Arithmetical Aspects of Power Series

The product of any $m$ consecutive integers is divisible by the product of the first $m$ positive integers.

The product of $m$ integers in arithmetic progression is divisible by $m!$ if the common difference of the progression is prime to $m!$.

The product of the differences of any $m$ integers is divisible by the product of the differences of the first $m$ positive integers. (The product of the differences of $m$ variables $x_1,x_2,\ldots ,x_m$, $\prod_{j<k}^{1,2,\ldots ,m}(x_k-x_j)$, consists of $\tfrac 12m(m-1)$ factors.)

What is the exponent of the highest power of a prime number $p$ that is a divisor of $n!$?

In how many zeros does 1000! end when expressed in the decimal system?

Let $a$ and $b$ be positive integers. Then the integer $2a!\, 2b!$ is divisible by $a!\, (a+b)!\, b!$.

If $h$ and $n$ are positive integers, then $(hn)!/[(h!)^nn!]$ is an integer.

Let $s$ be an integer, $t$ a positive integer, $(s,t)=1$. Show that only prime factors of $t$ divide the denominators of the coefficients of the power series with rational coefficients

Let $p$ be a prime factor of $t$ and let $p^\alpha ,p^{\alpha_0},p^{\alpha_1},\ldots ,p^{\alpha_n},\ldots$ respectively be the highest powers of $p$ that divide $t,t_0,t_1,\ldots ,t_n,\ldots$, where $t_n$ denotes the denominator of the $n$th coefficient in the power series expansion of $(1+z)^{s/t}$; $(s,t)=1$. Calculate $\lim_{n\to \infty }(\alpha_n/n)$.

Find the smallest integer $T$ with the property that all the coefficients of the power series expansion $(1+Tz)^{s/t}$ are integers.

Prove Eisenstein's theorem in the simplest special case, namely for rational functions.

We say that the power series

with rational coefficients ($s_n,t_n$ integers, $t_n\geq 1$, $(s_n,t_n)=1$) satisfies the Eisenstein condition, if there exists an integer $T$ such that $T^n$ is divisible by $t_n$ for $n=1,2,3,\ldots$. If two power series $f(z)$ and $g(z)$ with rational coefficients satisfy the Eisenstein condition, then the series $f(z)+g(z)$, $f(z)-g(z)$, $f(z)g(z)$, also satisfy the condition, and also $f(z)/g(z)$ (assuming $g(0)\gtreqless 0$) and $f(g(z))$ (assuming $g(0)=0$).

If the two power series with rational coefficients

satisfy the Eisenstein condition, then the series

also satisfies the condition.

Show that the Eisenstein condition is equivalent to the following pair of simultaneous conditions:

Expand the two functions

in powers of $z$. Do the power series obtained satisfy the two conditions (1), (2) in 144? Are the functions represented algebraic?

By a hypergeometric series we mean a series of the form

If $\alpha$ and $\beta$ are rational numbers but not both integers, and if $\gamma$ is a positive integer, then it is possible to choose a positive integer $T$ such that all the coefficients in the hypergeometric series $F(\alpha ,\beta ,\gamma ;Tz)$ become integers. (The function

for example is not an algebraic function [Problem 90 of Part I]: the Eisenstein condition is necessary but not sufficient for a function to be algebraic.)

Assume that $\alpha \neq \gamma$, $\beta \neq \gamma$ and $\alpha ,\beta ,\gamma$ are not all rational, but all the coefficients in the hypergeometric series $F(\alpha ,\beta ,\gamma ;z)$ are rational. Then $\alpha$ and $\beta$ are the roots of an irreducible quadratic equation with rational coefficients.

In the case discussed in 147, $F(\alpha ,\beta ,\gamma ;z)$ does not represent an algebraic function of $z$. Prove this from Eisenstein's theorem!

A rational function (quotient of two polynomials), whose expansion in ascending powers of $z$ has only rational numbers as coefficients, is the quotient of two polynomials with rational coefficients.

Let the power series $f(z)$ with rational coefficients satisfy an equation of the form

where $P_0(z),P_1(z),\ldots ,P_l(z)$ are polynomials, $P_0(z)\neq 0$. Show that $f(z)$ also satisfies an equation of the same form, where the polynomials occurring in it have coefficients that are rational integers.

Let the power series with rational coefficients $y=a_0+a_1z+a_2z^2+\cdots +a_nz^n+\cdots$ satisfy an algebraic differential equation, i.e. an equation of the form

where $R$ denotes a rational entire function (multinomial) of its $r+2$ arguments. Show that $y$ must satisfy an equation of the form

where $R^*$ denotes a rational entire function with integral coefficients.

Let the functions $f(z)=c_0+c_1z+c_2z^2+\cdots$, $F_0(z),F_1(z),F_2(z),\ldots ,F_l(z)$ ($F_0(z)\neq 0$) be regular in a neighbourhood of the origin and related by the equation

Then the power series

if $m$ is suitably chosen, satisfies an equation of the form

where $k\leq l$, $G_0(0)=G_1(0)=\cdots =G_{k-2}(0)=0$, $G_{k-1}(0)\neq 0$, $G_0(z)\not\equiv 0$ and $G_\kappa (z)$, $\kappa =0,1,2,\ldots ,k$, is a rational entire expression with integral coefficients in $F_0(z),F_1(z),\ldots ,F_l(z),z,c_0,c_1,\ldots ,c_{m-1}$, possibly divided by a power of $z$. (Function-theoretic preparation for 153, 154.)

If the power series $P_0(z),P_1(z),\ldots ,P_l(z)$ with rational coefficients satisfy the Eisenstein condition, and if the power series $f(z)$ with rational coefficients satisfies the equation

then $f(z)$ also satisfies the Eisenstein condition.

Let $s_n$ and $t_n$ be integers, $(s_n,t_n)=1$, $t_n\geq 1$, and denote the greatest prime factor that divides $t_n$ by $P_n$. We say that the power series

satisfies the Tchebychev condition if $P_n/n$ is bounded, and the Hurwitz condition if $\log P_n/\log n$ is bounded, $n=2,3,4,\ldots$. For the Tchebychev condition as well as for the Hurwitz condition analogous theorems to 142, 143, and 153 hold.

Let $P(z),Q(z),Q_2(z),Q_3(z),\ldots$ be power series with rational coefficients, and set $y=c_0+c_1z+c_2z^2+\cdots$. If one of the following six equations holds

then $c_n$ is a uniquely determined rational number, given as a rational and entire function of the coefficients of $1,z,z^2,\ldots ,z^n$ in the expansions of $P(z),Q(z),Q_2(z),\ldots$, which therefore can have only such prime numbers in its denominator as occur also in the denominators of the coefficients mentioned above. [Cf. op. cit. 144.]

Let $P(z),Q(z),Q_2(z),Q_3(z),\ldots$ be power series with rational coefficients, and set $y=c_0+c_1z+c_2z^2+\cdots$. If one of the following six equations holds

then $c_n$ is a uniquely determined rational number, given as a rational and entire function of the coefficients of $1,z,z^2,\ldots ,z^n$ in the expansions of $P(z),Q(z),Q_2(z),\ldots$, which therefore can have only such prime numbers in its denominator as occur also in the denominators of the coefficients mentioned above. [Cf. op. cit. 144.]

We term the power series with integral coefficients $a_0+a_1z+a_2z^2+\cdots +a_nz^n+\cdots$ primitive if the numbers $a_0,a_1,a_2,\ldots ,a_n,\ldots$ have no common divisor apart from $\pm 1$. Prove that the product of two primitive power series is a primitive power series.

If the power series with integral coefficients $a_0+a_1z+a_2z^2+\cdots +a_nz^n+\cdots$ represents a rational function, then it can be expressed in the form $P(z)/Q(z)$, where $P(z)$ and $Q(z)$ are polynomials with integral coefficients and $Q(0)=1$. [155.]

Let

be a positive number $\leq 1$ expressed in the decimal system, $0\leq a_\nu \leq 9$, $\nu =1,2,3,\ldots$. The power series with integral coefficients $a_1z+a_2z^2+a_3z^3+\cdots +a_nz^n+\cdots$ represents a rational function if $\vartheta$ is rational, and does not represent a rational function if $\vartheta$ is irrational.

Let there be only finitely many distinct numbers in the infinite sequence of numbers $a_0,a_1,a_2,\ldots ,a_n,\ldots$. Then the power series $a_0+a_1z+a_2z^2+\cdots +a_nz^n+\cdots$ represents a rational function if and only if the sequence of coefficients after a certain number of terms is periodic.

Let $l$ be an integer, $l\geq 0$, and $P(z)$ a polynomial with integral coefficients. In the power series expansion of $P(z)(1-z)^{-l-1}$ the coefficients are integers that form a periodic sequence modulo any prime number $p$; the length of the period is a power of $p$.

Let $D$ be an integer and not divisible by the odd prime number $p$. The coefficients in the power series expansion of $[(D-1)z]/[(1-Dz)(1-z)]$ are periodic mod $p$, where the length of the period is a proper or an improper divisor of $p-1$.

Let the integers $D$ and $p$ be as in 160. In the power series expansion of $(1-Dz^2)^{-1}$ the coefficients are periodic mod $p$. The length of the period is always a divisor of $2(p-1)$. It is or is not a divisor of $p-1$ according to $\left(\dfrac Dp\right)=+1$ or $-1$.

The sequence of Fibonacci numbers (cf. Problem 53 of Part VII) is periodic mod $m$, where $m$ is any integer. Calculate the length of the period for all prime numbers less than 30.

In a power series expansion with integral coefficients that represents a rational function the coefficients, after a certain number of terms, form a periodic sequence mod $m$ ($m$ arbitrary).

The algebraic function $1/\sqrt{1-4z}$ yields on expansion in powers of $z$ a power series with integral coefficients. These coefficients are for no odd prime number $p$ periodic mod $p$. (163 cannot be extended to algebraic functions.)

The radius of convergence of a non-terminating power series with integral coefficients is $\leq 1$.

A non-terminating power series with integral coefficients that is convergent in the interior of the unit circle cannot represent a bounded function there.

If a power series with integral coefficients represents an algebraic function that is not a rational function, then its radius of convergence is $<1$.

In 167 the upper bound 1 is the best possible. In other words: If $\varepsilon >0$, then there exists a power series with integral coefficients that represents an algebraic function that is not a rational function and whose radius of convergence is $>1-\varepsilon$.

Let the coefficients of the polynomial

be real. The analytic function defined by the power series

is

1. a rational function, if the $r$ numbers $a_0,a_1,\ldots ,a_{r-1}$ are all rational,

2. not a rational function, if the $r$ numbers $a_0,a_1,\ldots ,a_{r-1}$ are not all rational. (Cf. Problem 168 of Part II.) [Problem 85 of Part I.]

If the sequence of non-negative integers $a_1,a_2,a_3,\ldots ,a_n,\ldots$ is bounded and contains infinitely many terms that are different from 0, then the series

does not represent a rational function.

If the sequence $a_1,a_2,a_3,\ldots ,a_n,\ldots$ satisfies the same assumptions as in 170, then the series

does not represent a rational function.

Denote by $Q_n$ the ``cross sum'' of the number $n$, i.e. the sum of its digits. E.g. we have $Q_{137}=1+3+7=11$. The power series

has the unit circle as its natural boundary.

In an infinite table of logarithms let the logarithms of all the positive integers $1,2,3,\ldots$ be written underneath each other in their natural order, in such a fashion that decimal places of the same order appear in the same vertical column. If we interpret the digits in any vertical column as the coefficients of a power series, then we obtain a power series that cannot be analytically continued.

In a different formulation we are required to prove the following theorem: Let $j$ be an integer and in the decimal representation of $\log 1,\log 2,\log 3,\ldots ,\log n,\ldots$ let the $j$th digit after the decimal point be denoted by $d_1,d_2,d_3,\ldots ,d_n,\ldots$ respectively. Then the analytic function defined by the power series

has as its natural boundary the circle $\left|z\right|=1$.

If $f(z)$ is a power series with H-integral coefficients, then

are also power series with H-integral coefficients.

If $f(z)$ and $g(z)$ are power series with H-integral coefficients, then

are also power series with H-integral coefficients, and under the further assumption that $g(0)=\pm 1$ also

If $f(z)$ is a power series with H-integral coefficients and $f(0)=0$, then

is also a power series with H-integral coefficients, $n=1,2,3,\ldots$ [174].

If $f(z)$ and $g(z)$ are power series with H-integral coefficients and $f(0)=0$, then $g[f(z)]$ is also a power series with H-integral coefficients.

The function $\varphi (z)$, well defined by the differential equation

and the initial conditions $\varphi (0)=0$, $\varphi '(0)>0$, may be expanded in a power series with H-integral coefficients.

For every prime number $p$, we have

For every composite number $m$ greater than 4 we have

The Bernoulli numbers $B_n$ are defined as the coefficients in the power series expansion

(Problem 154 of Part I). We have

In general we have

where $G_n$ is an integer and $2,3,\alpha ,\beta ,\gamma ,\ldots$ denote those prime numbers that exceed by one the divisors of $2n$. [Expand first $z$ and then $z/(e^z-1)$ in ascending powers of $(e^z-1)$ and apply 179--181.]

The coefficients $C_n$ in the series

where $\varphi (z)$ denotes the function defined in 178, are rational numbers and moreover the denominator of $C_n$ is not divisible by any square other than 1 and is divisible only by such prime numbers as are $\equiv 1 \pmod4$. [Expand first $z$ and then $z/\varphi (z)$ in ascending powers of $\varphi (z)$.]

The differential equation

has a well-defined solution of the form

($\wp (z)$ is a particular elliptic function, termed a ``lemniscate function''; its period-parallelogram is a square.) The numbers $D_n$ are rational, and moreover the denominator of $D_n$ is not divisible by any square other than 1, and is divisible only by such prime numbers as are $\equiv 1 \pmod4$. [Show that $\wp (z)=[\varphi (z)]^{-2}$, cf. 183. Expand first $z^2$ and then $z^2[\varphi (z)]^{-2}$ in ascending powers of $\varphi (z)$; $z^2$ as a function of $\varphi (z)$ satisfies a linear differential equation of second order.]

If a power series with H-integral coefficients $\sum_{n=0}^\infty (a_n/n!)z^n$ satisfies a homogeneous linear differential equation with constant coefficients, then the sequence of coefficients $a_0,a_1,a_2,\ldots ,a_n,\ldots$ is periodic mod $m$ after a certain number of terms, where $m$ is an arbitrary integer. [Examples of this are found already in 179, 180.]

The power series

shows that theorem 185 cannot without further assumptions be extended to homogeneous linear differential equations whose coefficients are polynomials.

If a transcendental entire function $g(z)$ has a power series expansion with H-integral coefficients, then we have for the maximum modulus (Chapter 1 of Part IV) of $g(z)$

Let the power series in descending powers of $z$

converge outside some circle. Let the numbers $b_1,b_2,\ldots ,b_m$ be rational. If the series assumes integral values for infinitely many integral values of $z$, then $b_0$ is rational and

(Generalization of 93.) [Hurwitz--Courant, p. 30.]

The series

represents an integer for infinitely many integral values of $z$.

Deduce 114 from 188.

Let the power series

that is not everywhere divergent represent integers for all sufficiently large integral values of $z$. Then $F(z)$ is an integral-valued polynomial.

If two polynomials are such that they assume integral values at the same points of the complex plane, then either their sum or their difference is a constant. (This constant is naturally an integer.)

If $m$ is the degree of $y=f(x)$ and $n$ the degree of $z=g(x)$, then we have for sufficiently large $\left|y\right|$

where all $m$ determinations are admissible for $y^{\frac1m}$.