Decomposing Functions Into Even and Odd Parts

Pairing Inputs at Opposite Points

For a function defined at both \(x\) and \(-x\), the two values can be compared to separate what stays the same under reflection from what changes sign. A function that stays the same is even; one that changes sign is odd. The useful fact is that every real-valued function on a domain symmetric about zero can be written as the sum of one function of each kind.

Let \(D\subseteq\mathbb{R}\). The domain is symmetric about zero if \(x\in D\) implies \(-x\in D\). For a symmetric domain, a function \(g:D\to\mathbb{R}\) is even if \(g(-x)=g(x)\) for every \(x\in D\), and it is odd if \(g(-x)=-g(x)\) for every \(x\in D\). These equations apply at \(x=0\) as well: an odd function must satisfy \(g(0)=-g(0)\), so \(g(0)=0\).

In this tutorial, sums and scalar multiples of functions are pointwise: for functions \(g,h:D\to\mathbb{R}\), \((g+h)(x)=g(x)+h(x)\), and \((ag)(x)=ag(x)\). Thus writing \(f=e+o\) means \(f(x)=e(x)+o(x)\) at every point of the domain.

The Two Formulas

Suppose \(f:D\to\mathbb{R}\) and \(D\) is symmetric about zero. If \(f=e+o\) for an even function \(e\) and an odd function \(o\), then reflecting the input gives \(f(-x)=e(x)-o(x)\). Adding and subtracting these two equations suggests formulas for the parts.

Proof. Since \(D\) is symmetric, whenever \(x\in D\), both \(x\) and \(-x\) belong to \(D\). The formulas therefore define values at every point of \(D\). For the even part, replacing \(x\) by \(-x\) gives $$ f_{\mathrm{even}}(-x) =\frac{f(-x)+f(-(-x))}{2} =\frac{f(-x)+f(x)}{2} =f_{\mathrm{even}}(x). $$ So \(f_{\mathrm{even}}\) is even. For the odd part, $$ f_{\mathrm{odd}}(-x) =\frac{f(-x)-f(x)}{2} =-\frac{f(x)-f(-x)}{2} =-f_{\mathrm{odd}}(x), $$ so \(f_{\mathrm{odd}}\) is odd. Finally, at every \(x\in D\), $$ f_{\mathrm{even}}(x)+f_{\mathrm{odd}}(x) =\frac{f(x)+f(-x)+f(x)-f(-x)}{2} =f(x). $$ All three claims follow. \(\square\)

The denominator \(2\) is valid because the scalars are real. The formulas construct the parts directly from \(f\); there is no need to guess an even function and then adjust it. The proof also shows why the domain condition matters: \(f(-x)\) must be defined whenever \(f(x)\) is used.

A Decomposition Is Unique

The formulas produce a decomposition, but an existence statement alone does not rule out different even and odd functions summing to the same \(f\). Uniqueness follows because the only function that is both even and odd is the zero function. The field matters here: over \(\mathbb{R}\), the equation \(2a=0\) implies \(a=0\).

Proof. Existence has been proved by constructing \(f_{\mathrm{even}}\) and \(f_{\mathrm{odd}}\). To prove uniqueness, suppose \(f=e_1+o_1=e_2+o_2\), where \(e_1,e_2\) are even and \(o_1,o_2\) are odd. Define \(q=e_1-e_2\). Since both \(e_1\) and \(e_2\) are even, \(q(-x)=q(x)\). The equality of the two decompositions also gives \(q=o_2-o_1\). Since both \(o_1\) and \(o_2\) are odd, \(q(-x)=-q(x)\). Therefore \(q(x)=-q(x)\), so \(2q(x)=0\), and hence \(q(x)=0\) for every \(x\in D\). Thus \(e_1=e_2\); substituting into \(e_1+o_1=e_2+o_2\) gives \(o_1=o_2\). The decomposition is unique. \(\square\)

This proof also establishes the related fact directly: if a function \(q\) is both even and odd, then \(q(x)=q(-x)=-q(x)\) for each \(x\), so \(q(x)=0\) everywhere. The zero function is indeed both even and odd. Consequently, the even and odd parts are not merely one possible split; they are forced by the values of \(f\) at \(x\) and \(-x\).

Piecewise Functions and the Point at Zero

The formulas apply just as well when a function has different rules on the positive and negative parts of its domain. It is important to evaluate \(f(-x)\) from the branch that contains \(-x\), rather than reuse the branch for \(x\). The point \(0\) also deserves an explicit check, because it reflects to itself.

Common Pitfalls

Forgetting to divide by two changes the answer. For the polynomial example, adding \(f(x)\) and \(f(-x)\) without dividing by \(2\) gives \(4x^2+8\). This is not the even part: adding the correct parts would give \(2f_{\mathrm{even}}(x)=4x^2+8\), twice as much as the even contribution needed in \(f\). The correction is \(f_{\mathrm{even}}(x)=\frac{f(x)+f(-x)}{2}=2x^2+4\). The odd formula also requires division by \(2\).

Check that the reflected input is in the domain. If \(f\) is defined only on \([0,1]\), then \(f(-x)\) is not defined for most \(x\in[0,1]\). It is therefore not valid to apply the formulas on that domain as written. Use a domain symmetric about zero, or provide a function defined on a symmetric domain before attempting this decomposition.

Do not assume the original function is itself even or odd. The formulas separate its two contributions; either contribution may be zero, but neither has to be. For example, the polynomial above has both a nonzero even part and a nonzero odd part. Its decomposition is pointwise and unique, not a claim that the original function has one parity.