# odd/even function hybrid combination

Every continuous smooth real function f(x) is either odd , even, or sum/difference of an even and odd (mixed like e.g., exponential) functions. Is this generally correct?

Can the even part of a mixed f(x) and odd part of another mixed g(x) be combined to make a new combo h(x)? If so how so and if not why not?

EDIT1:

Do such hybrid h(x) satisfy mean value theorem? Will there be no problem with computing their curvatures and inflection points? I asked this as I somehow feel that a mixed Taylor series has some of its " chemistry" or nature altered.

• Are Taylor series allowed in the answer? – imranfat Oct 13 '15 at 18:15
• Yes, afik that is the only way such a function is constituted. – Narasimham Oct 13 '15 at 18:21
• Well, then it's a done deal... – imranfat Oct 13 '15 at 18:22

You don't need smooth, or even continuous. Any function $f(x)$ is the sum of the two functions $$\frac{f(x) - f(-x)}{2}\quad,\quad \frac{f(x) + f(-x)}{2}$$ where the first one is odd and the second one is even. If $f$ itself is odd or even, then one of the fractions above evaluates to $0$ and the other to $f(x)$. These parts can be freely added to the corresponding parts of other functions.
Write $$f(x) = \frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2} = f_e(x) + f_o(x)$$ This means \begin{align} (f + g)(x) &= f(x) + g(x) \\ &= (f_e(x) + f_o(x) )+ (g_e(x) + g_o(x) ) \\ &= (f_e + g_e)(x) + (f_o + g_o)(x) \\ &= (f+g)_e(x) + (f+g)_o(x) \end{align} If $f = F_e$ and $g = G_o$ then \begin{align} (f+g)_e &= f_e + g_e = F_e + 0 = F_e \\ (f+g)_o &= f_o + g_o = 0 + G_o= G_o \end{align}