If $f(x)=\frac{1}{1-x-x^2}$, find $\frac{f^{(10)}(0)}{10!}$ My math teacher posed this question to my calculus class. If $f(x)=\frac{1}{1-x-x^2}$, find $\frac{f^{(10)}(0)}{10!}$.
At first, I began by taking the first few derivatives, but it soon go out of hand with the repeated quotient rules. I'm sure that I could continue for $10$ derivatives, but I believe that there must be an easier solution.
 A: Use the fact that $$f(x)=\frac{1}{1-x-x^2}=\sum \limits_{n=0}^{\infty} F_{n+1} x^n.$$ Where $F_n$ is the $nth$ Fibonacci number. To prove this property, recall the recursive definition of the Fibonacci sequence.
Thus, we only need to find the $11th$ term in the series, which is of the form $F_{11} x^{10}$. The first $9$ terms vanish after $10$ repeated derivatives and the infinitude of terms after the $10th$ disappear when we substitute $x=0$. 
Interestingly, from this relation, you can extract an explicit formulafor the $nth$ Fibonacci number. You only need to use partial fractions on the LHS and the formula for a geometric sum.
EDIT:
Thanks to Brevan Ellefsen, I would recommend looking up generating functions, which describe sequences by treating them like the coefficients in an infinite series.
A: If you don't know the theory of generating functions, the trick is to get the Taylor expansion with partial fractions. Write
$$
\frac{1}{1-x-x^2}=\frac{a}{1-\alpha x}+\frac{b}{1-\beta x}
$$
that gives the relations
$$
\begin{cases}
a+b=1\\[4px]
a\beta+b\alpha=0\\[4px]
\alpha+\beta=1\\[4px]
\alpha\beta=-1
\end{cases}
$$
Now you can use
$$
\frac{a}{1-\alpha x}=a\sum_{n\ge0}\alpha^nx^n
$$
so
$$
\frac{f^{(10)}(0)}{10!}=a\alpha^{10}+b\beta^{10}
$$

Once you compute $a$, $b$, $\alpha$ and $\beta$, you'll notice that
$$
a\alpha^n+b\beta^n=\frac{\varphi^n-\bar{\varphi}^n}{\sqrt{5}}
$$
where
$$
\varphi=\frac{1+\sqrt{5}}{2},\qquad
\bar{\varphi}=\frac{1-\sqrt{5}}{2}
$$
which is the Bézout formula for the Fibonacci numbers.
A: Hint:
$$\frac1{1-u}=1+u+u^2+\dots+u^{10}+o(u^{10})$$
Set $u=xx+x^2$ and find the coefficient of $x^{10}$.
Another hint:
Perform the division of $1$ by $1-x-x^2$ along the increasing powers of $x$.
You should find the Fibonacci number $F_{11}$.
A: If we consider 
$$ g(z) = \frac{z}{1-z-z^2} = \sum_{n\geq 0} a_n z^n $$
we have an analytic function in a neighbourhood of zero whose coefficients fulfill $a_0=0, a_1=1$ and $a_{n+2}-a_{n+1}-a_{n}=0$ for any $n\geq 0$ (it is enough to compute the coefficient of $z^{n+2}$ in $z=(1-z-z^2)\sum_{n\geq 0}a_n z^n$). It follows that $a_n = F_n$, the $n$-th Fibonacci number, so:
$$ f(z) = \frac{g(z)}{z} = \frac{1}{1-z-z^2} = \sum_{n\geq 0}\frac{f^{(n)}(0)}{n!}\cdot z^n = \sum_{n\geq 0} F_{n+1}\cdot z^n $$
and the answer is given by $\color{red}{F_{11}=89}$.
