Odd and Even Functions

Recall that $\sin(-x) = -\sin(x)$ and $\cos(-x) = \cos(x)$ for any $x \in \mathbb R$ .

Definition 1. For any function $f : \mathbb R \to \mathbb R$ , define $\tilde f : \mathbb R \to \mathbb R$ by $\tilde f(x) := f(-x)$ .

Problem 1. Given functions $f, g: \mathbb R \to \mathbb R$ , define $h_1 := f + g$ and $h_2 := f \cdot g$ . Prove that $\tilde h_1 = \tilde f + \tilde g$ and $\tilde h_2 = \tilde f \cdot \tilde g$ .

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Solution. By definition,

$\begin{aligned} \tilde h_1(x) = h_1(-x) &= (f+g)(-x) \\ &= f(-x) + g(-x) \\ &= \tilde f(x) + \tilde g(x) \\ &= (\tilde f + \tilde g)(x), \end{aligned}$

and

$\begin{aligned} \tilde h_2(x) = h_2(-x) &= (f \cdot g)(-x) \\ &= f(-x) \cdot g(-x) \\ &= \tilde f(x) \cdot \tilde g(x) \\ &= (\tilde f \cdot \tilde g)(x). \end{aligned}$

Definition 2. A function $f : \mathbb R \to \mathbb R$ is odd (resp. even) if there exists an odd (resp. even) integer $n$ such that $\tilde f = (-1)^n \cdot f$ .

Example 1. For any integer $n$ , the function $f_n$ defined by $f_n(x) = x^n$ is an odd (resp. even) function if and only if $n$ is an odd (resp. even) number. Furthermore, $\sin$ is an odd function while $\cos$ is an even function.

Problem 2. For any function $f: \mathbb R \to \mathbb R$ and real number $\alpha \in \mathbb R$ , define the function $f_\alpha$ by $f_\alpha(x) := f(\alpha x)$ . In particular, $f_{-1} = \tilde f$ . Prove that $f_\alpha$ is odd (resp. even) if $f$ is odd (resp. even).

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Solution. We observe that if $f$ is odd or even, then $\tilde f = (-1)^n f$ , so that

$\begin{aligned} \tilde f_\alpha(x) &= f_\alpha(-x) = f(\alpha(-x))= f(-(\alpha x)) \\ &= \tilde f(\alpha x) = (-1)^n \cdot f(\alpha x) = (-1)^n \cdot f_\alpha(x), \end{aligned}$

hence $\tilde f_\alpha= (-1)^n \cdot f_\alpha$ .

Problem 3. Prove the following properties:

if $f, g$ are odd, then $f + g$ is odd and $f \cdot g$ is even,
if $f, g$ are even, then $f + g$ is even and $f \cdot g$ is even,
if $f$ is odd and $g$ is odd (resp. even), then $g \circ f$ is odd (resp. even) whenever it exists.
if $f$ is even, then $g \circ f$ is even.

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Solution. Suppose $\tilde f = (-1)^m \cdot f$ and $\tilde g = (-1)^n \cdot g$ , and assume $m \leq n$ for simplicity. Denote $h_1 = f+g$ and $h_2 = f \cdot g$ as per Problem 1. Then

$\tilde h_1 = \tilde f + \tilde g = (-1)^m \cdot f + (-1)^n \cdot g = (-1)^m \cdot (f + (-1)^{n-m} \cdot g).$

If $m$ and $n$ are both odd or both even, then $n-m$ is even, so that

$\tilde h_1 = (-1)^m \cdot (f + g) = (-1)^m \cdot h_1.$

Hence, $h_1$ is odd (resp. even) if and only if $f$ is odd (resp. even). Similarly,

$\tilde h_2 = \tilde f \cdot \tilde g = (-1)^m \cdot f \cdot (-1)^n \cdot g = (-1)^{m+n} \cdot h_2.$

If $m$ and $n$ are both odd or both even, then $m+n$ is always even, so that $\tilde h_2 = h_2$ unconditionally, i.e. $h_2 = f \cdot g$ is even. For the composite function, define $h_3 := g \circ f$ . Then

$\tilde h_3(x) = h_3(-x) = (g \circ f)(-x) = g(f(-x)) = g(\tilde f(x)) = (g \circ \tilde f)(x).$

Applying the odd or even condition on $f$ ,

$\begin{aligned} g \circ \tilde f &= g \circ (-(-1)^{m+1} \cdot f) \\ &= \tilde g \circ ((-1)^{m+1} \cdot f) \\ &= (-1)^n \cdot g \circ ((-1)^{m+1} \cdot f). \end{aligned}$

If $f$ is odd, then $m + 1$ is even, so that

$\tilde h_3 = g \circ \tilde f = (-1)^n \cdot (g \circ f) = (-1)^n \cdot h_3.$

Thus, $h_3$ is odd (resp. even) if and only if $g$ is odd (resp. even). If $f$ is even, then

$\tilde h_3 = g \circ \tilde f = g \circ f = h_3,$

so that $h_3$ is even.

Problem 4. For any function $f: \mathbb R \to \mathbb R$ , prove that there exist a unique odd function $f_1$ and a unique even function $f_2$ such that $f = f_1 + f_2$ .

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Solution. For existence, define the functions $f_1, f_2$ by

$f_1 := \frac 12 \cdot (f - \tilde f),\quad f_2 := \frac 12 \cdot (f + \tilde f).$

It is obvious that $f = f_1 + f_2$ . Since $\tilde{ \tilde f } = f$ ,

$\begin{aligned} \tilde f_1 &= \textstyle \frac 12 \cdot (\tilde f - \tilde{ \tilde f }) = \frac 12 \cdot (\tilde f - f) = -f_1, \\ \tilde f_2 &= \textstyle \frac 12 \cdot (\tilde f + \tilde{ \tilde f }) = \frac 12 \cdot (\tilde f + f) = f_2, \end{aligned}$

so that $f_1$ is odd and $f_2$ is even. For uniqueness, suppose $f = f_1 + f_2 = \hat f_1 + \hat f_2$ , where $\hat f_1$ is odd and $\hat f_2$ is even. Then $g := f_1 - \hat f_1 = \hat f_2 - f_2$ , so that $g$ is both odd and even. Hence,

$g = \tilde g = -g \quad \Rightarrow \quad g = 0\quad \Rightarrow \quad \hat f_1 = f_1\quad \Rightarrow \quad \hat f_2 = f_2.$

Problem 5. Fix $a > 0$ , any odd continuous function $f_1$ , and any even continuous function $f_2$ . Prove that

$\displaystyle \int_{-a}^a f_1 (x)\,\mathrm dx = 0, \quad \int_{-a}^a f_2 (x)\,\mathrm dx = 2 \cdot \int_0^a f_2 (x)\,\mathrm dx.$

In particular, for any continuous function $f : \mathbb R \to \mathbb R$ ,

$\displaystyle \int_{-a}^a f (x)\,\mathrm dx = 2 \cdot \int_0^a f_2 (x)\,\mathrm dx.$

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Solution. Using a substitution,

$\displaystyle \int_{-a}^0 f_1(x)\, \mathrm dx = \int_{a}^0 f_1(-t)\cdot (-1)\, \mathrm dt = \int_0^a -f_1(t)\, \mathrm dt = - \int_0^a f_1(t)\, \mathrm dt.$

Therefore,

$\displaystyle \begin{aligned} \int_{-a}^a f_1 (x)\,\mathrm dx &= \int_{-a}^0 f_1(x)\, \mathrm dx + \int_0^a f_1(x)\, \mathrm dx = 0.\end{aligned}$

Similarly, we can prove that

$\displaystyle \int_{-a}^0 f_2(x)\, \mathrm dx = \int_0^a f_2(t)\, \mathrm dt$

so that

$\displaystyle \begin{aligned} \int_{-a}^a f (x)\,\mathrm dx &= \int_{-a}^0 f_2(x)\, \mathrm dx + \int_0^a f_2(x)\, \mathrm dx = 2 \cdot \int_0^a f_2(x)\, \mathrm dx.\end{aligned}$

Problem 6. For any continuous function $f : \mathbb R \to \mathbb R$ and $T > 0$ , define

$\displaystyle a_n := \frac 2T \int_{-T/2}^{T/2} f(t)\, \mathrm \cos \left( \frac{2n \pi t}{T} \right)\, \mathrm dt,\quad b_n := \frac 2T \int_{-T/2}^{T/2} f(x)\, \mathrm \sin \left( \frac{2n \pi t}{T} \right)\, \mathrm dt.$

Prove that $a_n = 0$ (resp. $b_n = 0$ ) if $f$ is odd (resp. even).

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Solution. By Example 1 and Problem 2, $\cos_{2n\pi/T}$ is even and $\sin_{2n\pi/T}$ is odd. Thus, if $f$ is odd, then $f \cdot \cos_{2n\pi/T}$ is odd by Problem 3, so that $a_n = 0$ by Problem 5. Similarly, if $f$ is even, then $f \cdot \sin_{2n\pi/T}$ is odd by Problem 3, so that $b_n = 0$ by Problem 5.