Section 6

Power Series with Integral Coefficients in the Sense of Hurwitz

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Problem 8.174

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

df(z)dzand∫0zf(z) dz \frac{df(z)}{dz} \qquad \text{and} \qquad \int _0^z f(z)\, dz

are also power series with H-integral coefficients.

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Problem 8.175

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

f(z)+g(z),f(z)−g(z),f(z)g(z) f(z)+g(z) , \qquad f(z)-g(z) , \qquad f(z)g(z)

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

f(z)g(z). \frac{f(z)}{g(z)} .
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Problem 8.176

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

[f(z)]mm! \frac{[f(z)]^m}{m!}

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

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Problem 8.177

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

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Problem 8.178

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

(dφ(z)dz)2=1−[φ(z)]4 \left(\frac{d\varphi (z)}{dz}\right)^2 = 1-[\varphi (z)]^4

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

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Problem 8.179
(ez−1)3≡2(z33!+z55!+z77!+⋯ )(mod4). (e^z-1)^3 \equiv 2\left(\frac{z^3}{3!}+\frac{z^5}{5!}+\frac{z^7}{7!}+\cdots \right) \pmod4 .
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Problem 8.180

For every prime number pp, we have

(ez−1)p−1≡−(zp−1(p−1)!+z2(p−1)(2p−2)!+z3(p−1)(3p−3)!+⋯ )(modp). (e^z-1)^{p-1} \equiv -\left(\frac{z^{p-1}}{(p-1)!}+\frac{z^{2(p-1)}}{(2p-2)!}+\frac{z^{3(p-1)}}{(3p-3)!}+\cdots \right) \pmod p .
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Problem 8.181

For every composite number mm greater than 4 we have

(ez−1)m−1≡0(modm). (e^z-1)^{m-1} \equiv 0 \pmod m .
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Problem 8.182

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

zez−1=1−z2+∑n=1∞(−1)n−1Bn(2n)!z2n \frac{z}{e^z-1} = 1-\frac z2+\sum _{n=1}^\infty \frac{(-1)^{n-1}B_n}{(2n)!}z^{2n}

(Problem 154 of Part I). We have

B1=16=1−12−13,B2=130=−1+12+13+15, B_1=\frac16 = 1-\frac12-\frac13 , \qquad B_2=\frac{1}{30} = -1+\frac12+\frac13+\frac15 , B3=142=1−12−13−17,B4=130=−1+12+13+15, B_3=\frac{1}{42} = 1-\frac12-\frac13-\frac17 , \qquad B_4=\frac{1}{30} = -1+\frac12+\frac13+\frac15 , B5=566=1−12−13−111,B6=6912730=−1+12+13+15+17+113, B_5=\frac{5}{66} = 1-\frac12-\frac13-\frac{1}{11} , \qquad B_6=\frac{691}{2730} = -1+\frac12+\frac13+\frac15+\frac17+\frac{1}{13} , B7=76=2−12−13,B8=3617510=6+12+13+15+117. B_7=\frac76 = 2-\frac12-\frac13 , \qquad B_8=\frac{3617}{510} = 6+\frac12+\frac13+\frac15+\frac{1}{17} .

In general we have

Bn=Gn+(−1)n(12+13+1α+1β+1γ+⋯ ), B_n = G_n+(-1)^n\left(\frac12+\frac13+\frac1\alpha +\frac1\beta +\frac1\gamma +\cdots \right) ,

where GnG_n is an integer and 2,3,α,β,γ,…2,3,\alpha ,\beta ,\gamma ,\ldots denote those prime numbers that exceed by one the divisors of 2n2n. [Expand first zz and then z/(ez−1)z/(e^z-1) in ascending powers of (ez−1)(e^z-1) and apply 179--181.]

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Problem 8.183

The coefficients CnC_n in the series

zφ(z)=∑n=0∞Cn(4n)!z4n, \frac{z}{\varphi (z)} = \sum _{n=0}^\infty \frac{C_n}{(4n)!}z^{4n} ,

where φ(z)\varphi (z) denotes the function defined in 178, are rational numbers and moreover the denominator of CnC_n is not divisible by any square other than 1 and is divisible only by such prime numbers as are ≡1(mod4)\equiv 1 \pmod4. [Expand first zz and then z/φ(z)z/\varphi (z) in ascending powers of φ(z)\varphi (z).]

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Problem 8.184

The differential equation

(d℘(z)dz)2=4[℘(z)]3−4℘(z) \left(\frac{d\wp (z)}{dz}\right)^2 = 4[\wp (z)]^3-4\wp (z)

has a well-defined solution of the form

℘(z)=1z2+∑n=1∞Dn(4n)!z4n−2. \wp (z) = \frac{1}{z^2}+\sum _{n=1}^\infty \frac{D_n}{(4n)!}z^{4n-2} .

(℘(z)\wp (z) is a particular elliptic function, termed a ``lemniscate function''; its period-parallelogram is a square.) The numbers DnD_n are rational, and moreover the denominator of DnD_n is not divisible by any square other than 1, and is divisible only by such prime numbers as are ≡1(mod4)\equiv 1 \pmod4. [Show that ℘(z)=[φ(z)]−2\wp (z)=[\varphi (z)]^{-2}, cf. 183. Expand first z2z^2 and then z2[φ(z)]−2z^2[\varphi (z)]^{-2} in ascending powers of φ(z)\varphi (z); z2z^2 as a function of φ(z)\varphi (z) satisfies a linear differential equation of second order.]

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Problem 8.185

If a power series with H-integral coefficients ∑n=0∞(an/n!)zn\sum_{n=0}^\infty (a_n/n!)z^n satisfies a homogeneous linear differential equation with constant coefficients, then the sequence of coefficients a0,a1,a2,…,an,…a_0,a_1,a_2,\ldots ,a_n,\ldots is periodic mod mm after a certain number of terms, where mm is an arbitrary integer. [Examples of this are found already in 179, 180.]

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Problem 8.186

The power series

y=1+2x1!+6x22!+20x33!+⋯+(2nn)xnn!+⋯ y = 1+\frac{2x}{1!}+\frac{6x^2}{2!}+\frac{20x^3}{3!}+\cdots +\binom {2n}{n}\frac{x^n}{n!}+\cdots

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

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Problem 8.187

If a transcendental entire function g(z)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)g(z)

lim sup⁡r→∞M(r)e−rr≥12π. \limsup _{r\to \infty } M(r)e^{-r}\sqrt r \geq \frac{1}{\sqrt{2\pi }} .
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