# N roots?

1. Mar 10, 2010

### Mentallic

The function $$y=e^x$$ can be expanded using the power series, thus $$y=e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...$$
This is a polynomial of infinite degree, and the theorem that says a polynomial must have at least one root in the complex field, and thus this extends to a polynomial of nth degree having n roots (not necessarily distinct).

However, for $e^x=0$ there are zero roots. Why is this possible when, clearly by the power series for $e^x$ it should be an infinite degree polynomial with infinite roots (in the complex plane)?

2. Mar 10, 2010

### Staff: Mentor

According to wiki:

so perhaps that's where the problem lies.

3. Mar 10, 2010

### Mentallic

Was that definition placed before or after this exception was noticed?

4. Mar 10, 2010

### Staff: Mentor

No idea.

Actually, I have checked other source - Wolfram Mathworld - and their definition of polynomial doesn't state it has to be finite.

5. Mar 10, 2010

### HallsofIvy

Staff Emeritus
In any case, the statement "a polynomial of degree n has n zeros (counting multiplicity) over the complex field" is only true for n finite.

6. Mar 10, 2010

### g_edgar

The phrase "polynomial of infinite degree" is never never used by real mathematicians. In particular $$e^z$$ is not (repeat not) a polynomial in $$z$$.