Finding properties of operation defined by $x⊕y=\frac{1}{\frac{1}{x}+\frac{1}{y}}$? ("Reciprocal addition" common for parallel resistors)

Solution 1:

You're transporting usual addition in $\Bbb R^+$ by the bijection $s:\Bbb R^+\to\Bbb R^+$ that sends $x\longmapsto x^{-1}$. That is, to sum $x,y\in\Bbb R^+$, first transport them to $s(x),s(y)$, now sum them, to get $s(x)+s(y)$; now apply the inverse of $s$ (which is also $s$!) to get a "new" sum $x\oplus y=s(s(x)+s(y))$. This will satisfy all axioms that your original sum satisfies in $\Bbb R^+$. It won't have a zero, unless you adjoin say $+\infty$ and send $0\to +\infty$ (and yes, this works!).

In general, you can always transport any algebraic operation you have defined on a set onto another of the same cardinality using a given bijection, by $x\oplus y=f^{-1}(f(x)+f(y))$, just as above.

Solution 2:

This is not a name, but an exploration, since you asked for other properties. What you should instead look at is the "oplus-derivative"

$$f^{\oplus}(x) = \lim_{\Delta x \rightarrow 0} \left(f(x + \Delta x) \ominus f(x)\right) \Delta x$$

where $a \ominus b = a \oplus (-b)$ and is the analogue of subtraction, Then $(f \oplus g)^\oplus = f^\oplus \oplus g^\oplus$.

It is analogous to the "product derivative" that is sometimes seen which is a derivative defined using multiplication instead of addition:

$$f^*(x) = \lim_{\Delta x \rightarrow 0} \left(\frac{f(x + \Delta x)}{f(x)}\right)^{\frac{1}{\Delta x}}$$

because the inverse operation of multiplication is division as the inverse of addition is subtraction and the operation formed from repeating multiplication is exponentiation and its inverse is root-taking. Likewise, the inverse operation of oplus is ominus, the operation formed from repetition is division, and the inverse of that is multiplication.

For the "oplus-derivative", $f(x) = \frac{1}{x}$ has constant oplus-derivative 1, just as $f(x) = x$ has constant "plus-derivative" 1.

In fact, then we easily see the connection to the usual derivative. This "oplus derivative" turns out to just be the reciprocal of the derivative of $\frac{1}{f(x)}$ just as the product derivative equals $e^{f'(x)/f(x)}$:

$$\begin{align}f^{\oplus}(x) &= \lim_{\Delta x \rightarrow 0} \left(f(x + \Delta x) \ominus f(x)\right) \Delta x \\ &= \lim_{\Delta x \rightarrow 0} \left(\frac{1}{\frac{1}{f(x + \Delta x)} - \frac{1}{f(x)}}\right) \Delta x \\ &= \lim_{\Delta x \rightarrow 0} \frac{\Delta x}{\frac{1}{f(x + \Delta x)} - \frac{1}{f(x)}} \\ &= \left(\lim_{\Delta x \rightarrow 0} \left(\frac{\frac{1}{f(x + \Delta x)} - \frac{1}{f(x)}}{\Delta x}\right) \right)^{-1}\\ &= \left(\frac{d}{dx} \frac{1}{f(x)}\right)^{-1} \end{align} $$

Thus we can given an identity in terms of the usual derivative:

$$\left(\frac{d}{dx} \frac{1}{f(x) \oplus g(x)}\right)^{-1} = \left(\frac{d}{dx} \frac{1}{f(x)}\right)^{-1} \oplus \left(\frac{d}{dx} \frac{1}{g(x)}\right)^{-1}$$.

Or, since $\oplus$ of two reciprocals is just the reciprocal of the sum of the originals,

$$\frac{d}{dx} \frac{1}{f(x) \oplus g(x)} = \frac{d}{dx} \frac{1}{f(x)} + \frac{d}{dx} \frac{1}{g(x)}$$.

Note that this is actually just the derivative of a sum of reciprocals, but expressed using $\oplus$. Perhaps more interesting when expressed using the $\oplus$-derivative.