A contradiction involving derivative

Question

Possible Duplicate:
Where is the flaw in this argument of a proof that 1=2?

I am unable to find where the error is occurring in the following (I guess I can't take derivative, but why?):

$\underbrace{x+x+\dotsc}_{x \text{ times}}=x^2$

Differentiating both sides w.r.t. $x$, $\underbrace{1+1+\dotsc~~}_{x \text{ times}}=2x \implies x=2x$ !

Could anyone please explain the error.

Answer 1

This is ancient. I saw it in the '70s. You're saying $x$ is the number of terms. The number of terms can be $0,1,2,3,4,\ldots$ but it cannot be $3.14159\ldots$. The rules of differentiation apply to functions of a continuous variable, not to functions of a variable that varies discretely like this.

Answer 2

You have $\sum_{k=1}^x x = x^2$ so your summation depends on $x$. Thus as "$x$ changes" so does the length of the sum. Thus you'd need to figure out: What do you mean by summing $x=2.763$ number of $x=2.763$'s?

This is really the same issue that appears when taking the derivative of $x^x$ (which is not $x \cdot x^{x-1}$ since the exponent is changing).

Answer 3

Regarding this apparent paradox, it is often said that $x + x + ... + x = x^{2}$ is only meaningful when $x$ is a positive integer, and that differentiation rules for $x^2$ assume $x$ is a continuous variable. However, I don't think this gets to the heart of the matter. Consider $f(x) = x^2$ with domain equal to the set of positive integers. Then we can't take the derivative in the usual way, but we can still consider $\Delta = 1$ difference quotients:

$$ \frac{f(x+1) - f(x)}{1} \;\;= \;\;(x+1)^{2} - x^{2} \;\;=\;\; 2x + 1$$

Note that for large values of $x$, $2x + 1$ is asymptotically equal to the derivative of $x^2$, but $2x + 1$ is not asymptotically equal to $1$ added to itself $x$-times, which is $1 + 1 + ... + 1 = x.$

The factor of $2$ error comes about not because we're looking at functions defined only for the integers, but because we're incorrectly applying the sum rule for differentiating $x + x + ... + x$ ($x$-times). Specifically, this isn't a fixed-length sum. In order to correctly increment $x$ in the expression $x + x + ... + x$ ($x$-times), we have to increment each of the summands of $x$ AND increment the number of the $x$'s being added. Or, another way to look at this, in getting $1 + 1 + ... + 1 = x,$ we're incorrectly applying the product rule to $x^{2}=x \cdot x$ to get $1 \cdot x=x.$

Back in 2003 I posted (in the ap-calculus listserv) an ASCII diagram of what's going on, but I haven't been able to get a version to correctly display here. However, that diagram (along with some more explanation) can be found at the following URL. (The correct display of my original post at Math Forum has not survived their last two or three site modifications.)

https://web.archive.org/web/20040831204853/http://www.ncaapmt.org/calculus/newsletters/Winter2004/Vol12issue1.asp

Incidentally, there is a Letter to the Editor by David W. Erbach in The (Two-Year) College Mathematics Journal [Volume 6, Number 4, December 1975, pp. 2-3] that uses the greatest integer function in some way to get things to work out.

Answer 4

First recall the definition of a derivative: Let $X$ be a subset of the real numbers and $x_0\in X$ be such that $x_0$ is also a limit point of $X$. Let $f:X\rightarrow R$ be a function. If the limit:

$\lim_{x\to x_0;x\in X-\{x_0\}}\frac{f(x)-f(x_0)}{x-x_0}$

converges to some real number $L$ then we say that $f$ is differentiable at $x_0$ with derivative $L$.

Now in our case the moment we say $x+x+\cdots x$ times $=x^2$ we are implicitly assuming that $x$ is an element of the set of natural numbers. The function under consideration is given by $f(x)=x^2$ with domain the set of natural numbers. This function is not differentiable at any point $x_0$ as the set of natural numbers has no limit points. Hence we cannot take the derivative at all.