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Is anything interesting known about the binary operation $$ x\circ y = \exp_b((\log_b x)(\log_b y)) $$ where $0<b\ne 1$?

It's clearly commutatitive and associative, and satisfies $\forall x\in \mathbb R^+$, $x\circ b=x$. In one sense is obviously equivalant to multiplication so nothing interesting can be said after everything's been said about multiplication, but then we can wonder if there anything interesting to be said about this way of embedding a structure isomorphic to $(\mathbb R^+,\times)$ into the line. In particular, we have (as I noted yesterday in another thread in this forum) $$ \log_{{}\,p \,\circ\, q\,\circ\, r\,\circ\,\cdots} (w\circ x\circ y\circ\cdots) = (\log_{{}\,p} w)(\log_{{}\,q} x)(\log_{{}\,r} y)\cdots. $$

I know I've seen this function arising in routine stuff, but I can't remember any specifics. So a question is: in which contexts does this operation arise naturally?

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Your "multiplication" $x \circ y$ doesn't seem to interact well with addition -- certainly $\circ$ is not distributive over $+$. However, there are nice relationships with exponentiation: $(x \circ y)^z = x^z \circ y = x \circ y^z$. –  Shaun Ault Oct 3 '12 at 18:14
    
Under this operation, it's easy to show that $x^{-1}=b^{1/\log_b x}$. Look at the graph of $(x, x^{-1})$ and compare it with the same graph for ordinary multiplication. –  Rick Decker Oct 3 '12 at 18:17
    
There is distribution over ordinary multiplication: $x \circ(y \cdot z) = (x \circ y) \cdot (x \circ z)$. –  Shaun Ault Oct 3 '12 at 18:19
    
I would call this "conjugation of $\times$ with $\log$" (i.e. something like $x\circ y = \log^{-1}(\log(x)\times\log(y))$). –  Dirk Oct 3 '12 at 19:10
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This operation arises in homework assignments in group theory courses. You define $a*b=a^{\log b}$ on some appropriate subset of the reals, and then you ask the students, is it associative? is there an identity element? does each element have an inverse? is it commutative? It's a good exercise, since at first glance you might not expect it to have all these properties. –  Gerry Myerson Oct 4 '12 at 8:02

1 Answer 1

Ok, I think I have something interesting to say. I remembered an exercise I gave to my Linear Algebra course that asked students to prove that there is a linear transformation taking $(\mathbb{R}, +)$ to $(\mathbb{R}^+, \cdot)$, where the latter is considered a vector space under the binary operation $\cdot$, with scalar "multiplication" $c.x = x^c$. The linear transformation is $x \mapsto e^x$.

Now consider applying that same transformation $e^x$ from $\mathbb{R}^+$ to $\mathbb{R}^{>1}$. I claim it is a vector space isomorphism $(\mathbb{R}^+, \cdot) \to (\mathbb{R}^{>1}, \circ)$, where scalar multiplication is:

$$ c.x = \exp( (\log x)^c).$$

[Of course, I'm specializing to $b=e$, but the results are the same for any $b$]

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