5
$\begingroup$

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?

$\endgroup$
  • 5
    $\begingroup$ 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. $\endgroup$ – KCd Mar 30 '16 at 2:20
  • $\begingroup$ @KCd Can you mention the higher level context? I am not sure I see immediately what you mean. $\endgroup$ – Steven Gubkin Mar 30 '16 at 2:22
  • 6
    $\begingroup$ 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). $\endgroup$ – KCd Mar 30 '16 at 2:37
  • 1
    $\begingroup$ Ah okay. Also worth mentioning that the decomposition of complex differential $1$-forms into $(1,0)$ and $(0,1)$ also follows this pattern. $\endgroup$ – Steven Gubkin Mar 30 '16 at 3:01
11
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ It is a little bit fancy - they are the only pair of odd and even functions that add to give $f(x)$! $\endgroup$ – DavidButlerUofA Apr 4 '16 at 19:43
  • $\begingroup$ The hyperbolic sine (sinh x) and hyperbolic cosine (cosh x) are the result of applying this transformation to e^x. $\endgroup$ – Jasper Apr 5 '16 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.