dzyubam wrote:

I'll try to make this clearer by using our math syntax. See the short intro

here:

Given

P# = \frac{P}{P-1}, need to find

P## - original problem

We might simplify this problem by saying that this '#' sign is an operation that replaces each

P with

\frac{P}{P-1}.

P## is found by replacing all the

Ps with

\frac{P}{P-1} in this expression:

\frac{P}{P-1}, which equals

P#This is how it looks after these replacements:

\frac{\frac{P}{P-1}}{\frac{P}{P-1}-1}Now we simplify it gradually:

\frac{\frac{P}{P-1}}{\frac{P}{P-1}-\frac{P-1}{P-1}}=\frac{\frac{P}{P-1}}{\frac{P-P+1}{P-1}}=\frac{\frac{P}{P-1}}{\frac{1}{P-1}}=\frac{P}{P-1}*\frac{P-1}{1}=PHope this clears the doubts.

I think that's a baldy written question. Why not P## be a multiplication? Similarly to other questions like:

What is (2^3)(4^5)(6^7) ? Which straightly implies multiplication of the 3 numbers (as opposed to ask for a number where the 1st digit(s) would be 2^3, the next ones 4^5 and the last ones 6^7.

(Actually I got confused and ended up multiplying # * #).

In m opinion it should be written as a function (P(P#)) or something.