Real Analysis: The Real Numbers
Even and Odd Functions
Motivation - Symmetry is Useful
Symmetry is an ancient concept, praised in art as a source of beauty and in mathematics as a kind of elegance. The elegance of symmetry in mathematics derives from its power to save us time: prove something for one piece of the picture, and it will be true for the symmetric pieces of the picture. If we can analyze some simple kinds of symmetry for real functions, we'll be able to use these properties we discover later on to save us time.
Two Kinds of Symmetry - Even and Odd Functions
Real functions can have two kinds of symmetry about the $y$-axis. In one instance, the positive side can be mirrored across it on the negative side, as is the case for the function $f(x) = x^2.$ In the other case, the positive side can be rotated around the origin point $(0,0)$ by 180 degrees, as is the case for the function $f(x) = x^3.$ Note that the functions $f(x) = x^4$ and $f(x) = x^5$ also have these same kinds of symmetry, respectively. In fact, functions of the form $f(x) = x^{2n}$ and $f(x) = x^{2n+1}$ have these respective kinds of symmetry. Verily, these kinds of symmetry have names - the first is called even, the second is called odd.
Formally, a real function $f$ is even if $f(-x) = f(x).$ Alternatively, $f$ is odd if $f(-x) = -f(x).$ The names derive from the parity of the powers in the above examples.
Useful Properties
Even and odd functions have useful properties that come in two flavors. The first flavor concerns algebraic combinations. For example, the sum of two even functions is even, and the product of two odd functions is even. The second flavor is somewhat more powerful and concerns compositions. For example, the composition of two even functions is even, and the composition of two odd functions is odd, but mixing and matching the two or including functions that are neither even nor odd requires closer examination of the results.
Problems
Show that $f$ is both even and odd if and only if $f(x) = 0.$
Assume $f(x) = 0.$ Then $f(x) = 0 = f(-x),$ so $f$ is even. Likewise, $f(-x) = 0 = -0 = -f(x),$ so $f$ is odd.
Conversely, assume $f$ is both even and odd. Then $f(x) = f(-x) = -f(x).$ The only number that equals its own additive inverse is $0.$ Therefore $f(x) = 0.$
Show that the constant function $f(x) = c$ is even.
Let $x \in \mathbb{R}.$ Then $f(x) = c = f(-x),$ therefore $f$ is even.
Show that $f(x) = x$ is odd.
$f(x) = x$ and $f(-x) = -x,$ so $f(-x) = -f(x).$ Therefore $f$ is odd.
Show that if $f$ is even, then $-f$ is even, and if $f$ is odd, then $-f$ is also odd.
Assume $f$ is even. Then $f(x) = f(-x).$ Multiply both sides by $-1$ to show that $-f(x) = -f(-x).$ Therefore $-f$ is even. Next, assume $f$ is odd. Then $f(-x) = -f(x).$ Multiply both sides by $-1$ to show that $-f(-x) = -(-f(x)).$ Therefore $-f$ is odd.
Algebraic Properties of Even Functions. Let $f$ and $g$ be even. Show the following:
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$f + g$ is even.
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$f - g$ is even.
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$f \cdot g$ is even.
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$f / g$ is even.
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$(f + g)(x) = f(x) + g(x) = f(-x) + g(-x) = (f + g)(-x).$
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$(f - g)(x) = f(x) - g(x) = f(-x) - g(-x) = (f - g)(-x).$
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$(f \cdot g)(x) = f(x)g(x) = f(-x)g(-x) = (f \cdot g)(-x).$
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$(f / g)(x) = \dfrac{f(x)}{g(x)} = \dfrac{f(-x)}{g(-x)} = (f / g)(-x).$
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Algebraic Properties of Odd Functions. Let $f$ and $g$ be odd. Show the following:
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$f + g$ is odd.
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$f - g$ is odd.
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$f \cdot g$ is even.
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$f / g$ is even.
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$(f + g)(-x) = f(-x) + g(-x) = -f(x) - g(x) = -\left(f(x) + g(x)\right) = -(f + g)(x).$
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$(f - g)(-x) = f(-x) - g(-x) = -f(x) + g(x) = -(f - g)(x).$
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$(f \cdot g)(-x) = f(-x)g(-x) = (-f(x))(-g(x)) = f(x)g(x) = (f \cdot g)(x).$
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$(f / g)(-x) = \dfrac{f(-x)}{g(-x)} = \dfrac{-f(x)}{-g(x)} = \dfrac{f(x)}{g(x)} = (f / g)(x).$
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Algebraic Properties of Even and Odd Functions. Let $f$ be even and $g$ be odd. Show the following:
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$f + g$ is odd if and only if $f(x) = 0.$
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$f + g$ is even if and only if $g(x) = 0.$
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$f - g$ is odd if and only if $f(x) = 0.$
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$f - g$ is even if and only if $g(x) = 0.$
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$f \cdot g$ is odd.
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$f / g$ is odd.
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First, assume $f(x) = 0.$ Then $f + g = g,$ and since $g$ is odd, $f + g$ is odd. Conversely, assume $f + g$ is odd. Then $-(f + g)(x) = (f + g)(-x) = f(-x) + g(-x)$ and $-(f + g)(x) = -f(x) - g(x).$ Therefore $f(-x) + g(-x) = -f(x) - g(x).$ Since $g$ is odd, it follows that $f(-x) + g(-x) = -f(x) + g(-x).$ Therefore $f(-x) = -f(x),$ and so $f$ is odd. Since $f$ is both odd and even, it follows that $f(x) = 0.$
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First, assume $g(x) = 0.$ Then $f + g = f,$ and since $f$ is even, $f + g$ is even. Conversely, assume $f + g$ is even. Then $(f + g)(x) = (f + g)(-x) = f(-x) + g(-x)$ and $(f + g)(x) = f(x) + g(x).$ Then $f(-x) + g(-x) = f(x) + g(x).$ Since $f$ is even, it follows that $f(x) + g(-x) = f(x) + g(x).$ Therefore $g(-x) = g(x),$ and so $g$ is even. Since $g$ is both odd and even, it follows that $g(x) = 0.$
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First, assume $f(x) = 0.$ Then $(f - g)(-x) = f(-x) - g(-x) = g(-x) = g(x) = -f(x) + g(x) = -(f - g)(x).$ Conversely, assume $f - g$ is odd. Then $-(f - g)(x) = (f - g)(-x) = f(-x) - g(-x)$ and $-(f - g)(x) = -f(x) + g(x).$ Therefore $f(-x) - g(-x) = -f(x) + g(x).$ Since $g$ is odd, it follows that $f(-x) - g(-x) = -f(x) - g(-x).$ Therefore $f(-x) = -f(x),$ and so $f$ is odd. Since $f$ is both even and odd, it follows that $f(x) = 0.$
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First, assume $g(x) = 0.$ Then $f - g = f,$ and since $f$ is even, $f - g$ is even. Conversely, assume $f - g$ is even. Then $(f - g)(x) = (f - g)(-x) = f(-x) - g(-x)$ and $(f - g)(x) = f(x) - g(x).$ Then $f(-x) - g(-x) = f(x) - g(x).$ Since $f$ is even, it follows that $f(x) - g(-x) = f(x) - g(x).$ Therefore $g(-x) = g(x),$ and so $g$ is even. Since a $g$ is both even and odd,$ it follows that $g(x) = 0.$
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$(f \cdot g)(-x) = f(-x)g(-x) = f(x)(-g(x)) = -f(x)g(x) = -(f \cdot g)(x).$
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$(f / g)(-x) = \dfrac{f(-x)}{g(-x)} = \dfrac{f(x)}{(-g(x))} = -\dfrac{f(x)}{g(x)} = -(f / g)(x).$
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Compositions of Even and Odd Functions. Show the following
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If $f$ and $g$ are both even, then $f \circ g$ is also even.
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If $f$ and $g$ are both odd, then $f \circ g$ is also odd.
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If $f$ is even and $g$ is odd, then $f \circ g$ is even.
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If $g$ is even, then $f \circ g$ is even for any $f.$
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Assume $f$ and $g$ are even. Then
$(f \circ g)(x) = f(g(x)) \\ (f \circ g)(x) = f(g(-x)) \\ (f \circ g)(x) = (f \circ g)(-x).$
Therefore $f \circ g$ is even.
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Assume $f$ and $g$ are odd. Then
$(f \circ g)(-x) = f(g(-x)) \\ (f \circ g)(-x) = f(-g(x)) \\ (f \circ g)(-x) = -f(g(x)) \\ (f \circ g)(-x) = -(f \circ g)(x)$
Therefore $f \circ g$ is odd.
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Assume $f$ is even and $g$ is odd. Then
$(f \circ g)(-x) = f(g(-x)) \\ (f \circ g)(-x) = f(-g(x)) \\ (f \circ g)(-x) = f(g(x)) \\ (f \circ g)(-x) = (f \circ g)(x).$
Therefore $f \circ g$ is even.
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Assume $g$ is even. Then
$(f \circ g)(x) = f(g(x)) \\ (f \circ g)(x) = f(g(-x)) \\ (f \circ g)(x) = (f \circ g)(-x).$
Therefore $f \circ g$ is even.
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Show that $f(x) = x^2$ is even.
Direct proof: $f(x) = x^2 = 1(x \cdot x) = ((-1) \cdot (-1))(x \cdot x) = (-x)(-x) = (-x)^2 = f(-x).$
Proof using algebraic properties: $g(x) = x$ is odd, and $f(x) = g(x)g(x).$ Since the product of two odd functions is even, it follows that $f$ is even.
Show that monomials of even power are even.
Proof by induction. Base case: The function $f(x) = cx^0 = c$ is even. Inductive step: Assume $f(x) = cx^{2n}$ is even for some $n \in \mathbb{N}.$ Then $g(x) = cx^{2(n+1)} = cx^{2n+2} = cx^{2n}x^2$ is a product of two even functions $cx^{2n}$ and $x^2$ and is therefore even. The result follows by induction.
Show that monomials of odd power are odd.
Let $f$ be a monomial of odd power such that $f(x) = cx^{2n+1}.$ Note that $f(x) = cx^{2n} \cdot x.$ Since $f$ is a product of an even function $cx^{2n}$ and an odd function $x,$ it follows that $f$ is odd.