Exercise 1.3.71

Question

Let $f\item $ be a function with domain $\mathbb{R}$.
    \begin{enumerate}[label=(\alph*)]
\item Show that $E(x)=f(x)+f(-x)$ is an even function.
\item  Show that $O(x)=f(x)-f(-x)$ is an odd function.
\item  Prove that every function $f(x)$ can be written as a sum of an even function and an odd function.
\item  Express the function $f(x)=2^{x}+(x-3)^{2}$ as a sum of an even function and an odd function.
\end{enumerate}

Khagan Gulusoy · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item \begin{proof}We have:
$$
\forall x \in \mathbb{R}: E(-x)=f(-x)+f(x)=f(x)+f(-x)=E(x) 	ext {. }
$$
Therefore, $E$ is even.
\end{proof}
\item \begin{proof}
We have:
$$
\forall x \in \mathbb{R}: E(-x)=f(-x)-f(x)=-f(x)+f(-x)=-E(x
$$
Therefore, $E$ is odd.
\end{proof}

\item \begin{proof}Given an arbitrary function $f: \mathbb{R} 	o \mathbb{R}$, let $E$ be defined as un part (a) and $O$ be defined as in part (b). We have:
\begin{equation}
\forall x \in \mathbb{R}: \quad f(x)=\frac{E(x)+O(x)}{2}=\frac{1}{2} E(x)+\frac{1}{2} O(x) 	ext {. }
\end{equation}

Since multiplying an even/odd function by a positive factor $c>0$. the function remains even lad, above, $f$ is represented as the sum of an even and an odd function, as desired.\end{proof}

\item To solve this exercise, we should represent $f$ as in (1):

$$
\begin{aligned}
E(x) & =f(x)+f(-x) \
& \left.=2^{x}-12-3ight)^{2}+2^{-x}-(-x-3)^{2} \
& =2^{x}-x^{2}+62-9+2^{-x}-x^{2}-6 x-9= \
& =2^{x}+2^{-x}-2 x^{2}-18 . \
O(x) & =f(x)-f(-x) \
& =2^{x}-(x-3)^{2}-2^{-x}+(-x-3)^{2} \
& =2^{x}-2^{-x}-x^{2}+6 x-9+x^{2}+6 x+9 \
& =2^{x}-2^{-x}+12 x
\end{aligned}
$$
\end{enumerate}

Exercise 1.3.71

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Exercise 1.3.71

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