I notice that division distributes over addition Root extraction distributes over multiplication

What operator do logarithms distribute over: ie:

what non-constant function $H \in C^2 \rightarrow C$ is there such that:

$$\log_s(H(a,b)) = H(\log_s(a),\log_S(b))$$



We can notice the following:

$$\ln(H(x,y)) = H(\ln(x), \ln(y))$$


$$\frac{\partial }{\partial x}[\ln(H(x,y)] =\frac{\partial }{\partial x}[H(\ln(x), \ln(y)] $$

$$\frac{\partial }{\partial y}[\ln(H(x,y)] =\frac{\partial }{\partial y}[H(\ln(x), \ln(y)] $$


$$ \frac{1}{H(x,y)}\frac{\partial H(x,y)}{\partial x} = \frac{1}{x}\frac{\partial H(\ln(x),\ln(y))}{\partial x} $$

$$ \frac{1}{H(x,y)}\frac{\partial H(x,y)}{\partial y} = \frac{1}{y}\frac{\partial H(\ln(x),\ln(y))}{\partial y} $$

Solutions to that system of differential equations should correspond to functionals over which the natural logarithm distributes

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    $\begingroup$ I assume you want a symmetric non-constant H? Otherwise, if H ignores one parameter there are at least two solutions (identity and inverse of log). $\endgroup$ – user21820 Dec 11 '13 at 6:24
  • $\begingroup$ If symmetric means H(a,b) = H(b,a) then no, symmetric is not necessary, merely non constant H that are dependent on both parameters is all I seek. $\endgroup$ – frogeyedpeas Dec 11 '13 at 23:02
  • $\begingroup$ :/ anyone got an idea? $\endgroup$ – frogeyedpeas Dec 12 '13 at 14:53
  • $\begingroup$ I see that you've added some additional information, but in the first place you didn't require H to be differentiable. Do you now? $\endgroup$ – user21820 Dec 14 '13 at 2:24
  • $\begingroup$ How drastically would requiring and not requing differentiabulity change the nature if H? $\endgroup$ – frogeyedpeas Dec 14 '13 at 12:59

This is not an answer, but I feel that perhaps it may help others with similar questions understand the distributivity axioms and why they should be there. This would explain the distribution of division over addition and root extraction over multiplication, because in both cases the operation that distributes over the other is essentially the inverse of the repetition of the other.

$(a+b)/c = a/c + b/c$ is equivalent to $a*c + b*c = (a+b)*c$ for $c \ne 0$ (where "*" denotes multiplication)

$(a*b)^{1/c} = a^{1/c} b^{1/c}$ is equivalent to $a^c * b^c = (a*b)^c$ for $c \ne 0$ (using multi-valued functions for negative $a,b$)

These are obtained by using the inverse operations of the "outer" operation. For example, in the first case to get from the left to the right we can first substitute $(a,b)$ with $(a*c,b*c)$ to get $((a*c)+(b*c))/c = (a*c)/c + (b*c)/c$ and then cancel inverses to get $((a*c)+(b*c))/c = a+b$ and finally multiply both sides by $c$. The second is exactly the same but with different operations.

Now multiplication distributes over addition and powers distribute over multiplication because in each case the former is a repetition of the latter. To see this, we can recast them as:

$[+a] ^c [+b] ^c = ( [+a] [+b] ) ^c$

$[*a] ^c [*b] ^c = ( [*a] [*b] ) ^c$

This is a made-up notation where $[+a]$ means "add $a$" and $X^c$ means "repeat X $c$ times". Notice that $[+a] ^c = [+(a*c)]$ and $[*a] ^c = [*(a^c)]$. Also, notice that repetition distributes over any set of functions that commute with one another, hence the distributivity holds. Another way to see this is to note that the set $\{ [+x] : x \}$ is indeed a commutative ring with [+0] as identity and function composition as the binary operation, and likewise for $\{ [*x] : x \}$ with [*1] as identity.

To recover the original distributivity axioms we simply apply both sides of the above expressions to the respective identities in the corresponding ring, which is 0 for the first and 1 for the second. In other words, ( to 0 repeat ( ( adding a ) then ( adding b ) ) c times ) is the same as ( to 0 ( repeat ( adding a ) c times ) then ( repeat ( adding b ) c times ). If we took as axioms the distributivity of repetition over adding or multiplying, we see that the normal distributivity axioms become theorems. Of course, we only have integer $c$ but there is a natural extension to rationals, and then reals can be approximated by rationals in the usual way. Somehow they make complete sense to me this way, instead of just treating the axioms as arbitrary rules that work.

Back to the original question, $\log$ is the inverse of $\exp$ and neither are obvious repetitions of any sort. I'd be interested if anyone could recast either of them as such.

Anyway I would really like to see if anyone has a symmetric continuous solution to the functional equation using the exponential function: $e^{f(x,y)} = f(e^x,e^y)$ where f is a binary operator on $\mathbb{C}$.


Really, the business of $\ln$ is to move from the group $(\mathbb{R}_{>0},\times)$ to the group $(\mathbb{R},+)$. And it does this in a very natural way: $$\ln H_{\times}(a,b)=H_{+}(\ln a,\ln b)$$ where $H_{\times}(a,b)=a\,b$ and $H_{+}(a,b)=a+b$. That is, $$\ln(a\,b)=\ln a+\ln b$$ So if you think of $H(a,b)$ as just the group operation of a group, applied to $a$ and $b$, then we have $$\ln H(a,b)=H(\ln a,\ln b)$$ just with two different groups in play.

On the one hand, you may not find this very satisfying. On the other hand, there is something important here about what logarithms really are.

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    $\begingroup$ frogeyedpeas had already specified that H was to be a complex-valued binary operator on complex numbers, so this is not what he is looking for. $\endgroup$ – user21820 Dec 20 '13 at 15:32
  • $\begingroup$ @user21820 I think my answer is touching on why OPs original quest is not going to be easy, if even at all possible. Even the usual relation $\ln(ab)=\ln a+\ln b$ doesn't hold up in $\mathbb{C}$ (for any one particular branch cut of a logarithm.) $\endgroup$ – alex.jordan Dec 20 '13 at 15:51
  • $\begingroup$ Actually I presumed that he wanted to look for a solution to the corresponding functional equation using the exponential function instead of the logarithm, otherwise the question isn't very well defined due to the branch cuts as you also noticed. $\endgroup$ – user21820 Dec 20 '13 at 15:54

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