Show that if 0 is not a pole of $P/Q \in \mathbf{Q}(x)$, then there exists a unique power series with rational coefficients $g \in \mathbf{Q}\llbracket x \rrbracket$ such that $P = Q \cdot g$. Show that the map $P/Q \longmapsto g$ is compatible with addition and multiplication in $\mathbf{Q}(x)$ and in $\mathbf{Q}\llbracket x \rrbracket$, and that it sends the derivative $(P/Q)' = (P'Q - PQ')/Q^2$ to the derived power series $g'$.
Let $Q \in \mathbf{Q}[x]$ be a polynomial with rational coefficients whose constant term equals 1. Show that there exists an integer $b \geq 1$ such that $Q(bx)$ has integer coefficients.
Show that if $f \in \mathbf{Q}\llbracket x \rrbracket$ is the power series expansion of a rational function with rational coefficients, then $f$ is globally bounded. (A power series $f(x) = \sum_{n=0}^{\infty} c_n x^n \in \mathbf{Q}\llbracket x \rrbracket$ is globally bounded if there exist integers $A, B \geq 1$ such that $A f(Bx) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} (B^n A c_n) x^n$ is a power series with integer coefficients.)
Show that the power series $$\sum_{m=0}^{\infty} x^{m^2} = \sum_{n=0}^{\infty} c_n x^n$$ where $c_n = 1$ if $n$ is the square of an integer $m \geq 0$ and $c_n = 0$ otherwise, is not the power series expansion of a rational function.
Give an example of a power series expansion of a rational function whose antiderivative is not the expansion of a rational function. (The antiderivative of a power series $f(x) = \sum_{n=0}^{\infty} c_n x^n$ is defined as $\int_0^x f(t)\,dt \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} \frac{c_n}{n+1} x^{n+1}$.)
Q10
First order differential equations (integrating factor)View
(More difficult question) Let $f$ be the power series expansion of a rational function $P/Q \in \mathbf{Q}(x)$ all of whose poles are rational numbers. Suppose that the antiderivative $\int_0^x f(t)\,dt$ is globally bounded. Show that $\int_0^x f(t)\,dt$ is then the power series expansion of a rational function in $\mathbf{Q}(x)$.
Show that a power series $f(x) = \sum_{n=0}^{\infty} c_n x^n \in \mathbf{Q}\llbracket x \rrbracket$ is a solution of a differential equation if and only if there exist an integer $d \geq 0$ and polynomials not all zero $S_0, \ldots, S_d \in \mathbf{Z}[x]$ such that: for all $n \geq 0$, $$S_0(n) c_n + \cdots + S_d(n) c_{n+d} = 0.$$
Give a new proof, based on questions 12 and 13 above, of the fact that the power series $\sum_{m=0}^{\infty} x^{m^2}$ is not the expansion of a rational function.
Show that the power series $$h(x) = \sum_{n=0}^{\infty} \frac{(2n)!(3n)!}{(n!)^5} x^n$$ is a solution of a differential equation, then make one explicit.
Show that a power series $f(x) = \sum_{n=0}^{\infty} \frac{c_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ is a solution of a differential equation if and only if its Laplace transform $$\widehat{f}(x) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} c_n x^n$$ is a solution of a differential equation.
Let $r \geq 2$ be an integer and $a_1, \ldots, a_r \in \mathbf{Q}$ be distinct rationals. Let $b_1, \ldots, b_r \in \mathbf{Q}^{\times}$ be nonzero rationals. Set $e^{a_i x} \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} \frac{a_i^n}{n!} x^n$ and consider the power series $$f(x) = \sum_{n=0}^{\infty} \frac{u_n}{n!} x^n \stackrel{\text{def}}{=} b_1 e^{a_1 x} + \cdots + b_r e^{a_r x}.$$ Show that the Laplace transform $\widehat{f}(x) = \sum_{n=0}^{\infty} u_n x^n$ is the power series expansion of the rational function $$\sum_{i=1}^{r} \frac{b_i}{1 - a_i x}.$$ Deduce that $f$ is not the zero power series.
Consider the sequence $(v_n)_{n \geq 0}$ defined in terms of the coefficients $u_n$ by the formula $$v_n = n! \sum_{i=0}^{n} \frac{u_i}{i!}$$ and the power series $$v(x) = \sum_{n=0}^{\infty} v_n x^n \in \mathbf{Q}\llbracket x \rrbracket.$$ Show the equality of power series $$\sum_{n=0}^{\infty} (v_n - n v_{n-1}) x^n = \sum_{n=0}^{\infty} u_n x^n.$$
With the notation of question 18 and 19, show that the differential operator $L = -x^2 \left(\frac{d}{dx}\right) + (1-x)$ acts on $v(x)$ by $$(L \cdot v)(x) = \sum_{i=1}^{r} \frac{b_i}{1 - a_i x}.$$
With the notation of questions 18--20, deduce that if $v(x)$ is the power series expansion of a rational fraction $P/Q$, then every element of the non-empty set $\{1/a_i \mid a_i \neq 0\}$ is a pole of $P/Q$.
Show that Theorem 1 is equivalent to the following statement: Let $f(x) \in \mathbf{Q}\llbracket x \rrbracket$ be an exponential polynomial such that $f(1) = \sum_{i=1}^{s} P_i(1) e^{c_i}$ vanishes. Then $f(x)/(x-1)$ is still an exponential polynomial. (An exponential polynomial is any power series with rational coefficients of the form $f(x) = \sum_{i=1}^{s} P_i(x) e^{c_i x}$, where $c_1, \ldots, c_s \in \mathbf{Q}$ are rationals and $P_1, \ldots, P_s \in \mathbf{Q}[x]$ are polynomials.)
Let $D$ be a common denominator of the rational numbers $a_1, \ldots, a_r$ and let $A = \max(1, |a_1|, \ldots, |a_r|)$. Show that $D^n u_n \in \mathbf{Z}$ for all $n \geq 0$.
For all integers $n, k \geq 0$, define the rational number $v_n(k)$ as the coefficient of degree $n$ in the power series $$\left(1 - s_1 x - \cdots - s_r x^r\right)^k v(x) = \sum_{n=0}^{\infty} v_n(k) x^n.$$ Show that $v(x)$ is the power series expansion of a rational fraction if and only if there exists an integer $k \geq 0$ such that $\sum_{n=0}^{\infty} v_n(k) x^n$ is a polynomial.
For all integers $n, k \geq 0$, define the rational number $v_n(k)$ as the coefficient of degree $n$ in the power series $$\left(1 - s_1 x - \cdots - s_r x^r\right)^k v(x) = \sum_{n=0}^{\infty} v_n(k) x^n.$$ Observe the equality: for all $n \geq r$ and $k \geq 0$, $$v_n(k+1) = v_n(k) - s_1 v_{n-1}(k) - \cdots - s_r v_{n-r}(k).$$
Let $C = 1 + |s_1| + \cdots + |s_r|$. Show that $D^n v_n(k) \in \mathbf{Z}$ and that there exists a real number $c_2 > 0$ such that $$|v_n(k)| \leq c_2 A^n C^k \text{ for all } n \geq kr.$$
Let $\ell \geq 0$ be an integer. Show that there exists a polynomial $P_\ell \in \mathbf{Q}[x]$ of degree $< r(\ell+1)$ satisfying $$\sum_{n=0}^{\infty} n(n-1)\cdots(n-\ell+1)\, u_{n-\ell}\, x^n = \frac{P_\ell(x)}{\left(1 - s_1 x - \cdots - s_r x^r\right)^{\ell+1}}.$$
Define two sequences $(w_{n,k})_{n,k \geq 0}$ and $(w_n(k))_{n,k \geq 0}$ by the formulas $$w_{n,k} = n! \sum_{i=0}^{n-k} \frac{u_i}{i!} \quad \text{and} \quad \sum_{n=0}^{\infty} w_n(k) x^n = \left(1 - s_1 x - \cdots - s_r x^r\right)^k \sum_{n=0}^{\infty} w_{n,k}\, x^n.$$ Show the equality $w_n(k) = v_n(k)$ for all $n$ and $k$ such that $n \geq kr$.
A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$. Show that if $f$ is a function $E$, then the numerical series $\sum_{n=0}^{\infty} \frac{b_n}{n!} \alpha^n$ converges for every real number $\alpha$. We denote its value by $f(\alpha)$.
A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$. Let $f$ be a function $E$ that is not a polynomial. Show that there exists $R > 0$ such that the numerical series $\sum_{n=0}^{\infty} b_n \alpha^n$ diverges for every real number $\alpha$ with $|\alpha| > R$.
A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$. Recall that $\widehat{f}(x)$ denotes the Laplace transform $\widehat{f}(x) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} b_n x^n$. Which functions $E$ are such that $\widehat{f}$ is also a function $E$?
A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$. Prove that functions $E$ are closed under addition and multiplication.
A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$. Let $f$ be an exponential polynomial (i.e., $f(x) = \sum_{i=1}^{s} P_i(x) e^{c_i x}$ with $c_i \in \mathbf{Q}$ and $P_i \in \mathbf{Q}[x]$). Show that $f$ is a function $E$ such that $\widehat{f}$ is the power series expansion of a rational fraction with rational poles.
A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$. Show that if $\sum_{n=0}^{\infty} b_n x^n$ is the power series expansion of a rational fraction with rational poles, then $\sum_{n=0}^{\infty} \frac{b_n}{n!} x^n$ is a function $E$.
A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$. Show that the Bessel function $$J_0(x) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n!)^2} \left(\frac{x}{2}\right)^{2n}$$ is a function $E$ such that $\widehat{J}_0(x)$ satisfies the equation $(1 + x^2)\widehat{J}_0(x)^2 = 1$. Deduce that $J_0(x)$ is not an exponential polynomial.
Show that the real zeros of the Bessel function $J_0(x) \stackrel{\text{def}}{=} \sum_{n=0}^{\infty} \frac{(-1)^n}{(n!)^2} \left(\frac{x}{2}\right)^{2n}$ are simple, that is, if $J_0(\alpha) = 0$, then $J_0'(\alpha) \neq 0$.
A function $E$ (with rational coefficients) is a power series $f(x) = \sum_{n=0}^{\infty} \frac{b_n}{n!} x^n \in \mathbf{Q}\llbracket x \rrbracket$ satisfying: (a) $f$ is a solution of a differential equation; (b) there exists a real number $C > 0$ such that $|b_n| \leq C^n$ and $\operatorname{denom}(b_0, \ldots, b_n) \leq C^n$ for all $n \geq 1$. Let $f(x)$ be a function $E$ such that $f(1) = 0$. Show that the power series $f(x)/(x-1)$ is still a function $E$.