Does the order of the Fibonacci sequence's initial values matter?

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.

In the particular case of the Fibonacci number sequence OEIS A000045 (or series) there is some difference of opinion as amply evidenced by the Wikipedia article and OEIS entry. The most common convention is that $\,F_0=0\,$ and $\,F_1=1\,$ and this choice has much in its favor. This choice implies that its generating function is $$ \sum_{n=0}^\infty F_n\,x^n = \frac{x}{1-x-x^2}. $$ Notice the $\,x\,$ in the numerator. For this and a few other reasons, some prefer the choice $\,F_0=F_1=1\,$ instead, which implies that the generating function is now $$ \sum_{n=0}^\infty F_n\,x^n = \frac{1}{1-x-x^2}. $$ Both choices are valid and useful in some cases and not as useful in other cases. Thus, it is a good idea to be clear on which choice is being used in any given context since the choice will usually matter.

The situation with negative index Fibonacci sequence elements is that the recurrence relation for the sequence can be used to uniquely extend the sequence in the negative index direction. For the common convention this implies that $$ F_{-n} = (-1)^{n-1}F_n \quad\text{ for all integer }n. $$ The result for the other convention it is that $$ F_{-n} = (-1)^n F_{n-2} \quad\text{ for all integer }n. $$

What if we choose arbitrary starting values with the same recurrence relation? That is, if we choose $\,F_0=a\,$ and $\,F_1=b,\,$ what do we get? The recurrence relation is linear and this implies that the sequence of numbers we get is a linear combination of the two initial value choice sequences. The generating function now is $$ (a) \!+\! (b)x \!+\! (a+b)x^2 \!+\! (a+2b)x^3 \!+\! \cdots \!=\! \frac{a\!+\!(b\!-\!a)x}{1-x-x^2}. $$

Note that initial values of Fibonacci and related sequences can be given for any two consecutive index values--not just for $\,a_0\,$ and $\,a_1\,$ since the recursion going forward and backward can reach any other integer index value. Again, this is just a matter of convenience and convention. In the case you cite, specifying $\,a_{-1}\,$ and $\,a_0\,$ is fine. The initial values $\,a_{-1}=1,\,a_0=0\,$ gives the same sequence as $\,a_0=0,\,a_1=1.$ This special case may be misleading though. The initial values $\,a_{-1}=u,\,a_0=v\,$ gives the same sequence as $\,a_0=v,\,a_1=u+v\,$ and $\,u+v\,$ is equal to $\,u\,$ if and only if $\,v=0.$

Note that the Wikipedia article Generalizations of Fibonacci numbers mentions some of the issues such as extension to negative integers. It mentions a vector space of number sequences:

Since the set of sequences satisfying the relation $\,S(n)=S(n-1)+S(n-2)\,$ is closed under termwise addition and under termwise multiplication by a constant, it can be viewed as a vector space. Any such sequence is uniquely determined by a choice of two elements, so the vector space is two-dimensional. If we abbreviate such a sequence as $\,(S(0),S(1)),\,$ the Fibonacci sequence $\,F(n)=(0,1)\,$ and the shifted Fibonacci sequence $\,F(n-1)=(1,0)\,$ are seen to form a canonical basis for this space, yielding the identity: $$ S(n)=S(0)F(n-1)+S(1)F(n) $$ for all such sequences $\,S$.

The recursion $a_n=a_{n-1}+a_{n-2}$ is "encoded" in the denominator of the generating function: $1-x-x^2$. The numerator is determined by the values of $a_0$ and $a_1$. In particular, $$ \overbrace{\left(1-x-x^2\right)\vphantom{\sum^\infty}}^\text{denominator}\overbrace{\sum_{n=0}^\infty a_kx^k}^{\substack{\text{generating}\\\text{function}}}=\overbrace{a_0+(a_1-a_0)\,x\vphantom{\sum^\infty}}^\text{numerator}+\sum_{n=2}^\infty\overbrace{(a_n-a_{n-1}-a_{n-2})\vphantom{\sum^\infty}}^0\,x^n $$ Thus, the generating function is $$ \frac{a_0+(a_1-a_0)x}{1-x-x^2} $$

I prefer the definition of the Fibonacci Numbers that starts with $F_0=0$ and $F_1=1$, whose generating function is $$ \frac{x}{1-x-x^2} $$ because, among other things, we have $$ n\mid m\implies F_n\mid F_m $$ Furthermore, $$ F_{-n}=(-1)^{n-1}F_n $$ Lucas Numbers also obey the same recursion as Fibonacci Numbers, and their generating function is $$ \frac{2-x}{1-x-x^2} $$