Section 3

Multiples of an Irrational Number

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Problem 2.166

Let θ\theta be an irrational number. The numbers

xn=θn−[θn] x_{n} = \theta n - \left[\theta n\right]

are equidistributed on the interval [0,1]\left[0,1\right].

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Problem 2.167

Let θ\theta denote an irrational number. Put εn=1\varepsilon_{n}=1 or εn=0\varepsilon_{n}=0 according as the integer next to nθn\theta is larger or smaller than nθn\theta. Let aa and dd, a≥0a \geq 0, d>0d > 0, stand for two integers. Then we find that

lim⁡n→∞∑k=0n−1εa+kdn=12. \lim _{n \to \infty } \frac{\sum _{k=0}^{n-1} \varepsilon _{a+kd}}{n} = \frac{1}{2} .
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Problem 2.168

Let θ\theta be an irrational number and α\alpha be defined as α=qθ\alpha = q\theta, qq integer, q≠0q \neq 0. As zz converges to e2πiαe^{2\pi i \alpha } along the ray arg⁡z=2πα\arg z = 2\pi \alpha, the function

f(z)=∑n=1∞(nθ−[nθ])zn,z arbitrary complex, ∣z∣<1, f(z) = \sum _{n=1}^{\infty } (n\theta - \left[n\theta \right]) z^{n}, \qquad z \text{ arbitrary complex}, \ \left|z\right| < 1,

increases to ∞\infty in such a way that

lim⁡r→1−0(1−r)f(re2πiα)=12πiq \lim _{r \to 1-0} (1-r) f(re^{2\pi i \alpha }) = \frac{1}{2\pi i q}
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Remark.
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Problem 88 of Part I.

Problem 2.169

Determine for real xx the function

f(x)=lim⁡n→∞cos⁡2πx+cos⁡42πx+cos⁡63πx+⋯+cos⁡2nnπxn. f(x) = \lim _{n \to \infty } \frac{\cos ^{2}\pi x + \cos ^{4}2\pi x + \cos ^{6}3\pi x + \cdots + \cos ^{2n} n\pi x}{n} .
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Problem 2.170

The decimal fraction

θ=0.12345678910111213‾… \theta = 0.123456789101112\underline{13}\ldots

(the natural numbers listed consecutively) represents an irrational number. According to Problem 166 the numbers

nθ−[nθ],n=1,2,3,… n\theta - \left[n\theta \right], \qquad n = 1, 2, 3, \ldots

are everywhere dense on the interval [0,1]\left[0,1\right]. Show that this is already the case for the subset

10nθ−[10nθ],n=0,1,2,3,… 10^{n}\theta - \left[10^{n}\theta \right], \qquad n = 0, 1, 2, 3, \ldots
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Problem 2.171

The number

e=1+11!+12!+13!+⋯+1n!+⋯ e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!} + \cdots

is irrational (see Problem 258 of Part VIII). Prove that the only limit point of the set

n! e−[n! e],n=1,2,3,… n! \, e - \left[n! \, e\right], \qquad n = 1, 2, 3, \ldots

is zero.

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Problem 2.172

Suppose that the polynomial P(x)=a1x+a2x2+⋯+arxrP(x) = a_{1}x + a_{2}x^{2} + \cdots + a_{r}x^{r} has at least one irrational coefficient. Then the numbers

P(n)−[P(n)],n=1,2,3,… P(n) - \left[P(n)\right], \qquad n = 1, 2, 3, \ldots

have infinitely many limit points.

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Problem 2.173

Let θ\theta be an irrational number, xn=nθ−[nθ]x_{n} = n\theta - \left[n\theta \right], n=1,2,3,…n = 1, 2, 3, \ldots and let α1,α2,α3,…,αn,…\alpha_{1}, \alpha_{2}, \alpha_{3}, \ldots , \alpha_{n}, \ldots be a monotone decreasing sequence of positive numbers whose sum diverges. Then we find for any properly integrable function f(x)f(x) on [0,1]\left[0,1\right] that

lim⁡n→∞α1f(x1)+α2f(x2)+⋯+αnf(xn)α1+α2+⋯+αn=∫01f(x) dx. \lim _{n \to \infty } \frac{\alpha _{1}f(x_{1}) + \alpha _{2}f(x_{2}) + \cdots + \alpha _{n}f(x_{n})}{\alpha _{1}+\alpha _{2}+\cdots +\alpha _{n}} = \int _{0}^{1} f(x) \, dx .
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