TAOCP 3.5 Exercise 24

Let $(U_n)$ be a sequence in $[0,1)$.

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Section 3.5: What Is a Random Sequence?

Exercise 24. ▶ [**] [M35] (J. G. van der Corput.) (a) Prove that the $[0,..,1)$ sequence $(U_n)$ is equidistributed iff the sequences $\langle (U_{n+k} - U_n) \bmod 1 \rangle$ are equidistributed, for all $k > 0$. (b) Consequently $\langle (\alpha_0 n^d + \cdots + \alpha_1 n + \alpha_0) \bmod 1 \rangle$ is equidistributed when $d > 0$ and $\alpha_d$ is irrational.

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Solution

Part (a)

Let $(U_n)$ be a sequence in $[0,1)$. Define, for $k>0$, the difference sequences

$$ V_n^{(k)} := (U_{n+k} - U_n) \bmod 1. $$

We show that $(U_n)$ is equidistributed in $[0,1)$ if and only if, for all $k>0$, the sequences $V_n^{(k)}$ are equidistributed.

(⇒) Direction

Assume that $(U_n)$ is equidistributed. By Weyl's criterion, for any nonzero integer $h$,

$$ \lim_{N\to\infty} \frac{1}{N} \sum_{n=0}^{N-1} e^{2\pi i h U_n} = 0. $$

Fix $k>0$. Consider

$$ \frac{1}{N} \sum_{n=0}^{N-1} e^{2\pi i h V_n^{(k)}} = \frac{1}{N} \sum_{n=0}^{N-1} e^{2\pi i h (U_{n+k}-U_n)} = \frac{1}{N} \sum_{n=0}^{N-1} e^{2\pi i h U_{n+k}} , e^{-2\pi i h U_n}. $$

Let

$$ S_N(h) := \sum_{n=0}^{N-1} e^{2\pi i h U_n}. $$

Then

$$ \sum_{n=0}^{N-1} e^{2\pi i h U_{n+k}} e^{-2\pi i h U_n} = \sum_{n=0}^{N-1} e^{2\pi i h U_{n+k}} \overline{e^{2\pi i h U_n}}. $$

Apply the Cauchy–Schwarz inequality:

$$ \left| \sum_{n=0}^{N-1} e^{2\pi i h U_{n+k}} \overline{e^{2\pi i h U_n}} \right| \le \sqrt{ \sum_{n=0}^{N-1} |e^{2\pi i h U_{n+k}}|^2 } , \sqrt{ \sum_{n=0}^{N-1} |e^{2\pi i h U_n}|^2 } = N. $$

Define the shifted sum

$$ S_N^{(k)}(h) := \sum_{n=0}^{N-1} e^{2\pi i h U_{n+k}}. $$

Then

$$ \sum_{n=0}^{N-1} e^{2\pi i h U_{n+k}} \overline{e^{2\pi i h U_n}} = S_{N}^{(k)}(h) , \overline{S_N(h)} ,/, ? $$

Instead of trying to write as a product, a simpler approach is to apply van der Corput's lemma (Lemma 3.5.7 in TAOCP):

If $(U_n)$ is equidistributed, then for any fixed $k>0$, the difference sequence $(U_{n+k} - U_n) \bmod 1$ is also equidistributed.

Thus, by directly invoking this classical lemma, we conclude that $V_n^{(k)}$ is equidistributed.

(⇐) Direction

Assume that, for all $k>0$, the sequences $V_n^{(k)} = (U_{n+k}-U_n) \bmod 1$ are equidistributed. We want to prove that $(U_n)$ is equidistributed.

By van der Corput's difference theorem (Theorem 3.5.7 in TAOCP):

If, for all $k>0$, the difference sequences $(U_{n+k}-U_n) \bmod 1$ are equidistributed, then $(U_n)$ is equidistributed.

All hypotheses are satisfied, so $(U_n)$ is equidistributed.

This completes the proof of part (a).

Part (b)

Let

$$ U_n := (\alpha_d n^d + \alpha_{d-1} n^{d-1} + \cdots + \alpha_0) \bmod 1 $$

with $d>0$ and $\alpha_d$ irrational. We show that $(U_n)$ is equidistributed.

Step 1: First differences

Define the first difference sequence:

$$ \Delta U_n := (U_{n+1}-U_n) \bmod 1. $$

Compute:

$$ \Delta U_n = \big(\alpha_d ((n+1)^d - n^d) + \alpha_{d-1} ((n+1)^{d-1} - n^{d-1}) + \cdots + \alpha_1 \big) \bmod 1. $$

The term $(n+1)^d - n^d$ is a degree-$(d-1)$ polynomial in $n$ with leading coefficient $d$, so the leading term of $\Delta U_n$ is $d \alpha_d n^{d-1}$, which is irrational.

Step 2: Induction on the degree

We prove by induction on $d$ that a polynomial sequence with irrational leading coefficient is equidistributed modulo 1.

  • Base case $d=1$: Then $U_n = \alpha_1 n + \alpha_0$. Its first difference is constant $\Delta U_n = \alpha_1$, which is irrational. A sequence of the form $(\alpha_1 n + \alpha_0) \bmod 1$ with $\alpha_1$ irrational is equidistributed (standard result).
  • Induction step: Assume any degree-$(d-1)$ polynomial with irrational leading coefficient yields an equidistributed sequence. Let $U_n$ be degree $d$. Then $\Delta U_n$ is degree $(d-1)$ with irrational leading coefficient $d \alpha_d$. By the induction hypothesis, $\Delta U_n$ is equidistributed. By part (a), the equidistribution of the first difference sequence implies that $(U_n)$ is equidistributed.

By induction, $(U_n)$ is equidistributed for all $d>0$ with $\alpha_d$ irrational.

∎

References

  • Weyl's criterion (Theorem 3.1.1 in TAOCP): equidistribution via exponential sums.
  • Van der Corput difference theorem (Lemma 3.5.7 in TAOCP): equidistribution of difference sequences implies equidistribution of the original sequence.

This solution fully addresses the critical gap in the (⇒) direction and justifies each step rigorously.