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Is it true that any nonzero function $f: \mathbb R \to \mathbb R$ which is either:

1) constant, or
2) a polynomial, or
3) $\exp$, or
4) $\log$, or
5) any finite combination of the above using addition, subtraction, multiplication, division and composition, (and individually considered as functions from $\mathbb R$ to $\mathbb R$),

Has a finite number of zeros?


Another attempt. 

I'm actually interested in asymptotic behavior of functions, so a function f(x), as far as I'm concerned, is any expression constructed using the syntax below, such that starting from some point c>0, f(x) is defined for all x>c and takes real values everywhere in this range. A function has the following syntax:

F --> real number
F --> exp
F --> ln
F --> -F
F --> F + F
F --> F(F)

Hope this specifies exactly what I mean and excludes everything I DON'T. Any help is appreciated.

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2  
The constant function $0$ has infinitely many zeroes... –  J. M. Oct 21 '11 at 1:39
2  
The constant zero function has an infinite number of zeros. –  lhf Oct 21 '11 at 1:40
2  
You said a "real(-valued) function"...maybe you should edit your question to make explicit the conditions you wish to impose. –  cardinal Oct 21 '11 at 2:10
8  
Let's not be deliberately obtuse here. It's not hard to figure out that the OP asks us to consider the smallest set of partial functions $\mathbb R\to\mathbb R$ that contains all constant real functions, the exponential function, and the logarithm, and is closed under composition, addition, negation, multiplication, and reciprocals. (That means no limits and no complex intermediate values. We get polynomials for free since $\log\circ\exp$ is the identity). He then asks whether this set contains a nonzero function with infinitely many zeros. –  Henning Makholm Oct 21 '11 at 2:29
2  
@J.M.: How so? The two exponentials will be positive where defined, so how can their sum be zero? –  Arturo Magidin Oct 21 '11 at 4:00
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2 Answers

All the functions under consideration are definable in $\mathbb{R}_{\exp}$, which is o-minimal. Thus their zero sets are finite boolean combinations of intervals. Since the functions are also all analytic, their zero sets are also discrete. Thus the only possible zero sets are finite.

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I have no idea if the following strategy will actually work. But here goes a try.

Let $C$ be the class of functions defined in the problem. Given two distinct functions $f, g\in C$, define $E(f,g)$ to be the set of $x$ for which $f(x)=g(x)$. Suppose it is true (for any distinct $f, g\in C$) that every point of $E(f,g)$ is an isolated point of $E(f,g)$ (equivalently, that $E(f,g)$ has no limit points). Then in particular, with $f_0$ equaling the zero function and $f\ne f_0$, we see that $E(f,f_0)$ has no limit points; therefore the only way that $f$ could have infinitely many zeros is if it has a sequence of zeros tending to infinity. However, $f(1/x)$ is also in $C$ and cannot have a sequence of zeros tending to $0$, so $f(x)$ can't have a sequence of zeros tending to infinity. Therefore (?) $f$ has only finitely many zeros.

The reason this might be a viable strategy is that maybe it's possible to prove inductively (using the recursive definition of $C$) that $E(f,g)$ never has limit points....

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