Create odd function from arbitrary function I have a product of two arbitrary functions $f$ and $g$,
$$y(x) = f(x)g(x)$$
and I want to make the product $y$ odd. I know $f$, e.g. it's a typical Lorentzian function
$$f(x) = \dfrac{b}{(x-a)+b^2},$$
but I want to deduce $g$.
What strategy can I use to find $g$?
Context: I want to start by assuming that these functions are well behaved, defined in real space, and differentiable everywhere. How would I find $g$ so that the product $fg$ becomes odd? Then my goal is to make $f$ more complex (e.g. multiplying the Lorentzian above with the discontinuous Bose-Einstein distribution, for example) then see if it's still possible to make $fg$ an odd function.
This reference summarises odd/even functions but I didn't find it too helpful for my problem.
 A: Okay, so, of course there are lots of options here. The simplest, but unsatisfying, would be to make $g$ be $\frac{1}{f(x)}x$, so that $y$ is just $x$. What that example  demonstrates is that there's just way too little information here to pin down one specific $g$.
But one that might be more satisfying would be to take $g(x) = xf(-x)$. This has the advantage of retaining some of the "character" of $f$, but without knowing more about your goal here I can't tell if this is what you're looking for.
A: Odd and even functions behave exactly the same way under multiplication as odd and even numbers (hence the terminology). In particular,


*

*an odd function times an odd function is even

*an even function times an even function is even

*an odd function times an even function is odd


It might help to know a little bit more about the function $f$. If it's a Lorentzian centred on zero, then it's even, and any odd function $g$ will suffice to make $y$ odd. If it never takes the value zero, then we can take $g$ to be $1/f$ multiplied by any odd function. Trivially we can always just take $g = 0$ and $y$ will be odd (and, incidentally, even).
A: Let $e(x) = \frac{1}{2}(f(x) + f(-x))$ and write
$$  o(x) = f(x) - e(x)  \text{.}  $$
That is, let $e$ be the even part of $f$ and $o$ be the odd part of $f$.
Then let $g(x) = \frac{o(x)}{f(x)}$ so that $y(x) = f(x) \frac{o(x)}{f(x)}$.  This is undefined anywhere $f$ is zero, but $f(x) = \frac{b}{(x-a)^2+b^2}$ (which only differs from the Cauchy/Lorentzian distribution by a normalizing factor) is always positive, as long as $b \neq 0$ (which seems likely).   
Following the above prescription, 
$$  o(x) = \frac{2 a b x}{a^4 + 2a^2(b-x)(b+x) + (b^2 + x^2)^2} $$
and
$$  g(x) = \frac{2 a x}{b^2 + (x+a)^2}  \text{.}  $$
As others have noted, there are many other ways to select $g$, but this one picks out an intrinsic property of $f$ (its odd part), so may capture some other property that you want but have not specifically called out.
