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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.

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  • $\begingroup$ Please edit your title to make it more clear which fake proof you are asking about. $\endgroup$
    – Larry Wang
    Jul 29 '10 at 2:47
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    $\begingroup$ Right before taking the derivative, x was an integer, and what you did before that point only makes sense for x integer. Then you computed the derivative, and then that does not make sense. $\endgroup$ Jul 29 '10 at 2:48
  • $\begingroup$ @Kaestur I am not sure what you want me to do. Is there a specific name for this fake proof? $\endgroup$
    – user116
    Jul 29 '10 at 2:51
  • $\begingroup$ @Srikant: To clarify, the reason I asked that is so that people about to ask a similar question in the future may see yours come up as a suggested duplicate. Something including "derivative of 1+...+1 (x times)" would be uniquely identifying, I think. If there is a canonical name for this one, chances are people about to ask it won't know it, but will recognize that line. $\endgroup$
    – Larry Wang
    Jul 29 '10 at 3:04
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    $\begingroup$ @Kaestur I actually think retaining the title as it is written now is better as most people are more likely to write 'proof of 1 = 2' rather than some description of their proof in the title. $\endgroup$
    – user116
    Jul 29 '10 at 4:22

11 Answers 11

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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 $\underset{\lfloor x\rfloor\mathrm{\ addends}}{\underbrace{x+x+\cdots+x}}+(x-\lfloor x\rfloor)\cdot x$.)

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    $\begingroup$ A simpler version: 1+1+...+1 repeated x times = x. Now you can see that the left hand side is obviously not constant with respect to x. $\endgroup$
    – Jules
    Feb 22 '11 at 19:44
  • $\begingroup$ Of course, the derivative of $(x-\lfloor x\rfloor)\cdot x$ is $2x-\lfloor x\rfloor$ when the derivative is defined, so the value you'd get is $2x$ if you took the continuous version. $\endgroup$ May 16 '17 at 20:30
  • $\begingroup$ "You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x."-@issac i really loved this statement...but could you please explain about it a bit more? $\endgroup$ Mar 25 '19 at 4:19
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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.

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    $\begingroup$ This is a nice explanation. $\endgroup$
    – Matt E
    Aug 11 '10 at 4:35
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    $\begingroup$ I side with you on this one. I answered a similar thing here $\endgroup$
    – Pedro Tamaroff
    Jul 27 '12 at 2:41
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    $\begingroup$ any one please could help me? what are $F_u$ and $F_v$ ? $\endgroup$ Sep 6 '20 at 3:19
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    $\begingroup$ @JoséMarín: Partial derivatives. The partial derivative of $F$ with respect to $u$ is $F_u(u,v) = v$, and likewise $F_v(u,v) = u$. $\endgroup$ Sep 6 '20 at 3:26
  • $\begingroup$ @NateEldredge Thank you very much! now I understand and agree. In the original argument ( fallacy) the problem is with the abuse of notation ( en.wikipedia.org/wiki/Abuse_of_notation ) with the sum operation under the derivative. $\endgroup$ Sep 6 '20 at 3:36
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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 !

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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: $$ \Delta(\underbrace{n + \ldots + n}_{n \textrm{ times}})= \underbrace{(n+1) + \ldots + (n+1)}_{\color{Green}{(n+1) \textrm{ times}}}-(\underbrace{n + \ldots + n}_{n \textrm{ times}})=n+(n+1). $$

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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.

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    $\begingroup$ The problem is not so much that the LHS is not continuous: it is defined on a discrete set (so in particular it is continuous!). $\endgroup$ Jul 29 '10 at 3:00
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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

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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.

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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 $$ \frac{d}{dx}(x^2) = x + x = 2x, $$

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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 $G(u,v)=\int_0^u v\,dt$.

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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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Even if x were to be an integer, once you have x=2x, a possible value for is 0 which would make that equation true. Thus if x can be zero, then you are simply dividing by a variable that equals 0. And you cannot divide by zero!!!!

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  • $\begingroup$ The derivative of $f(x) = x^2$ at $x = 0$ is $0$. No division by zero necessary. -1 $\endgroup$
    – Thomas
    Mar 3 '14 at 4:40