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

I have recently found some interesting properties of the function/operation:

$x⊕y = \frac{1}{\frac{1}{x}+\frac{1}{y}} = \frac{xy}{x+y}$ where $x,y\ne0$.

and similarly, its inverse operation:

$x⊖y = \frac{1}{\frac{1}{x}-\frac{1}{y}} = \frac{xy}{y-x} = x⊕(-y)$

This is a formula often used in physics, for calculating equivalent resistance/capacitance in circuits. I've heard this referred to by my teacher as "reciprocal addition" but have not been able to find much significant mention of this operation outside of its direct connection to circuits and an isolated youtube video.

Since there is no common name/use for this operation, I'm asking for help in finding properties of this function.

to denote that the operations ("o-plus" and "o-minus" -- name from the youtube channel 3Blue1Brown) and proven a few simple properties of it, including many similarities, but also differences, with addition.

For example, the ⊕ ("o-plus") operation is:

1. Associative (EDIT: this only seems fully true if we extend the domain to the projective reals, and say $\infty$ is the identity and $a⊕0=0$ while $a⊕-a=\infty$)

2. Commutative

3. Distributive with multiplication

However, it does not have an identity (such as zero for addition or 1 for multiplication) at least among the real numbers. EDIT: adding $\infty$ to the domain and working under the projective reals gives $\infty$ as the identity.

I've done a lot of playing around with simple properties of ⊕ and found many interesting things about it, such as ways to do arithmetic using it - for instance:

$\frac{a}{x}⊕\frac{a}{y} = \frac{a}{x+y}$

Which means one would "o-plus" fractions by finding a common numerator instead of a common denominator.

Also, just as repeated addition is a form of multiplication($a+a+a+a = 4a$), repeated "o-plussing" is a form of division.

$a⊕a⊕a⊕a = \frac{a}{4}$

There are a multitude of cool properties that come out of the definition of this operation that I've defined here, including one the the aforementioned 3Blue1Brown explains in this video (https://www.youtube.com/watch?v=EOtduunD9hA):

$\sqrt[a] x \sqrt[b] x = \sqrt[a⊕b]{x}$

One other thing I found was the derivative of two "o-plussed" functions:

$\frac{d}{dx}(f(x)⊕g(x)) = \frac{f'(x)g(x)^2 + g'(x)f(x)^2}{(f(x)+g(x))^2}$

which looks over-complicated, and I think it is. I'd like to find a way to describe this function that actually includes the ⊕ operation itself, but have been unable to find one. However, there is a nice parallel which I found which makes more sense of this derivative rule:

$f⊕g = \frac{1}{\frac{1}{f}+\frac{1}{g}} = \frac{fg}{f+g} = \frac{fg}{f+g}\frac{(f+g)}{(f+g)} = \frac{fg^2+gf^2}{(f+g)^2}$

which parallels perfectly with what I said before,

$(f⊕g)' = \frac{f'g^2+g'f^2}{(f+g)^2}$.

Again, my question is if there exists a generally accepted notation or mathematical discussion of this operation, or if it is used or mathematically discussed outside of a few very specific locations. If not, could anybody just offer comments on my work or prove other interesting properties of this operation? Thanks to anybody for your help!

EDIT: Thanks to dls's comment about the Harmonic Mean, I've also found that a⊕b is the solution to the "crossed ladder problem"

Which asks to find h, given A and B as heights to "cross ladders" over. PHOTO: Wikipedia.

This leads me to believe it would be natural to define a⊕0 to be zero, since the lines given those heights would intersect at a height of 0. This is not a formal 'definition', just pattern-continuation. One might also define a⊕-a to be 'infinite', since that seems to solve the associativity problem in some cases, though this may not rigorously work...

Edit: Since oplus is so closely related to the Harmonic Mean, it's also interesting to view how the definition of the oplus operation makes the connection between the three means (arithmetic, geometric and harmonic) so much more natural. Starting with the arithmetic mean,

$$A = \frac{1}{n}\sum_{i=0}^n{x_i}$$

we see it involves repeated addition, followed by division, which is the inverse of multiplication (the operation which repeated addition of the same number defines). Similarly, for the geometric mean,

$$G = ({\prod_{i=0}^n{x_i}})^{1/n}$$

it involves repeated multiplication, followed by a root, which is the inverse of exponentiation (the operation which repeated multiplication of the same number defines). And finally, the harmonic mean can be written,

(there is no easy way to write this in a similar notation to the above, but the idea is the same)

$$H = n(x_i⊕x_2⊕x_3⊕...⊕x_n)$$

which involves repeated oplussing, followed by multiplication, which is the inverse of division (the operation which repeated oplussing of the same number defines - as described above). I find this parallel to be an extremely natural way to define the Harmonic mean, and any mean for that matter. In fact, one might imagine a mean defined by any operation of any order (for example, an exponentiation mean, which would use repeated exponentiation followed by a super-root, the inverse of tetration).

UPDATE: I have found a really cool connection between this operation and another area: Since the oplussing of two real numbers creates an output which is less than both of the originals, I wanted to see what conditions were necessary for the case to be that this is true:

$x⊕y=x-y$

and the result was fantastic (derivation below)!

$x⊕y=x-y$

$\frac{1}{\frac{1}{x}+\frac{1}{y}} = x-y$

$\frac{xy}{x+y} = x-y$

$xy = x^2-y^2$

$0 = x^2-xy-y^2$

$x = \frac{y±\sqrt{y^2+4y^2}}{2}$

$x = y\frac{1±\sqrt{5}}{2}$

which means that the ratio between x and y is φ≈1.618, the golden ratio!

$x = φy$

And through this, I saw that only the quadratic formula was necessary to find this ratio, and wondered if there was a number (clearly complex) which might satisfy:

$x⊕y=x+y$

and there is!

$\frac{1}{\frac{1}{x}+\frac{1}{y}} = x+y$

$\frac{xy}{x+y} = x+y$

$xy = x^2+2xy+y^2$

$0 = x^2+xy+y^2$

$x = \frac{-y±\sqrt{y^2-4y^2}}{2}$

$x = y\frac{1±\sqrt{-3}}{2}$

$x = y\frac{1±\sqrt{3}i}{2}$

And this is an extremely interesting number - I call it (and I found one internet forum from 2011 which has also named it this) the "imaginary golden ratio".

$φ_i = \frac{1+\sqrt{3}i}{2}$

It has a ton of interesting properties similar to the real golden ratio, and some of its own. I was wondering if anybody could help me uncover a bit of this mystery. Once I am able to write a question which asks more in depth about this number, because I'd like to reignite a discussion that seems to have been left unmentioned since the other forum in 2011.

Here are some properties of the imaginary golden ratio.

$φ_i = cos(π/3)+ isin(π/3) = e^{iπ/3}$

$|φ_i|=1$

$φ_i^2=φ_i-1$

which parallels

$φ^2=φ+1$

Here is the link to the other forum: http://mymathforum.com/number-theory/17605-imaginary-golden-ratio.html

• Your expression is (almost) the harmonic mean of the two inputs. – dls May 15 '16 at 3:19
• I love 3Blue1Brown's videos! Hope this gets a good answer – Omnomnomnom May 15 '16 at 3:38
• It's not entirely associative: $(1 \oplus 1) \oplus -1 = 1$, but what about $1 \oplus ( 1 \oplus -1 )$? I think the reason this isn't studied very widely is that it's just the conjugation of addition with the almost-bijection $x \mapsto 1/x$. So pretty much all of its properties are inherited from simple addition. – Erick Wong May 15 '16 at 4:48
• @ErickWong Using imallett's application of division by zero, we have $1 \oplus (1 \oplus -1) = 1 \oplus \infty = 1$. – lastresort May 15 '16 at 5:22
• So in general you would have $a \oplus 0 = 0$ and $a \oplus \infty = a$. – lastresort May 15 '16 at 5:28

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.

• Thank you, that's quite helpful! You only mention the positive reals in this case, can this extend to both positive and negative reals, and in this case would negative infinity be separate and behave differently from positive infinity? – MathTrain May 15 '16 at 5:48
• I'm very sure you meant "addition" instead of multiplication in your first sentence. Also, I hope you don't mind that I added brackets to ensure that readers understand your function application. =) – user21820 May 15 '16 at 6:29

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.

• It can be shown using simple derivative rules from what you came up with that the oplus derivative $f^⊕(x) = -\frac{f^2}{f'}$ and that this, I have realized, is equivalent to another operator I was curious about in my more recent post: the operator $\frac{dx}{d}f$. You can find that question here if you're still interested: math.stackexchange.com/questions/1878230/what-is-fracdxd/… – MathTrain Aug 2 '16 at 0:29