Chapter 3.2, Problem 77E

(a)

To determine

To show: that a polynomial $P\left(x\right)$ that contains only off powers of $x$ is an odd function.

Explanation of Solution

Given information:

Recall that a function $f$ is odd if $f\left(-x\right)=-f\left(x\right)$ or even if $f\left(-x\right)=f\left(x\right)$ for all real $x$ .

Calculation:

Need to show that a polynomial $P\left(x\right)$ that contains only off powers of $x$ is an odd function.

Let us consider,

  $P_0\left(x\right)=a_{2n-1}{x}^{2n-1}+a_{2n-3}{x}^{2n-3}+a_{2n-5}{x}^{2n-5}+\mathrm{...}+a_3{x}^3+a_{1}x$

Observe that all the powers of $x$ are odd integer only.

Also, there is no constant term while a constant is an even power of $x$ .

To check the even or odd, replace $x$ by $-x$ .

  $\begin{array}{c}{P}_0\left(-x\right)=a_{2n-1}{\left(-x\right)}^{2n-1}+a_{2n-3}{\left(-x\right)}^{2n-3}+a_{2n-5}{\left(-x\right)}^{2n-5}+\mathrm{...}+a_3{\left(-x\right)}^3+a_1\left(-x\right)\\ =a_{2n-1}{\left(-1\right)}^{2n-1}{\left(x\right)}^{2n-1}+a_{2n-3}{\left(-1\right)}^{2n-3}{\left(x\right)}^{2n-3}+a_{2n-5}{\left(-1\right)}^{2n-5}{\left(x\right)}^{2n-5}+\mathrm{...}+a_3{\left(-1\right)}^3{\left(x\right)}^3+a_1{\left(-1\right)}^1\left(x\right)\\ =-\left[a_{2n-1}{x}^{2n-1}+a_{2n-3}{x}^{2n-3}+a_{2n-5}{x}^{2n-5}+\mathrm{...}+a_3{x}^3+a_{1}x\right]\\ =-P_0\left(x\right)\end{array}$

So, $P_0\left(-x\right)=-P_0\left(x\right)$ confirms that $P_0\left(x\right)$ is an odd function.

Therefore, a polynomial $P\left(x\right)$ that contains only odd powers of $x$ is an odd function.

(b)

To determine

To show: that a polynomial $P\left(x\right)$ that contains only even powers of $x$ is an even function.

Explanation of Solution

Given information:

Recall that a function $f$ is odd if $f\left(-x\right)=-f\left(x\right)$ or even if $f\left(-x\right)=f\left(x\right)$ for all real $x$ .

Calculation:

Need to show that a polynomial $P\left(x\right)$ that contains only even powers of $x$ is an even function.

Let us consider,

  $P_E\left(x\right)=a_{2n}{x}^{2n}+a_{2n-2}{x}^{2n-2}+a_{2n-4}{x}^{2n-4}+\mathrm{...}+a_2{x}^2+a_0$

To check the even or odd, replace $x$ by $-x$ .

  $\begin{array}{c}{P}_E\left(-x\right)=a_{2n}{\left(-x\right)}^{2n}+a_{2n-2}{\left(-x\right)}^{2n-2}+a_{2n-4}{\left(-x\right)}^{2n-4}+\mathrm{...}+a_2{\left(-x\right)}^2+a_0\\ =a_{2n}{\left(-1\right)}^{2n}{\left(x\right)}^{2n}+a_{2n-2}{\left(-1\right)}^{2n-2}{\left(x\right)}^{2n-2}+a_{2n-4}{\left(-1\right)}^{2n-4}{\left(x\right)}^{2n-4}+\mathrm{...}+a_2{\left(-1\right)}^2{\left(x\right)}^2+a_1\\ =\left[a_{2n}{x}^{2n}+a_{2n-2}{x}^{2n-2}+a_{2n-4}{x}^{2n-4}+\mathrm{...}+a_2{x}^2+a_1\right]\\ =P_E\left(x\right)\end{array}$

So, $P_E\left(-x\right)=P_E\left(x\right)$ confirms that $P_E\left(x\right)$ is an even function.

Therefore, a polynomial $P_E\left(x\right)$ that contains only even powers of $x$ is an even function.

(c)

To determine

To show: that a polynomial $P\left(x\right)$ that contains both odd and even powers of $x$ .

Explanation of Solution

Given information:

Recall that a function $f$ is odd if $f\left(-x\right)=-f\left(x\right)$ or even if $f\left(-x\right)=f\left(x\right)$ for all real $x$ .

Calculation:

Need to show that a polynomial $P\left(x\right)$ that contains both odd and even powers of $x$ .

Let us consider,

  $P\left(x\right)=a_{2n}{x}^{2n}+a_{2n-1}{x}^{2n-1}+a_{2n-2}{x}^{2n-2}+a_{2n-3}{x}^{2n-3}+a_{2n-4}{x}^{2n-4}+\mathrm{...}+a_3{x}^3+a_2{x}^2+a_{1}x+a_0$

To check the even or odd, replace $x$ by $-x$ .

  $\begin{array}{c}P\left(-x\right)=a_{2n}{\left(-x\right)}^{2n}+a_{2n-1}{\left(-x\right)}^{2n-1}+a_{2n-2}{\left(-x\right)}^{2n-2}+a_{2n-3}{\left(-x\right)}^{2n-3}+a_{2n-4}{\left(-x\right)}^{2n-4}+\mathrm{...}+a_3{\left(-x\right)}^3+a_2{\left(-x\right)}^2+a_1\left(-x\right)+a_0\\ =a_{2n}{\left(x\right)}^{2n}-a_{2n-1}{\left(x\right)}^{2n-1}+a_{2n-2}{\left(x\right)}^{2n-2}-a_{2n-3}{\left(x\right)}^{2n-3}+a_{2n-4}{\left(x\right)}^{2n-4}+\mathrm{...}-a_3{\left(x\right)}^3+a_2{\left(x\right)}^2-a_{1}x+a_0\\ \ne P\left(x\right)\end{array}$

So, $P\left(-x\right)\ne P\left(x\right)$ and $P\left(-x\right)\ne -P\left(x\right)$

Therefore, a polynomial $P_E\left(x\right)$ that contains both odd and even powers of $x$ .

(d)

To determine

To express: the given function.

Answer to Problem 77E

  $P\left(x\right)=P_0\left(x\right)+P_E\left(x\right)$

Where $P_0\left(x\right)=x^5+6x^3-2x$ and $P_E\left(x\right)=-x^2+5$ .

Explanation of Solution

Given information:

Recall that a function $f$ is odd if $f\left(-x\right)=-f\left(x\right)$ or even if $f\left(-x\right)=f\left(x\right)$ for all real $x$ .

Given function:

  $P\left(x\right)=x^5+6x^3-x^2-2x+5$

Calculation:

Consider the function,

  $P\left(x\right)=x^5+6x^3-x^2-2x+5$

Need to express the function as the sum of odd function and even function.

Rewrite $P\left(x\right)$ as the sum of odd powers of $x$ and the sum of even powers of $x$ as,

  $\begin{array}{c}P\left(x\right)=x^5+6x^3-x^2-2x+5\\ =\left(x^5+6x^3-2x\right)+\left(-x^2+5\right)\\ =P_0\left(x\right)+P_E\left(x\right)\end{array}$

Where $P_0\left(x\right)=x^5+6x^3-2x$ and $P_E\left(x\right)=-x^2+5$ .