Chapter 1.8, Problem 3E

a.

To determine

To prove: that g is even and h is odd.

Explanation of Solution

Given Information:

  $\begin{array}{l}g\left(x\right)=\frac{1}{2}\left[f\left(x\right)+f\left(-x\right)\right]\cdots \left(i\right)\\ h\left(x\right)=\frac{1}{2}\left[f\left(x\right)-f\left(-x\right)\right]\cdots \left(ii\right)\end{array}$

Formula Used:

If a function g is even then $g\left(x\right)=g\left(-x\right)$

If a function h is odd then $h\left(-x\right)=-h\left(x\right)$

Calculation:

First find $g\left(-x\right)$

For $g\left(-x\right)$ just replace x by $-x$ in equation $\left(i\right)$ ,

  $\begin{array}{c}g\left(-x\right)=\frac{1}{2}\left[f\left(-x\right)+f\left(-\left(-x\right)\right)\right]\\ g\left(-x\right)=\frac{1}{2}\left[f\left(-x\right)+f\left(x\right)\right]\\ g\left(-x\right)=g\left(x\right)\end{array}$

Hence g is even.

Now for $h\left(x\right)$

First we find $h\left(-x\right)$

To find $h\left(-x\right)$ replace x by $-x$ in equation $\left(ii\right)$

  $\begin{array}{c}h\left(-x\right)=\frac{1}{2}\left[f\left(-x\right)-f\left(-\left(-x\right)\right)\right]\\ h\left(-x\right)=\frac{1}{2}\left[f\left(-x\right)-f\left(x\right)\right]\\ h\left(-x\right)=-\frac{1}{2}\left[f\left(x\right)-f\left(-x\right)\right]\\ =-h\left(x\right)\\ h\left(-x\right)=-h\left(x\right)\end{array}$

Hence h is odd.

Proved

b.

To determine

To prove: that any function can be written as a sum of even and odd function.

Answer to Problem 3E

  $f\left(x\right)=g\left(x\right)+h\left(x\right)$

Explanation of Solution

Given Information:

From part (a), $g\left(x\right)$ is even function and $h\left(x\right)$ is odd function.

  $\begin{array}{l}g\left(x\right)=\frac{1}{2}\left[f\left(x\right)+f\left(-x\right)\right]\cdots \left(i\right)\\ h\left(x\right)=\frac{1}{2}\left[f\left(x\right)-f\left(-x\right)\right]\cdots \left(ii\right)\end{array}$

Calculation:

Adding $g\left(x\right)\&h\left(x\right)$

  $\begin{array}{c}g\left(x\right)+h\left(x\right)=\frac{1}{2}\left[f\left(x\right)+f\left(-x\right)\right]+\frac{1}{2}\left[f\left(x\right)-f\left(-x\right)\right]\\ =\frac{1}{2}\left[f\left(x\right)+f\left(-x\right)+f\left(x\right)-f\left(-x\right)\right]\\ =\frac{1}{2}\left[2f\left(x\right)\right]\\ =f\left(x\right)\end{array}$

So, $f\left(x\right)=g\left(x\right)+h\left(x\right)$

c.

To determine

To write: each function as a sum of even and odd functions using the result of part (b).

Answer to Problem 3E

Even function for $f\left(x\right)=x^2-2x+1$ is x² +1 and odd function is $-2x$

Even function for $k\left(x\right)=\frac{1}{x+1}$ is $\frac{1}{1-x^2}$ and odd function is

Explanation of Solution

Given Information:

  $f\left(x\right)=x^2-2x+1$ and $k\left(x\right)=\frac{1}{x+1}$

Calculation:

First find the even function $g\left(x\right)$ for $f\left(x\right)=x^2-2x+1$

  $\begin{array}{c}g\left(x\right)=\frac{1}{2}\left(f\left(x\right)+f\left(-x\right)\right)\\ =\frac{1}{2}\left(x^2-2x+1+{\left(-x\right)}^2-2\left(-x\right)+1\right)\\ =\frac{1}{2}\left(x^2-2x+1+x^2+2x+1\right)\\ =\frac{1}{2}\left(2x^2+2\right)\\ =x^2+1\end{array}$

Now we will find the odd function $h\left(x\right)$ for $f\left(x\right)=x^2-2x+1$

  $\begin{array}{c}h\left(x\right)=\frac{1}{2}\left(f\left(x\right)-f\left(-x\right)\right)\\ =\frac{1}{2}\left(x^2-2x+1-{\left(-x\right)}^2+2\left(-x\right)-1\right)\\ =\frac{1}{2}\left(x^2-2x+1-x^2-2x-1\right)\\ =\frac{1}{2}\left(-4x\right)\\ =-2x\end{array}$

So even function for $f\left(x\right)=x^2-2x+1$ is x² +1 and odd function is $-2x$

Now $k\left(x\right)=\frac{1}{x+1}$

First find the even function $g\left(x\right)$ for $k\left(x\right)=\frac{1}{x+1}$

  $\begin{array}{c}g\left(x\right)=\frac{1}{2}\left(k\left(x\right)+k\left(-x\right)\right)\\ =\frac{1}{2}\left(\frac{1}{x+1}+\frac{1}{-x+1}\right)\\ =\frac{1}{2}\left(-\frac{2}{x^2-1}\right)\\ =\frac{1}{x^2-1}\end{array}$

Now we will find the odd function $h\left(x\right)$ for $k\left(x\right)=\frac{1}{x+1}$

  $\begin{array}{c}h\left(x\right)=\frac{1}{2}\left(k\left(x\right)-k\left(-x\right)\right)\\ =\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{-x+1}\right)\\ =\frac{1}{2}\left(\frac{2x}{x^2-1}\right)\\ =\frac{x}{x^2-1}\end{array}$

So even function for $k\left(x\right)=\frac{1}{x+1}$ is $\frac{1}{1-x^2}$ and odd function is $\frac{2x}{x^2-1}$ .