TITLE: Does the order of the Fibonacci sequence's initial values matter?
QUESTION [15 upvotes]: I am reading about generating functions in this reputable engineering textbook and the author uses the Fibonacci sequence as an example:

The Fibonacci sequence is defined by the initial conditions $a_0=0$,
$a_{-1}=1$, and the recursion relation $a_n=a_{n-1}+a_{n-2}$ for
$n>0$.

Using this definition, I am able to work through the author's example to get the correct generating function:
$$G(x)=\frac{1}{1-x-x^2}$$
However, everywhere I read online states the initial conditions of the Fibonacci sequence as (0,1) and not (1,0). So I tried that instead, but I get a different result:
$$G(x)=\frac{1+x}{1-x-x^2}$$
This generating function seems to produce the Fibonacci sequence, but starting from the second number instead of the first $(1,2,3,5,8,13,...)$. Where did I go wrong? I would have thought that the sequence should be identical in both cases.
Many thanks for any help!
My calculations:
The author starts with a general expression for a generating function for a sequence $\{a_n\}$:
$$G(x)=\sum_{n=0}^{\infty}a_nx^n$$
He then looks at the special case where $\{a_n\}$ is a linear recurrence of range $r$:
$$a_n=\sum_{i=1}^rc_ia_{n-i}$$
and proves that this leads to:
$$G(x)=\frac{g(x)}{f(x)}=\frac{\sum_{i=1}^rc_ix^i\left(a_{-i}x^{-i}+\cdots+a_{-1}x^{-1}\right)}{1-\sum_{i=1}^rc_ix^i}$$
So, for the Fibonacci sequence, I guess we have $r=2$ and $c_1=c_2=1$. So we get:
$$g(x)=a_{-1}+a_{-2}+a_{-1}x$$
Now I'm a bit confused because the subscripts don't seem to match up. In all the author's discussion, the "initial conditions" have strictly negative subscripts ($a_{-i}$ for $i=1,2,\dots,r$). But just for the Fibonacci example, we have initial conditions $a_0=0$ and $a_{-1}=1$. So I assume these are just off by 1. Therefore, substituting $a_{-1}=0$ and $a_{-2}=1$ gives the author's result above and using $a_{-2}=0$ and $a_{-1}=1$ gives my result above.

REPLY [2 votes]: Although one does not always make this explicit, giving a sequence of numbers really amounts to giving a set of (index,value) pairs; for instance an infinite sequence of real numbers is really a function $\Bbb N\to\Bbb R$, whose graph is an infinite set of such pairs. The generating series for a set $S$ of such pairs $(i,a_i)$ is $\sum_{(i,a_i)\in S}a_iX^i$, so for a series of numbers corresponding to $f:\Bbb N\to\Bbb R$ this is $\sum_{i=0}^\infty f(i)X^i$.
Let me first put aside the easy answer to the question in your title: of course order of initial terms matters, and permuting them in general gives an entirely different sequence. But that is not what your question was really about, which seems to be more about whether a shift in indexing produces a different generating series. Here too the answer is yes: a pure shift in indexing amounts for the generating series in multiplication by a power of $X$. If the shift is accompanied by addition or removal of nonzero values in the sequence, then this amounts to addition of terms to the generating series.
Note that normally a generating series is a formal power series in $X$, with no negative powers of $X$ at all. In this sense there is no generating series at all for a sequence with a nonzero value at some negative index, as your book apparently wants to do by starting at $a_{-1}=1$. If you insist on including that, the generating series would have a term $X^{-1}$, and thereby become a Laurent series rather than a formal power series. But usually those who entertain the idea of Fibonacci numbers at negative indices want them for all negative integer indices, but for such bi-infinite sequences the idea of generating series simply does not work: there is no appropriate algebraic framework. Think of it: if there were such a thing as $S=\sum_{i\in\Bbb Z}a_iX^i$ for a sequence $(a_i)_{i\in\Bbb Z}$ satisfying $a_{i+2}=a_i+a_{i+1}$ everywhere, then one would want to express this by saying $S(1-X-X^2)=0$, but $S=0/(1-X-X^2)$ in no way describes a generating series for the Fibonacci numbers (for one thing that expression gives no information about any "initial terms" that were used, for another it just simplifies to $0$ in any reasonable algebraic setting).