Calculation with infinitely many operands

Question

Recently I have come across the equation

$$x^{x^{x^{x^{...}}}} = 2$$

which I found out to be solvable using substitution:

$$u = x^{x^{x^{x^{...}}}} = 2$$ $$x^u = 2$$ $$x^2 = 2$$ $$x = \sqrt{2}$$

Thinking about that, I tried to apply the same concept to other operations

$$ x \cdot x \cdot x \cdot x\space \cdot \space ...\space= 2 $$ $$u = x \cdot x \cdot x \cdot x\space \cdot \space ...\space= 2$$ $$x \cdot u = 2$$ $$2x = 2$$ $$x = 1$$

This solution is rather strange, as $1 \cdot 1 \cdot 1 ...$ should of course always equal $1$, but the equation says something else. Same thing goes for the addition:

$$x + x + x + x + \space ...\space= 2$$ $$u = x + x + x + \space...\space= 2$$ $$x + u = 2$$ $$x + 2 = 2$$ $$x = 0 $$

Which basically says that if you add $0$ up often enough you get $2$. And the next problem is, if you define the sum to equal $1$ instead of $2$ you end up with $1 = 2$. So obviously there must be a mistake here.

Is there any reason why this approach can be used for infinite powers but not sums and products?

Answer 1

The problem is when you make those kinds of substitutions you are assuming that the associated sequence is convergent. That is true for the first case but it is not for the second.

In the second case you have:

$$x\cdot x\cdot x ...=2$$

It is equivalent to

$$\lim_{n\to \infty}x^n=2$$

But if you have $|x|>1$ then we have two cases: if $x>1$ then the limit will be $\infty$ but if $x<-1$ then $x^n$ will oscillate for $+\infty$ or $-\infty$ and then we don't have limit is this case:

$$\lim_{n\to \infty}x^n=\infty \text{ or } \nexists \lim_{n\to \infty}x^n \quad (1)$$

which means that it is not convergent. Otherwise, if $|x|<1$ then

$$\lim_{n\to \infty}x^n=0 \quad (2)$$

For, if $x=1$ we get

$$\lim_{n\to \infty}x^n=1 \quad (3)$$

and, finally, for $x=-1$ we have that $x^n$ will oscillate. It will be $1$ when $n$ is even and $-1$ when $n$ is odd so $\lim_{n\to \infty}x^n$ doesn't exist. $(4)$

From $(1)$, $(2)$, $(3)$ and $(4)$ we conclude that you will never have

$$x\cdot x\cdot x ...=2$$

Answer 2

There are three issues here.

You did it backwards

You have done what is basically the common high school algebra mistake of doing problems backwards — what you have proven is:

(Assuming that the product satisfies that algebraic identity) if $x$ is a real number solution to $x \cdot x \cdot x \cdot \ldots = 2$, then $x = 1$

In particular, you have completely forgotten to put any work towards showing

If $x=1$, then $x$ is a solution to $x \cdot x \cdot x \cdot \ldots = 2$

You haven't defined the problem

Multiplication is a binary operation. By induction, you can extend multiplication to any finite number of operands.

If you want to talk about multiplying infinitely many operands, then the above is insufficient. You have to find some way to make sense of such an operation.

You might do this by considering the "limit of partial products" approach from introductory calculus; e.g.

$$ x \cdot x \cdot x \cdot \ldots := \lim_{n \to \infty} \underbrace{x \cdot x \cdot \ldots \cdot x}_{n \text{ factors}} $$

You probably misunderstand infinitary operations

I infer this because you write as if:

• Your product has a first factor
• Your product has a last factor
• Your product has infinitely many factors

While it is possible to arrange for such a thing, it is unusual and contrary to the common meaning of notation. One who actually understands how to arrange for this would understand the need to actually explain more precisely what they actually mean.

Thus, I imagine you either made a mistake (e.g. you did not mean to put a final term), or simply misunderstand infinitary algebra.

Answer 3

Perhaps someone will chime in with a deeper reason, but let me explain "what you're doing" more formally. An infinite expression like $$ x^{x^{x^{x^\cdots}}} $$ is typically taken to mean the limit of the sequence $$ x, x^x, x^{x^x}, x^{x^{x^x}}, \ldots. $$ That is, the sequence $s$ is defined by $s_0 = x$ and $s_{n+1} = x^{s_n}$; and the number we are looking for is the limit of this sequence, if it exists. Now note that the operation $r \mapsto x^r$ is continuous for positive $r$ (let's say that $x$ is also positive), and thus we have that $$ \lim_{n \to \infty} s_n = \lim_{n \to \infty} s_{n+1} = \lim_{n \to \infty} x^{s_n} = x^{\lim_{n \to \infty} s_n}. $$ Now, solving this equation is what gave you $x = \sqrt 2$. Thus, this tells you that if the sequence converges, then $x = \sqrt 2$.

In the other examples, you know the same thing. However, because you don't know if the sequences converge (and in fact you can easily check that they don't) the knowledge is vacuous.

Answer 4

Well, convergence issues. You have to carefully define what your infinite power is and check for convergence. Your method only works if we indeed have a convergent sequence.

Answer 5

The problem is in the convergence. you can only get an value of a function if it convergences. Look at $x\times x\times x\times\cdots$: $$x\times x\times x\times\cdots=\lim_{n\to\infty}x^n=\begin{cases}\infty & |x|>1\\0 & |x|<1\end{cases}$$ so the equation $x\times x\times x\times\cdots=2\,$ has no answer $$$$and for $x+x+x+\cdots$: $$x+x+x+\cdots=\lim_{n\to\infty}nx=\begin{cases}\infty&x>0\\0&x=0\\-\infty&x<0\end{cases}$$so $x+x+x+\cdots=2$ also has no answer