User:Thecmad

Topology: an application to number theory
 Statement  There exist infinitely many primes in $$\mathbb{Z}$$.

 Proof  Consider the collection $$\mathcal{O}$$ of subsets of $$\mathbb{Z}$$ that consists of all sets of the form $$\alpha\mathbb{Z}+\beta, \; \; \alpha, \beta \in \mathbb{Z}, \; \; \alpha \neq 0$$, as well as of any (possibly trivial) union of such sets. Then, it is trivial to see that $$\mathcal{O}$$ induces a topology on $$\mathbb{Z}$$. Furthermore, observe that for all such $$\alpha, \beta: \; \; \alpha \mathbb{Z} + \beta = \mathbb{Z} \setminus \bigcup_{\kappa=1}^{|\alpha|-1} \left( \alpha \mathbb{Z} + \beta + \kappa \right)$$, which implies that $$\alpha \mathbb{Z} + \beta$$ is closed, as the complement of a union of open sets.

Suppose, for the sake of contradiction, that there were finitely many primes $$\mathbb{P}$$ in $$\mathbb{Z}$$. Then $$\mathbb{Z} \setminus \{\pm 1\} = \bigcup_{p \in \mathbb{P}} p\mathbb{Z}$$ would be closed, as a finite union of closed sets. However, $$\{\pm 1\} \not \in \mathcal{O}$$, so its complement is not open, a contradiction. Consequently, there exist infinitely many primes.

Counting in two ways: an application to trigonometry
Statement For any $$\{ a_i \}_{i=1}^\infty \subseteq [0,1]$$ we have $$\sum_{i=1}^\infty \left( a_i \prod_{j=1}^{i-1} (1-a_j) \right) = 1 - \prod_{i=1}^\infty (1-a_i)$$.

Proof Consider a sequence $$\{\mathcal{C}_i\}_{i=1}^\infty$$ of unfair coins, each having probability $$a_i$$, $$i = 1, 2, \ldots$$, of turning up heads when tossed. Denote by $$E$$ the event that, upon sequentially tossing the coins, a head eventually turns up; this can happen in one of countably many ways: a head comes up immediately, or a tail comes up first followed by a head, or two tails come up followed by a head, and so on. For a head to first come up on the $$i$$-th toss, all previous tosses must have resulted in tails, which event happens with probability $$a_i \prod_{j=1}^{i-1} (1-a_j)$$. Therefore, $$P(E) = \sum_{i=1}^\infty \left( a_i \prod_{j=1}^{i-1} (1-a_j) \right)$$. We may, however, calculate $$P(E)$$ with another method. For $$E$$ not to happen, each toss must result in a tail, for $$i=1,2,\ldots$$. The associated probability is $$P(E^c) = \prod_{i=1}^\infty (1-a_i)$$. Since $$P(E) = 1- P(E^c)$$, the result follows.

Application This can be applied whenever we can construct a sequence whose range lies in the unit interval $$[0,1]$$. One particularly interesting application is obtained by setting $$a_i = \cos^2 t_i$$, for $$t_i \in \mathbb{R}, i = 1,2,\ldots$$, thus establishing the trigonometric identity $$\sum_{i=1}^\infty \left( \cos^2 t_i \prod_{j=1}^{i-1} \sin^2 t_j \right) + \prod_{i=1}^\infty \sin^2 t_j = 1$$.

Counting in two ways: an application to number theory
Lemma If two positive integers $$\mathcal{X}, \mathcal{Y}$$ are chosen at random, each number being equally likely to have been selected, then the probability that $$\mathcal{X}, \mathcal{Y}$$ are coprime is $$\frac{6}{\pi^2}$$

Proof Let us adopt the notion that $$\mathbb{N}$$ is the set of positive integers. Recalling that $$(\cdot,\cdot) : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$$ denotes the greatest common divisor, we wish to show that $$\mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = 1\right] = \frac{6}{\pi^2}$$.

First, we claim that $$\mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = k\right] = \frac{1}{k^2}\mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = 1\right]$$, for any positive integer $$k$$. Indeed, it is easy to see that $$\mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = k\right] = \mathbb{P}\left[\left(\frac{1}{k}\mathcal{X}, \frac{1}{k}\mathcal{Y}\right) = 1 \mbox{ and } \mathcal{X}, \mathcal{Y} \in k\mathbb{N} \right] = \mathbb{P}\left[\mathcal{X}, \mathcal{Y} \in k\mathbb{N}\right] \; \mathbb{P}\left[\left(\frac{1}{k}\mathcal{X}, \frac{1}{k}\mathcal{Y}\right) = 1 \; \vert \; \mathcal{X}, \mathcal{Y} \in k\mathbb{N} \right]$$. If $$\mathcal{X}, \mathcal{Y}$$, being uniformly distributed over $$\mathbb{N}$$, are both restricted to $$k\mathbb{N}$$, then $$\frac{1}{k}\mathcal{X}, \frac{1}{k}\mathcal{Y}$$ are uniformly distributed over $$\mathbb{N}$$, so the right factor equals $$\mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = 1\right]$$, and thus: $$\mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = k\right] = \mathbb{P}\left[\mathcal{X}, \mathcal{Y} \in k\mathbb{N}\right] \; \mathbb{P}\left[\left(\frac{1}{k}\mathcal{X}, \frac{1}{k}\mathcal{Y}\right) = 1 \mbox{ and } \mathcal{X}, \mathcal{Y} \in k\mathbb{N} \right] = \mathbb{P}\left[\mathcal{X} \in k\mathbb{N}\right] \; \mathbb{P}\left[\mathcal{Y} \in k\mathbb{N}\right] \; \mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = 1\right] = \frac{1}{k^2} \mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = 1\right]$$, and the claim follows.

Now, observe that $$1 = \sum_{k=1}^\infty \mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = k\right] = \sum_{k=1}^\infty \frac{1}{k^2} \mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = 1\right]$$, so $$\mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = 1\right] = \frac{1}{\sum_{k=1}^\infty \frac{1}{k^2}} = \frac{1}{\zeta(2)} = \frac{6}{\pi^2}$$, and the result follows.

Statement If $$\{p_i\}_{i=1}^\infty$$ denotes the sequence of primes in $$\mathbb{Z}$$, then $$\prod_{n=1}^\infty \frac{p_n^2 - 1}{p_n^2} = \frac{6}{\pi^2}$$

Proof Due to the lemma, it suffices to show that $$\prod_{n=1}^\infty \frac{p_n^2 - 1}{p_n^2} = \mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = 1\right]$$, where $$\mathcal{X}, \mathcal{Y}$$ are random variables uniformly distributed over $$\mathbb{N}$$. Let us find yet another (easier) way of computing $$\mathbb{P}\left[(\mathcal{X}, \mathcal{Y}) = 1\right]$$. Two integers are coprime whenever they share no prime factors. Therefore, $$\mathcal{X}, \mathcal{Y}$$ are coprime if, for every $$n = 1,2,\ldots$$, $$\mathcal{X}, \mathcal{Y}$$ are not both in $$p_n\mathbb{N}$$, an event whose probability is $$1 - \frac{1}{p_n^2} = \frac{p_n^2 - 1}{p_n^2}$$. After multiplying the expressions for $$n=1,2,\ldots$$, since the events are independent, the result follows.

Isometric embeddings: an application to computational geometry
Lemma Consider the map $$\sigma : \mathbb{R}^k \to \mathbb{R}^{2^k}$$ defined as $$\sigma(\mathbf{x}) := \oplus_{\mathbf{q} \in \{\pm 1\}^k} \langle \mathbf{x}, \mathbf{q} \rangle$$, with $$\langle \cdot, \cdot \rangle$$ being the normal dot product in $$\mathbb{R}^k$$. Then for any two $$\mathbf{x}, \mathbf{y} \in \mathbb{R}^k$$ we have $$\Vert \mathbf{x} - \mathbf{y} \Vert_1 = \Vert \sigma(\mathbf{x}) - \sigma(\mathbf{y}) \Vert_{\text{sup}}$$

Proof Indeed, observe that

$$\Vert \sigma(\mathbf{x}) - \sigma(\mathbf{y}) \Vert_{\text{sup}} = \max_{\mathbf{q} \in \{\pm 1\}^k} \left( \langle \mathbf{x}, \mathbf{q} \rangle - \langle \mathbf{y}, \mathbf{q} \rangle \right) = \max_{\mathbf{q} \in \{\pm 1\}^k} \langle \mathbf{x} - \mathbf{y}, \mathbf{q} \rangle = \max_{\mathbf{q} \in \{\pm 1\}^k} \sum_{i=1}^k (x_i - y_i)q_i \leq \sum_{i=1}^k \vert x_i - y_i \vert = \Vert \mathbf{x} - \mathbf{y} \Vert_1$$

In fact, equality is achieved for $$\mathbf{q} = \oplus_{i=1}^k \text{sgn} (x_i - y_i)$$ (in this case, we adopt the notion that $$\text{sgn}(0) = 1$$), and the lemma follows.

Statement Given $$N$$ points $$\mathbf{p}_1, \ldots, \mathbf{p}_N \in \mathbb{R}^k, 1 \leq k < \infty$$, it is possible to compute the points' Manhattan diameter, $$\max_{1 \leq i, j \leq N} \Vert \mathbf{p}_i - \mathbf{p}_j \Vert_1$$, in runtime that is linear in $$N$$; specifically, we can achieve a runtime of $$\Theta(k \; 2^k \; N)$$.

Proof Let $$\sigma$$ and $$\langle \cdot, \cdot \rangle$$ be as in the lemma. Then

$$\max_{1 \leq i, j \leq N} \Vert \mathbf{p}_i - \mathbf{p}_j \Vert_1 = \max_{1 \leq i, j \leq N} \Vert \sigma(\mathbf{p}_i) - \sigma(\mathbf{p}_j) \Vert_{\text{sup}} = \max_{1 \leq i, j \leq N} \max_{\mathbf{q} \in \{\pm 1\}^k}\left( \langle \mathbf{p}_i, \mathbf{q} \rangle - \langle \mathbf{p}_j, \mathbf{q} \rangle \right) =$$

$$= \max_{\mathbf{q} \in \{\pm 1\}^k} \max_{1 \leq i, j \leq N} \left( \langle \mathbf{p}_i, \mathbf{q} \rangle - \langle \mathbf{p}_j, \mathbf{q} \rangle \right) = \max_{\mathbf{q} \in \{\pm 1\}^k} \left( \max_{1 \leq i \leq N}  \langle \mathbf{p}_i, \mathbf{q} \rangle - \min_{1 \leq j \leq N} \langle \mathbf{p}_j, \mathbf{q} \rangle \right)$$

which is easily computed in $$\Theta(k \; 2^k \; N)$$.

Uncategorized Problems
Prove that if $$\mu$$ is singular with respect to Lebesgue measure on $$\mathbb{S}^1$$ and $$d\mu(x + \alpha \pi) = d \mu(x)$$ on $$\mathbb{S}^1$$ for some $$\alpha$$ irrational, then $$\mu = 0$$. Stanford Qualifying Exams.