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So a friend shows me this :

$x^4= x^2+x^2+ \cdots +x^2 $ ( i.e. $x^2$ added $x^2$ times)

Now take the derivative of both side;

$4x^3 = 2x + 2x + \cdots + 2x $;

So $4x^3 = 2x^3 \cdots $(1)

And so dividing by $x^3$ gives $2=1 \cdots $(2).

I know we can't divide by 0 so that makes (2) false, but to show that (1) is false too ?

SchrodingersCat
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Sudhanshu
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    The derivative on the RHS is incorrect. The number of $x^2$'s is dependent on $x$, so you would have to factor this into the derivative. Namely $x^2 + \cdots + x^2 = (x^2)(x^2) = (2x)(x^2) + (x^2)(2x) = 4x^3$. – Trevor Norton Feb 16 '16 at 14:37
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    Essentially, one side is treated like a constant while the other is treated like a function when the derivative is taken. – zz20s Feb 16 '16 at 14:39
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    See: http://math.stackexchange.com/questions/1096/where-is-the-flaw-in-this-proof-that-1-2-derivative-of-repeated-addition?rq=1 . – nukeguy Feb 16 '16 at 14:39

2 Answers2

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  1. First of all, the statement

$x^2$ added $x^2$ times

makes sense only if $x^2$ is a positive integer. Else if $x^2$ is not a positive integer, then the statement is meaningless.

  1. Moreover, from $(1)$, we have that $4x^3=2x^3 \Rightarrow 4x^3-2x^3=0 \Rightarrow 2x^3=0 \Rightarrow x^3=0$

And hence division by $x^3$ in the next step is meaningless.

SchrodingersCat
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You can only do the first step if $x^2$ is an positive integer. So since it doesn't hold for all $x$ you can't take the derivative of both sides like that. It doesn't even make sense to talk about "$x^2$ times" if $x^2$ is not a natural number. To take the derivative of both sides you'd need equality in an entire interval.

Gregory Grant
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