Generating Function for Fibonacci Sequence

We seek closed form for the power series $\sum_{n=0}^\infty F_nz^n$, where $F_0 = 1$, $F_1 = 1$ and $F_{n+2} = F_{n+1} + F_n$. First we note that the ratio test yields

$\lim_{n\rightarrow\infty}\frac{F_{n+1}}{F_n}|z| = \varphi |z|$, where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. Thus the series converges on the region of the complex plane where $|z| < \frac{1}{\varphi}$. Now let the power series be denoted by $F(z)$, and note that

$F(z) - zF(z) - z^2F(z) = 1 + \sum_{n=0}^\infty (F_{n+2}-F_{n+1}-F_n)z^{n+2} = 1$

and so $F(z) = \frac{1}{1-z-z^2}$. It seems so unlikely, before you know better at least, that things like that can be done. Why should it be that the nth Fibonacci number is given by $\frac{1}{2\pi i}\oint\frac{1}{z^{n+1}(1-z-z^2)}dz$, the integral along some closed contour in the complex plane of a function that otherwise seems unrelated. That’s maths!

On Complex vs Real Differentiability

Once upon a time I was confused about what really made a holomorphic function different from a function on $\mathbb{R}^2$ which is differentiable in the real sense. I mean obviously I knew how they were different but I wanted to put it in a context that was more meaningful for me, and after some thinking I thought the following and was satisfied:

Let’s cover some definitions; it’s likely I’ll write these classic definitions in more detail in later posts. Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a complex function and let $z_0$ be a complex number. $f$ is called (complex) differentiable at $z_0$ if the limit

$f'(z_0) = \lim_{z\rightarrow z_0}\frac{f(z) - f(z_0)}{z-z_0}$

exists, where the limit is taken along whatever path you like. The function $f$ is differentiable on a subset $A\subset\mathbb{C}$ if it is differentiable for all $z_0\in A$. Now a complex number can be written as $z = x+iy$, where $x,y\in\mathbb{R}$, and a complex function can be written as $f(z=x+iy) = u(x,y)+iv(x,y)$ where $u$ and $v$ are both functions from $\mathbb{R}^2$ to $\mathbb{R}$. As $\mathbb{R}$-vector spaces, $\mathbb{C}$ and $\mathbb{R}^2$ are isomorphic and we can think of $f$ as a function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ given by

$(x,y)\mapsto (u(x,y),v(x,y))$.

Now we it can be said that a complex function is complex differentiable if and only if it is differentiable in the real sense (i.e. all partial derivatives of $u$ and $v$ exist and are continuous) and $u$ and $v$ satisfy the Cauchy-Riemann equations ($\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}$).

This observation is well known and does a good job of contrasting the two senses of differentiability, but I wanted to know the answer to “why is the derivative of the real version of the function a linear transformation (given by the Jacobian matrix) but the derivative of the complex version another complex function?” The answer to that is obvious in hindsight, but I was happy when I went from not knowing to knowing so I thought I better recount it here. The trick is, of course, that we construct a field isomorphism (which I’ll call $\varphi$) from $\mathbb{C}$ into $\mathrm{M}_2(\mathbb{R})$ (the ring of 2 by 2 matrices with real entries) given by

$\varphi(z) = \begin{pmatrix} \mathrm{Re}z & -\mathrm{Im}z \\ \mathrm{Im}z& \mathrm{Re}z \end{pmatrix}$.

Now taking the derivative  of $f$ gives us a function $\mathrm{D}f:\mathbb{R}^2\rightarrow\mathrm{M}_2(\mathbb{R})$, but the Cauchy-Riemann equations mean that if $f$ is complex differentiable, then $\mathrm{D}f(\mathbb{R}^2) \subset \varphi(\mathbb{C})\subset\mathrm{M}_2(\mathbb{R})$. So in that case we can think of $\mathrm{D}f$ as a function from $\mathbb{R}^2\cong\mathbb{C}$ to $\varphi(\mathbb{C})\cong\mathbb{C}$, and so my question is resolved! The real version of the function is complex differentiable precisely when the Jacobian matrix can be identified with a complex function through the isomorphism $\varphi$.