Power Series

By Sébastien Boisgérault, Mines ParisTech, under CC BY-NC-SA 4.0

September 30, 2019

Contents

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 \(a_n\), \(n \in \mathbb{N}\).

Answers


  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 \(p\).

Answer

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.

Answer

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.

Answer

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.

Answer

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, \[ \sum_{n=1}^{+\infty} na_n (z-c)^{n-1} = \lim_{m \to +\infty} \sum_{n=1}^m na_n (z-c)^{n-1} = f'(z). \]