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?

Parcly Taxel
  • 89,705
  • 18
  • 102
  • 171
  • 603
  • 5
  • 8
  • 52
    If you assume a solution exists then you can find it or come upon a contradiction. If you come across a contradiction, that implies that one of your assumptions was incorrect, which in this case would be the assumption that a solution exists in the first place. – JMoravitz Oct 18 '17 at 19:10
  • This reminds me of the non-measurable set example. A set $A$ is non-measurable if countably infinite copies of it constitute a set $B$ with finite measure. The set $A$ cannot have measure $0$, because otherwise the measure of $B$ would be $0$. Neither can $A$ have a positive measure, because otherwise the measure of $B$ would be infinite. – Zhuoran He Oct 18 '17 at 20:15
  • 6
    Always suspect of infinity, weird things happen over there :-) Also "often enough" is very, very, really, really! very far from infinity. – Rolazaro Azeveires Oct 18 '17 at 22:05
  • [I ran into a similar problem with continued fractions.](https://math.stackexchange.com/q/1681993/50421) The issue at heart is the same (i.e. the assumption that the sequence converges at all), so you might find the answers there to be useful as well. – Martin Ender Oct 19 '17 at 09:46
  • 2
    @JMoravitz Why didn't you put that as an answer instead of a comment? Comments are not for answers, and you would have received 320 rep (at the time of writing this comment) :P – Pedro A Oct 19 '17 at 23:06

5 Answers5


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

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

  • 20,741
  • 4
  • 26
  • 52
  • 1
    Thanks for the quick answer. When thinking about the problem I was already considering that the problem might be the convergence. However I dumped that thought immediately. After all, the concept of infinity his rather hard to grasp. After that explanation it seems pretty obvious to me though. – TheCodingDamian Oct 18 '17 at 19:19
  • 5
    You'd need to handle the $|x|=1$ case too. – hmakholm left over Monica Oct 18 '17 at 19:21
  • To add to this answer, here's a graph: https://imgur.com/GmucBJA . For each case, we can look at the loci of points such that $y=x+y,\;y=x\cdot y,\;y=x^y$. For $y=x+y$ (in red), the only place points that satisfy the equation are where $x=0$ (snore). For $y=x\cdot y$ (in black), the only points satisfy the equation have either $x=0$ or $x=1$ (still snore). But for $y=x^y$, we get a whole funky curve in green - all of these points are limits of the sequence $x^{x^{x^\ldots}}$. – Jam Oct 18 '17 at 19:32
  • ^ I think I should correct my comment - *some* of the points on the graph (for $0 – Jam Oct 18 '17 at 19:56
  • @HenningMakholm: Thanks. Updated. – Arnaldo Oct 18 '17 at 20:19
  • @HenningMakholm The x = 1 case works fine. If we say x*x*x.... = 1, Set u equal to the infinite product, we gt u*x = 1 or 1*x = 1, so x = 1. – RothX Oct 19 '17 at 00:38
  • Are these results the same for non-real $x$? – r12 Oct 20 '17 at 05:48
  • Is it true to say (1), i.e. that the limit as $n$ approaches infinity of $x^n$ where $|x|>1$ is equal to infinity? Because if $x$ is negative, as $n$ goes to infinity, $x^n$ bounces between positive and negative values (not just large positive). This isn't intended as critique in any way, I'm just curious if this is a mistake or perhaps an accepted convention I'm not aware of. – Questions about math Oct 25 '17 at 11:44
  • @Questionsaboutmath: Thanks for the tips. You are right, it does not change the solution that much but it is a valid observation. Solution updated. – Arnaldo Oct 25 '17 at 12:27

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.

  • 1
    You are right, I don't know much about infinitary algebra. As I said, I found out about the infinite power equation and tried to use that knowledge to learn even more things. I have one question regarding your answer though, if it is not too complex for a comment: In the third part, you say, I write as if my product had a first factor. I do understand that there is no last factor, but why isn't there any first factor? – TheCodingDamian Oct 18 '17 at 19:30
  • @zockDoc Infinity stretches in both directions, and we treat each end as the same point. See after the two series [here](https://en.wikipedia.org/wiki/Infinity#Real_Analysis). – DonielF Oct 19 '17 at 14:03
  • 1
    @zockDoc I disagree with DonielF and agree with you. It is completely OK for it to have a first factor. For Hurkyl's answer, although I can't speak for him, I believe that what he meant is that you can't normally have those three bullet points *at the same time*. But you can have a first factor if you don't have a last factor. Also, you could have a last factor if you don't have a first factor (as in a product from minus infinity to zero, for example). And you could have both if you have a finite amount of factors. "Infinity stretches in both directions" is incorrect in general. – Pedro A Oct 19 '17 at 23:04
  • @Hamsteriffic - You are also making an assumption. Actually, you can have a first and last factor and infinitely many between them. The ordinals from $0$ to $\omega + 1$ is an example. What you can't have is (1) a first and last element (2) every other element has both an immediate predecessor and immediate successor and (3) an infinite number of elements. – Paul Sinclair Oct 20 '17 at 03:46
  • @PaulSinclair I know that, hence I said "normally" (and Hurkyl also said it's possible). But I see that perhaps I could have been more precise in my comment. I just didn't want to insert this complicated stuff when I see that OP's question is about the simpler situation. Thanks :) – Pedro A Oct 20 '17 at 08:40
  • @PaulSinclair No, this too is possible: $\omega+\omega^*$ (where $\omega^*$ is the order type of the negative natural numbers) has the three listed properties. What is true is that if an order is well-ordered and converse well-ordered then it is finite. – Mario Carneiro Oct 24 '17 at 19:10
  • @MarioCarneiro - I knew at the time I should have thought harder about that comment. Thanks. – Paul Sinclair Oct 24 '17 at 23:01

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.

Mees de Vries
  • 24,889
  • 1
  • 40
  • 73
  • 5
    That proviso "*if* the sequence converges" is needed. If the 2 in the OP's equation is replaced by 4, that gives $x^4=4$, yielding $x=\sqrt{2}$ again! (Supposing that a real $x>0$ is required.) If $x=\sqrt{2}$, $x^{x^{x^{\dots}}}$ can't possibly be both 2 and 4. – Rosie F Oct 19 '17 at 11:16
  • +1 for actually going into why the approach works for the power tower case. – R.M. Oct 20 '17 at 16:02

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.

Qi Zhu
  • 6,822
  • 1
  • 14
  • 34

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

  • 8,659
  • 2
  • 14
  • 35
  • 3
    Why would the second limit be "indeterminate" for $x=0$? The limit of the sequence $0,0,0,\ldots$ is obviously $0$. – hmakholm left over Monica Oct 18 '17 at 19:20
  • if $0\times n=0$ and $\infty\times n=\infty$ then how can you know what $0\times \infty$ is? you can get a result only if there are $2$ functions that $f(x_0)=0,g(x_0)=\infty$ so $$\lim_{x\to x_0}f(x)g(x)$$ because $f(x)g(x)=\frac{1}{\frac{1}{f(x)}}g(x)=\frac{g(x)}{\frac{1}{f(x)}}$ and when you take the limit you will get $\frac{\infty}{\infty}$ and you can solve that limit, but here you have $0\times\infty$ without any way to get the limit – ℋolo Oct 18 '17 at 19:27
  • 2
    @Holo Limits (as $x\to c$) aren't substitutions of $c$ in place of $x$; they're what happens to sequences when $x$ gets near $c$. Clearly you can't substitute $\infty$ in place of $x$ so Henning's point is that the sequence $n\cdot x$ for $n=1,2,3,4,\ldots$ and $x=0$ is the sequence $(0\cdot 1), (0\cdot 2), (0\cdot 3), \ldots$, which is $0,0,0\ldots$. Clearly no matter how many terms you've got, the elements of the sequence are still $0$ (so the limit is definitely *not* indeterminate) - we're not finding $0\times \infty$, we're just showing what the sequence approaches :) – Jam Oct 18 '17 at 19:37
  • @Jam oh, okay i changed it, btw, what is the behavior of $x\times x\times x\cdots$ at $x=-1$? – ℋolo Oct 18 '17 at 19:43
  • @Holo Oscillates between $1$ and $-1$ (so it's not got a limit) – Jam Oct 18 '17 at 19:45