Does a polynomial have to have a finite number of terms?

[morson]

Does a polynomial:

\(\displaystyle \L\ \sum_{k=0}^n\ A_k x^{n - k}\)

have to have a finite number of terms? If n tends to infinity, does it just become your regular power series? I was wondering if the FTA applies to:

\(\displaystyle \L\ x - \frac{x^3}{3!}\ + \frac{x^5}{5!}\ - \frac{x^7}{7!}\ + ... = 0\)

... which of course has solutions \(\displaystyle x = n\pi\\), where n is some integer. I was wondering whether the left hand side of the equation has a "degree" of infinity, and so by the FTA, it has an infinite number of solutions. But then that would be giving a trigonometric function a "polynomial degree," which doesn't seem to make sense. At the moment I think that a polynomial must have a finite # of terms but I'd like to hear from you guys.

 

[tkhunny]

You probably do not have a definition of "infinity" that makes it a number.

 

[galactus]

Remember that pi is transcendental and can not be the root of a polynomial. Actually, this is more than likely a good field theory question.

*Formal definition of a polynomial:
Actually, Let R be a ring, then a polynomial with coefficients in R is an infinite formal sum \(\displaystyle \L\\\sum_{n=0}^{\infty}A_{n}x^{n}=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+............\)
where \(\displaystyle \L\\a_{i}\in{R}, \;\ a_{i}=0\) for all but a finite number of values of i. If for some \(\displaystyle i\geq{0}\) it is true that \(\displaystyle a_{i}\neq{0}\), the largest such values of i is the degree of f(x). If all \(\displaystyle a_{i}=0\), then the degree of f(x) is undefined. Which tells us that there is a highest coefficient that does not have zero in front and everything else is zero. So in that case polynomials are always finite.

*This being the definition from field theory.

In analysis a polynomial in x is of the form:

\(\displaystyle \L\\a_{n}x^{n}=a_{n-1}x^{n-1}+.......+a_{1}x+a_{0}\)
where n is a nonnegative integer and each coefficient \(\displaystyle a_{i}\) is a real number. If \(\displaystyle a_{n}\neq{0}\), then the polynomial has degree n.

Are you asking if a power series can be considered a polynomial?.
I suppose it can if one thinks of it as some limit. Such as in a Taylor or MacLaurin series.

A polynomial in calc and a polynomial in field theory are two different animals. Polynomials in calc are finite by definition.

pka will probably have something for you. This would be a nice proof, if there is one. Now you have me wondering. I am not as well-versed in group theory as I would like to be.

 

[morson]

Thanks galactus for your response. I guess I was wanting to make sense of:

\(\displaystyle \L\ \lim_{n \to \infty}\ \sum_{k = 0}^{n} \ A_k x^{n - k}\)

And I understand now that it can't really be a polynomial. Thanks for that definition, that cleared things up.

As for your first sentence, you must mean a polynomial with rational coefficients, right? Because, consider:

\(\displaystyle \L\ x^2 + 2x\pi - 3\pi^2 = 0\)

I was taught that a transcedental number was a number which can not be the root of a polynomial with integral coefficients.

 

[galactus]

Yes, morson, that's what I meant. I should've elaborated.