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
- $1 = 2$.
Clearly something is wrong but I am unable pinpoint my mistake.
$\endgroup$ 710 Answers
$\begingroup$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.
$\endgroup$ 5 $\begingroup$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 $\underset{\lfloor x\rfloor\mathrm{\ addends}}{\underbrace{x+x+\cdots+x}}+(x-\lfloor x\rfloor)\cdot x$.)
$\endgroup$ 3 $\begingroup$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 botharguments of the product $\rm\: x \ f(x)\:$ when applying the chain rule.
The source of the error becomes clearer if we consider adiscrete 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)nBut 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 ≠ nSThis 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 !
$\endgroup$ $\begingroup$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:
$$ \underbrace{n + n + n + \ldots + n}_{n \textrm{ times}}= n^2 $$
and we apply $\Delta$ on both sides taking advantage from the "theorems" above: we get $$ \underbrace{1 + 1 + 1 + \ldots + 1}_{n \textrm{ times}} = 2n+1 $$ 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: $$ \Delta(\underbrace{n + \ldots + n}_{n \textrm{ times}})= \underbrace{(n+1) + \ldots + (n+1)}_{\color{Red}{n \textrm{ times}}}-(\underbrace{n + \ldots + n}_{n \textrm{ times}})=n $$ that is wrong, the right way being this: $$ \Delta(\underbrace{n + \ldots + n}_{n \textrm{ times}})= \underbrace{(n+1) + \ldots + (n+1)}_{\color{Green}{(n+1) \textrm{ times}}}-(\underbrace{n + \ldots + n}_{n \textrm{ times}})=n+(n+1). $$
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$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
Applying the sum rule in differentiation we get:$$\forall k \in \mathbb {N} \,(k\,f(x))' = k\,f'(x).$$The symbol $x$ does not represent any value, it is simply a placeholder meaning that $$\forall x \,(kf)'(x)=(kf')(x).$$I skip the mention that $x\in X$, where $X \in dom(f)$ to make it look more simple.
So we get$$\forall k \in \mathbb {N} \,\bigl( \forall x \,(kf)'(x)=(kf')(x)\bigr).$$
As we see now, we can't substitute $k$ with $x$, because $x$ is from the scope of the quantifier $\forall x$.
But we can substitute $k$ and $x$ with the same number, for example, with 4, and get the right state. We get$$(4f)'(4)=(4f')(4),$$ so $$(4id)'(4)=\textbf{4} (4)=4= (4\,\mathbf{1})(4)=(4\, id')(4),$$ where $\textbf{1}$ and $\textbf{4}$ are constant functions.
$\endgroup$ $\begingroup$Here is a nice way to write this joke ...
$$ x^2 = \underbrace{\color{red}{x} + \color{red}{x} + \color{red}{x} + \dots + \color{red}{x}}_{\text{repeated $\color{blue}{x}$ times}} $$ To differentiate the right side, use the chain rule for partial derivatives. We get two terms. One where we differentiate the red $x$ in the top line, the other when we differentiate the blue $x$ in the bottom line:$$ \underbrace{1 + 1 + 1 + \dots + 1}_{\text{repeated $x$ times}}\quad +\quad \underbrace{x + x + x + \dots + x}_{\text{repeated $1$ times}} $$and this is of course the right answer$$ \frac{d}{dx}(x^2) = x + x = 2x, $$
$\endgroup$ $\begingroup$It has already been pointed out that “$x$ times” makes sense only for integer $x$, and several ways of salvaging the example for continuous $x$ have been given. Here is another informal interpretation that seems fitting for first-semester calculus. The continuous analog of a discrete sum is a definite integral. So let’s define adding up a constant quantity $f(x)$ “$x$ times” by the integral$$\int_0^x f(x) \; dt \,.$$Differentiating* this yields$$\int_0^x f’(x) \; dt + f(x) \,.$$For $f(x)=x$, we get$${d \over dx} \int_0^x x \; dt = \int_0^x 1 \; dt + x \,.$$In words, the derivative equals $1$ added $x$ times plus $x$ added once, which is exactly what @GEdgar derived and similar to @arena-ru.
*There are two ways to explain the differentiation. One is through the product rule applied to $\int_0^x f(x)\,dt=f(x) \int_0^x dt$, and the other is through the multivariate chain rule applied to $G(u,v)=\int_0^u v\,dt$.
$\endgroup$ $\begingroup$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,
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