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Textbook by: Lawrence Baggett. E-mail the author

# Polynomial Functions

Module by: Lawrence Baggett. E-mail the author

Summary: A module about polynomial functions, with a theorem establishing some elementary properties of polynomial functions, such as the uniqueness of coefficients and the behavior at infinity.

If p(z)=k=0nakzkp(z)=k=0nakzk and q(z)=j=0mbjzjq(z)=j=0mbjzj are two polynomials, it certainly seems clear that they determine the same function only if they have identical coefficients. This is true, but by no means an obvious fact. Also, it seems clear that, as |z||z| gets larger and larger, a polynomial function is more and more comparable to its leading term anzn.anzn. We collect in the next theorem some elementary properties of polynomial functions, and in particular we verify the above “uniqueness of coefficients” result and the “behavior at infinity” result.

## Theorem 1

1. Suppose p(z)=k=0nakzkp(z)=k=0nakzk is a nonconstant polynomial of degree n>0.n>0. Then p(z)=0p(z)=0 for at most nn distinct complex numbers.
2. If rr is a polynomial for which r(z)=0r(z)=0 for an infinite number of distinct points, then rr is the zero polynomial. That is, all of its coefficients are 0.
3. Suppose pp and qq are nonzero polynomials, and assume that p(z)=q(z)p(z)=q(z) for an infinite number of distinct points. Then p(z)=q(z)p(z)=q(z) for all z,z, and pp and qq have the same coefficients. That is, they are the same polynomial.
4. Let p(z)=j=0ncjzjp(z)=j=0ncjzj be a polynomial of degree n>0.n>0. Then there exist positive constants mm and BB such that
|cn|2|z|n|p(z)|M|z|n|cn|2|z|n|p(z)|M|z|n
(1)
for all complex numbers zz for which |z|B.|z|B. That is, For all complex numbers zz with |z|B,|z|B, the numbers |p(z)||p(z)| and |z|n|z|n are “comparable.”
5. If f:[0,)Cf:[0,)C is defined by f(x)=x,f(x)=x, then there is no polynomial pp for which f(x)=p(x)f(x)=p(x) for all x0.x0. That is, the square root function does not agree with any polynomial function.

### Proof

We prove part (1) using an argument by contradiction. Thus, suppose there does exist a counterexample to the claim, i.e., a nonzero polynomial pp of degree nn and n+1n+1 distinct points {c1,c2,...,cn+1}{c1,c2,...,cn+1} for which p(cj)=0p(cj)=0 for all 1jn+1.1jn+1. From the set of all such counterexamples, let p0p0 be one with minimum degree n0.n0. That is, the claim in part (1) is true for any polynomial whose degree is smaller than n0.n0. We write

p 0 ( z ) = k = 0 n 0 a k z k , p 0 ( z ) = k = 0 n 0 a k z k ,
(2)

and we suppose that p0(cj)=0p0(cj)=0 for j=1j=1 to n0+1,n0+1, where these ckck's are distinct complex numbers. We use next the Root Theorem (part (d) of (Reference)) to write p0(z)=(z-cn0+1)q(z),p0(z)=(z-cn0+1)q(z), where q(z)=k=0n0-1bkzk.q(z)=k=0n0-1bkzk. We have that qq is a polynomial of degree n0-1n0-1 and the leading coefficient an0an0 of p0p0 equals the leading coefficient bn0-1bn0-1 of q.q. Note that for 1jn01jn0 we have

0 = p 0 ( c j ) = ( c j - c n 0 + 1 ) q ( c j ) , 0 = p 0 ( c j ) = ( c j - c n 0 + 1 ) q ( c j ) ,
(3)

which implies that q(cj)=0q(cj)=0 for 1jn0,1jn0, since cj-cn0+10.cj-cn0+10. But, since deg(q)<n0,deg(q)<n0, the nonzero polynomial qq can not be a counterexample to part (1), implying that q(z)=0q(z)=0 for at most n0-1n0-1 distinct points. We have arrived at a contradiction, and part (1) is proved.

Next, let rr be a polynomial for which r(z)=0r(z)=0 for an infinite number of distinct points. It follows from part (1) that rr cannot be a nonzero polynomial, for in that case it would have a degree n0n0 and could be 0 for at most nn distinct points. Hence, rr is the zero polynomial, and part (2) is proved.

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

To prove the first inequality in part (4), suppose that |z|>1,|z|>1, and from the backwards triangle inequality, note that

| p ( z ) | = | k = 0 n c k z k | = | z | n | k = 0 n c k z n - k | = | z | n | ( k = 0 n - 1 c k z n - k ) + c n | | z | n ( | c n | - | k = 0 n - 1 c k z n - k | ) | z | n ( | c n | - k = 0 n - 1 | c k | | z | n - k ) | z | n ( | c n | - k = 0 n - 1 | c k | | z | ) | z | n ( | c n | - 1 | z | k = 0 n - 1 | c k | ) . | p ( z ) | = | k = 0 n c k z k | = | z | n | k = 0 n c k z n - k | = | z | n | ( k = 0 n - 1 c k z n - k ) + c n | | z | n ( | c n | - | k = 0 n - 1 c k z n - k | ) | z | n ( | c n | - k = 0 n - 1 | c k | | z | n - k ) | z | n ( | c n | - k = 0 n - 1 | c k | | z | ) | z | n ( | c n | - 1 | z | k = 0 n - 1 | c k | ) .
(4)

Set BB equal to the constant (2/|cn|)j=0n-1|cj|.(2/|cn|)j=0n-1|cj|. Then, replacing the 1/|z|1/|z| in the preceding calculation by 1/B,1/B, we obtain

| p ( z ) | m | z | n | p ( z ) | m | z | n
(5)

for every zz for which |z|B.|z|B. This proves the first half of part (4).

To get the other half of part (4), suppose again that |z|>1.|z|>1. We have

| p ( z ) | k = 0 n | c k | | z | k k = 0 n | c k | | z | n , | p ( z ) | k = 0 n | c k | | z | k k = 0 n | c k | | z | n ,
(6)

so that we get the other half of part (4) by setting M=k=0n|ck|.M=k=0n|ck|.

Finally, to see part (5), suppose that there does exist a polynomial pp of degree nn such that x=p(x)x=p(x) for all x0.x0. Then x=(p(x))2x=(p(x))2 for all x0.x0. Now p2p2 is a polynomial of degree 2n.2n. By part (2), the two polynomials q(x)=xq(x)=x and (p(x))2(p(x))2 must be the same, implying that they have the same degree. However, the degree of qq is 1, which is odd, and the degree of p2p2 is 2n,2n, which is even. Hence, we have arrived at a contradiction.

## Exercise 1

1. Let r(z)=p(z)/q(z)r(z)=p(z)/q(z) and r'(z)=p'(z)/q'(z)r'(z)=p'(z)/q'(z) be two rational functions. Suppose r(z)=r'(z)r(z)=r'(z) for infinitely many zz's. Prove that r(z)=r'(z)r(z)=r'(z) for all zz in the intersection of their domains. Is it true that p=p'p=p' and q=q'?q=q'?
2. Let pp and qq be polynomials of degree nn and mm respectively, and define a rational function rr by r=p/q.r=p/q. Prove that there exist positive constants CC and BB such that |r(z)|<C|z|n-m|r(z)|<C|z|n-m for all complex numbers zz for which |z|>B.|z|>B.
3. Define f:[0,)Rf:[0,)R by f(x)=x.f(x)=x. Show that there is no rational function rr such that f(x)=r(x)f(x)=r(x) for all x0.x0. That is, the square root function does not agree with a rational function.
4. Define the real-valued function rr on RR by r(x)=1/(1+x2).r(x)=1/(1+x2). Prove that there is no polynomial pp such that p(x)=r(x)p(x)=r(x) for infinitely many real numbers xx.
5. If ff is the real-valued function of a real variable given by f(x)=|x|,f(x)=|x|, show that ff is not a rational function. HINT: Suppose |x|=p(x)/q(x).|x|=p(x)/q(x). Then |x|q(x)=p(x)|x|q(x)=p(x) implying that |x|q(x)|x|q(x) is a polynomial s(x).s(x). Now use Theorem 3.1 to conclude that p(x)=xq(x)p(x)=xq(x) for all xx and that p(x)=-xq(x)p(x)=-xq(x) for all x.x.
6. Let ff be any complex-valued function of a complex variable, and let c1,...,cnc1,...,cn be nn distinct complex numbers that belong to the domain of f.f. Show that there does exist a polynomial pp of degree nn such that p(cj)=f(cj)p(cj)=f(cj) for all 1jn.1jn. HINT: Describe pp in factored form.
7. Give examples to show that the maximum and minimum of two polynomials need not be a polynomial or even a rational function.

Very important is the definition of the compositiongfgf of two functions ff and g.g.

Definition 1:

Let f:STf:ST and g:TUg:TU be functions. We define a function gf,gf, with domain SS and codomain U,U, by (gf)(x)=g(f(x)).(gf)(x)=g(f(x)).

If f:ST,f:ST,g:TS,g:TS, and gf(x)=xgf(x)=x for all xS,xS, then gg is called a left inverse of f.f. If fg(y)=yfg(y)=y for all yT,yT, then gg is called a right inverse for f.f. If gg is both a left inverse and a right inverse, then gg is called an inverse for f,f,ff is called invertible, and we denote gg by f-1.f-1.

## Exercise 2

1. Suppose f:STf:ST has a left inverse. Prove that ff is 1-1.
2. Suppose f:STf:ST has a right inverse. Prove that ff is onto.
3. Show that the composition of two polynomials is a polynomial and that the composition of two rational functions is a rational function. HINT: If pp is a polynomial, show by induction that pnpn is a polynomial. Now use (Reference).
4. Find formulas for gfgf and fgfg for the following. What are the domains of these compositions?
1. f(x)=1+x2f(x)=1+x2 and g(x)=1/(1+x)1/2.g(x)=1/(1+x)1/2.
2. f(x)=x/(x+1)f(x)=x/(x+1) and g(x)=x/(1-x).g(x)=x/(1-x).
3. f(x)=ax+bf(x)=ax+b and g(x)=cx+d.g(x)=cx+d.

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