Generating Function for even and odd Fibonacci Numbers, Missing Steps

Question

I am working through this post:https://math.stackexchange.com/a/787841/283262 and its connected to this one too: Recurrences for even and odd indexed of Fibonacci Numbers

I did not get all parts so I have added some parts, and the parts I did not get, maybe someone could ad the missing steps.

Definition of Fibonacci Numbers: $$ F_{n } = F_{n - 1} + F_{n-2} \qquad F_0 = 0, F_1 = 1 $$ The sequence :(0,1,1,2,3,5,8,13,21,34,55,...)

One way to find the generating Function could be: $$ F(x):=\sum_{n \ge 0}F_n x^n=_{since F_0=0}\sum_{n \ge 1}F_n x^n $$ $$ F(x):= x + \sum_{n \ge 2}(F_{n-1}-F_{n-2}) x^n=x+x\sum_{n \ge 2}F_{n-1} x^{n-1}+x^2\sum_{n \ge 2}F_{n-2} x^{n-2} $$ (Q1) One Question would be why this: $$ F(x)=x+x*F(x)+x^2*F(x) $$ My assumption is that since each of the sums converge towards F(x).

Now once could solve for F(x) and get: $$ F(x)=\frac{x}{1-x-x^2} $$

Then one would need to proove: $$ \frac{1}{2}(A(x)+A(-x))=\sum_{n \ge 0} a_{2 n} x^{2 n} and \frac{1}{2}(A(x)-A(-x))=\sum_{n \ge 0} a_{2 n + 1} x^{2 n + 1} $$ With $ A(z) = \sum_{n \ge 0} a_n x^n$. This part is clear, open summation to even index and odd and the put it back together.

(Q2) I dont know where this is coming from? And how to prrove it. $$ \sum_{n \ge 0} \left( \sum_{0 \le k \le n} a_k \right) z^n = \frac{A(z)}{1 - z} $$ (Q3) This part I dont get at all: why is $z^{1/2}$, and how to get the generating function: So, for even Fibonacci numbers: \begin{align} F_e(z) &= \sum_{n \ge 0} F_{2 n} z^n \\ &= \frac{F(z^{1/2}) + F(- z^{1/2})}{2} \\ &= \frac{z}{1 - 3 z + z^2} \\ \end{align}

Maybe someone could add the missing steps. Thank you in advance.

update (Q2) solved. $$ F(x)=\frac{z}{1 - x} =1+z+z^2+... $$ The coefficients are (1,1,1,1,...) When multiplied with G(z) and apply the basic property of multiplcation for generating functions, you get this. Basic properties: Concrete Mathmatics p335

Answer 1

(Q1) Here there's no need for analytic solutions, just manipulate the series algebraically changing indices. Since

$$ F(x) = x+\sum_{n \ge 2}(F_{n-1}-F_{n-2})X^n = X+X\sum_{n \ge 2}F_{n-1} X^{n-1}+X^2\sum_{n \ge 2}F_{n-2} X^{n-2} $$

it suffices to note that

$$ \sum_{n \ge 2}F_{n-1} X^{n-1} = \sum_{n \ge 1}F_{n} X^{n} = \sum_{n \ge 0}F_{n} X^{n} $$

and

$$ \sum_{n \ge 2}F_{n-2} X^{n-2} = \sum_{n \ge 0}F_{n} X^{n} $$

If you stare at this long enough, you'll be convinced that in effect these are the same expressions. We're just changing the offset from the expression towards the index.

(Q2) Given two formal series $f(X) = \sum_{n \geq 1}c_nX^n$, $g(X) = \sum_{n\geq 1}d_nX^n$, their product is defined as

$$ fg(X) = \sum_{n\geq 1}\big(\sum_{k = 0}^na_kb_{n-k}\big)X^n $$

if you take $A = f$ and $g = \sum_{n\geq 1}X^n$, we have

$$ \sum_{n \ge 0} \left( \sum_{0 \le k \le n} a_k \right) X^n = A(X)\big(\sum_{n \geq 1}X^n\big) $$

it suffices then to note that $\sum_{n \geq 1}X^n = (1-X)^{-1}$. To prove this, multiply the series by $(1-X)$ and conclude that their product is $1$ (once again, we only need algebraic manipulations of formal series).

(Q3) As you noted, summing $\frac{F(X) + F(-X)}{2}$ will cancel out odd exponents, leaving you with the series $\sum_{n \geq 1}F_{2n}X^{2n}$. However, you'd like to have $\sum_{n \geq 1}F_{2n}X^n$. From the previous expression, it's quite clear that composing by $X^{\frac{1}{2}}$ does the job. Since you have a closed expression for $F$ in terms of a rational function, you can calculate $\frac{F(X^{\frac{1}{2}}) + F(-X^{\frac{1}{2}})}{2}$ without dealing with series, which produces the aforementioned result, that is, $\frac{X}{1-3X+X^2}$.