algebraic binary operations that are associative and nice in a way.

Question

It is well known that $$ a*b = ab + a + b$$ and $$ x*y = \dfrac{x+y}{1-xy} $$ are commutative associative binary operations. What rational function $f$ is a commutative associative binary operations that satisfy (informally writing) $$ a_1*a_2*\cdots*a_n = f(e_0,e_1,e_2,\cdots,)$$ where $e_i$s are elementary symmetric polynomials and thought of as $e_i=0$ if $i\geq n+1$ and $e_0=1$.

Or more formally, There is a sequence of rational functions $(f_i)_i$ that satisfy $$ f_i(a_1,a_2,\ldots,a_i) = f_{i+1}(a_1,a_2,\ldots,a_i,a_{i+1})$$ and $$ a_1*a_2*\cdots a_n = f_n(a_1,a_2,\ldots,a_n).$$

Answer 1

The operations can be represented by bijection of the real numbers, applying an operation there, and then transforming back again.

For example, in the case of

$$a*b=ab+a+b$$

first apply $f(x) = x+1$ to both $a$ and $b$, then apply multiplication as usual, and then apply $f^{-1} = x-1$:

$$a*b=f^{-1}(f(a)f(b)) = (a+1)(b+1) - 1 = ab+a+b$$

That is we have the commutative diagram

$$\require{AMScd}\begin{CD} (\Bbb R^*,*) @>>> (\Bbb R^*,*) \\ @VV{f}V @V{f}VV \\ (\Bbb R^\times,\cdot) @>>> (\Bbb R^\times,\cdot) \\ \end{CD}$$

which describes a bijective homomorphism (ispmorphism) of two (infinite, abelian, Lie) groups that can be written

$$f(a*b) = f(a)\cdot f(b)$$

In particular, the mapping carries over group properties like:

Neutral Element: $f$ maps $0$ to $1$, and hence $0$ is the neutral element of the group $(\Bbb R^*,*)$ since $a*0 = a\cdot0 +a+0 = a$.

Inverse Elements: $f$ maps $-1$ to $0$, a real number that has no multiplicative inverse, thus only $\Bbb R^\times=\Bbb R\backslash\{0\}$ is a group under multiplication but $\Bbb R$ isn't. Consequently, $\Bbb R$ is not a group under $*$ but only $\Bbb R^*=\Bbb R\backslash\{-1\}$, because $a*(-1) = -1$ for all $a$, which has no inverse element. The inverse of $a\in(\Bbb R^*,*)$ is $a^{-1}=\dfrac1{a+1}-1=-\dfrac a{a+1}$ since $$a*a^{-1}= a\left(-\frac a{a+1}\right)+a-\frac a{a+1}=0$$ and $0$ is the neutral element.

Associativity: $$\begin{align} (a*b)*c &= f^{-1}(f(a*b)f(c)) \\ &= f^{-1}(f(a)f(b)f(c)) \\ &= f^{-1}(f(a)f(b*c)) = a*(b*c) \end{align}$$

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The second mapping $$x*y = \frac{x+y}{1-xy}$$

is realized by $f(x) = \arctan x$ due to addition theorem of arctan

$$\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}$$ resp.

$$\tan(x+y) = \frac{\tan x+ \tan y}{1-\tan x \tan y}$$

This means you can first apply arctan and then addition, or alternatively first apply $*$ and then arctan. This also establishes an isomorphism or groups, one is $(\Bbb R,+)$ and the other is $((-\pi/2,\pi/2),*)$.

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The $n$-fold operations can then be expressed by the associated operation. For the 1st case:

$$\begin{align} {\Huge*}_{k=1}^n a_k &= f^{-1}\left(\prod_{k=1}^nf(a_k)\right) \\ &= -1+ \prod_{k=1}^n(1+a_k) \\ \end{align}$$

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One of the most prominent and useful such isomorphism is the the isomorphism between $(\Bbb R^+,\cdot)$ and $(\Bbb R,+)$ mediated by $\ln$ and $\exp$ (or any log-exp pair to some other basis like 10)

$$\require{AMScd}\begin{CD} (\Bbb R^+,\cdot) @>>> (\Bbb R^+,\cdot) \\ @VV{\ln}V @A{\exp}AA \\ (\Bbb R,+) @>>> (\Bbb R,+) \\ \end{CD}$$

using

$$\log_b(xy) = \log_b x + \log_b y \quad\text{ resp. }\quad b^{x+y}=b^xb^y$$

that is, instead of multiplication (expensive) you can first apply log then add the values, then inverse log. The latter three operations can be performed using a slide-rule, or log and exp can be performed using lookup-tables. This method lost significance with the advent of modern computes, but was very important in earlier times.

Answer 2

Another example is $$x*y=\frac{x+y}{1+xy}\tag{1}$$ which, like your $$x*y=\frac{x+y}{1-xy}\tag{2}$$ has the property $h(x*y)=h(x)+h(y)$ for some invertible function $h$. As the earlier answer pointed out, with $(2)$, $h$ is arctan; with $(1)$, $h$ is arctanh.

With $(2)$, $\infty$ is in the range (when the denominator is 0) and so must be in the domain and codomain, to keep $*$ closed. With $(1)$, a similar approach is possible; alternatively, you could set the domain to be $(-1, 1)$, as $*$ is closed on $(-1, 1)$.