1 Even and odd functions

1.1 Constructing even and odd functions

A given function $f\left(x\right)$ can always be split into two parts, one of which is even and one of which is odd. To do this write $f\left(x\right)$ as $\frac{1}{2}\left[f\left(x\right)+f\left(x\right)\right]$ and then simply add and subtract $\frac{1}{2}f\left(-x\right)$ to this to give

$f\left(x\right)=\frac{1}{2}\left[f\left(x\right)+f\left(-x\right)\right]+\frac{1}{2}\left[f\left(x\right)-f\left(-x\right)\right]$

The term $\frac{1}{2}\left[f\left(x\right)+f\left(-x\right)\right]$ is even because when $x$ is replaced by $-x$ we have $\frac{1}{2}\left[f\left(-x\right)+f\left(x\right)\right]$ which is the same as the original. However, the term $\frac{1}{2}\left[f\left(x\right)-f\left(-x\right)\right]$ is odd since, on replacing $x$ by $-x$ we have $\frac{1}{2}\left[f\left(-x\right)-f\left(x\right)\right]=-\frac{1}{2}\left[f\left(x\right)-f\left(-x\right)\right]$ which is the negative of the original.

Example 2

Separate $x^3+2^x$ into odd and even parts.

Solution

$f\left(x\right)=x^3+2^x$

$f\left(-x\right)={\left(-x\right)}^3+2^{-x}=-x^3+2^{-x}$

Even part:

$\frac{1}{2}\left(f\left(x\right)+f\left(-x\right)\right)=\frac{1}{2}\left(x^3+2^x-x^3+2^{-x}\right)=\frac{1}{2}\left(2^x+2^{-x}\right)$

Odd part:

$\frac{1}{2}\left(f\left(x\right)-f\left(-x\right)\right)=\frac{1}{2}\left(x^3+2^x+x^3-2^{-x}\right)=\frac{1}{2}\left(2x^3+2^x-2^{-x}\right)$

Task!

Separate the function $x^2-3^x$ into odd and even parts.

First, define $f\left(x\right)$ and find $f\left(-x\right)$ :

Answer

Answer

1.2 The odd and even parts of the exponential function

Using the approach outlined above we see that the even part of ${\text{e}}^x$ is

$\frac{1}{2}\left({\text{e}}^x+{\text{e}}^{-x}\right)$

and the odd part of ${\text{e}}^x$ is

$\frac{1}{2}\left({\text{e}}^x-{\text{e}}^{-x}\right)$

We give these new functions special names: $coshx$ (pronounced ‘cosh’ $x$ ) and $sinhx$ (pronounced ‘shine’ $x$ ).

These two functions, when added and subtracted, give

$coshx+sinhx\equiv {\text{e}}^x\text{and}coshx-sinhx\equiv {\text{e}}^{-x}$

The graphs of $coshx$ and $sinhx$ are shown in Figure 4.

Figure 4 :

Note that $coshx>0$ for all values of $x$ and that $sinhx$ is zero only when $x=0$ .