Dgdrxrt

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.

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

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 !

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.

We can create the same "paradox" with finite differences over integers.

Given $f: mathbb Z to mathbb Z$ define the "discrete derivative"
$$
Delta f (n)=f(n+1)-f(n)
$$
we have the following obvious "theorems":

• $Delta(n)=n+1-n=1$


• $Delta(n^2)=(n+1)^2-n^2=2n+1$


• $Delta (f_1+cdots +f_k)=Delta f_1 + cdots +Delta f_k$


• $f(n)=g(n) ; forall n quad implies quad Delta f(n)=Delta g(n) ; forall n$

So we can start with the correct equality:

$$
underbracen + n + n + ldots + n_n textrm times= n^2
$$

and we apply $Delta$ on both sides taking advantage from the "theorems" above: we get
$$
underbrace1 + 1 + 1 + ldots + 1_n textrm times = 2n+1
$$
so we conclude $n=2n+1$ and we have the paradox.

Here maybe the mistake is more clear: the rule $Delta (f_1+cdots +f_k)=Delta f_1 + cdots +Delta f_k$ doesn't work when $k$ is a function (of the same variable of the $f_i$), in fact it amounts to do a computation like this:
$$
Delta(underbracen + ldots + n_n textrm times)=
underbrace(n+1) + ldots + (n+1)_colorRedn textrm times-(underbracen + ldots + n_n textrm times)=n
$$
that is wrong, the right way being this:
$$
Delta(underbracen + ldots + n_n textrm times)=
underbrace(n+1) + ldots + (n+1)_colorGreen(n+1) textrm times-(underbracen + ldots + n_n textrm times)=n+(n+1).
$$

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

There is the sum rule in differentiation:
$$(f_1(x)+f_2(x)+...+f_k(x))'=f_1'(x)+f_2'(x)+...+f_k'(x),$$ where $k$ is any positive integer number.

We can not use this rule to take the derivative of LHS, because $x$ in "$x$ times" is not a number, it is a variable, like in "function $f(x)$".

The problem is that the equation $x + cdots + x = x^2$ only holds for two values of $x$, namely if you added $x$ with itself $n$ times, it holds at $x=0$ and $x=n$. Therefore your equation becomes $nx = x^2$, the differential is $n = 2x$ (the number of terms in the LHS of the equation should not depend on the real/complex parameter $x$) and the only thing you can deduce from this is that the first equation is equivalent to $x^2 - nx = x(x-n) = 0$ and the second equation is $x=n/2$, so when $x=0$ or $x=n$ then the sum of the $x$'s and $x^2$ are equal, and when $x=n/2$ the derivative of the sum and $x^2$ are equal. The deduction that $1=2$ is simply not true. The equality $nx = x^2$ is not comparable to an equality like $sin x^2 = 1 - cos x^2$ : the equation $nx = x^2$ holds for only two values of $x$, where as the trigonometric equation holds for any real/complex value of $x$.

Hope that helps,