What do logarithms distribute over?

Question

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:

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

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ADDITIONAL INFO:

We can notice the following:

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

Thus:

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

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

Thus:

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

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

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

Answer 1

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.

Answer 2

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 $\mathbb{C}$.

Answer 3

I have asked the same question here: A functional equation $\log f(x,y) = f(\log x, \log y) $ This is not a complete answer, but the function $$f(x,y) = \lim_{n\rightarrow \infty}\underbrace{\exp\Bigg(\exp\Bigg(\cdots\exp\Bigg(}_n\sqrt{(\underbrace{\log\log\cdots\log}_n\,x)(\underbrace{\log\log\cdots\log}_n\,y)}\Bigg)\Bigg)\Bigg)$$ solves the functional equation.

Answer 4

Since you specifically asked for a function $H: \mathbb{C}^2 \to \mathbb{C}$, the answer should technically be that there is no such function for the simple reason that $H(0, 0)$ would be undefined, since the functional equation in this case would be $$\ln\left[H(0, 0)\right] = H(\ln 0, \ln 0)$$

Which is meaningless since $\ln 0$ itself is undefined. Even in the sense of limits, $\lim\limits_{z\to 0}\ln z$ is at best defined as "the point at infinity" in the sense that the real part of $\ln z$ diverges to $-\infty$, but its imaginary part can be anything depending on the direction of approach.

So at best you would have to specify what the equation means in this case. And even if you exclude $(0, 0)$ from the domain, you will still run into the issue when looking at $H(1, 1)$, which will relate to $H(0, 0)$ through the functional equation.

In addition, since $\ln$ is multivalued on $\mathbb{C}^2 \setminus \{(0, 0)\}$, the functional equation would also require a choice of branch for $\ln$ otherwise $H(\ln a, \ln b)$ would itself be multivalued.