Exercises

The Fibonacci Sequence

We search for a closed form of the Fibonacci sequence $$a_n$$, defined by $a_0 = 0, \; a_1 = 1, \; \forall \, n \in \mathbb{N}, \, a_{n+2} = a_n + a_{n+1}.$

Questions

1. Show that the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}$ is the largest solution of the equation $$x^2 = x + 1$$ and that the other solution is $$\psi = - {1}/ {\phi}.$$

2. Establish that for any $$n \in \mathbb{N}$$, $$a_n \leq \phi^n$$.

3. Show that the radius of convergence of the generating function $f(z) = \sum_{n=0}^{+\infty} a_n z^n$ is at least $$1/\phi$$.

4. Compute $$f(z)$$ when $$|z| < 1/\phi$$.

5. Find a closed form for $$a_n$$, $$n \in \mathbb{N}$$.

1. The discriminant $$\Delta$$ of the quadratic equation $$x^2 - x - 1 = 0$$ is $\Delta = (-1)^2 - 4 \times 1 \times (-1) = 5,$ therefore the solutions are $x = \frac{1 \pm \sqrt{5}}{2}.$ The golden ratio $$\phi$$, equal to $$(1 + \sqrt{5}) / 2$$, is the largest of the two. The fact that the other root $$\psi$$ of the equation is equal to $$-1/\phi$$ can be demonstrated directly; we have indeed $\psi = \frac{1-\sqrt{5}}{2} = \frac{1+\sqrt{5}}{1+\sqrt{5}} \frac{1 - \sqrt{5}}{2} = \frac{1^2 - \sqrt{5}^2}{2(1+\sqrt{5})} = - \frac{2}{1 + \sqrt{5}}.$ Alternatively, we know that $x^2 - x - 1 = (x-\phi)(x-\psi) = x^2 - (\phi+\psi) x + \phi \psi,$ hence $$\phi \psi = -1$$.

2. It is clear that $$a_0 = 0 \leq 1 = \phi^0$$ and $$a_1 = 1 \leq \phi = \phi^1$$. If we assume that the inequality $$a_n \leq \phi^n$$ holds for $$n=0, 1, \dots, m+1$$, the recursive definition of the Fibonacci sequence yields $a_{m+2} = a_m + a_{m+1} \leq \phi^m + \phi^{m+1} = \phi^m(1 + \phi) = \phi^{m+2}.$ Hence, by induction, the inequality holds for every $$n \in \mathbb{N}$$.

3. The inequality $$a_n \leq \phi^n$$ provides $\limsup_{n \to +\infty} \sqrt[n]{|a_n|} \leq \phi,$ and hence, by the Cauchy-Hadamard formula, the radius of convergence of the series $$\sum_{n \geq 0} a_n z^n$$ is at least $$1/\phi$$.

4. If $$|z| < 1/\phi$$, we can write the expansion of $$f(z)$$ as $f(z) = \sum_{n=0}^{+\infty} a_n z^n = a_0 + a_1 z + \sum_{n=0}^{+\infty} a_{n+2} z^{n+2} = z + \sum_{n=0}^{+\infty} a_{n+2} z^{n+2}.$ Using $$a_{n+2} = a_n + a_{n+1}$$, we deduce that $f(z) = z + z^2 \sum_{n=0}^{+\infty} a_n z^n + z \sum_{n=0}^{+\infty} a_{n+1} z^{n+1} = z + z^2 f(z) + z f(z),$ hence $f(z) = \frac{z}{1 - z - z^2}.$

5. The roots of the polynomial $$1 - z - z^2$$ are $$-\phi$$ and $$-\psi$$, hence $- z^2 - z + 1= - (z + \phi)(z + \psi).$ Thus, for any $$|z| < 1/\phi$$, we have $f(z) = \frac{-z}{(z+\phi)(z+\psi)} = \frac{1}{\phi- \psi} \left[\frac{-\phi}{z+\phi} + \frac{\psi}{z+\psi} \right],$ or equivalently, using $$\psi = -1/\phi$$, $f(z) = \frac{1}{\phi - \psi} \left[ \frac{-1}{1 - \psi z} + \frac{1}{1 - \phi z} \right].$ If $$|z| < 1/\phi$$, then $$|\phi z| < 1$$ and $$|\psi z| < 1$$ and consequently $\frac{1}{1 - \phi z} = \sum_{n=0}^{+\infty} \phi^n z^n,\; \frac{1}{1 - \psi z} = \sum_{n=0}^{+\infty} \psi^n z^n.$ Thus, $$f(z)$$ can be expanded as $f(z) = \sum_{n=0}^{+\infty} \frac{1}{\phi - \psi} \left[\phi^n - \psi^n \right]z^n.$ The power series expansion of $$f(z)$$ in the disk centered on the origin with radius $$1/\phi$$ is unique, therefore $a_n = \frac{1}{\phi - \psi} \left[ \phi^n - \psi^n \right] = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n \right]$ for every $$n \in \mathbb{N}$$.

Entire Functions Dominated By Polynomials

Question

Show that if a holomorphic function $$f:\mathbb{C} \to \mathbb{C}$$ is dominated by a polynomial $$P$$ of order $$p$$ $\forall \, z \in \mathbb{C}, \; |f(z)| \leq |P(z)|$ then it is a polynomial whose degree is at most $$p$$.

Let $$\sum_{n=0}^{+\infty} a_n z^n$$ be the power series expansion of $$f$$ in $$\mathbb{C}$$. For any $$r>0$$, we have $a_n = \frac{1}{i 2\pi} \int_{r[\circlearrowleft]} \frac{f(z)}{z^{n+1}} dz,$ hence by the M-L estimation lemma, $|a_n| \leq \frac{\sup \, \{|P(re^{i2\pi t})| \; | \; t \in [0,1]\}}{r^n}.$ For any $$n> p$$, letting $$r \to +\infty$$ provides $$a_n=0$$. Hence, the function $$f$$ is a polynomial of degree at most $$p$$.

Existence of Primitives

Question

Show that the function $f: z \in \mathbb{C} \setminus [-1, 1] \mapsto \frac{\pi}{z} \frac{1}{\sin \pi/z}$ has a primitive.

The function $$f$$ is defined and holomorphic in $$\mathbb{C} \setminus [-1,1]$$ (the zeros of $$\sin \pi/z$$ are $$z=1/k$$ for $$k \in \mathbb{N}^*$$).

We first consider the restriction of $$f$$ to the annulus $$A(0,1, +\infty)$$. For any $$z$$ in this annulus, $$-z$$ also belong to it and $$f(-z) = f(z)$$. Hence, if $$\sum_{n=-\infty}^{+\infty} a_n z^n$$ is a Laurent series expansion of $$f$$, $$\sum_{n=-\infty}^{+\infty} (-1)^n a_n z^n$$ is another valid one. The uniqueness of the expansion yields that $$a_n = 0$$ if $$n$$ is odd; in particular, $$a_{-1}=0$$ and the sum $\sum_{p=-\infty}^{+\infty} \frac{a_{2p}}{2p+1} z^{2p+1}$ provide a primitive of $$f$$ on the annulus.

Now, let $$\gamma$$ be an arbitrary closed rectifiable path of $$\mathbb{C} \setminus [-1, 1]$$. Let $$n=\mathrm{ind}(\gamma, 0)$$; define the path $$\mu:t \in [0,1] \mapsto 2 e^{i2\pi n t}$$ and the sequence of paths $$\nu = (\gamma, \mu^{\leftarrow})$$. As $$[-1,1]$$ is a connected subset of $$\mathbb{C} \setminus \nu ([0,1])$$, for any $$z\in[-1,1]$$, $$\mathrm{ind}(\nu, z) = \mathrm{ind}(\nu, 0)= 0$$. Consequently, $$\mathrm{Int} \, \nu \subset \mathbb{C} \setminus [-1,1]$$ and Cauchy’s integral theorem provides $\int_{\gamma} f(z) \, dz = \int_{\mu} f(z) \, dz.$ As $$f$$ has a primitive on the annulus $$A(0,1, +\infty)$$, the integral in the right-hand side of this equation is equal to zero. The classic criteria therefore proves that primitives of $$f$$ exist in $$\mathbb{C} \setminus [-1, 1]$$.

A Removable Set

Let $$f: \mathbb{C} \to \mathbb{C}$$ be a continuous function which is holomorphic on $$\mathbb{C} \setminus \mathbb{U}$$ (where $$\mathbb{U} = \{z \in \mathbb{C} \; | \; |z|=1\}$$).

Question

Prove that $$f$$ is an entire function.

Let $$\sum_{n=0}^{+\infty} a_n z^n$$ be the Taylor series expansion of $$f$$ in $$D(0,1)$$; we are going to prove that this expansion is actually a valid expansion of $$f$$ in $$\mathbb{C}$$. Consider the Laurent expansion $$\sum_{n=-\infty}^{+\infty} b_n z^n$$ of $$f$$ in $$A(0,1,+\infty)$$. For any $$n \in \mathbb{Z}$$ and any $$r > 1$$, we have $b_n = \frac{1}{i2\pi}\int_{r[\circlearrowleft]} \frac{f(z)}{z^{n+1}} dz,$ thus, by continuity of $$f$$ $\begin{split} b_n &= \lim_{r\to 1^+} \frac{1}{i2\pi}\int_{r[\circlearrowleft]} \frac{f(z)}{z^{n+1}} dz\\ &= \lim_{r\to 1^-} \frac{1}{i2\pi}\int_{r[\circlearrowleft]} \frac{f(z)}{z^{n+1}} dz \end{split}$ and consequently, $$b_n= a_n$$ if $$n$$ is non-negative and zero otherwise. The sum $$\sum_{n=0}^{+\infty} a_n z^n$$ is defined for any $$|z| > 1$$, thus its open disk of convergence is the full complex plane. It is equal to $$f$$ on $$\mathbb{C} \setminus \mathbb{U}$$ and both functions are continuous on $$\mathbb{C}$$, hence they are equal on $$\mathbb{C}$$: the function $$f$$ is entire.

Derivative of Power Series

Question

Provide an alternate proof of the existence and value of the derivative of the sum $$\sum_{n=0}^{+\infty} a_n (z-c)^n$$ in its open disk of convergence.

Hint: a locally uniform limit of a sequence of holomorphic functions is holomorphic.

Let $$f_m(z) = \sum_{n=0}^m a_n (z-c)^n$$. Every polynomial $$f_m$$ is holomorphic and the sequence converges locally uniformly to $$f(z) = \sum_{n=0}^{+\infty} a_n (z-c)^n$$ in the open disk of convergence $$D(c, r)$$ of the series, thus $$f$$ is holomorphic.
For any holomorphic function $$\phi$$ in $$D(c,r)$$ and any $$\rho \in \left]0,r\right[$$ $\phi'(z) = \frac{1}{i2\pi} \int_{c+\rho[\circlearrowleft]} \frac{\phi(w)}{(w-z)^2}.$ Thus, for any $$m \in \mathbb{N}$$, $f'_m(z) = \sum_{n=1}^m na_n (z-c)^{n-1} = \frac{1}{i2\pi} \int_{c+\rho[\circlearrowleft]} \frac{f_m(w)}{(w-z)^2}.$ The integrand above converges locally uniformly in $$D(c,r)$$, hence $\lim_{m \to +\infty} \frac{1}{i2\pi} \int_{c+\rho[\circlearrowleft]} \frac{f_m(w)}{(w-z)^2} = \frac{1}{i2\pi} \int_{c+\rho[\circlearrowleft]} \frac{f(w)}{(w-z)^2} = f'(z).$ Finally, $\sum_{n=1}^{+\infty} na_n (z-c)^{n-1} = \lim_{m \to +\infty} \sum_{n=1}^m na_n (z-c)^{n-1} = f'(z).$