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

Question

Consider the following:

• $1 = 1^2$
• $2 + 2 = 2^2$
• $3 + 3 + 3 = 3^2$

Therefore,

• $\underbrace{x + x + x + \ldots + x}_{x \textrm{ times}}= x^2$

Take the derivative of lhs and rhs and we get:

• $\underbrace{1 + 1 + 1 + \ldots + 1}_{x \textrm{ times}} = 2x$

Which simplifies to:

• $x = 2x$

and hence

• $1 = 2$.

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

Answer 1

You cannot take the derivative of $\underbrace{x + x + x + \dots + x}_{\text{repeated $x$ times}}$ with respect to $x$ one term at a time because the number of terms depends on $x$.

Even beyond that, if you can express $x^2$ as $\underbrace{x + x + x + \dots + x}_{\text{repeated $x$ times}}$, then $x$ must be an integer, and if the domain of the expression is the integers, (continuous) differentiation does not make sense and/or the derivatives do not exist.

(edit: I gave my first reason first because the second reason can be smoothed over by taking "repeated $x$ times" to mean something like $\underset{\lfloor x\rfloor\mathrm{\ addends}}{\underbrace{x+x+\cdots+x}}+(x-\lfloor x\rfloor)\cdot x$.)

Answer 2

You cannot differentiate the LHS of your equation

$x + x + x + \cdots$ (repeated $x$ times) = $x^2$

This is because the LHS is not a continuous function; the number of terms depends on $x$ so the LHS is not well defined when $x$ is not an integer. We can only differentiate continuous functions, so this is not valid.

Answer 3

Here's my explanation from an old sci.math post:

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Zachary Turner wrote on 26 Jul 2002:

Let D = d/dx = derivative wrt x. Then

D[x^2] = D[x  +   x  + ... +   x  (x times)]
       = D[x] + D[x] + ... + D[x] (x times)
       =   1  +   1  + ... +   1  (x times)
       =   x

An obvious analogous fallacious argument proves both

• $ $ D[x f(x)] = Df(x) (x times) = x Df(x)

• $ $ D[x f(x)] = Dx (f(x) times) = f(x), via Dx = 1

vs. the correct result: their sum $\rm\:f(x) + x\, Df(x)\:$ as given by the Leibniz product rule (= chain rule for times). The error arises from overlooking the dependence upon x in both arguments of the product $\rm\: x \ f(x)\:$ when applying the chain rule.

The source of the error becomes clearer if we consider a discrete analog. This will also eliminate any tangential concerns on the meaning of "(x times)" for non-integer x. Namely, we consider the shift operator $\rm\ S:\, n \to n+1\ $ on polynomials $\rm\:p(n)\:$ with integer coefficients, where $\rm\:S p(n) = p(n+1).\:$ Here is a similar fallacy

  S[n^2] =  S[n  +   n  + ... +   n  (n times)]
         =  S[n] + S[n] + ... + S[n] (n times)
         =  1+n  + 1+n  + ... + 1+n  (n times)
         = (1+n)n

But correct is $\rm\ S[n^2] = (n+1)^2.\:$ Here the "product rule" is $\rm\ S[fg] = S[f]\, S[g],\ $ not $\rm\: S[f] g\:$ as above.

The fallacy actually boils down to operator noncommutativity. On the space of functions $\rm\:f(x),\:$ consider "x" as the linear operator of multiplication by x, so $\rm\ x:\, f(x) \to x f(x).\:$ Then the linear operators $\rm\:D\:$ and $\rm\:x\:$ generate an operator algebra of polynomials $\rm\:p(x,D)\:$ in NON-commutative indeterminates $\rm\:x,D\:$ since we have

  (Dx)[f] = D[xf] = xD[f] + f = (xD+1)[f], so  Dx = xD + 1 ≠ xD

  (Sn)[f] = S[nf] = (n+1)S[f], so  Sn = (n+1)S ≠ nS

This view reveals the error as mistakenly assuming commutativity of the operators $\rm\:x,D\:$ or $\rm\:n,S.$

Perhaps something to ponder on boring commutes !

Answer 4

Lets define what is x+x+x+... x times for x - real. Natural definition is x+x+x.. := x*x (note - just the same as Isaac has wrote in his edit).

Suppose we want left our initial definition as is. We don't know what is x+x+.. repeat x times for x - real (and note we don't have rule how to obtain derivative from such func). So lets use definition of derivative. f(x):=x+x+x.. repeat x times, Df(x)=(f(x+h)-f(x))/h, h->0. Df(x)=((x+h+x+h+x+h.. repeat x+h times) - (x+x+x.. repeat x times))/h, h->0. Suppose x+x+... repeat a+b times := (x+x+.. repeat a times) + (x+x+.. repeat b times) we have Df(x)=((x+h+x+h+x+h.. repeat x times) - (x+x+x.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h->0, Df(x)=((h+h+h.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h->0, or Df(x)=(1+1+1.. repeat x times) + (x+h+x+h+x+h.. repeat 1 times), h->0 and at last Df(x)=x + x+h, h->0 = 2x

Answer 5

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 $u + \dots + u$, $v$ times. Then $x^2 = F(x,x)$. Differentiating both sides gives $2x = F_u(x,x) + F_v(x,x)$, which is perfectly true. In the fallacious example, the problem is essentially that the $F_v$ 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.

Answer 6

That differentiation step is invalid, because that equation is not actually differentiable, because its domain is comprised of isolated points, because $x\in\mathbb Z_0^+;$ to wit:

In general, if a theorem's condition(s) have not been fulfilled, then its consequence is not guaranteed.

P.S. A note on the above notation: What does $\mathbb{R}_0^+$ mean?