# Functions can be divided into odd and even components - name of theorem?

I'm explaining to a student that all functions can be divided into odd and even (symmetric and anti-symmetric) components. It is easy to prove (basic algebra or Taylor series) but is not referenced in the text. I'm using it in relation to the definition of trigonometric functions.

Does this theorem have a name? Are there any good online resources for this that cover this at a High School/Calc I level (i.e. no Fourier series). Or is this considered simply too obvious to cover formally?

• You can't prove this with Taylor series. Not all functions (not even all differentiable functions) have a Taylor series representation, yet the theorem is true for all functions $\mathbf R \rightarrow \mathbf R$. This theorem is a special case of a general theorem in linear algebra about eigenspace decompositions, but at the high school level that will not make sense. I would not say the theorem is too obvious to have a name, but the context into which this theorem naturally fits as a special case is beyond the level of high school math or calculus. – KCd Mar 30 '16 at 2:20
• @KCd Can you mention the higher level context? I am not sure I see immediately what you mean. – Steven Gubkin Mar 30 '16 at 2:22
• Even functions and odd functions are eigenfunctions for the operator $T$ where $(Tf)(x) = f(-x)$. The formulas for the even and odd parts of a function are projections onto the two eigenspaces. If a linear operator on a vector space has finite order $n$ and all the (distinct!) $n$th roots of unity are in the ground field then the space decomposes into a direct sum of eigenspaces for each $n$th root of unity as eigenvalue. Or think of this as a special case of decomposing a representation of a finite abelian group into irreducible subrepresentations (finite Fourier analysis). – KCd Mar 30 '16 at 2:37
• Ah okay. Also worth mentioning that the decomposition of complex differential $1$-forms into $(1,0)$ and $(0,1)$ also follows this pattern. – Steven Gubkin Mar 30 '16 at 3:01

Take the function $f(x)$ and write the functions $\frac{1}{2}(f(x)-f(-x))$ and $\frac{1}{2}(f(x)+f(-x))$. They are odd and even, respectively, and their sum is $f(x)$. Nothing fancy required so I doubt it is named.
Edit: If you wanted to generalize this to the action of $n$th roots of unity as @KCd suggests, you could decompose the function $f(z)$ into functions $f_k(z)$ of the form $$f_k(z) = \frac{1}{n} \sum_{j=0}^n \omega^{kj} f(\omega^j z)$$ where $\omega$ is a primitive $n$th root of unity. It is easy to check that $f= \sum_k f_k$ and that $f_k(\omega z) = \omega^k f(z)$. The odd/even case then corresponds to $n=2$, $\omega=-1$. You might even be able to explain that to advanced high school students if you can get them to understand why $\sum_{j=0}^n \omega^j = 0$. However, at that stage, the potential for confusion is probably not worth the reward.
• It is a little bit fancy - they are the only pair of odd and even functions that add to give $f(x)$! – DavidButlerUofA Apr 4 '16 at 19:43