# Challenge 3b: What's in a polynomial?

1. Oct 10, 2013

### Office_Shredder

Staff Emeritus
In order to challenge a broader section of the forum, there is a part a and a part b to this challenge - if you feel that part b is an appropriate challenge, then I request you do not post a solution to part a as part a is a strictly easier question than part b.

The new challenge: Are there any infinite subsets of ℂ on which ex is equal to a non-constant polynomial?

The old challenge: Prove there is no infinite subset of ℂ on which ex is equal to a polynomial (or make me look foolish and produce a counterexample)

Last edited: Oct 12, 2013
2. Oct 11, 2013

### Citan Uzuki

1 is a polynomial, and on the set $2\pi i \mathbb{Z}$, e^x is identically equal to 1.

3. Oct 12, 2013

### Office_Shredder

Staff Emeritus
Ack! Somehow in constructing this question I convinced myself that constant functions weren't polynomials. Congratulations on making me look very foolish and scoring a solution haha. OK, I'll re-open the challenge in this thread with the more interesting question.. are there any examples with non-constant polynomials?

4. Oct 12, 2013

### D H

Staff Emeritus
Let z be a solution to ze-z=c, where c is some constant. There are an infinite number of such solutions, given by the Lambert W function and it's analytic continuations. For each element z in the set of all z's that are solutions to this equality, we have ez = z/c, which is a non-constant polynomial.

5. Oct 12, 2013

### Citan Uzuki

Actually, in a certain sense there are solutions for almost any non-constant polynomial. Let $f(x)$ be an arbitrary polynomial. The function $x \mapsto e^x - f(x)$ has an essential singularity at $\infty$. By Picard's theorem, it follows that for every value of c except possibly one, we will have $e^x - f(x) = c$ at least once in every neighborhood of infinity (hence infinitely many times), and thus $e^x = f(x) + c$ will have infinitely many solutions for all but possibly one value of c.

So the interesting question isn't whether there are any polynomials for which $e^x = f(x)$ has infinitely many solutions, but whether there are any nonzero polynomials for which it does not.

6. Oct 13, 2013

### D H

Staff Emeritus
Rather than asking us to prove that there exists no infinite subset on which ez-p(z)=0, perhaps you meant to ask use to prove that for any polynomial p(z), ez-p(z)=0 is only true on a countable subset of the complex plane.

7. Oct 13, 2013

### Citan Uzuki

That's still a pretty easy problem though isn't it? After all any uncountable subset of the plane would have an accumulation point in the plane, and the only way the zeros of holomorphic function can have an accumulation point in the domain is if it is identically zero. Since ez is not a polynomial, ez - p(z) is not identically zero, and hence can only be zero on a countable subset. Q.E.D.

8. Oct 13, 2013

### Citan Uzuki

Well, it looks like Office Shredder is as foolish as possible -- because for every nonzero polynomial $f(z)$, the function $e^z - f(z)$ has infinitely many zeros in the complex plane.

Proof: Suppose that $e^z - f(z)$ has only finitely many zeros, say, $z_1, z_2, \ldots, z_n$ (where the zeros are listed with their multiplicity). Let $g(z) = \prod_{j=1}^{n} (z-z_j)$. Then after removing the removable singularities, we have that:

$$\frac{e^z - f(z)}{g(z)}$$

is an entire function which is never zero, and hence is equal to $e^{h(z)}$ for some entire function h. Now, observe that for all sufficiently large z, $|g(z)| \geq 1$ and $|f(z)| \leq e^{|z|}$. Therefore for all sufficiently large z:

$$e^{\Re (h(z))} = |e^{h(z)}| = \left| \frac{e^z - f(z)}{g(z)} \right| \leq |e^z - f(z)| \leq |e^z| + |f(z)| \leq 2e^{|z|} \leq e^{2|z|}$$

And hence for all sufficiently large z:

$$\Re(h(z)) \leq 2|z|$$

By applying the Borel-Caratheordory theorem, it follows that |h(z)| grows at most linearly with |z|, which in turn implies that h(z) must actually be a linear polynomial. That is, for some complex constants a and b, we have $h(z) = az + b$.

Now, recall that we defined h(z) such that:

$$\frac{e^z - f(z)}{g(z)} = e^{h(z)} = e^{az+b}$$

So:

$$e^z = f(z) + e^{az+b}g(z)$$

And thus:

$$1 = f(z)e^{-z} + e^{(a-1)z + b}g(z)$$

Take limits of both sides as $z \rightarrow \infty$ along the positive real axis. Clearly $f(z) e^{-z} \rightarrow 0$. Since the left hand side is a constant 1, this implies that $e^{(a-1)z + b}g(z) \rightarrow 1$. Now, if $\Re(a-1) < 0$, then as $z \rightarrow \infty$ along the positive real axis, we have that $e^{(a-1)z + b}g(z) \rightarrow 0$, and if $\Re(a-1) > 0$, then $e^{(a-1)z + b}g(z) \rightarrow \infty$. Therefore $\Re(a-1) = 0$, and so $|e^{(a-1)z + b}|$ is constant on the real line. Since $|e^{(a-1)z + b}g(z)| \rightarrow |1| = 1$ as $z \rightarrow \infty$ along the positive real axis, this implies that $|g(z)|$ approaches a nonzero constant as $z \rightarrow \infty$ along the positive real axis. This is only possible if g has degree zero -- i.e., g is a constant. Letting $c = e^b g$, we can rewrite our equation as:

$$1 = f(z)e^{-z} + ce^{(a-1)z}$$

We still have that the second term on the RHS approaches 1 as $z \rightarrow \infty$ along the positive real axis. We also know that $\Re(a-1) = 0$ -- i.e. a-1 is pure imaginary. Now, if a-1 is a nonzero imaginary number, $e^{(a-1)z}$ will oscillate and fail to approach a limit as $z \rightarrow \infty$ along the positive real axis. Therefore, a-1 = 0. Which means the second term on the RHS is just the constant c, and so we must have c=1. Hence:

$$1 = f(z)e^{-z} + 1$$

Which by simple algebra, implies that $f(z) = 0$. Q.E.D.

9. Oct 13, 2013

### verty

What about bounded infinite subsets?

10. Oct 13, 2013

### Office_Shredder

Staff Emeritus
I had a painful construction of a counterexample I just thought this would be a fun way to ask the challenge... I didn't realize it was possible to solve the problem for so many polynomials at the same time! (letting the constant polynomial solve the problem was an oversight on my part)

Last edited: Oct 13, 2013