Problems

Solution:

1. Claim: If \(n \in \mathbb{Z}\) then \(\lfloor x + n \rfloor = \lfloor x \rfloor + n\) Proof: Since \(\lfloor x \rfloor \leq x\) then \(\lfloor x \rfloor + n \leq x + n\) and \(\lfloor x \rfloor + n \in \mathbb{Z}\) we must have that \(\lfloor x \rfloor +n \leq \lfloor x + n \rfloor\). However, since \(\lfloor x \rfloor + 1 > x\) we must also have that \(\lfloor x \rfloor + 1 + n > x + n\), therefore \(\lfloor x \rfloor + n\) is the largest integer less than \(x + n\) as required.
2. 1. Claim: \(f_n\left(x + \frac{1}{n}\right) = f_n(x)\) Proof: \begin{align*} f_n\left(x + \frac{1}{n}\right) &=\left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{1}{n}+ \frac{1}{n} \right\rfloor + \left\lfloor x+ \frac{1}{n} + \frac{2}{n} \right\rfloor + \ldots + \left\lfloor x+ \frac{1}{n} + \frac{n-1}{n} \right\rfloor - \left \lfloor n\left ( x + \frac{1}{n} \right) \right \rfloor \\ &= \left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{2}{n}\right\rfloor + \left\lfloor x+ \frac{3}{n} \right\rfloor + \ldots + \left\lfloor x+ \frac{n}{n} \right\rfloor - \left \lfloor nx + 1 \right \rfloor \\ &= \left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{2}{n}\right\rfloor + \left\lfloor x+ \frac{3}{n} \right\rfloor + \ldots + \left\lfloor x+ 1 \right\rfloor - \left \lfloor nx + 1 \right \rfloor \\ &= \left \lfloor x+ \frac{1}{n} \right \rfloor + \left\lfloor x + \frac{2}{n}\right\rfloor + \left\lfloor x+ \frac{3}{n} \right\rfloor + \ldots + \lfloor x \rfloor + 1 - \left ( \lfloor nx \rfloor + 1 \right) \\ &= \lfloor x \rfloor + \left\lfloor x + \frac{1}{n} \right\rfloor + \left\lfloor x + \frac{2}{n} \right\rfloor + \ldots + \left\lfloor x + \frac{n-1}{n} \right\rfloor - \lfloor nx \rfloor \\ &= f_n(x) \end{align*}
   2. Suppose \(0 \leq t < \frac1n\), then note that \(\left \lfloor t + \frac{k}{n} \right \rfloor = 0\) for \(0 \leq k \leq n - 1\) and \(\lfloor n t \rfloor = 0\) since \(nt < 1\)
   3. Since \(f_n(x)\) is zero on \([0, \tfrac1n)\) and periodic with period \(\tfrac1n\) it must be constantly zero
3. 1. Claim: \(\left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor = \lfloor x \rfloor\) Proof: Suppose \(x = n + \epsilon\) where \(0 \leq \epsilon < 1\), ie \(n = \lfloor x \rfloor\), then consider two cases: Case 1: \(n = 2k\) \begin{align*} \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor &= \left\lfloor \frac{n + \epsilon}{2} \right\rfloor + \left\lfloor \frac{n + \epsilon+1}{2} \right\rfloor \\ &= \left\lfloor \frac{2k + \epsilon}{2} \right\rfloor + \left\lfloor \frac{2k + \epsilon+1}{2} \right\rfloor \\ &= k + \left\lfloor \frac{\epsilon}{2} \right\rfloor + k + \left\lfloor \frac{\epsilon+1}{2} \right\rfloor \\ &= 2k \\ &= n \end{align*} Case 2: \(n = 2k + 1\) \begin{align*} \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x+1}{2} \right\rfloor &= \left\lfloor \frac{n + \epsilon}{2} \right\rfloor + \left\lfloor \frac{n + \epsilon+1}{2} \right\rfloor \\ &= \left\lfloor \frac{2k +1+ \epsilon}{2} \right\rfloor + \left\lfloor \frac{2k +1+ \epsilon+1}{2} \right\rfloor \\ &= k + \left\lfloor \frac{\epsilon+1}{2} \right\rfloor + k +1+ \left\lfloor \frac{\epsilon}{2} \right\rfloor \\ &= 2k +1\\ &= n \end{align*} as required.
   2. Since \(\left \lfloor \frac{x+1}{2} \right \rfloor = \lfloor x \rfloor - \lfloor \frac{x}{2} \rfloor\) and in general, \(\left \lfloor \frac{x+2^k}{2^{k+1}} \right \rfloor = \lfloor \frac{x}{2^k} \rfloor - \lfloor \frac{x}{2^{k+1}} \rfloor\) and so in general: \begin{align*} \sum_{k=0}^\infty \left \lfloor \frac{x+2^k}{2^{k+1}} \right \rfloor &= \sum_{k=0}^\infty \left ( \left \lfloor \frac{x}{2^k} \right \rfloor -\left \lfloor \frac{x}{2^{k+1}} \right \rfloor \right) \\ &= \lfloor x \rfloor \end{align*}