# Notation: ln_2(k)?

1. Oct 29, 2005

### CRGreathouse

I started looking at a proof I had set aside for myself to read some time ago* and wondered if I was missing something in the notation.
In several of the author's formulas, he writes $$\ln_2k$$. Is this just $$\log_2k=\lg k$$?

On a related note, since I see this as well: I've seen notation of the form $$f^k$$ used in two different ways -- $$f^k(x)=f(x)\cdot f^{k-1}(x)$$ (usually trig functions or logs) and $$f^k(x)=f(f^{k-1}(x))$$ (usually number-theoretic functions like $$\sigma$$, but I suppose function inverses are like this in a sense). Is there any rhyme or reason behind the choice to use one or the other as convention? Is one gaining popularity with respect to the other over time?

* Pierre Dusart, "The kth Prime is Greater than $$k(\ln k+\ln\ln k-1)$$ for $$k\geq2$$", Math. Comp. 68/225 Jan 1999, pp. 411-415.

Edit: I'm such a fool. The author does define it after all, I just missed it in my carelessness. $$\ln_2k=\ln\ln k$$ in this paper.

Last edited: Oct 30, 2005
2. Oct 29, 2005

### mathman

Notation can be a funny thing, particularly when talking about superscipts. It is important for the author to define precisely what is meant in a given context. Sometimes it could mean the kth derivative as opposed to the two definitions you described.

For the log notation, once you have the subscript 2, it doesn't matter whether you write ln or log. However it is customary to use ln without a subscript to refer to base e (natural) log.

3. Oct 30, 2005

### CRGreathouse

I've seen $$f^{(n)}$$ for that, but never just $$f^n$$ for the nth derivative.

I wish the author was more explicit -- a base-2 log is a base-2 log, but when you write ln ("natural log"), it seems like a way to disclaim that interpretation for another. You're probably right on that, though; it is the most sensible possibility I see.