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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.

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marked as duplicate by Pedro Tamaroff, Jasper Loy, Tom Oldfield, vonbrand, rschwieb Mar 27 '13 at 14:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Differentiate it again, you get $ 0 =2$. That's more question. –  Inceptio Mar 27 '13 at 13:17
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This is a duplicate –  Pedro Tamaroff Mar 27 '13 at 13:18
    
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 $x+x=2x$. –  Pedro Tamaroff Mar 27 '13 at 13:23
    
Many other fake proofs here –  MathGeek Mar 27 '13 at 13:38
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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. –  Karl Kronenfeld Mar 27 '13 at 13:43

4 Answers 4

up vote 9 down vote accepted

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.

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What happens when $x$ is an integer? –  Inceptio Mar 27 '13 at 13:19
    
@Inceptio: Huh? –  Stefan Hansen Mar 27 '13 at 13:20
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@Inceptio: I don't understand where you are going. –  Stefan Hansen Mar 27 '13 at 13:22
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@Inceptio: What goes wrong is encapsulated in "Furthermore, you cannot differentiate term-by-term since the number of terms also depends on x." –  Stefan Hansen Mar 27 '13 at 13:27
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@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 '$\text{x times}$'. –  xylon97 Mar 27 '13 at 13:48

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.

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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 $0=2$.

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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 )

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