3.2 Polynomial functions  (Page 3/2)

If $p\left(z\right)={\sum }_{k=0}^n{a}_k{z}^k$ and $q\left(z\right)={\sum }_{j=0}^m{b}_j{z}^j$ are two polynomials, it certainly seems clear that they determine the same functiononly if they have identical coefficients. This is true, but by no means an obvious fact.Also, it seems clear that, as $\left|z\right|$ gets larger and larger, a polynomial function is more and more comparable to its leading term $a_n{z}^n.$ We collect in the next theorem some elementary properties of polynomial functions, and in particular we verify the above “uniqueness of coefficients” resultand the “behavior at infinity” result.

1. Suppose $p\left(z\right)={\sum }_{k=0}^n{a}_k{z}^k$ is a nonconstant polynomial of degree $n>0.$ Then $p\left(z\right)=0$ for at most $n$ distinct complex numbers.
2. If $r$ is a polynomial for which $r\left(z\right)=0$ for an infinite number of distinct points, then $r$ is the zero polynomial. That is, all of its coefficients are 0.
3. Suppose $p$ and $q$ are nonzero polynomials, and assume that $p\left(z\right)=q\left(z\right)$ for an infinite number of distinct points. Then $p\left(z\right)=q\left(z\right)$ for all $z,$ and $p$ and $q$ have the same coefficients. That is, they are the same polynomial.
4. Let $p\left(z\right)={\sum }_{j=0}^n{c}_j{z}^j$ be a polynomial of degree $n>0.$ Then there exist positive constants $m$ and $B$ such that
   $$\frac{|c_n|}{2}{\left|z\right|}^n\le \left|p\left(z\right)\right|\le {M\left|z\right|}^n$$
   for all complex numbers $z$ for which $\left|z\right|\ge B.$ That is, For all complex numbers $z$ with $\left|z\right|\ge B,$ the numbers $\left|p\right(z\left)\right|$ and ${\left|z\right|}^n$ are “comparable.”
5. If $f:[0,\infty )\to C$ is defined by $f\left(x\right)=\sqrt{x},$ then there is no polynomial $p$ for which $f\left(x\right)=p\left(x\right)$ for all $x\ge 0.$ That is, the square root function does not agree with any polynomial function.

We prove part (1) using an argument by contradiction. Thus, suppose there does exist a counterexample to the claim, i.e., a nonzero polynomial $p$ of degree $n$ and $n+1$ distinct points $\{c_1,c_2,...,c_{n+1}\}$ for which $p\left(c_j\right)=0$ for all $1\le j\le n+1.$ From the set of all such counterexamples, let $p_0$ be one with minimum degree $n_0.$ That is, the claim in part (1) is true for any polynomial whose degree is smaller than $n_0.$ We write

and we suppose that $p_0\left(c_j\right)=0$ for $j=1$ to $n_0+1,$ where these $c_k$ 's are distinct complex numbers. We use next the Root Theorem (part (d) of [link] ) to write $p_0\left(z\right)=(z-c_{n_0+1})q\left(z\right),$ where $q\left(z\right)={\sum }_{k=0}^{n_0-1}{b}_k{z}^k.$ We have that $q$ is a polynomial of degree $n_0-1$ and the leading coefficient $a_{n_0}$ of $p_0$ equals the leading coefficient $b_{n_0-1}$ of $q.$ Note that for $1\le j\le n_0$ we have

which implies that $q\left(c_j\right)=0$ for $1\le j\le n_0,$ since $c_j-c_{n_0+1}\ne 0.$ But, since $\text{deg}\left(q\right)<n_0,$ the nonzero polynomial $q$ can not be a counterexample to part (1), implying that $q\left(z\right)=0$ for at most $n_0-1$ distinct points. We have arrived at a contradiction, and part (1) is proved.

Next, let $r$ be a polynomial for which $r\left(z\right)=0$ for an infinite number of distinct points.It follows from part (1) that $r$ cannot be a nonzero polynomial, for in that case it would have a degree $n\ge 0$ and could be 0 for at most $n$ distinct points. Hence, $r$ is the zero polynomial, and part (2) is proved.

Now, to see part (3), set $r=p-q.$ Then $r$ is a polynomial for which $r\left(z\right)=0$ for infinitely many $z$ 's. By part (2), it follows then that $r\left(z\right)=0$ for all $z,$ whence $p\left(z\right)=q\left(z\right)$ for all $z.$ Moreover, $p-q$ is the zero polynomial, all of whose coefficients are 0, and this implies that the coefficients for $p$ and $q$ are identical.