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February 15

Solve for x

Recently, I stumbled upon this equation:

$x^x=x+1$

What is the exact value of x?

I've tried to solve it, but I'm stumped. WolframAlpha gives an approximate value of 1.77, though requesting an exact form gives an inadequate answer. If it turns out the equation cannot be solved, how come? – Poml (talk) 16:32, 15 February 2019 (UTC)

$x=0$ is an exact solution. There is also a solution with $2>x>1$ (this follows from the intermediate value theorem). So the equation can be solved (a solution exists), just maybe there is no closed formula for doing so. Most equations are like that -- it is easy to show that solutions exist, and they can be approximated numerically, but there does not need to be a nice way to express them in terms of known numbers and functions. —Kusma (t·c) 16:48, 15 February 2019 (UTC)

Template:Ping, in my view, $x=0$ cannot be a solution, because the pseudo identity $0^0=1$ is controversial, and we have also an article about that. HOTmag (talk) 19:10, 16 February 2019 (UTC)

In fact the true ambiguity here is that it is not stated where the unknown has to be. What is x? The complete statement should be (e.g.) solve $x^x=x+1$ for real x>0.

Further, in my view, $0^0=0$, because this is what directly follows from the well known axiomatic definition of the Power function: $a^x=a^{x-1}a$. Yes, I don't see any reason, why this definition, which is agreed to be true - for every non-zero integer $a$ - when applied to every integer $x$, shouldn't be true also for every integer $a$ (including $a=0$) - when applied to every integer $x$, so that for every integer $x$ one consistently receives: $0^x=0^{x-1}0=0$. Hence, for $x=0$ one consistently receives: $0^0=0$. By the way, also for $x=-1$ one consistently receives: $0^{-1}=0$. HOTmag (talk) 19:10, 16 February 2019 (UTC)

Well, among integers, $0^0=1$ is the only choice consistent with set theory. $a^b$ is the number of different functions from a set with $b$ elements into one with $a$ elements. There are no functions from a nonempty set into the empty set, so $0^b=0$ for nonzero $b$, but there is exactly one function from the empty set into any other set, the empty function, so $a^0=1$. This also works for $a=0$. In any case, in the question at hand, both $x^x$ and $x+1$ have the limit 1 at 0, so you can consider 0 as a solution if you extend by continuity. —Kusma (t·c) 19:39, 16 February 2019 (UTC)

Of course, there is one (and only one) function from the empty set (having zero elements) into the empty set (having zero elements). However, please notice, that your assertion: "$a^b$ is the number of different functions from a set with $b$ elements into one with $a$ elements", has never been proved for $(a,b)=(0,0);$ Unless you define $0^0=1$, but this very definition is controversial, as pointed out in our article 0^0.

As for continuity: Please notice that the function $f=a^0$ has never been proved to be continuous at $a=0;$ Unless you define $0^0=1$, but this very definition is controversial, as pointed out in our article 0^0. Further, your consistent assumption of continuity of the function $f=a^0$ at every real $a$, contradicts the analogous consistent assumption of continuity of the function $f=0^x$ at every real $x$, which consistently entails that $0^0=0$, so why should one prefer - your consistent assumption of continuity - to the second consistent assumption of continuity?

On the other hand, the well known axiomatic definition of the Power function: $a^x=a^{x-1}a$ (along with $a^1=a$), assumes nothing: Neither continuity, nor the very existence of non-integers (needed for continuity). That axiomatic definition just defines (rather than assumes) what being a "power function" means, so this definition should be applied for every pair of integers $(a,x)$, and not only for pairs of integers $(a,x)$ other than $(0,0)$, at least according to Frege who demanded that every function should be applied in the whole domain of discourse, wherever possible consistently. In the question at hand, it's possible to consistently apply the axiomatic definition of the Power function: $a^x=a^{x-1}a$ (along with $a^1=a$), in the whole domain of integers. Consequently, for every integer $x$ one consistently receives: $0^x=0^{x-1}0=0$. Hence, for $x=0$ one consistently receives: $0^0=0$. By the way, also for $x=-1$ one consistently receives: $0^{-1}=0.$ HOTmag (talk) 21:17, 16 February 2019 (UTC)

We are rather offtopic here. My assertion above about cardinalities is what I use as the definition of the power function for integers. Your $0^{-1}=0$ is "consistent" only with all powers of zero being zero, not with the much more useful definition of setting $a^{-1}$ to be the multiplicative inverse of $a$, so you won't convince me that this is of any use. The continuity I refer to above is that $0^0=1$ makes $x^x$ continuous at 1, which is relevant to the original question. —Kusma (t·c) 07:50, 17 February 2019 (UTC)

As for your comment that "We are rather offtopic here". Please notice that I have been referring to your original claim that 0 can be a solution as well, so I don't see we are offtopic (or rather offtopic).

As for your claim that the identity $0^{-1}=0$ is "consistent only with all powers of zero being zero": Please notice that the universal identity $0^x=0$ for every integer $x$, is not only a "consistent" assumption; As I have already pointed out, one consistently "receives" it, i.e. concludes it as a necessary result, from the very axiomatic definition of the Power function: $a^x=a^{x-1}a,$ because once $a$ is set to be $0$ - while $x$ is unlimited, one receives: $0^x=0^{x-1}0=0.$ As I have already pointed out, that axiomatic definition $a^x=a^{x-1}a$ (along with $a^1=a$), just defines (rather than assumes) what being a "power function" means, so this definition should be applied for every pair of integers $(a,x)$, and not only for pairs of integers $(a,x)$ other than $(0,0)$, at least according to Frege who demanded that every function should be applied in the whole domain of discourse, wherever possible consistently.

As for (what you call): "setting $a^{-1}$ to be the multiplicative inverse of $a$": Please notice, that before you "set" anything, you must make sure that your "setting" mentioned above does not contradict the very axiomatic definition of the Power function: $a^x=a^{x-1}a$ (along with $a^1=a$), but it does! Further, your "setting" has only been proved for every non-zero $a$, but has never been proved for $a=0$. As opposed to the very axiomatic definition of the Power function: $a^x=a^{x-1}a$ (along with $a^1=a$), which doesn't have to be "proved", because it's just an axiomatic definition of the power function.

As for your comment that I "won't convince" you that the identity $0^x=0$ for every integer $x$ "is of any use". First, I have never claimed it's useful, I have only claimed it's a necessary result from the very axiomatic definition of the Power function: $a^x=a^{x-1}a$. Second, your avoidance of having any value of $0^x$ for any negative $x$, is not more useful.

As for your assumption that the function $x^x$ is continuous at 0 (I guess you meant "0" rather than "1"): Please notice that for this function to be continuous at zero, it must be defined also for some neighborhood (around zero) - which necessarily contains negative numbers, but this function can't be defined for any neighborhood around zero - assuming the domain of discourse is of the real numbers, and can't have any limit at zero - assuming the domain of discourse is of the complex numbers. HOTmag (talk) 14:20, 17 February 2019 (UTC)

The only Plouffe's inverter-like tool that seems to be up has no other suggestions for the numerical value: [1]. So it may not be a number that has other nice forms. —Kusma (t·c) 16:52, 15 February 2019 (UTC)

Thank you very much for the helpful answer. While I personally do not consider $x=0$ as a solution, your clarification of the second solution was informative. – Poml (talk) 17:09, 15 February 2019 (UTC)

It doesn't work in this case, as far as I can see, but the Lambert W function tends to crop up in this type of equation. HTH, Robinh (talk) 09:56, 16 February 2019 (UTC)

We may argue what is the value of $0^0$ in Theoretic Philosophy, or in Theology, or in Voodoo rituals if you like. But there is no discussion about Mathematics; in Mathematics $0^0=1$, of course. pma 18:02, 21 February 2019 (UTC)

No, that is not correct, or at least is not the generally agreed view. There is indeed discussion. We have a whole article on it, at zero to the power of zero. --Trovatore (talk) 21:21, 21 February 2019 (UTC)

Yes, it's a very nice article. As I see it, the only reason to leave $0^0$ undefined is some timidity about discontinuity situations, which is why some calculus textbooks prefer to do it. But if there was a general tendency to do so in mathematics, then we would better introduce a new name and notation for exponentials, covering $0^0=1$. pma 07:38, 22 February 2019 (UTC)

I'll just briefly state my view in response, without inviting debate on this page; this has all been gone over many times and if you want clarification I'll point you to stuff in the archives. My view is that 0^0=1 makes perfect sense when thinking of the exponent as a natural number, but not when thinking of it as a real number. The two sorts of exponentiation are not the same function; they have different meanings, and there is no reason that they have to make a commutative diagram with the canonical embedding from N to R wherever they possibly could. In C language terms, pown(0,0)==1, but powr(0.0, 0.0) is undefined. --Trovatore (talk) 18:30, 22 February 2019 (UTC)

Would "mathematical homophone (or homonym)" cover it? -- Jack of Oz ^{[pleasantries]} 18:39, 22 February 2019 (UTC)

The softwarey term is "overloading". --Trovatore (talk) 18:48, 22 February 2019 (UTC)