. This indicates that the problem is misunderstanding a notation. $\endgroup$
    – Karl Kroningfeld
    Mar 27, 2013 at 13:43

4 Answers 4

Reset to default
9
$\begingroup$

This is simply wrong because the expression $$ x+x+\cdots+x\quad(x\text{ times}) $$ does not make sense unless $x$ is an integer. For instance if $x=2.5$, then what is $2.5+\cdots+2.5$ ($2.5$ times)? Furthermore, you cannot differentiate term-by-term since the number of terms also depends on $x$.

Note that also $$x^2=1+1+\cdots + 1\quad (x^2 \text{ times})$$ where the right-hand side has derivative zero. This means that everything has derivative $0$ because $f(x)=1+\cdots + 1$ ($f(x)$ times) for any function $f$ according to this logic.

$\endgroup$
7
  • 1
    $\begingroup$ What happens when $x$ is an integer? $\endgroup$
    – Inceptio
    Mar 27, 2013 at 13:19
  • $\begingroup$ @Inceptio: Huh? $\endgroup$
    – Stefan Hansen
    Mar 27, 2013 at 13:20
  • 1
    $\begingroup$ @Inceptio: I don't understand where you are going. $\endgroup$
    – Stefan Hansen
    Mar 27, 2013 at 13:22
  • 1
    $\begingroup$ @Inceptio: What goes wrong is encapsulated in "Furthermore, you cannot differentiate term-by-term since the number of terms also depends on x." $\endgroup$
    – Stefan Hansen
    Mar 27, 2013 at 13:27
  • 1
    $\begingroup$ @StefanHansen, your answer is completely correct. Inceptio we know nothing about $x$. It is not mentioned that $x$ is an integer, or anything. Stefan makes a very valid point about that, by demonstrating the case of 2.5. This is the problem in the proof, we cannot say '$\text{x times}$'. $\endgroup$
    – xylon97
    Mar 27, 2013 at 13:48
6
$\begingroup$

In your second line, you are treating $x$ as a constant, which is where you go wrong.

If you really want to take multiplication as repeated addition, then $x^2$ should be written as $x \times x$, and then differentiated using the chain rule.

$\frac{d}{dx} (x \times x) = \big(\frac{d}{dx} x \big) \times x + x \times \big(\frac{d}{dx}x\big) = x + x = 2x$.

You miss one of them when you say $\text{x times}$, where you are assuming $x$ to be constant, whereas it is a variable.

$\endgroup$
6
$\begingroup$

You cannot differentiate since the number of summands depends on $x$. Moreover, if you differentiate in such a manner the second equation, you will get $0=2$.

$\endgroup$
1
$\begingroup$

your first equal is true only for integer X.and so with attention to define of differentiate we can not differentiate form this equal(your equal must be true at an interval at least , it must be true at a neighberhood )

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.