Where is the flaw in this “proof” that 1=2? (Derivative of repeated addition)

Consider the following:

  • 1=12
  • 2+2=22
  • 3+3+3=32

Therefore,

  • x+x+x++xx times=x2

Take the derivative of lhs and rhs and we get:

  • 1+1+1++1x times=2x

Which simplifies to:

  • x=2x

and hence

  • 1=2.

Clearly something is wrong but I am unable pinpoint my mistake.

Answer

I think the discrete/continuous issue is sort of a red herring. To me, the problem is forgetting to use the chain rule!

To un-discretize, think of the function F(u,v)=uv, which we could think of as u++u, v times. Then x2=F(x,x). Differentiating both sides gives 2x=Fu(x,x)+Fv(x,x), which is perfectly true. In the fallacious example, the problem is essentially that the Fv term has been omitted. In some sense, one has forgotten to differentiate the operation “x times” with respect to x! Of course, the notation makes this easier to do.

Attribution
Source : Link , Question Author : Community , Answer Author : Nate Eldredge

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