A question about $[c_0,c_1,\ldots,c_n]$ notation for continued fractions

Question

I try to understand why by definition

1. $[c_0,c_1,\ldots,c_n]=[c_0,[c_1,\ldots,c_n]]$ and also
2. $[c_0,c_1,\ldots,c_n]=[c_0,c_1,\ldots,c_{n-2},[c_{n-1},c_n]]$ .

Those are continued fractions, and $1$ and $2$ are notes I have in the lecture summery.

But we can add bracket where we we want. for example:
$[c_0,c_1,\ldots,c_n]=[c_0,c_1,[c_2,\ldots,c_n]]$

Thanks!

Answer 1

It's a fairly easy inductive proof that you can write $$[c_0, c_1, c_2, \dots, c_{n-1}, c_n] = [c_0, [c_1, [c_2, \cdots [c_{n-1}, c_n] \cdots ]]]$$ and you can add or remove any of the sets of square brackets that appear on the right-hand side as you please. That is, you can stick a $[$ where you like as long as the closing $]$ is right at the end, not somewhere in the middle

What you can't do is add square brackets into the left-hand side willy-nilly. For instance, in general, $$c_0+\dfrac{1}{c_1+\frac{1}{c_2}} = [c_0, [c_1, c_2]] \ne [[c_0, c_1], c_2] = c_0+\dfrac{1}{c_1}+\dfrac{1}{c_2}$$

Answer 2

Usually one wouldn't write such a thing as $[c_0; [c_1, \ldots, c_n]]$, because elements appearing in a continued fraction representation are supposed to be positive integers (except the one before the semicolon, if there is one, which may be a negative integer).

But it doesn't contradict anything to write this. Well, from the definition of a continued fraction, you have

$$[c_0; c_1, \ldots, c_n] = c_0 + \frac 1 {[c_1, \ldots, c_n]}$$

And also

$$[c_0; c_1] = c_0 + \frac 1 {c_1}$$

Can you see how claim 1 follows from this? For claim 2, try using induction with claim 1.

Note: I added a semicolon after $c_0$ because the index "$0$" seems to indicate that is what you were looking for.

Answer 3

By definition:

$$[c_0,c_1,\ldots,c_n]:=c_0+\cfrac{1}{c_1+\cfrac{1}{c_2+\cfrac{1}{\ddots+\cfrac{1}{c_n}}}}=c_0+\cfrac{1}{[c_1,c_2,\ldots,c_n]}=[c_0,[c_1,\ldots,c_n]]\;,\;etc.$$

So yes: you can put the brackets where you want on the right.