3.2 Polynomial functions (Page 2/2)

Analysis of functions of a single Page 2 / 2

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

\begin{matrix} | p (z) | & = & | \sum_{k = 0}^{n} c_{k} z^{k} | \\ = & {| z |}^{n} | \sum_{k = 0}^{n} \frac{c_{k}}{z^{n - k}} | \\ = & {| z |}^{n} | (\sum_{k = 0}^{n - 1} \frac{c_{k}}{z^{n - k}}) + c_{n} | \\ \geq & {| z |}^{n} (| c_{n} | - | \sum_{k = 0}^{n - 1} \frac{c_{k}}{z^{n - k}} |) \\ \geq & {| z |}^{n} (| c_{n} | - \sum_{k = 0}^{n - 1} \frac{| c_{k} |}{{| z |}^{n - k}}) \\ \geq & {| z |}^{n} (| c_{n} | - \sum_{k = 0}^{n - 1} \frac{| c_{k} |}{| z |}) \\ \geq & {| z |}^{n} (| c_{n} | - \frac{1}{| z |} \sum_{k = 0}^{n - 1} | c_{k} |) . \end{matrix}

Set $B$ equal to the constant $(2 / | c_{n} |) \sum_{j = 0}^{n - 1} | c_{j} | .$ Then, replacing the $1 / | z |$ in the preceding calculation by $1 / B,$ we obtain

| p (z) | \geq {m | z |}^{n}

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

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

| p (z) | \leq \sum_{k = 0}^{n} | c_{k} {| | z |}^{k} \leq \sum_{k = 0}^{n} | c_{k} {| | z |}^{n},

so that we get the other half of part (4) by setting $M = \sum_{k = 0}^{n} | c_{k} | .$

Finally, to see part (5), suppose that there does exist a polynomial $p$ of degree $n$ such that $\sqrt{x} = p (x)$ for all $x \geq 0 .$ Then $x = {(p (x))}^{2}$ for all $x \geq 0 .$ Now $p^{2}$ is a polynomial of degree $2 n .$ By part (2), the two polynomials $q (x) = x$ and ${(p (x))}^{2}$ must be the same, implying that they have the same degree. However, the degree of $q$ is 1, which is odd, and the degree of $p^{2}$ is $2 n,$ which is even. Hence, we have arrived at a contradiction.

Let $r (z) = p (z) / q (z)$ and $r^{'} (z) = p^{'} (z) / q^{'} (z)$ be two rational functions. Suppose $r (z) = r^{'} (z)$ for infinitely many $z$ 's. Prove that $r (z) = r^{'} (z)$ for all $z$ in the intersection of their domains. Is it true that $p = p^{'}$ and $q = q^{'} ?$
Let $p$ and $q$ be polynomials of degree $n$ and $m$ respectively, and define a rational function $r$ by $r = p / q .$ Prove that there exist positive constants $C$ and $B$ such that ${| r (z) | < C | z |}^{n - m}$ for all complex numbers $z$ for which $| z | > B .$
Define $f : [0, \infty) \to R$ by $f (x) = \sqrt{x} .$ Show that there is no rational function $r$ such that $f (x) = r (x)$ for all $x \geq 0 .$ That is, the square root function does not agree with a rational function.
Define the real-valued function $r$ on $R$ by $r (x) = 1 / (1 + x^{2}) .$ Prove that there is no polynomial $p$ such that $p (x) = r (x)$ for infinitely many real numbers $x$ .
If $f$ is the real-valued function of a real variable given by $f (x) = | x |,$ show that $f$ is not a rational function. HINT: Suppose $| x | = p (x) / q (x) .$ Then $| x | q (x) = p (x)$ implying that $| x | q (x)$ is a polynomial $s (x) .$ Now use Theorem 3.1 to conclude that $p (x) = x q (x)$ for all $x$ and that $p (x) = - x q (x)$ for all $x .$
Let $f$ be any complex-valued function of a complex variable, and let $c_{1}, ..., c_{n}$ be $n$ distinct complex numbers that belong to the domain of $f .$ Show that there does exist a polynomial $p$ of degree $n$ such that $p (c_{j}) = f (c_{j})$ for all $1 \leq j \leq n .$ HINT: Describe $p$ in factored form.
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 composition $g \circ f$ of two functions $f$ and $g .$

Let $f : S \to T$ and $g : T \to U$ be functions. We define a function $g \circ f,$ with domain $S$ and codomain $U,$ by $(g \circ f) (x) = g (f (x)) .$

If $f : S \to T,$ $g : T \to S,$ and $g \circ f (x) = x$ for all $x \in S,$ then $g$ is called a left inverse of $f .$ If $f \circ g (y) = y$ for all $y \in T,$ then $g$ is called a right inverse for $f .$ If $g$ is both a left inverse and a right inverse, then $g$ is called an inverse for $f,$ $f$ is called invertible , and we denote $g$ by $f^{- 1} .$

Suppose $f : S \to T$ has a left inverse. Prove that $f$ is 1-1.
Suppose $f : S \to T$ has a right inverse. Prove that $f$ is onto.
Show that the composition of two polynomials is a polynomial and that the composition of two rational functions is a rational function.HINT: If $p$ is a polynomial, show by induction that $p^{n}$ is a polynomial. Now use [link] .
Find formulas for g ∘ f and f ∘ g for the following. What are the domains of these compositions?
1. $f (x) = 1 + x^{2}$ and $g (x) = 1 / {(1 + x)}^{1 / 2} .$
2. $f (x) = x / (x + 1)$ and $g (x) = x / (1 - x) .$
3. $f (x) = a x + b$ and $g (x) = c x + d .$

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Source: OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1

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