Chapter A6, Problem 65E

(a)

To determine

To show: $\frac{f\left(x\right)+f\left(-x\right)}{2}$ is even and $\frac{f\left(x\right)-f\left(-x\right)}{2}$ is odd.

Answer to Problem 65E

Since $E\left(x\right)=E\left(-x\right)$ ,itis an Even Function, and since $O\left(x\right)=-O\left(-x\right)$ , itis an Odd Function.

Explanation of Solution

Given:

  $f\left(x\right)=\frac{f\left(x\right)+f\left(-x\right)}{2}+\frac{f\left(x\right)-f\left(-x\right)}{2}$

Calculation:

Let us assume two functions $E\left(x\right)andO\left(x\right)$ such that,

  $\begin{array}{l}E\left(x\right)=\frac{f\left(x\right)+f\left(-x\right)}{2}\\ O\left(x\right)=\frac{f\left(x\right)-f\left(-x\right)}{2}\end{array}$

Now as we can calculate,

  $\begin{array}{l}E\left(x\right)+O\left(x\right)=\frac{f\left(x\right)+f\left(-x\right)}{2}+\frac{f\left(x\right)-f\left(-x\right)}{2}\\ =\frac{2f\left(x\right)}{2}\\ =f\left(x\right)\end{array}$

Also now we calculate $E\left(-x\right)andO\left(-x\right)$

  $\begin{array}{l}E\left(-x\right)=\frac{f\left(-x\right)+f\left(-\left(-x\right)\right)}{2}\\ \boxed{E\left(-x\right)=\frac{f\left(x\right)+f\left(-x\right)}{2}}\\ O\left(-x\right)=\frac{f\left(-x\right)-f\left(-\left(-x\right)\right)}{2}\\ O\left(-x\right)=\frac{f\left(-x\right)-f\left(x\right)}{2}\end{array}$

Since $E\left(x\right)=E\left(-x\right)$ ,itis an Even Function, and since $O\left(x\right)=-O\left(-x\right)$ , itis an Odd Function.

Hence proved

Conclusion:

Here, $E\left(x\right)=E\left(-x\right)$ ,itis an Even Function, and since $O\left(x\right)=-O\left(-x\right)$ , itis an Odd Function.

(b)

To determine

To identify: the even and odd parts of $f$ .

Answer to Problem 65E

Even part is $\cosh x$

Odd part is $\sinh x$

Explanation of Solution

Given:

  $f\left(x\right)=e^x$

Calculation:

When $f\left(x\right)=e^x$

Even part

= $\begin{array}{l}\frac{e^x+e^{-x}}{2}\\ =\cosh x\end{array}$

Odd part

= $\begin{array}{l}\frac{e^x-e^{-x}}{2}\\ =\sinh x\end{array}$

Conclusion:

Even part is $\cosh x$

Odd part is $\sinh x$