40

When I was a child, I wanted to be a mathematician so I asked my parents to buy me a computer to make super complex calculations. Of course, they were not crazy enough to buy an expensive super computer, so they bought me a way cheaper Stupid Computer™. In the TV ad, they said that «Stupid Computer™ can perform any operations a Super Computer™ can do !». As trusting as a 8 years old kid can be toward marketing, I trusted them.

In fact, and I realized that years later, Stupid Computer™ was just a Super Computer™ with a production defect.

With a Super Computer™ you can compute every polynomial functions like $ x^7-42x^3+3x$.

A bug in Stupid Computer™ prevents you to use the $x$ key more than once. For example, you can't enter $x²+x$ but you can enter $(x+1/2)^2-1/4$ instead.

Super and Stupid computer can only use operations like addition, multiplication, exponent... and their opposite : subtraction, division, roots, logarithm...

The question is : Can really a Stupid Computer do everything a Super Computer can ?

Pyrofoux
  • 1,548
  • 12
  • 21
  • 36
    If any polynomial could be written with just one $x$, you could solve any polynomial by radicals, which we know is false. –  Nov 16 '14 at 19:26
  • Can you do $x^5+x$ with this? (Also: are you French?) – Akiva Weinberger Nov 16 '14 at 19:30
  • @columbus8myhw You can't enter $x^5+x$ in a Stupid Computer because it has 2 $x$. And yes, I'm French. I guess you guessed right because of a French-typical English grammar error ? – Pyrofoux Nov 16 '14 at 19:33
  • @StevenTaschuk Can you please elaborate on what is "solving polynomials by radicals" and how do we know it's false ? Thank you ! – Pyrofoux Nov 16 '14 at 19:37
  • 3
    If you have an equation like "$\text{formula}=0$" and there's only one $x$ on the LHS, you can solve for $x$ by reversing all the operations, ending up with a formula for the solutions of your equation in terms of reverse operations. Just like how you get the quadratic formula from completing the square (as in your example). But such formulas do not exist for the roots of high-degree polynomials; this was proven by Galois (but note that as [wiki](http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem) says, this is stronger than what we usually mean by "not soluble by radicals"). –  Nov 16 '14 at 19:55
  • 2
    (I guess it's not entirely clear that solvability by radicals answers your question, since you have additional operations available.) –  Nov 16 '14 at 20:00
  • Sorry for being unclear. While you couldn't do $x^2+x$ directly, you _could_ do $(x+\frac12)^2-\frac14$, right? Similarly, you can't do $x^5+x$ directly, but I was wondering if you could do something that equalled it. – Akiva Weinberger Nov 16 '14 at 22:25
  • (I suppose @columbus8myhw noticed your guillemets.) –  Nov 16 '14 at 23:12
  • 2
    @StevenTaschuk I think indeed that the fact that there is every additions and higher operations with their inverse available may surpass the radical solvability, but I can't see a way to prove it. – Pyrofoux Nov 17 '14 at 20:51
  • @Pyrofoux: Please note, that your example is *not* a polynomial due to the summand $\frac{1}{3x}$ – epi163sqrt Nov 19 '14 at 14:50
  • @MarkusScheuer Corrected, thank you. – Pyrofoux Nov 19 '14 at 18:52
  • @Pyrofoux: You're welcome! – epi163sqrt Nov 19 '14 at 19:58
  • @Steven: Unsolvable quintics don't become solvable even if we allow exponentiation and logarithms in the solution, do they? –  Nov 22 '14 at 06:25
  • @Rahul: I would guess not, but I don't know. –  Nov 22 '14 at 13:02
  • @Steven: I think it makes sense to allow arbitrary (non-radical) real constants in the expression, in this case unsolvability by radicals does not solve the problem. – sdcvvc Nov 23 '14 at 10:59
  • 2
    A fun problem would be to find the minimal number of x's allowed in an equation so that all possible polynomials can be formed. Is it 2? Is it finite/infinite? – Ryan Nov 23 '14 at 16:15

3 Answers3

7

First, let's assume that Schanuel's conjecture is true and speak very loosely. :)

Timothy Chow's 1999 article What is a Closed-Form Number? proves that the exponential and logarithm functions don't really help us to express algebraic numbers. Any algebraic number that can be expressed using those functions can also be expressed using only radicals. This is stated as Corollary 1 at the top of page 444.

So the expression $x^5-x$ can't be rewritten with a single $x$ using exponentials and logarithms. If it could, we would be able to solve the equation $x^5-x=1$ by inverting those functions, which means we could solve it using radicals, which is impossible.

Can we do without Schanuel's conjecture? Maybe. Chow hints at partial results that don't need the conjecture:

It is folklore that general polynomial equations (i.e., those with variable coefficients) cannot be solved in terms of the exponential and logarithmic functions, although nobody seems to have written down a complete proof; partial proofs may be found in [C. Jordan, Traité des Substitutions et des Équations Algébriques, Gauthier-Villars, 1870, paragraph 513] and [V. B. Alekseev, Abel's Theorem in Problems and Solutions, Izdat. "Nauka," 1976 (Russian), p. 114].

I haven't tracked down those references. Maybe there's an equation where only the 0th coefficient is a variable, like $x^5-x+C=0$, that can't be solved in the relevant way. I'm not necessarily gunning for the bounty, so perhaps someone else can pick up the story from there?

Chris Culter
  • 26,935
  • 4
  • 46
  • 102
2

$f(x) = x^{3} - x$ is a counter-example. Assume for contradiction that there exist $\alpha, \beta \in\mathbb{R}, n \in \mathbb{Z}$ such that $f(x) = (x + \alpha)^{n} + \beta$. Since $f(x)$ has degree $3$, we know $n = 3$. So $f(x) = x^{3} + 3 \alpha x^{2} + 3 \alpha^{2} x + ( \alpha^{3} + \beta) $. If $\alpha \neq 0$, then we have an $x^{2}$ term, and if $\alpha = 0$, then we have $x^{3} + \beta$. Neither works, so it cannot be written as such.

AJY
  • 7,909
  • 2
  • 15
  • 39
  • 3
    You are allowed to use more operations that that (like exponentiation), and also iterate them (like $((x+\alpha)^n+\beta)^m+\gamma$.) – Cheerful Parsnip Nov 18 '14 at 21:19
  • 1
    I see, but I think the argument would still work, as $3$ is prime, so either $m = 1, n = 3$, or $m = 3, n = 1$ (still need a third-degree), so you're still curbing a first-degree factor. – AJY Nov 18 '14 at 21:52
  • 1
    that was just an example. And you also need to consider $e^u$, $\ln(u)$ and perhaps others. The OP is a little open-ended. – Cheerful Parsnip Nov 18 '14 at 22:32
  • 1
    If I'm not mistaken, we need only concern ourselves with algebraic functions (since we are, after all, trying to construct a polynomial, which is algebraic), which would exclude the exponential and logarithmic functions, as well as any trig business. – AJY Nov 18 '14 at 22:35
  • 1
    I agree that's likely, but the main point I think, is to make that precise. – Cheerful Parsnip Nov 18 '14 at 22:42
  • I think that if we're limited to only using elementary functions, then it could be shown that you can't construct any more polynomials by showing that there are limit behaviors you couldn't get. That is, a polynomials limits at $\pm \infty$ are infinite, and I believe it could be shown (rather tediously) that you wouldn't be able to get certain desired end behaviors by composition with elementary functions. I can try to do it later. The limitation to elementary functions seems to be the most sensible one. – AJY Nov 25 '14 at 16:31
1

Can really a Stupid Computer do everything a Super Computer can ?

This is sort of an open-ended question. Here are some interpretations I have of it.

Can the ring of polynomials over $\mathbb{R}$ be represented as a power of a polynomial of degree one plus some constant.

Any expression using the field operations on $\mathbb{R}$ and writing $x$ only once can be reduced to some polynomial of the form

$$ (ax^m + b)^n + c $$

This is pretty straightforward to check, so I'll let you do that. The answer to this question is then no, as pointed out by user AJY in another answer.

edit: I made an error above as pointed out by user David Speyer in the comments below.

Are we limited in what calculations we can do with Stupid Computer™?

Well, polynomials can be represented just as a sequence of coefficients, so there's actually no need to write $x$ at all!

Each polynomial might have a real root. If we look at the set of all polynomials and the collection of all possible roots, is this root set the same for both Stupid Computer™ and Super Computer™?

Yes. If our polynomial coefficients are in $\mathbb{R}$ then it's not a realistic computer and trivially true since for any desired root $a$, the degree 1 polynomial $x-a$ has precisely the root we want.

However, this is even true if our coefficients are in $\mathbb{Q}$! The set of all real roots for all polynomials is just a linear combination of elements of $\mathbb{Q}$ and roots thereof.

Community
  • 1
Ncat
  • 336
  • 1
  • 8